A Student's Guide to the Schrödinger Equation

Begin by writing the wavefunction \(\psi(x)\) as the sum of component wavefunctions \(\psi_1(x)\) and \(\psi_2(x)\):
\begin{equation*}
\psi(x)=\psi_1(x)+\psi_2(x)
\end{equation*}

and then write another wavefunction \(\psi'(x)\) as the sum of component wavefunctions \(\psi_1(x)\) and \(\psi_2(x)\) to which a universal phase factor \(e^{i\theta}\) has been applied (in this context, “universal” means that the same phase factor is applied to each component wavefunction). So

\begin{equation*}
\psi'(x)=e^{i\theta}[\psi_1(x)+\psi_2(x)]=\psi_1e^{i\theta}+\psi_2e^{i\theta}.
\end{equation*}

To show that the universal phase factor does not affect the probability density \(Pden(x)\), write the probability density for \(\psi(x)\) and for \(\psi'(x)\):
\begin{equation*}
Pden(x)=[\psi(x)]^*[\psi(x)]=[\psi_1+\psi_2]^*[\psi_1+\psi_2]
\end{equation*}

and

\begin{align*}
Pden'(x)&=[\psi'(x)]^*[\psi'(x)]=[\psi_1e^{i\theta}+\psi_2e^{i\theta}]^*[\psi_1e^{i\theta}+\psi_2e^{i\theta}]\\
&=[\psi_1^*e^{-i\theta}+\psi_2^*e^{-i\theta}][\psi_1e^{i\theta}+\psi_2e^{i\theta}].
\end{align*}

Carrying out the multiplications in the previous hint gives
\begin{equation*}
Pden(x)=[\psi_1+\psi_2]^*[\psi_1+\psi_2]=\psi_1^*\psi_1+\psi_1^*\psi_2+\psi_2^*\psi_1+\psi_2^*\psi_2
\end{equation*}

and

\begin{align*}
Pden^{\prime}(x)&=\psi_1^*\psi_1e^{-i\theta+i\theta}+\psi_1^*\psi_2e^{-i\theta+i\theta}+\psi_2^*\psi_1e^{-i\theta+i\theta}+\psi_2^*\psi_2e^{-i\theta+i\theta}\\
&=\psi_1^*\psi_1+\psi_1^*\psi_2+\psi_2^*\psi_1+\psi_2^*\psi_2.
\end{align*}

So the presence of the universal phase factor does not alter the probability density.

To show that the relative phase of the component wavefunctions does have an effect on the probability density, write another wavefunction \(\psi^{\prime\prime}(x)\) in which the component wavefunctions have (possibly different) phase factors \(e^{i\theta_1}\) and \(e^{i\theta_2}\):
\begin{equation*}
\psi^{\prime\prime}(x)=\psi_1(x)e^{i\theta_1}+\psi_2(x)e^{i\theta_2}.
\end{equation*}

Now determine the probability density of \(\psi”(x)\):
\begin{align*}
Pden^{\prime\prime}(x)&=[\psi^{\prime\prime}(x)]^*[\psi^{\prime\prime}(x)]=[\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}]^*[\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}]\\
&=[\psi_1^*e^{-i\theta_1}+\psi_2^*e^{-i\theta_2}][\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}].
\end{align*}

Performing the multiplications in the previous hint gives
\begin{align*}
Pden^{\prime\prime}(x)&=\psi_1^*\psi_1e^{-i\theta_1+i\theta_1}+\psi_1^*\psi_2e^{-i\theta_1+i\theta_2}\\
&\hspace {0.25cm}+\psi_2^*\psi_1e^{-i\theta_2+i\theta_1}+\psi_2^*\psi_2e^{-i\theta_2+i\theta_2}\\
&=\psi_1^*\psi_1+\psi_1^*\psi_2e^{-i(\theta_1-\theta_2)}\\
&\hspace {0.25cm}+\psi_2^*\psi_1e^{-i(\theta_2-\theta_1)}+\psi_2^*\psi_2
\end{align*}

which shows that the probability density depends on the difference between the individual phase factors (note that this expression reduces to the previous cases \(Pden(x)\) and \(Pden'(x)\) if \(\theta_1=\theta_2\)).

Full Solution: Begin by writing the wavefunction \(\psi(x)\) as the sum of component wavefunctions \(\psi_1(x)\) and \(\psi_2(x)\):
\begin{equation*}
\psi(x)=\psi_1(x)+\psi_2(x)
\end{equation*}

and then write another wavefunction \(\psi'(x)\) as the sum of component wavefunctions \(\psi_1(x)\) and \(\psi_2(x)\) to which a universal phase factor \(e^{i\theta}\) has been applied (in this context, “universal” means that the same phase factor is applied to each component wavefunction). So

\begin{equation*}
\psi'(x)=e^{i\theta}[\psi_1(x)+\psi_2(x)]=\psi_1e^{i\theta}+\psi_2e^{i\theta}.
\end{equation*}

To show that the universal phase factor does not affect the probability density \(Pden(x)\), write the probability density for \(\psi(x)\) and for \(\psi'(x)\):
\begin{equation*}
Pden(x)=[\psi(x)]^*[\psi(x)]=[\psi_1+\psi_2]^*[\psi_1+\psi_2]
\end{equation*}

and

\begin{align*}
Pden'(x)&=[\psi'(x)]^*[\psi'(x)]=[\psi_1e^{i\theta}+\psi_2e^{i\theta}]^*[\psi_1e^{i\theta}+\psi_2e^{i\theta}]\\
&=[\psi_1^*e^{-i\theta}+\psi_2^*e^{-i\theta}][\psi_1e^{i\theta}+\psi_2e^{i\theta}].
\end{align*}

Carrying out these multiplications gives
\begin{equation*}
Pden(x)=[\psi_1+\psi_2]^*[\psi_1+\psi_2]=\psi_1^*\psi_1+\psi_1^*\psi_2+\psi_2^*\psi_1+\psi_2^*\psi_2
\end{equation*}

and

\begin{align*}
Pden'(x)&=\psi_1^*\psi_1e^{-i\theta+i\theta}+\psi_1^*\psi_2e^{-i\theta+i\theta}+\psi_2^*\psi_1e^{-i\theta+i\theta}+\psi_2^*\psi_2e^{-i\theta+i\theta}\\
&=\psi_1^*\psi_1+\psi_1^*\psi_2+\psi_2^*\psi_1+\psi_2^*\psi_2.
\end{align*}

So the presence of the universal phase factor does not alter the probability density.

To show that the relative phase of the component wavefunctions does have an effect on the probability density, write another wavefunction \(\psi^{\prime\prime}(x)\) in which the component wavefunctions have (possibly different) phase factors \(e^{i\theta_1}\) and \(e^{i\theta_2}\):
\begin{equation*}
\psi^{\prime\prime}(x)=\psi_1(x)e^{i\theta_1}+\psi_2(x)e^{i\theta_2}.
\end{equation*}

Now determine the probability density of \(\psi^{\prime\prime}(x)\):
\begin{align*}
Pden^{\prime\prime}(x)&=[\psi^{\prime\prime}(x)]^*[\psi^{\prime\prime}(x)]=[\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}]^*[\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}]\\
&=[\psi_1^*e^{-i\theta_1}+\psi_2^*e^{-i\theta_2}][\psi_1e^{i\theta_1}+\psi_2e^{i\theta_2}].
\end{align*}

Performing these multiplications gives
\begin{align*}
Pden^{\prime\prime}(x)&=\psi_1^*\psi_1e^{-i\theta_1+i\theta_1}+\psi_1^*\psi_2e^{-i\theta_1+i\theta_2}\\
&\hspace {0.25cm}+\psi_2^*\psi_1e^{-i\theta_2+i\theta_1}+\psi_2^*\psi_2e^{-i\theta_2+i\theta_2}\\
&=\psi_1^*\psi_1+\psi_1^*\psi_2e^{-i(\theta_1-\theta_2)}\\
&\hspace {0.25cm}+\psi_2^*\psi_1e^{-i(\theta_2-\theta_1)}+\psi_2^*\psi_2
\end{align*}

which shows that the probability density depends on the difference between the individual phase factors (note that this expression reduces to the previous cases \(Pden(x)\) and \(Pden'(x)\) if \(\theta_1=\theta_2\)).

For an infinite rectangular potential well extending from \(x=0\) to \(x=a\), the wavefunctions are given by Eq. 5.9 as
\begin{equation*}
\psi_n(x)=\sqrt{\frac{2}{a}}\sin{\left(\frac{n\pi x}{a}\right)}
\end{equation*}

so the ground-state wavefunction (\(n=1\)) is
\begin{equation*}
\psi_1(x)=\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}.
\end{equation*}

The expectation value of position \(x\) is given by Eq. 2.60 as
\begin{equation*}
\braket{x}=\bra{\psi}\widehat{X}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X}\) is the position operator.

Plugging the expression for \(\psi_1(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{x}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \widehat{X}\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx
\end{equation*}

and since the eigenvalue equation for the position operator in position space is \(\widehat{X}\psi(x)=x\psi(x)\), this is
\begin{align*}
\braket{x}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\int_{0}^a x\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int x \sin^2{(cx)}dx=\frac{x^2}{4}-\frac{x\sin{(2cx)}}{4c}-\frac{\cos{(2cx)}}{8c^2}
\end{equation*}

with \(c=\pi/a\).

Evaluating the integral in the previous hint gives
\begin{align*}
\braket{x}&=\frac{2}{a}\int_{0}^a x\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\left[\frac{x^2}{4}-\frac{x\sin{\left(\frac{2\pi x}{a}\right)}}{4\left(\frac{\pi}{a}\right)}-\frac{\cos{\left(\frac{2\pi x}{a}\right)}}{8\left(\frac{\pi}{a}\right)^2}\right]\bigg|_0^a\\
&=\frac{2}{a}\left[\frac{a^2}{4}-\frac{\cos{(2\pi)}}{8\left(\frac{\pi}{a}\right)^2}+\frac{\cos{(0)}}{8\left(\frac{\pi}{a}\right)^2}\right]\\
&=\frac{2}{a}\left(\frac{a^2}{4}\right)=\frac{a}{2}.
\end{align*}

To find the expectation value of momentum \(\braket{p}\), use Eq. 4.60:
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P}_x\) represents the position-basis version of the momentum operator.

Plugging the expression for \(\psi_1(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{p}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\widehat{P}_x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx
\end{equation*}

and since the momentum operator in position space is given by Eq. 4.64 as
\begin{equation*}
\widehat{P}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}=-i\hbar\frac{\partial}{\partial x},
\end{equation*}

\(\braket{p}\) can be written as
\begin{equation*}
\braket{p}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\left(-i\hbar\frac{\partial}{\partial x}\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx.
\end{equation*}

Taking the derivative in the previous hint gives
\begin{align*}
\braket{p}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \left(-i\hbar \frac{\pi}{a}\right)\left[\sqrt{\frac{2}{a}}\cos{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{-2\pi i\hbar}{a^2}\int_{0}^a \left[\sin{\left(\frac{\pi x}{a}\right)}\cos{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int \sin{(cx)}\cos{(cx)}dx=\frac{\sin^2{(cx)}}{2c}
\end{equation*}

with \(c=\pi/a\).

Evaluating the integral in the previous hint gives
\begin{align*}
\braket{p}&=\frac{-2\pi i\hbar}{a^2}\int_{0}^a \left[\sin{\left(\frac{\pi x}{a}\right)}\cos{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{-2\pi i\hbar}{a^2}\left[\frac{\sin^2{\left(\frac{\pi}{a}x\right)}}{2\frac{\pi}{a}}\right]\bigg|_0^a\\
&=\frac{-i\hbar}{a}\left[\sin^2{\pi}-\sin^2{0}\right]=0.
\end{align*}

The expectation value of the square of position \(x^2\) is given by
\begin{equation*}
\braket{x^2}=\bra{\psi}\widehat{X^2}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X^2}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X^2}\) is the position-squared operator.

Plugging the expression for \(\psi_1(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{x^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \widehat{X^2}\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx
\end{equation*}

and since the eigenvalue equation for the position-squared operator in position space is \(\widehat{X^2}\phi(x)=x^2\phi(x)\), this is
\begin{align*}
\braket{x}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* x^2\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\int_{0}^a x^2\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int x^2 \sin^2{(cx)}dx=\frac{x^3}{6}-\left(\frac{x^2}{4c}-\frac{1}{8c^3}\right)\sin{(2cx)}-\frac{x\cos{(2cx)}}{4c^2}
\end{equation*}

with \(c=\pi/a\).

Evaluating the integral in the previous hint gives
\begin{align*}
\braket{x^2}&=\frac{2}{a}\int_{0}^a x^2\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\left[\frac{x^3}{6}-\left(\frac{x^2}{4\left(\frac{\pi}{a}\right)}-\frac{1}{8\left(\frac{\pi}{a}\right)^3}\right)\sin{\left(\frac{2\pi x}{a}\right)}-\frac{x\cos{\left(\frac{2\pi x}{a}\right)}}{4\left(\frac{\pi}{a}\right)^2} \right]\bigg|_0^a\\
&=\frac{2}{a}\left[\frac{a^3}{6}-\frac{a\cos{(2\pi)}}{4\left(\frac{\pi}{a}\right)^2}+\frac{(0)\cos{(0)}}{4\left(\frac{\pi}{a}\right)^2} \right]\\
&=\frac{2}{a}\left(\frac{a^3}{6}-\frac{a^3}{4\pi^2}\right)=a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right).
\end{align*}

To find the momentum-squared expectation value \(\braket{p^2}\), use:
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P^2}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P^2}_x\) represents the position-basis version of the momentum-squared operator.

Plugging the expression for \(\psi_1(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{p^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\widehat{P^2}_x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx
\end{equation*}

and the momentum-squared operator in position space is
\begin{equation*}
\widehat{P^2}_x=\left[\frac{\hbar}{i}\frac{\partial}{\partial x}\right]^2=-\hbar^2\frac{\partial^2}{\partial x^2},
\end{equation*}

so \(\braket{p^2}\) can be written as
\begin{equation*}
\braket{p^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}
\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}
\right]dx.
\end{equation*}

Taking the second derivative in the previous hint gives
\begin{align*}
\braket{p^2}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \left(\hbar^2 \frac{\pi^2}{a^2}\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx\\
&=\frac{2\pi^2 \hbar^2}{a^3}\int_0^a \left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int \sin^2{(cx)}dx=\frac{x}{2}-\frac{\sin{(2cx)}}{4c}
\end{equation*}

with \(c=\pi/a\).

Evaluating the integral in the previous hint gives
\begin{align*}
\braket{p^2}&=\frac{2\pi^2 \hbar^2}{a^3}\int_{0}^a \left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2\pi^2 \hbar^2}{a^3}\left[\frac{x}{2}-\frac{\sin{(\left(\frac{2\pi x}{a}\right))}}{4\left(\frac{\pi}{a}\right)}\right]\bigg|_0^a=\frac{2\pi^2 \hbar^2}{a^3}\left(\frac{a}{2}\right)\\
&=\left(\frac{\pi \hbar}{a}\right)^2.
\end{align*}

Part (a):

For an infinite rectangular potential well extending from \(x=0\) to \(x=a\), the wavefunctions are given by Eq. 5.9 as
\begin{equation*}
\psi_n(x)=\sqrt{\frac{2}{a}}\sin{\left(\frac{n\pi x}{a}\right)}
\end{equation*}

so the ground-state wavefunction (\(n=1\)) is
\begin{equation*}
\psi_1(x)=\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}.
\end{equation*}

The expectation value of position \(x\) is given by Eq. 2.60 as
\begin{equation*}
\braket{x}=\bra{\psi}\widehat{X}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X}\) is the position operator.

Plugging the expression for \(\psi_1(x)\) into this equation for expectation value yields
\begin{equation*}
\braket{x}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \widehat{X}\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx,
\end{equation*}

and since the eigenvalue equation for the position operator in position space is \(\widehat{X}\psi(x)=x\psi(x)\), this is
\begin{align*}
\braket{x}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\int_{0}^a x\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx,
\end{align*}

which can be evaluated using
\begin{equation*}
\int x \sin^2{(cx)}dx=\frac{x^2}{4}-\frac{x\sin{(2cx)}}{4c}-\frac{\cos{(2cx)}}{8c^2}
\end{equation*}

with \(c=\pi/a\).

Evaluating this integral gives
\begin{align*}
\braket{x}&=\frac{2}{a}\int_{0}^a x\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\left[\frac{x^2}{4}-\frac{x\sin{\left(\frac{2\pi x}{a}\right)}}{4\left(\frac{\pi}{a}\right)}-\frac{\cos{\left(\frac{2\pi x}{a}\right)}}{8\left(\frac{\pi}{a}\right)^2}\right]\bigg|_0^a\\
&=\frac{2}{a}\left[\frac{a^2}{4}-\frac{\cos{(2\pi)}}{8\left(\frac{\pi}{a}\right)^2}+\frac{\cos{(0)}}{8\left(\frac{\pi}{a}\right)^2}\right]\\
&=\frac{2}{a}\left(\frac{a^2}{4}\right)=\frac{a}{2}.
\end{align*}

Part (b):

To find the expectation value of momentum \(\braket{p}\), use Eq. 4.60:
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P}_x\) represents the position-basis version of the momentum operator.

Plugging the expression for \(\psi_1(x)\) into this equation for expectation value yields
\begin{equation*}
\braket{p}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\widehat{P}_x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx
\end{equation*}

and since the momentum operator in position space is given by Eq. 4.64 as
\begin{equation*}
\widehat{P}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}=-i\hbar\frac{\partial}{\partial x},
\end{equation*}

\(\braket{p}\) can be written as
\begin{equation*}
\braket{p}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\left(-i\hbar\frac{\partial}{\partial x}\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx.
\end{equation*}

Taking these derivatives gives
\begin{align*}
\braket{p}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \left(-i\hbar \frac{\pi}{a}\right)\left[\sqrt{\frac{2}{a}}\cos{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{-2\pi i\hbar}{a^2}\int_{0}^a \left[\sin{\left(\frac{\pi x}{a}\right)}\cos{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int \sin{(cx)}\cos{(cx)}dx=\frac{\sin^2{(cx)}}{2c}
\end{equation*}

with \(c=\pi/a\).

Evaluating this integral gives
\begin{align*}
\braket{p}&=\frac{-2\pi i\hbar}{a^2}\int_{0}^a \left[\sin{\left(\frac{\pi x}{a}\right)}\cos{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{-2\pi i\hbar}{a^2}\left[\frac{\sin^2{\left(\frac{\pi}{a}x\right)}}{2\frac{\pi}{a}}\right]\bigg|_0^a\\
&=\frac{-i\hbar}{a}\left[\sin^2{\pi}-\sin^2{0}\right]=0.
\end{align*}

Part (c):

The expectation value of the square of position \(x^2\) is given by
\begin{equation*}
\braket{x^2}=\bra{\psi}\widehat{X^2}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X^2}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X^2}\) is the position-squared operator.

Plugging the expression for \(\psi_1(x)\) into this equation for expectation value yields
\begin{equation*}
\braket{x^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \widehat{X^2}\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx
\end{equation*}

and since the eigenvalue equation for the position-squared operator in position space is \(\widehat{X^2}\phi(x)=x^2\phi(x)\), this is
\begin{align*}
\braket{x}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* x^2\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\int_{0}^a x^2\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int x^2 \sin^2{(cx)}dx=\frac{x^3}{6}-\left(\frac{x^2}{4c}-\frac{1}{8c^3}\right)\sin{(2cx)}-\frac{x\cos{(2cx)}}{4c^2}
\end{equation*}

with \(c=\pi/a\).\\

Evaluating this integral gives
\begin{align*}
\braket{x^2}&=\frac{2}{a}\int_{0}^a x^2\left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2}{a}\left[\frac{x^3}{6}-\left(\frac{x^2}{4\left(\frac{\pi}{a}\right)}-\frac{1}{8\left(\frac{\pi}{a}\right)^3}\right)\sin{\left(\frac{2\pi x}{a}\right)}-\frac{x\cos{\left(\frac{2\pi x}{a}\right)}}{4\left(\frac{\pi}{a}\right)^2} \right]\bigg|_0^a\\
&=\frac{2}{a}\left[\frac{a^3}{6}-\frac{a\cos{(2\pi)}}{4\left(\frac{\pi}{a}\right)^2}+\frac{(0)\cos{(0)}}{4\left(\frac{\pi}{a}\right)^2} \right]\\
&=\frac{2}{a}\left(\frac{a^3}{6}-\frac{a^3}{4\pi^2}\right)=a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right).
\end{align*}

Part (d):

To find the expectation value of momentum squared \(\braket{p^2}\), use:
\begin{equation*}
\braket{p^2}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P^2}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P^2}_x\) represents the position-basis version of the momentum-squared operator.

Plugging the expression for \(\psi_1(x)\) into this equation for expectation value yields
\begin{equation*}
\braket{p^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\widehat{P^2}_x\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx
\end{equation*}

since the momentum-squared operator in position space is
\begin{equation*}
\widehat{P^2}_x=\left[\frac{\hbar}{i}\frac{\partial}{\partial x}\right]^2=-\hbar^2\frac{\partial^2}{\partial x^2},
\end{equation*}

so \(\braket{p^2}\) can be written as
\begin{equation*}
\braket{p^2}=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}
\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}
\right]dx.
\end{equation*}

Taking this second derivative gives
\begin{align*}
\braket{p^2}&=\int_{0}^a \left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]^* \left(\hbar^2 \frac{\pi^2}{a^2}\right)\left[\sqrt{\frac{2}{a}}\sin{\left(\frac{\pi x}{a}\right)}\right]dx\\
&=\frac{2\pi^2 \hbar^2}{a^3}\int_0^a \left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx
\end{align*}

which can be evaluated using
\begin{equation*}
\int \sin^2{(cx)}dx=\frac{x}{2}-\frac{\sin{(2cx)}}{4c}
\end{equation*}

with \(c=\pi/a\).

Evaluating this integral gives
\begin{align*}
\braket{p^2}&=\frac{2\pi^2 \hbar^2}{a^3}\int_{0}^a \left[\sin^2{\left(\frac{\pi x}{a}\right)}\right] dx\\
&=\frac{2\pi^2 \hbar^2}{a^3}\left[\frac{x}{2}-\frac{\sin{(\left(\frac{2\pi x}{a}\right))}}{4\left(\frac{\pi}{a}\right)}\right]\bigg|_0^a=\frac{2\pi^2 \hbar^2}{a^3}\left(\frac{a}{2}\right)\\
&=\left(\frac{\pi \hbar}{a}\right)^2.
\end{align*}

The uncertainty in position is given by Eq. 2.62:

\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}
\end{equation*}

and the uncertainty in momentum is given by Eq. 2.64:

\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}.
\end{equation*}

The values for \(\braket{x}\) and \(\braket{x^2}\) from the previous problem are
\begin{equation*}
\braket{x}=\frac{a}{2}
\end{equation*}

and

\begin{equation*}
\braket{x^2}=a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right).
\end{equation*}

Plugging in the values for \(\braket{x^2}\) and \(\braket{x}\) gives the position uncertainty:
\begin{align*}
\Delta x&=\sqrt{\braket{x^2}-\braket{x}^2}=\sqrt{a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right)-\left(\frac{a}{2}\right)^2}\\
&=\sqrt{0.0327a^2}=0.181a.
\end{align*}

This is the reason that 18% of the well width was taken as the uncertainty in position for the ground state of the infinite rectangular well in Section 5.1.

The values for \(\braket{p}\) and \(\braket{p^2}\) from the previous problem are
\begin{equation*}
\braket{p}=0
\end{equation*}

and

\begin{equation*}
\braket{p^2}=\left(\frac{\pi \hbar}{a}\right)^2.
\end{equation*}

Plugging in the values for \(\braket{p^2}\) and \(\braket{p}\) gives the momentum uncertainty:
\begin{align*}
\Delta p&=\sqrt{\braket{p^2}-\braket{p}^2}\\
&=\sqrt{\left(\frac{\pi \hbar}{a}\right)^2-(0)^2}=\frac{\pi \hbar}{a}.
\end{align*}

Using the expressions for \(\Delta x\) and \(\Delta p\) from Hints 3 and 5 to form the product \(\Delta x \Delta p\) yields
\begin{equation*}
\Delta x \Delta p=0.181a\left(\frac{\pi \hbar}{a}\right)=0.57\hbar
\end{equation*}

which is greater than \(\hbar/2\), in accordance with the Heisenberg Uncertainty principle.

The uncertainty in position is given by Eq. 2.62:

\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}
\end{equation*}

and the uncertainty in momentum is given by Eq. 2.64:

\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}.
\end{equation*}

The values for \(\braket{x}\) and \(\braket{x^2}\) from the previous problem are
\begin{equation*}
\braket{x}=\frac{a}{2}
\end{equation*}

and

\begin{equation*}
\braket{x^2}=a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right).
\end{equation*}

Plugging in the values for \(\braket{x^2}\) and \(\braket{x}\) gives the position uncertainty:
\begin{align*}
\Delta x&=\sqrt{\braket{x^2}-\braket{x}^2}=\sqrt{a^2\left(\frac{1}{3}-\frac{1}{2\pi^2}\right)-\left(\frac{a}{2}\right)^2}\\
&=\sqrt{0.0327a^2}=0.181a.
\end{align*}

This is the reason that 18% of the well width was taken as the uncertainty in position for the ground state of the infinite rectangular well in Section 5.1.

The values for \(\braket{p}\) and \(\braket{p^2}\) from the previous problem are
\begin{equation*}
\braket{p}=0
\end{equation*}

and

\begin{equation*}
\braket{p^2}=\left(\frac{\pi \hbar}{a}\right)^2.
\end{equation*}

Plugging in the values for \(\braket{p^2}\) and \(\braket{p}\) gives the momentum uncertainty:
\begin{align*}
\Delta p&=\sqrt{\braket{p^2}-\braket{p}^2}\\
&=\sqrt{\left(\frac{\pi \hbar}{a}\right)^2-(0)^2}=\frac{\pi \hbar}{a}.
\end{align*}

Using the expressions for \(\Delta x\) and \(\Delta p\) from above to form the product \(\Delta x \Delta p\) yields
\begin{equation*}
\Delta x \Delta p=0.181a\left(\frac{\pi \hbar}{a}\right)=0.57\hbar
\end{equation*}

which is greater than \(\hbar/2\), in accordance with the Heisenberg Uncertainty principle.

As explained in Section 4.2, the possible energy measurement outcomes are the eigenvalues of the energy operator, and the probability of each outcome is proportional to the square of the magnitude of the wavefunction value \(c_n\).

The energy eigenvalues are given in Eq. 5.7:
\begin{equation*}
E_n=\frac{k_n^2\hbar^2}{2m}=\frac{n^2\pi^2\hbar^2}{2ma^2}
\end{equation*}

which means that the possible outcomes of an energy measurement are
\begin{align*}
E_1&=\frac{k_1^2\hbar^2}{2m}=\frac{(1)^2\pi^2\hbar^2}{2ma^2}\\
E_2&=\frac{k_2^2\hbar^2}{2m}=\frac{(2)^2\pi^2\hbar^2}{2ma^2}\\
E_3&=\frac{k_3^2\hbar^2}{2m}=\frac{(3)^2\pi^2\hbar^2}{2ma^2}.
\end{align*}

The \(c_n\) values for the component wavefunctions \(\psi_1(x)\), \(\psi_2(x)\), and \(\psi_3(x)\) can be read directly from the expression for \(\psi(x)\) given in the problem statement:
\begin{align*}
c_1&=\frac{1}{2}\\
c_2&=\frac{3i}{4}\\
c_3&=\frac{\sqrt{3}}{4}
\end{align*}

which makes the probability of each outcome
\begin{align*}
|c_1|^2&=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\\
|c_2|^2&=\left(\frac{3i}{4}\right)\left(\frac{-3i}{4}\right)=\frac{9}{16}\\
|c_3|^2&=\left(\frac{\sqrt{3}}{4}\right)\left(\frac{\sqrt{3}}{4}\right)=\frac{3}{16}.
\end{align*}

Note that the total probability is
\begin{equation*}
\frac{1}{4}+\frac{9}{16}+\frac{3}{16}=1.
\end{equation*}

The expectation value for energy is the sum of each possible outcome \(E_n\) times the probability of that outcome \(|c_n|^2\), as given by Eq. 5.13:
\begin{equation*}
\braket{E}=\sum_n |c_n|^2 E_n.
\end{equation*}

Plugging in the values for \(E_n\) and \(|c_n|^2\) gives
\begin{align*}
\braket{E}&=\sum_n |c_n|^2 E_n=E_1|c_1|^2+E_2|c_2|^2+E_3|c_3|^2\\
&=\frac{\pi^2\hbar^2}{2ma^2}\left(\frac{1}{4}\right)+\frac{4\pi^2\hbar^2}{2ma^2}\left(\frac{9}{16}\right)+\frac{9\pi^2\hbar^2}{2ma^2}\left(\frac{3}{16}\right)\\
&=\frac{\pi^2\hbar^2}{2ma^2}\left(\frac{1}{4}+\frac{36}{16}+\frac{27}{16}\right)=4.19\frac{\pi^2\hbar^2}{2ma^2}\\
&=4.19E_1.
\end{align*}

Part (a):

As explained in Section 4.2, the possible energy measurement outcomes are the eigenvalues of the energy operator, and the probability of each outcome is proportional to the square of the magnitude of the wavefunction value \(c_n\).

The energy eigenvalues are given in Eq. 5.7:
\begin{equation*}
E_n=\frac{k_n^2\hbar^2}{2m}=\frac{n^2\pi^2\hbar^2}{2ma^2}
\end{equation*}

which means that the possible outcomes of an energy measurement are
\begin{align*}
E_1&=\frac{k_1^2\hbar^2}{2m}=\frac{(1)^2\pi^2\hbar^2}{2ma^2}\\
E_2&=\frac{k_2^2\hbar^2}{2m}=\frac{(2)^2\pi^2\hbar^2}{2ma^2}\\
E_3&=\frac{k_3^2\hbar^2}{2m}=\frac{(3)^2\pi^2\hbar^2}{2ma^2}.
\end{align*}

The \(c_n\) values for the component wavefunctions \(\psi_1(x)\), \(\psi_2(x)\), and \(\psi_3(x)\) can be read directly from the expression for \(\psi(x)\) given in the problem statement:
\begin{align*}
c_1&=\frac{1}{2}\\
c_2&=\frac{3i}{4}\\
c_3&=\frac{\sqrt{3}}{4}
\end{align*}

which makes the probability of each outcome
\begin{align*}
|c_1|^2&=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\\
|c_2|^2&=\left(\frac{3i}{4}\right)\left(\frac{-3i}{4}\right)=\frac{9}{16}\\
|c_3|^2&=\left(\frac{\sqrt{3}}{4}\right)\left(\frac{\sqrt{3}}{4}\right)=\frac{3}{16}.
\end{align*}

Note that the total probability is
\begin{equation*}
\frac{1}{4}+\frac{9}{16}+\frac{3}{16}=1.
\end{equation*}

Part (b):

The expectation value for energy is the sum of each possible outcome \(E_n\) times the probability of that outcome \(|c_n|^2\), as given by Eq. 5.13:
\begin{equation*}
\braket{E}=\sum_n |c_n|^2 E_n
\end{equation*}

and plugging in the values for \(E_n\) and \(|c_n|^2\) gives
\begin{align*}
\braket{E}&=\sum_n |c_n|^2 E_n=E_1|c_1|^2+E_2|c_2|^2+E_3|c_3|^2\\
&=\frac{\pi^2\hbar^2}{2ma^2}\left(\frac{1}{4}\right)+\frac{4\pi^2\hbar^2}{2ma^2}\left(\frac{9}{16}\right)+\frac{9\pi^2\hbar^2}{2ma^2}\left(\frac{3}{16}\right)\\
&=\frac{\pi^2\hbar^2}{2ma^2}\left(\frac{1}{4}+\frac{36}{16}+\frac{27}{16}\right)=4.19\frac{\pi^2\hbar^2}{2ma^2}\\
&=4.19E_1.
\end{align*}

Recall from Section 5.1 that you can determine the probability of measuring the particle’s position to be within a specified region by integrating the probability density over that region. Specifically, to determine the probability of measuring the particle to be within a region of width \(\Delta x\) centered on position \(x_0\), you can use Eq. 5.15:
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}\frac{2}{a}\sin^2{\left(\frac{n\pi x}{a}\right)}dx.
\end{equation*}

For the first excited state (\(n=2\)) of the infinite rectangular well of width \(a\) centered on \(x=a/2\), the integral for determining the probability of finding a particle in the region between \(x=0.25a\) and \(x=0.75a\) (so \(x_0=0.5a\) and \(\Delta x=0.5a\)) is
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{2\pi x}{a}\right)}dx.
\end{equation*}

which can be evaluated using
\begin{equation*}
\int \sin^2{(cx)}dx=\frac{x}{2}-\frac{\sin{(2cx)}}{4c}
\end{equation*}

with \(c=2\pi/a\).

The probability integral is therefore
\begin{align*}
\textrm{Prob}&=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{2\pi x}{a}\right)}dx\\
&=\frac{2}{a}\left[\frac{x}{2}-\frac{\sin{\left(\frac{4\pi x}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\bigg|_{0.25a}^{0.75a}\\
&=\frac{2}{a}\left[\frac{(0.75a-0.25a}{2}-\frac{\sin{\left(\frac{4\pi(0.75a)}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}+\frac{\sin{\left(\frac{4\pi(0.25a)}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{\sin{\left(3\pi\right)}}{4\left(\frac{2\pi}{a}\right)}+\frac{\sin{\left(\pi\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left(\frac{0.5a}{2}\right)=0.5.
\end{align*}

For the second excited state \(n=3\), the probability integral is
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{3\pi x}{a}\right)}dx.
\end{equation*}

The same integral relation as in the \(n=2\) case can be used, but in this case \(c=3\pi/a\), which makes the probability integral
\begin{align*}
\textrm{Prob}&=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{3\pi x}{a}\right)}dx=\frac{2}{a}\left[\frac{x}{2}-\frac{\sin{\left(\frac{6\pi x}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\bigg|_{0.25a}^{0.75a}\\
&=\frac{2}{a}\left[\frac{(0.75a-0.25a}{2}-\frac{\sin{\left(\frac{6\pi(0.75a)}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}+\frac{\sin{\left(\frac{6\pi(0.25a)}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{\sin{\left(4.5\pi\right)}}{4\left(\frac{3\pi}{a}\right)}+\frac{\sin{\left(1.5\pi\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{1}{\left(\frac{12\pi}{a}\right)}+\frac{-1}{\left(\frac{12\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{2a}{12\pi}\right]\\
&=\frac{2}{a}\left(\frac{0.5a}{2}-\frac{a}{6\pi}\right)=0.5-0.106=0.394.
\end{align*}

Recall from Section 5.1 that you can determine the probability of measuring the particle’s position to be within a specified region by integrating the probability density over that region. Specifically, to determine the probability of measuring the particle to be within a region of width \(\Delta x\) centered on position \(x_0\), you can use Eq. 5.15:
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}\frac{2}{a}\sin^2{\left(\frac{n\pi x}{a}\right)}dx.
\end{equation*}

For the first excited state (\)n=2\)) of the infinite rectangular well of width \(a\) centered on \(x=a/2\), the integral for determining the probability of finding a particle in the region between \(x=0.25a\) and \(x=0.75a\) (so \(x_0=0.5a\) and \(\Delta x=0.5a\)) is
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{2\pi x}{a}\right)}dx.
\end{equation*}

which can be evaluated using
\begin{equation*}
\int \sin^2{(cx)}dx=\frac{x}{2}-\frac{\sin{(2cx)}}{4c}
\end{equation*}

with \(c=2\pi/a\).

The probability integral is therefore
\begin{align*}
\textrm{Prob}&=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{2\pi x}{a}\right)}dx\\
&=\frac{2}{a}\left[\frac{x}{2}-\frac{\sin{\left(\frac{4\pi x}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\bigg|_{0.25a}^{0.75a}\\
&=\frac{2}{a}\left[\frac{(0.75a-0.25a}{2}-\frac{\sin{\left(\frac{4\pi(0.75a)}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}+\frac{\sin{\left(\frac{4\pi(0.25a)}{a}\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{\sin{\left(3\pi\right)}}{4\left(\frac{2\pi}{a}\right)}+\frac{\sin{\left(\pi\right)}}{4\left(\frac{2\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left(\frac{0.5a}{2}\right)=0.5.
\end{align*}

For the second excited state \(n=3\), the probability integral is
\begin{equation*}
\int_{x_0-\Delta x/2}^{x_0+\Delta x/2}[\psi(x)]^*[\psi(x)]dx=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{3\pi x}{a}\right)}dx.
\end{equation*}

The same integral relation as in the \(n=2\) case can be used, but in this case \(c=3\pi/a\), which makes the probability \begin{align*}
\textrm{Prob}&=\int_{0.5a-0.25a}^{0.5a+0.25a}\frac{2}{a}\sin^2{\left(\frac{3\pi x}{a}\right)}dx=\frac{2}{a}\left[\frac{x}{2}-\frac{\sin{\left(\frac{6\pi x}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\bigg|_{0.25a}^{0.75a}\\
&=\frac{2}{a}\left[\frac{(0.75a-0.25a}{2}-\frac{\sin{\left(\frac{6\pi(0.75a)}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}+\frac{\sin{\left(\frac{6\pi(0.25a)}{a}\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{\sin{\left(4.5\pi\right)}}{4\left(\frac{3\pi}{a}\right)}+\frac{\sin{\left(1.5\pi\right)}}{4\left(\frac{3\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{1}{\left(\frac{12\pi}{a}\right)}+\frac{-1}{\left(\frac{12\pi}{a}\right)}\right]\\
&=\frac{2}{a}\left[\frac{0.5a}{2}-\frac{2a}{12\pi}\right]\\
&=\frac{2}{a}\left(\frac{0.5a}{2}-\frac{a}{6\pi}\right)=0.5-0.106=0.394.
\end{align*}

Eq. 5.16 says
\begin{equation*}
\tilde{\phi}(p)=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{equation*}
in which \(p_n=\hbar k_n\).

To get to this equation, start with the expression for \(\tilde{\phi}(p)\) given just above Eq. 5.16:
\begin{equation*}
\tilde{\phi}(p)=\frac{1}{\sqrt{\pi a \hbar}}\int_{0}^a \sin{\left(\frac{n\pi x}{a}\right)}e^{-ipx/\hbar}dx.
\end{equation*}

and use Euler’s relation to write the \(e^{-ipx/\hbar}\) term as \(\cos{(px/\hbar)}-i\sin{(px/\hbar)}\).

Since \(k_n=n\pi/a\) and \(p_n=\hbar k_n\), you can write the term \(\sin{\left(\frac{n\pi x}{a}\right)}\) as \(\sin{(k_n x)}=\sin{(p_n x/\hbar)}\).

The integral
\begin{equation*}
\tilde{\phi}(p)=\frac{1}{\sqrt{\pi a \hbar}}\int_{0}^a \sin{(p_n x/\hbar)}[\cos{(px/\hbar)}-i\sin{(px/\hbar)}]dx.
\end{equation*}
can be evaluated using
\begin{equation*}
\sin{(ax)}\cos{(bx)}=\frac{1}{2}\{\sin{[(a+b)x]}+\sin{[(a-b)x]}\}
\end{equation*}
and
\begin{equation*}
\sin{(ax)}\sin{(bx)}=\frac{1}{2}\{\cos{[(a-b)x]}-\cos{[(a+b)x]}\}.
\end{equation*}

The integrals
\begin{align*}
\tilde{\phi}(p)&=\frac{1}{\sqrt{\pi a \hbar}}\left[\int_{0}^a \sin{(p_n x/\hbar)}\cos{(px/\hbar)}dx\right.\\
&\hspace{2cm}\left.-i\int_{0}^a \sin{(p_n x/\hbar)}\sin{(px/\hbar)}dx\right]\\
&=\frac{1}{\sqrt{\pi a \hbar}}\left(\frac{1}{2}\right)\left[\int_{0}^a \sin{[(p_n/\hbar+p/\hbar)]x}dx\right.\\
&\hspace{2.5cm}\left.+\int_{0}^a \sin{[(p_n/\hbar-p/\hbar)]x}dx\right.\\
&\hspace{3cm}\left.-i\int_{0}^a \cos{[(p_n/\hbar-p/\hbar)]x}dx\right.\\
&\hspace{3.5cm}\left.+i\int_{0}^a \cos{[(p_n/\hbar+p/\hbar)]x}dx\right]
\end{align*}
evaluate to
\begin{align*}
\tilde{\phi}(p)&=\frac{1}{2\sqrt{\pi a \hbar}}\left[-\frac{\cos{[(p_n+p)x/\hbar]}}{(p_n+p)/\hbar}\right|_0^a-\left.\frac{\cos{[(p_n-p)x/\hbar]}}{(p_n-p)/\hbar}\right|_0^a\\
&\left.\left.-i\frac{\sin{[(p_n-p)x/\hbar]}}{(p_n-p)/\hbar}\right|_0^a+\left.i\frac{\sin{[(p_n+p)x/\hbar]}}{(p_n+p)/\hbar}\right|_0^a\right].
\end{align*}
or, plugging in the limits,
\begin{align*}
\tilde{\phi}(p)&=\frac{\hbar}{2\sqrt{\pi a \hbar}}\left[-\frac{\cos{[(p_n+p)a/\hbar]}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\left.-\frac{\cos{[(p_n-p)a/\hbar]}}{p_n-p}+\frac{1}{p_n-p}\right.\\
&\left.-i\frac{\sin{[(p_n-p)a/\hbar]}}{p_n-p}+i\frac{\sin{[(p_n+p)a/\hbar]}}{p_n+p}\right].
\end{align*}

Group the terms of the final expression of the previous hint as
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{\cos{[(p_n+p)a/\hbar]}}{p_n+p}+i\frac{\sin{[(p_n+p)a/\hbar]}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\hspace{1cm}\left.-\frac{\cos{[(p_n-p)a/\hbar]}}{p_n-p}-i\frac{\sin{[(p_n-p)a/\hbar]}}{p_n-p}+\frac{1}{p_n-p}\right]
\end{align*}
and use Euler’s relation to make this
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\hspace{1cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}+\frac{1}{p_n-p}\right].
\end{align*}

Write the two non-exponential terms in the final expression of the previous hint over a common denominator:
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{p_n-p}{p_n^2-p^2}\right.\\
&\hspace{2cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}+\frac{p_n+p}{p_n^2-p^2}\right].\\
&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{2p_n}{p_n^2-p^2}\right.\\
&\hspace{2cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}-\right]
\end{align*}
So
\begin{equation*}
\tilde{\phi}(p)=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{equation*}

in agreement with Eq. 5.16.

The probability density in momentum space \(P_{den}(p)\) is given by Eq. 5.17 as
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{2p_n^2}{(p_n^2-p^2)^2}[1-(-1)^n\cos{(pa/\hbar)}].
\end{equation*}
As indicated in the text, you can get to this expression by multiplying the expression for the momentum-basis wavefunction \(\tilde{\phi}(p)\) given in Eq. 5.16 by its complex conjugate.

The complex conjugate of the momentum-basis wavefunction \(\tilde{\phi}(p)\) given in Eq. 5.16 is
\begin{equation*}
[\tilde{\phi}(p)]^*=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right].
\end{equation*}

Multiplying \(\tilde{\phi}(p)\) by \([\tilde{\phi}(p)]^*\) looks like this:
\begin{align*}
P_{den}(p)&=\tilde{\phi}(p)*[\tilde{\phi}(p)]^*\\
&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]\\
&\hspace{2 cm}*\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{align*}
which can be multiplied term by term.

Multiplying the three-term expressions shown in the previous hint gives nine terms:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left[\frac{4p_n^2}{(p_n^2-p^2)^2}-\left(\frac{2p_n}{p_n^2-p^2}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left. -\left(\frac{2p_n}{p_n^2-p^2}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right) \right.\\
&\left.+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{2p_n}{p_n^2-p^2}\right)+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left.+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right)\right.\\
&\left.+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{2p_n}{p_n^2-p^2}\right)+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left.+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right)\right.
\end{align*}

To simplify the nine-term expression shown in the previous hint, multiply the exponentials and gather terms with common factors:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{2cm}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}+\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{e^{-i(2p_n)a/\hbar}}{p_n^2-p^2}+\frac{e^{i(2p_n)a/\hbar}}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\}.
\end{align*}

Use the inverse Euler relation for cosines to convert the four terms inside the square brackets into two cosines and the two exponential terms outside the square brackets into another cosine:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{2\cos{[(p_n+p)a/\hbar]}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{6cm}+\frac{2\cos{[(p_n-p)a/\hbar]}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{2\cos{[(2p_n)a/\hbar]}}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\}.
\end{align*}

Recall that \(\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}\) and that \(p_n=\hbar k_n\) and \(k_n=n\pi/a\), which means that
\begin{align*}
\cos{\left(\frac{p_n a}{\hbar}\right)}&=\cos{\left(\frac{\hbar k_n a}{\hbar}\right)}=\cos{\left[\left(\frac{\hbar n\pi}{a}\right)\left(\frac{a}{\hbar}\right)\right]}=\cos{(n\pi)}\\
&=(-1)^n
\end{align*}
and
\begin{align*}
\sin{\left(\frac{p_n a}{\hbar}\right)}&=\sin{\left(\frac{\hbar k_n a}{\hbar}\right)}=\sin{\left[\left(\frac{\hbar n\pi}{a}\right)\left(\frac{a}{\hbar}\right)\right]}=\sin{(n\pi)}\\
&=0.
\end{align*}
So
\begin{align*}
\cos{\left[\frac{(p_n-p)a}{\hbar}\right]}&=\cos{\left(\frac{p_na}{\hbar}\right)}\cos{\left(-\frac{pa}{\hbar}\right)}-\sin{\left(\frac{p_na}{\hbar}\right)}\sin{\left(-\frac{pa}{\hbar}\right)}\\
&=(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}
\end{align*}
and
\begin{equation*}
\cos{\left(\frac{2p_n a}{\hbar}\right)}=\cos{(2n\pi)}=1.
\end{equation*}

One way to simplify the expression
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{2(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{6cm}+\frac{2(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{2}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\},
\end{align*}
is to pull a factor of \(\frac{4p_n^2}{(p_n^2-p^2)^2}\) out of each term, which gives
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\frac{4p_n^2}{(p_n^2-p^2)^2}\left\{1-\frac{p_n^2-p^2}{p_n}\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{1}{p_n+p}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{1}{p_n-p}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{p_n^2-p^2}{(p_n+p)^2}+\frac{p_n^2-p^2}{(p_n-p)^2}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}.
\end{align*}

Writing \(p_n^2-p^2\) as \((p_n-p)(p_n+p)\) makes the expression from the previous hint look like this:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{p_n-p}{p_n}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{p_n+p}{p_n}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{p_n-p}{p_n+p}+\frac{p_n+p}{p_n-p}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}
\end{align*}
or
\begin{align*}
P_{den}(p)&=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{p_n-p}{p_n}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{p_n+p}{p_n}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{(p_n-p)^2}{p_n^2-p^2}+\frac{(p_n+p)^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}.
\end{align*}

The expression in the previous hint may be simplified using
\begin{equation*}
\frac{p_n-p}{p_n}+\frac{p_n+p}{p_n}=\frac{p_n+p_n}{p_n}=2,
\end{equation*}
and the terms
\begin{equation*}
\frac{p_n^2-p^2}{4p_n^2}\left[\frac{(p_n-p)^2}{p_n^2-p^2}+\frac{(p_n+p)^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}
\end{equation*}
simplify to
\begin{equation*}
\frac{p_n^2-p^2}{4p_n^2}\left[\frac{2p_n^2+2p^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}=\frac{p_n^2+p^2}{2p_n^2}+\frac{p_n^2-p^2}{2p_n^2}=1,
\end{equation*}
so
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)(2)}\right]+1\right\}.
\end{equation*}
or
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{2p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\right]\right\}.
\end{equation*}

in agreement with Eq. 5.17.

Part (a):

Eq. 5.16 says
\begin{equation*}
\tilde{\phi}(p)=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{equation*}
in which \(p_n=\hbar k_n\).

To get to this equation, start with the expression for \(\tilde{\phi}(p)\) given just above Eq. 5.16:
\begin{equation*}
\tilde{\phi}(p)=\frac{1}{\sqrt{\pi a \hbar}}\int_{0}^a \sin{\left(\frac{n\pi x}{a}\right)}e^{-ipx/\hbar}dx.
\end{equation*}

and use Euler’s relation to write the \(e^{-ipx/\hbar}\) term as \(\cos{(px/\hbar)}-i\sin{(px/\hbar)}\).

Since \(k_n=n\pi/a\) and \(p_n=\hbar k_n\), you can write the term \(\sin{\left(\frac{n\pi x}{a}\right)}\) as \(\sin{(k_n x)}=\sin{(p_n x/\hbar)}\).

The integral
\begin{equation*}
\tilde{\phi}(p)=\frac{1}{\sqrt{\pi a \hbar}}\int_{0}^a \sin{(p_n x/\hbar)}[\cos{(px/\hbar)}-i\sin{(px/\hbar)}]dx.
\end{equation*}
can be evaluated using
\begin{equation*}
\sin{(ax)}\cos{(bx)}=\frac{1}{2}\{\sin{[(a+b)x]}+\sin{[(a-b)x]}\}
\end{equation*}
and
\begin{equation*}
\sin{(ax)}\sin{(bx)}=\frac{1}{2}\{\cos{[(a-b)x]}-\cos{[(a+b)x]}\}.
\end{equation*}

The integrals
\begin{align*}
\tilde{\phi}(p)&=\frac{1}{\sqrt{\pi a \hbar}}\left[\int_{0}^a \sin{(p_n x/\hbar)}\cos{(px/\hbar)}dx\right.\\
&\hspace{2cm}\left.-i\int_{0}^a \sin{(p_n x/\hbar)}\sin{(px/\hbar)}dx\right]\\
&=\frac{1}{\sqrt{\pi a \hbar}}\left(\frac{1}{2}\right)\left[\int_{0}^a \sin{[(p_n/\hbar+p/\hbar)]x}dx\right.\\
&\hspace{2.5cm}\left.+\int_{0}^a \sin{[(p_n/\hbar-p/\hbar)]x}dx\right.\\
&\hspace{3cm}\left.-i\int_{0}^a \cos{[(p_n/\hbar-p/\hbar)]x}dx\right.\\
&\hspace{3.5cm}\left.+i\int_{0}^a \cos{[(p_n/\hbar+p/\hbar)]x}dx\right]
\end{align*}
evaluate to
\begin{align*}
\tilde{\phi}(p)&=\frac{1}{2\sqrt{\pi a \hbar}}\left[-\frac{\cos{[(p_n+p)x/\hbar]}}{(p_n+p)/\hbar}\right|_0^a-\left.\frac{\cos{[(p_n-p)x/\hbar]}}{(p_n-p)/\hbar}\right|_0^a\\
&\left.\left.-i\frac{\sin{[(p_n-p)x/\hbar]}}{(p_n-p)/\hbar}\right|_0^a+\left.i\frac{\sin{[(p_n+p)x/\hbar]}}{(p_n+p)/\hbar}\right|_0^a\right].
\end{align*}
or, plugging in the limits,
\begin{align*}
\tilde{\phi}(p)&=\frac{\hbar}{2\sqrt{\pi a \hbar}}\left[-\frac{\cos{[(p_n+p)a/\hbar]}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\left.-\frac{\cos{[(p_n-p)a/\hbar]}}{p_n-p}+\frac{1}{p_n-p}\right.\\
&\left.-i\frac{\sin{[(p_n-p)a/\hbar]}}{p_n-p}+i\frac{\sin{[(p_n+p)a/\hbar]}}{p_n+p}\right].
\end{align*}

Now group the terms of this expression as
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{\cos{[(p_n+p)a/\hbar]}}{p_n+p}+i\frac{\sin{[(p_n+p)a/\hbar]}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\hspace{1cm}\left.-\frac{\cos{[(p_n-p)a/\hbar]}}{p_n-p}-i\frac{\sin{[(p_n-p)a/\hbar]}}{p_n-p}+\frac{1}{p_n-p}\right]
\end{align*}
and use Euler’s relation to make this
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{1}{p_n+p}\right.\\
&\hspace{1cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}+\frac{1}{p_n-p}\right].
\end{align*}

Next, write the two non-exponential terms in the final expression of the previous hint over a common denominator:
\begin{align*}
\tilde{\phi}(p)&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{p_n-p}{p_n^2-p^2}\right.\\
&\hspace{2cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}+\frac{p_n+p}{p_n^2-p^2}\right].\\
&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{2p_n}{p_n^2-p^2}\right.\\
&\hspace{2cm}\left.-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}-\right].
\end{align*}
So
\begin{equation*}
\tilde{\phi}(p)=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{equation*}

in agreement with Eq. 5.16.

Part (b):

The probability density in momentum space \(P_{den}(p)\) is given by Eq. 5.17 as
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{2p_n^2}{(p_n^2-p^2)^2}[1-(-1)^n\cos{(pa/\hbar)}].
\end{equation*}
As indicated in the text, you can get to this expression by multiplying the expression for the momentum-basis wavefunction \(\tilde{\phi}(p)\) given in Eq. 5.16 by its complex conjugate.

The complex conjugate of the momentum-basis wavefunction \(\tilde{\phi}(p)\) given in Eq. 5.16 is
\begin{equation*}
[\tilde{\phi}(p)]^*=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right].
\end{equation*}
Multiplying \(\tilde{\phi}(p)\) by \([\tilde{\phi}(p)]^*\) looks like this:
\begin{align*}
P_{den}(p)&=\tilde{\phi}(p)*[\tilde{\phi}(p)]^*\\
&=\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]\\
&\hspace{2 cm}*\frac{\sqrt{\hbar}}{2\sqrt{\pi a}}\left[\frac{2p_n}{p_n^2-p^2}-\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right]
\end{align*}
which can be multiplied term by term to give nine terms:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left[\frac{4p_n^2}{(p_n^2-p^2)^2}-\left(\frac{2p_n}{p_n^2-p^2}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left. -\left(\frac{2p_n}{p_n^2-p^2}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right) \right.\\
&\left.+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{2p_n}{p_n^2-p^2}\right)+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left.+\left(\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right)\right.\\
&\left.+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{2p_n}{p_n^2-p^2}\right)+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right)\right.\\
&\hspace{4cm}\left.+\left(\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right)\left(\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}\right)\right.
\end{align*}

To simplify this nine-term expression, multiply the exponentials and gather terms with common factors:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{e^{i(p_n+p)a/\hbar}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{2cm}-\frac{e^{-i(p_n-p)a/\hbar}}{p_n-p}+\frac{e^{-i(p_n+p)a/\hbar}}{p_n+p}+\frac{e^{i(p_n-p)a/\hbar}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{e^{-i(2p_n)a/\hbar}}{p_n^2-p^2}+\frac{e^{i(2p_n)a/\hbar}}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\}.
\end{align*}

Now use the inverse Euler relation for cosines to convert the four terms inside the square brackets into two cosines and the two exponential terms outside the square brackets into another cosine:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{2\cos{[(p_n+p)a/\hbar]}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{6cm}+\frac{2\cos{[(p_n-p)a/\hbar]}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{2\cos{[(2p_n)a/\hbar]}}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\}.
\end{align*}

Recall that \(\cos{(a+b)}=\cos{a}\cos{b}-\sin{a}\sin{b}\) and that \(p_n=\hbar k_n\) and \(k_n=n\pi/a\), which means that
\begin{align*}
\cos{\left(\frac{p_n a}{\hbar}\right)}&=\cos{\left(\frac{\hbar k_n a}{\hbar}\right)}=\cos{\left[\left(\frac{\hbar n\pi}{a}\right)\left(\frac{a}{\hbar}\right)\right]}=\cos{(n\pi)}\\
&=(-1)^n
\end{align*}
and
\begin{align*}
\sin{\left(\frac{p_n a}{\hbar}\right)}&=\sin{\left(\frac{\hbar k_n a}{\hbar}\right)}=\sin{\left[\left(\frac{\hbar n\pi}{a}\right)\left(\frac{a}{\hbar}\right)\right]}=\sin{(n\pi)}\\
&=0.
\end{align*}
So
\begin{align*}
\cos{\left[\frac{(p_n-p)a}{\hbar}\right]}&=\cos{\left(\frac{p_na}{\hbar}\right)}\cos{\left(-\frac{pa}{\hbar}\right)}-\sin{\left(\frac{p_na}{\hbar}\right)}\sin{\left(-\frac{pa}{\hbar}\right)}\\
&=(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}
\end{align*}
and
\begin{equation*}
\cos{\left(\frac{2p_n a}{\hbar}\right)}=\cos{(2n\pi)}=1.
\end{equation*}
One way to simplify the expression
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\left\{\frac{4p_n^2}{(p_n^2-p^2)^2}
-\left(\frac{2p_n}{p_n^2-p^2}\right)\left[\frac{2(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}}{p_n+p}\right.\right.\\
&\left.\left.\hspace{6cm}+\frac{2(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}}{p_n-p}\right]\right.\\
&\left.\hspace{1.5cm}+\frac{1}{(p_n+p)^2}+\frac{2}{p_n^2-p^2}+\frac{1}{(p_n-p)^2}\right\},
\end{align*}
is to pull a factor of \(\frac{4p_n^2}{(p_n^2-p^2)^2}\) out of each term, which gives
\begin{align*}
P_{den}(p)&=\frac{\hbar}{4\pi a}\frac{4p_n^2}{(p_n^2-p^2)^2}\left\{1-\frac{p_n^2-p^2}{p_n}\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{1}{p_n+p}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{1}{p_n-p}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{p_n^2-p^2}{(p_n+p)^2}+\frac{p_n^2-p^2}{(p_n-p)^2}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}.
\end{align*}

Writing \(p_n^2-p^2\) as \((p_n-p)(p_n+p)\) makes this expression look like this:
\begin{align*}
P_{den}(p)&=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{p_n-p}{p_n}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{p_n+p}{p_n}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{p_n-p}{p_n+p}+\frac{p_n+p}{p_n-p}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}
\end{align*}
or
\begin{align*}
P_{den}(p)&=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\left(\frac{p_n-p}{p_n}\right.\right.\right.\\
&\hspace{8cm}\left.\left.\left.+\frac{p_n+p}{p_n}\right)\right]\right.\\
&\left.\hspace{1.5cm}+\frac{p_n^2-p^2}{4p_n^2}\left[\frac{(p_n-p)^2}{p_n^2-p^2}+\frac{(p_n+p)^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}\right\}.
\end{align*}

This expression may be simplified using
\begin{equation*}
\frac{p_n-p}{p_n}+\frac{p_n+p}{p_n}=\frac{p_n+p_n}{p_n}=2,
\end{equation*}
and the terms
\begin{equation*}
\frac{p_n^2-p^2}{4p_n^2}\left[\frac{(p_n-p)^2}{p_n^2-p^2}+\frac{(p_n+p)^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}
\end{equation*}
simplify to
\begin{equation*}
\frac{p_n^2-p^2}{4p_n^2}\left[\frac{2p_n^2+2p^2}{p_n^2-p^2}\right]+\frac{p_n^2-p^2}{2p_n^2}=\frac{p_n^2+p^2}{2p_n^2}+\frac{p_n^2-p^2}{2p_n^2}=1,
\end{equation*}
so
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)(2)}\right]+1\right\}.
\end{equation*}
or
\begin{equation*}
P_{den}(p)=\frac{\hbar}{\pi a}\frac{2p_n^2}{(p_n^2-p^2)^2}\left\{1-\left[(-1)^n\cos{\left(\frac{pa}{\hbar}\right)}\right]\right\}.
\end{equation*}

in agreement with Eq. 5.17.

The wavefunctions for a quantum harmonic oscillator are given by Eq. 5.97 as
\begin{equation*}
\psi_n(\xi)=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{2^n n!}}\right)H_n(\xi)e^{-\frac{\xi^2}{2}}.
\end{equation*}

so the ground-state wavefunction \(n=0\) is
\begin{align*}
\psi_0(x)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(1)(1)}}\right)(1)e^{-\frac{m\omega}{2\hbar}x^2}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}.
\end{align*}

since \(0!=1\) and \(H_0(\xi)=1\), and \(\xi^2=\frac{m\omega}{\hbar}x^2\).

The expectation value of position \(x\) is given by Eq. 2.60 as
\begin{equation*}
\braket{x}=\bra{\psi}\widehat{X}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X}\) is the position operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{x}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{X}\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the eigenvalue equation for the position operator in position space is \(\widehat{X}\psi(x)=x\psi(x)\), this is
\begin{align*}
\braket{x}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* x\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x\left[e^{-\frac{m\omega}{\hbar}x^2}\right]dx
\end{align*}

which can be evaluated by noting that \(x\) is an odd function and \(e^{-\frac{m\omega}{\hbar}x^2}\) is an even function, so the product of these two functions is odd, and integrating an odd function over symmetric limits (\(-\infty\) to \(\infty\) in this case) gives zero. Hence
\begin{equation*}
\braket{x}=0.
\end{equation*}

To find the expectation value of momentum \(\braket{p}\), use Eq. 4.60:
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P}_x\) represents the position-basis version of the momentum operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{P}_x\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the momentum operator in position space is given by Eq. 4.64 as
\begin{equation*}
\widehat{P}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}=-i\hbar\frac{\partial}{\partial x},
\end{equation*}

\(\braket{p}\) can be written as
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-i\hbar\frac{\partial}{\partial x}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx.
\end{equation*}

Taking the derivative in the previous hint gives
\begin{align*}
\braket{p}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-i\hbar\right)\left(-\frac{2xm\omega}{2\hbar}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(i\hbar\right)\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\left(\frac{m\omega}{\hbar}\right)\int_{-\infty}^\infty xe^{-\frac{m\omega}{\hbar}x^2} dx\\
&=\left(\frac{i\hbar}{\sqrt{\pi}}\right)\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty xe^{-\frac{m\omega}{\hbar}x^2} dx.
\end{align*}

As in the case of \(\braket{x}\), this can be evaluated by noting that \(x\) is an odd function and \(e^{-\frac{m\omega}{\hbar}x^2}\) is an even function, so the product of these two functions is odd, and integrating an odd function over symmetric limits (\(-\infty\) to \(\infty\) in this case) gives zero. Hence
\begin{equation*}
\braket{p}=0.
\end{equation*}

The expectation value of the square of position \(x^2\) is given by
\begin{equation*}
\braket{x^2}=\bra{\psi}\widehat{X^2}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X^2}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X^2}\) is the position-squared operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{x^2}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{X^2}\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the eigenvalue equation for the position-squared operator in position space is \(\widehat{X^2}\psi(x)=x^2\psi(x)\), this is
\begin{align*}
\braket{x^2}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*x^2\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx
\end{align*}

which can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty x^2e^{-cx^2}dx=\frac{1}{2c}\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\).

Evaluating the integral in the previous hint gives
\begin{align*}
\braket{x^2}&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\frac{1}{2\left(\frac{m\omega}{\hbar}\right)}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}\\
&=\frac{1}{2}\frac{\hbar}{m\omega}.
\end{align*}

The expectation value of momentum squared \(\braket{p^2}\) can be found using:
\begin{equation*}
\braket{p^2}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P^2}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P^2}_x\) represents the position-basis version of the momentum-squared operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given in the previous hint yields
\begin{equation*}
\braket{p^2}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\widehat{P^2}_x\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]dx
\end{equation*}

and the momentum-squared operator in position space is
\begin{equation*}
\widehat{P^2}_x=\left[\frac{\hbar}{i}\frac{\partial}{\partial x}\right]^2=-\hbar^2\frac{\partial^2}{\partial x^2},
\end{equation*}

so \(\braket{p^2}\) can be written as
\begin{align*}
\braket{p^2}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty \left[e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\left[e^{-\frac{m\omega}{2\hbar}x^2}\right]dx.
\end{align*}

The second derivative in the previous hint is
\begin{align*}
\frac{\partial^2}{\partial x^2}\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)&=\frac{\partial}{\partial x}\left[\left(-\frac{m\omega}{2\hbar}\right)(2x)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\right]\\
&=\left(-\frac{m\omega}{2\hbar}\right)\left[(2)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)-\left(\frac{m\omega}{2\hbar}\right)(2x)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\right]\\
&=\left(-\frac{m\omega}{2\hbar}\right)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[2-\left(\frac{m\omega}{2\hbar}\right)(2x)^2\right]\\
&=\left(-\frac{m\omega}{\hbar}\right)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[1-\left(\frac{m\omega}{\hbar}\right)x^2\right]\\
&=\left(\frac{m\omega}{\hbar}\right)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[-\frac{\hbar}{m\omega}+x^2\right].
\end{align*}

Inserting the expression for the second derivative from the previous hint into the integral in Hint 2d gives
\begin{align*}
\braket{p^2}&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{2\hbar}x^2}\left[-\hbar^2\left(\frac{m\omega}{\hbar}\right)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left(x^2-\frac{\hbar}{m\omega}\right)\right]dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}\left[-\hbar^2\left(\frac{m\omega}{\hbar}\right)^2\left(x^2-\frac{\hbar}{m\omega}\right)\right]dx\\
&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}\left(x^2-\frac{\hbar}{m\omega}\right)dx\\
&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx+\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx.
\end{align*}

The first integral in the previous hint can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty x^2e^{-cx^2}dx=\frac{1}{2c}\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\), which gives
\begin{align*}
\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\frac{1}{2\left(\frac{m\omega}{\hbar}\right)}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}\\
&=-m\omega \hbar/2.
\end{align*}

The second integral in the previous hint can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty e^{-cx^2}dx=\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\), which gives
\begin{equation*}
\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx=\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}=m\omega \hbar.
\end{equation*}

Thus \(\braket{p^2}=-\frac{m\omega\hbar}{2}+m\omega\hbar=\frac{m\omega\hbar}{2}\).

The wavefunctions for a quantum harmonic oscillator are given by Eq. 5.97 as
\begin{equation*}
\psi_n(\xi)=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{2^n n!}}\right)H_n(\xi)e^{-\frac{\xi^2}{2}}.
\end{equation*}

so the ground-state wavefunction \(n=0\) is
\begin{align*}
\psi_0(x)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(1)(1)}}\right)(1)e^{-\frac{m\omega}{2\hbar}x^2}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}
\end{align*}

since \(0!=1\) and \(H_0(\xi)=1\), and \(\xi^2=\frac{m\omega}{\hbar}x^2\).

The expectation value of position \(x\) is given by Eq. 2.60 as
\begin{equation*}
\braket{x}=\bra{\psi}\widehat{X}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X}\) is the position operator.

Plugging the expression for \(\psi_0(x)\) into this equation for the expectation value yields
\begin{equation*}
\braket{x}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{X}\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the eigenvalue equation for the position operator in position space is \(\widehat{X}\psi(x)=x\psi(x)\), this is
\begin{align*}
\braket{x}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{X}\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x\left[e^{-\frac{m\omega}{\hbar}x^2}\right]dx
\end{align*}

which can be evaluated by noting that \(x\) is an odd function and \(e^{-\frac{m\omega}{\hbar}x^2}\) is an even function, so the product of these two functions is odd, and integrating an odd function over symmetric limits (\(-\infty\) to \(\infty\) in this case) gives zero. Hence
\begin{equation*}
\braket{x}=0.
\end{equation*}

To find the expectation value of momentum \(\braket{p}\), use Eq. 4.60:
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P}_x\) represents the position-basis version of the momentum operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given above yields
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{P}_x\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the momentum operator in position space is given by Eq. 4.64 as
\begin{equation*}
\widehat{P}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}=-i\hbar\frac{\partial}{\partial x},
\end{equation*}

\(\braket{p}\) can be written as
\begin{equation*}
\braket{p}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-i\hbar\frac{\partial}{\partial x}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx.
\end{equation*}

Taking this derivative gives
\begin{align*}
\braket{p}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-i\hbar\right)\left(-\frac{2xm\omega}{2\hbar}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(i\hbar\right)\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\left(\frac{m\omega}{\hbar}\right)\int_{-\infty}^\infty xe^{-\frac{m\omega}{\hbar}x^2} dx\\
&=\left(\frac{i\hbar}{\sqrt{\pi}}\right)\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty xe^{-\frac{m\omega}{\hbar}x^2} dx.
\end{align*}

As in the case of \(\braket{x}\), this can be evaluated by noting that \(x\) is an odd function and \(e^{-\frac{m\omega}{\hbar}x^2}\) is an even function, so the product of these two functions is odd, and integrating an odd function over symmetric limits \(-\infty\) to \(\infty\) in this case) gives zero. Hence
\begin{equation*}
\braket{p}=0.
\end{equation*}

The expectation value of the square of position \(x^2\) is given by
\begin{equation*}
\braket{x^2}=\bra{\psi}\widehat{X^2}\ket{\psi}=\int_{-\infty}^\infty [\psi(x)]^* \widehat{X^2}[\psi(x)] dx
\end{equation*}

in which \(\widehat{X^2}\) is the position-squared operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given above yields
\begin{equation*}
\braket{x^2}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^* \widehat{X^2}\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx
\end{equation*}

and since the eigenvalue equation for the position-squared operator in position space is \(\widehat{X^2}\psi(x)=x^2\psi(x)\), this is
\begin{align*}
\braket{x^2}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*x^2\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right] dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx
\end{align*}

which can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty x^2e^{-cx^2}dx=\frac{1}{2c}\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\).

Evaluating the integral for \(\braket{x^2}\) gives
\begin{align*}
\braket{x^2}&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\frac{1}{2\left(\frac{m\omega}{\hbar}\right)}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}\\
&=\frac{1}{2}\frac{\hbar}{m\omega}.
\end{align*}

To find the expectation value of momentum squared \(\braket{p^2}\), use:
\begin{equation*}
\braket{p^2}=\int_{-\infty}^\infty [\psi(x)]^*\widehat{P^2}_x[\psi(x)]dx
\end{equation*}
in which \(\widehat{P^2}_x\) represents the position-basis version of the momentum-squared operator.

Plugging the expression for \(\psi_0(x)\) into the equation for expectation value given above yields
\begin{equation*}
\braket{p^2}=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\widehat{P^2}_x\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]dx
\end{equation*}

and the momentum-squared operator in position space is
\begin{equation*}
\widehat{P^2}_x=\left[\frac{\hbar}{i}\frac{\partial}{\partial x}\right]^2=-\hbar^2\frac{\partial^2}{\partial x^2},
\end{equation*}

so \(\braket{p^2}\) can be written as
\begin{align*}
\braket{p^2}&=\int_{-\infty}^\infty \left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{m\omega}{2\hbar}x^2}\right]dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty \left[e^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\left[e^{-\frac{m\omega}{2\hbar}x^2}\right]dx.
\end{align*}

This second derivative is
\begin{align*}
\frac{\partial^2}{\partial x^2}\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)&=\frac{\partial}{\partial x}\left[\left(-\frac{m\omega}{2\hbar}\right)(2x)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\right]\\
&=\left(-\frac{m\omega}{2\hbar}\right)\left[(2)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)-\left(\frac{m\omega}{2\hbar}\right)(2x)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\right]\\
&=\left(-\frac{m\omega}{2\hbar}\right)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[2-\left(\frac{m\omega}{2\hbar}\right)(2x)^2\right]\\
&=\left(-\frac{m\omega}{\hbar}\right)\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[1-\left(\frac{m\omega}{\hbar}\right)x^2\right]\\
&=\left(\frac{m\omega}{\hbar}\right)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left[-\frac{\hbar}{m\omega}+x^2\right].
\end{align*}

Inserting this expression for the second derivative into the integral for \(\braket{p^2}\) gives
\begin{align*}
\braket{p^2}&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{2\hbar}x^2}\left[-\hbar^2\left(\frac{m\omega}{\hbar}\right)^2\left(e^{-\frac{m\omega}{2\hbar}x^2}\right)\left(x^2-\frac{\hbar}{m\omega}\right)\right]dx\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}\left[-\hbar^2\left(\frac{m\omega}{\hbar}\right)^2\left(x^2-\frac{\hbar}{m\omega}\right)\right]dx\\
&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}\left(x^2-\frac{\hbar}{m\omega}\right)dx\\
&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx+\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx.
\end{align*}

The first integral in this expression can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty x^2e^{-cx^2}dx=\frac{1}{2c}\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\), which gives
\begin{align*}
\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\int_{-\infty}^\infty x^2e^{-\frac{m\omega}{\hbar}x^2}dx&=\frac{-\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{5}{2}\frac{1}{2\left(\frac{m\omega}{\hbar}\right)}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}\\
&=-m\omega \hbar/2.
\end{align*}

The second integral the expression for \(\braket{p^2}\) can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty e^{-cx^2}dx=\left(\frac{\pi}{c}\right)^\frac{1}{2}
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\), which gives
\begin{equation*}
\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx=\frac{\hbar^2}{\sqrt{\pi}}\left(\frac{m\omega}{\hbar}\right)^\frac{3}{2}\left(\frac{\pi}{\frac{m\omega}{\hbar}}\right)^\frac{1}{2}=m\omega \hbar.
\end{equation*}

Thus \(\braket{p^2}=-\frac{m\omega\hbar}{2}+m\omega\hbar=\frac{m\omega\hbar}{2}\).

The uncertainty in position is given by Eq. 2.62:

\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}
\end{equation*}

and the uncertainty in momentum is given by Eq. 2.64:

\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}.
\end{equation*}

The values for \(\braket{x}\) and \(\braket{p}\) from the previous problem are
\begin{equation*}
\braket{x}=\braket{p}=0
\end{equation*}

and the values of \(\braket{x^2}\) and \(\braket{p^2}\) are

\begin{equation*}
\braket{x^2}=\frac{1}{2}\frac{\hbar}{m\omega}
\end{equation*}

and

\begin{equation*}
\braket{p^2}=\frac{m\omega\hbar}{2}.
\end{equation*}

Plugging in the values for \(\braket{x^2}\) and \(\braket{x}\) gives the position uncertainty:
\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}=\sqrt{\frac{1}{2}\frac{\hbar}{m\omega}-(0)^2}=\frac{1}{\sqrt{2}}\left(\frac{\hbar}{m\omega}\right)^{\frac{1}{2}}
\end{equation*}

and plugging in the values for \(\braket{p^2}\) and \(\braket{p}\) gives the momentum uncertainty:
\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}=\sqrt{\frac{m\omega\hbar}{2}-(0)^2}=\frac{1}{\sqrt{2}}\left(m\omega\hbar\right)^{\frac{1}{2}}.
\end{equation*}

Using the expressions for \(\Delta x\) and \(\Delta p\) from the previous hint to form the product \(\Delta x \Delta p\) yields
\begin{equation*}
\Delta x \Delta p=\frac{1}{\sqrt{2}}\left(\frac{\hbar}{m\omega}\right)^{\frac{1}{2}}\frac{1}{\sqrt{2}}\left(m\omega\hbar\right)^{\frac{1}{2}}=\frac{\hbar}{2}
\end{equation*}

in accordance with the Heisenberg Uncertainty principle.

The uncertainty in position is given by Eq. 2.62:

\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}
\end{equation*}

and the uncertainty in momentum is given by Eq. 2.64:

\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}.
\end{equation*}

The values for \(\braket{x}\) and \(\braket{p}\) from the previous problem are
\begin{equation*}
\braket{x}=\braket{p}=0
\end{equation*}

and the values of \(\braket{x^2}\) and \(\braket{p^2}\) are

\begin{equation*}
\braket{x^2}=\frac{1}{2}\frac{\hbar}{m\omega}
\end{equation*}

and

\begin{equation*}
\braket{p^2}=\frac{m\omega\hbar}{2}.
\end{equation*}

Plugging in the values for \(\braket{x^2}\) and \(\braket{x}\) gives the position uncertainty:
\begin{equation*}
\Delta x=\sqrt{\braket{x^2}-\braket{x}^2}=\sqrt{\frac{1}{2}\frac{\hbar}{m\omega}-(0)^2}=\frac{1}{\sqrt{2}}\left(\frac{\hbar}{m\omega}\right)^{\frac{1}{2}}
\end{equation*}

and plugging in the values for \(\braket{p^2}\) and \(\braket{p}\) gives the momentum uncertainty:
\begin{equation*}
\Delta p=\sqrt{\braket{p^2}-\braket{p}^2}=\sqrt{\frac{m\omega\hbar}{2}-(0)^2}=\frac{1}{\sqrt{2}}\left(m\omega\hbar\right)^{\frac{1}{2}}.
\end{equation*}

Using the expressions for \(\Delta x\) and \(\Delta p\) from the previous hint to form the product \(\Delta x \Delta p\) yields
\begin{equation*}
\Delta x \Delta p=\frac{1}{\sqrt{2}}\left(\frac{\hbar}{m\omega}\right)^{\frac{1}{2}}\frac{1}{\sqrt{2}}\left(m\omega\hbar\right)^{\frac{1}{2}}=\frac{\hbar}{2}
\end{equation*}

in accordance with the Heisenberg Uncertainty principle.

Recall that normalization of a quantum wavefunction requires that the probability density \([\psi(x)]^*[\psi(x)]\) integrated over all \(x\) must result in unity:
\begin{equation*}
1=\int_{-\infty}^\infty [\psi(x)]^*[\psi(x)] dx.
\end{equation*}

The relation between \(\xi\) and \(x\) is given by Eq. 5.63:
\begin{equation*}
\xi=\frac{x}{\sqrt{\frac{\hbar}{m\omega}}}
\end{equation*}

so the wavefunction for the ground state of the quantum harmonic oscillator may be written as
\begin{equation*}
\psi(x)=Ae^{\frac{1}{2}\left(\frac{x}{\sqrt{\frac{\hbar}{m\omega}}}\right)^2}=Ae^{-\frac{m\omega}{2\hbar}x^2}.
\end{equation*}

Using the expression for \(\psi(x)\) from the previous hint makes the normalization integral look like this:
\begin{align*}
1=\int_{-\infty}^\infty [\psi(x)]^*[\psi(x)] dx&=\int_{-\infty}^\infty \left[Ae^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left[Ae^{-\frac{m\omega}{2\hbar}x^2}\right]dx\\
&=|A|^2\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx.
\end{align*}

The integral in the previous hint can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty e^{-cx^2}dx=\sqrt{\frac{\pi}{c}}.
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\).

Applying the relation in the previous hint gives
\begin{align*}
1=|A|^2\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx&=|A|^2\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}\\
&=|A|^2\sqrt{\frac{\pi\hbar}{m\omega}},
\end{align*}

and solving for \(|A|\) gives
\begin{align*}
|A|^2&=\sqrt{\frac{m\omega}{\pi\hbar}}\\
|A|&=\left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}.
\end{align*}

Recall that normalization of a quantum wavefunction requires that the probability density \([\psi(x)]^*[\psi(x)]\) integrated over all \(x\) must result in unity:
\begin{equation*}
1=\int_{-\infty}^\infty [\psi(x)]^*[\psi(x)] dx.
\end{equation*}

The relation between \(\xi\) and \(x\) is given by Eq. 5.63:
\begin{equation*}
\xi=\frac{x}{\sqrt{\frac{\hbar}{m\omega}}}
\end{equation*}

so the wavefunction for the ground state of the quantum harmonic oscillator my be written as
\begin{equation*}
\psi(x)=Ae^{\frac{1}{2}\left(\frac{x}{\sqrt{\frac{\hbar}{m\omega}}}\right)^2}=Ae^{-\frac{m\omega}{2\hbar}x^2}.
\end{equation*}

Using this expression for \(\psi(x)\) makes the normalization integral look like this:
\begin{align*}
1=\int_{-\infty}^\infty [\psi(x)]^*[\psi(x)] dx&=\int_{-\infty}^\infty \left[Ae^{-\frac{m\omega}{2\hbar}x^2}\right]^*\left[Ae^{-\frac{m\omega}{2\hbar}x^2}\right]dx\\
&=|A|^2\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx.
\end{align*}

This integral can be evaluated using the relation
\begin{equation*}
\int_{-\infty}^\infty e^{-cx^2}dx=\sqrt{\frac{\pi}{c}}.
\end{equation*}

with \(c=\frac{m\omega}{\hbar}\).

Applying this relation gives
\begin{align*}
1=|A|^2\int_{-\infty}^\infty e^{-\frac{m\omega}{\hbar}x^2}dx&=|A|^2\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}\\
&=|A|^2\sqrt{\frac{\pi\hbar}{m\omega}},
\end{align*}

and solving for \(|A|\) gives
\begin{align*}
|A|^2&=\sqrt{\frac{m\omega}{\pi\hbar}}\\
|A|&=\left(\frac{m\omega}{\pi\hbar}\right)^\frac{1}{4}.
\end{align*}

The wavefunctions for a quantum harmonic oscillator are given by Eq. 5.97 as
\begin{equation*}
\psi_n(\xi)=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{2^n n!}}\right)H_n(\xi)e^{-\frac{\xi^2}{2}},
\end{equation*}

so the first excited-state wavefunction \(n=1\) is
\begin{align*}
\psi_1(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(2^1)(1!)}}\right)(2\xi)e^{-\frac{\xi^2}{2}}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\sqrt{2}\xi e^{-\frac{\xi^2}{2}}\right)
\end{align*}

since \(1!=1\) and \(H_1(\xi)=2\xi\). The second excited-state wavefunction \(n=2\) is
\begin{align*}
\psi_2(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(2^2)(2!)}}\right)(4\xi^2-2)e^{-\frac{\xi^2}{2}}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}}e^{-\frac{\xi^2}{2}}\right)
\end{align*}

since \(2!=2\) and \(H_2(\xi)=4\xi^2-2\).

The lowering operator \(\hat{a}\) is given by Eq. 5.103:
\begin{equation*}
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}}).
\end{equation*}

in which \(\hat{\mathcal{P}}\) is the dimensionless momentum operator \(\hat{\mathcal{P}}=\widehat{P}/p_{ref}\) and \(\hat{\xi}\) is the dimensionless position operator \(\hat{\xi}=\widehat{X}/x_{ref}\).

Applying the lowering operator to \(\psi_2(\xi)\) gives
\begin{align*}
\hat{a}^{\dagger}\psi_2(\xi)&=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\frac{1}{\sqrt{2}}\xi\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right]\\
&\hspace{0.3cm}+i\left(-i\frac{\partial}{\partial \xi}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right].
\end{align*}

Multiplying through, taking the derivatives and collecting terms in the expression in the previous hint gives
\begin{align*}
\hat{a}^{\dagger}\psi_1(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\left(\frac{\partial}{\partial \xi}\right)\left(\frac{2\xi^2-1}{2} e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\frac{\partial \left(\frac{2\xi^2-1}{2}\right)}{\partial \xi}e^{-\frac{\xi^2}{2}}\right.\\
&\hspace{5.5cm}\left.+\left(\frac{2\xi^2-1}{2}\right) \frac{\partial \left(e^{-\frac{\xi^2}{2}}\right)}{\partial \xi}\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\left(\frac{4\xi}{2}\right)e^{-\frac{\xi^2}{2}}\right.\\
&\hspace{5.5cm}\left.+\left(\frac{2\xi^2-1}{2}\right) \left(-\xi e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{\xi^2}{2}}\left[\left(\frac{2\xi^3-\xi}{2} \right)+\left(2\xi \right)+\left(\frac{2\xi^2-1}{2}\right) \left(-\xi \right)\right]\\
&=2\xi\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{\xi^2}{2}}=\sqrt{2}\psi_1(\xi)
\end{align*}

in accordance with Eq. 5.111.

Full Solution: The wavefunctions for a quantum harmonic oscillator are given by Eq. 5.97 as
\begin{equation*}
\psi_n(\xi)=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{2^n n!}}\right)H_n(\xi)e^{-\frac{\xi^2}{2}},
\end{equation*}

so the first excited-state wavefunction \(n=1\) is
\begin{align*}
\psi_1(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(2^1)(1!)}}\right)(2\xi)e^{-\frac{\xi^2}{2}}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\sqrt{2}\xi e^{-\frac{\xi^2}{2}}\right)
\end{align*}

since \(1!=1\) and \(H_1(\xi)=2\xi\). The second excited-state wavefunction \(n=2\) is
\begin{align*}
\psi_2(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{1}{\sqrt{(2^2)(2!)}}\right)(4\xi^2-2)e^{-\frac{\xi^2}{2}}\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}}e^{-\frac{\xi^2}{2}}\right)
\end{align*}

since \(2!=2\) and \(H_2(\xi)=4\xi^2-2\).

The lowering operator \(\hat{a}\) is given by Eq. 5.103:
\begin{equation*}
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}}).
\end{equation*}

in which \(\hat{\mathcal{P}}\) is the dimensionless momentum operator \(\hat{\mathcal{P}}=\widehat{P}/p_{ref}\) and \(\hat{\xi}\) is the dimensionless position operator \(\hat{\xi}=\widehat{X}/x_{ref}\),

and applying the lowering operator to \(\psi_2(\xi)\) gives
\begin{align*}
\hat{a}^{\dagger}\psi_2(\xi)&=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\frac{1}{\sqrt{2}}\xi\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right]\\
&\hspace{0.3cm}+i\left(-i\frac{\partial}{\partial \xi}\right)\left[\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left(\frac{2\xi^2-1}{\sqrt{2}} e^{-\frac{\xi^2}{2}}\right)\right].
\end{align*}

Multiplying through, taking the derivatives and collecting terms in this expression gives
\begin{align*}
\hat{a}^{\dagger}\psi_1(\xi)&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\left(\frac{\partial}{\partial \xi}\right)\left(\frac{2\xi^2-1}{2} e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\frac{\partial \left(\frac{2\xi^2-1}{2}\right)}{\partial \xi}e^{-\frac{\xi^2}{2}}\right.\\
&\hspace{5.5cm}\left.+\left(\frac{2\xi^2-1}{2}\right) \frac{\partial \left(e^{-\frac{\xi^2}{2}}\right)}{\partial \xi}\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}\left[\left(\frac{2\xi^3-\xi}{2} e^{-\frac{\xi^2}{2}}\right)+\left(\frac{4\xi}{2}\right)e^{-\frac{\xi^2}{2}}\right.\\
&\hspace{5.5cm}\left.+\left(\frac{2\xi^2-1}{2}\right) \left(-\xi e^{-\frac{\xi^2}{2}}\right)\right]\\
&=\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{\xi^2}{2}}\left[\left(\frac{2\xi^3-\xi}{2} \right)+\left(2\xi \right)+\left(\frac{2\xi^2-1}{2}\right) \left(-\xi \right)\right]\\
&=2\xi\left(\frac{m\omega}{\pi \hbar}\right)^\frac{1}{4}e^{-\frac{\xi^2}{2}}=\sqrt{2}\psi_1(\xi)
\end{align*}

in accordance with Eq. 5.111.

To show that the position operator \(\widehat{X}\) can be written in terms of the ladder operators \(\hat{a}^{\dagger}\) and \(\hat{a}\) as
\begin{equation*}
\widehat{X}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}^{\dagger}+\hat{a}),
\end{equation*}

begin by adding the equations for the ladder operators \(\hat{a}^{\dagger}\) (Eq. 5.102) and \(\hat{a}\) (5.103):
\begin{equation*}
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})
\end{equation*}

and

\begin{equation*}
\hat{a}=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}}),
\end{equation*}

which gives

\begin{align*}
\hat{a}^{\dagger}+\hat{a}&=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})+\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\\
&=\frac{2}{\sqrt{2}}\hat{\xi}=\sqrt{2}\hat{\xi}.
\end{align*}

Recall that \(\hat{\xi}=\widehat{X}/x_{ref}=\sqrt{\frac{m\omega}{\hbar}}\widehat{X}\), so
\begin{equation*}
\hat{a}^{\dagger}+\hat{a}=\sqrt{2}\hat{\xi}=\sqrt{2}\sqrt{\frac{m\omega}{\hbar}}\widehat{X}
\end{equation*}

which means
\begin{equation*}
\widehat{X}=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}^{\dagger}+\hat{a}\right).
\end{equation*}

To show that the momentum operator \(\widehat{P}\) can be written in terms of the ladder operators \(\hat{a}^{\dagger}\) and \(\hat{a}\) as
\begin{equation*}
\widehat{P}=i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a}^{\dagger}-\hat{a}),
\end{equation*}

begin by subtracting the equation for \(\hat{a}\) (5.103) from the equation for \(\hat{a}^{\dagger}\) (Eq. 5.102):
\begin{align*}
\hat{a}^{\dagger}-\hat{a}&=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})-\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\\
&=\frac{-2i}{\sqrt{2}}\hat{\mathcal{P}}=-i\sqrt{2}\hat{\mathcal{P}}.
\end{align*}

Recall that \(\hat{\mathcal{P}}=\widehat{P}/p_{ref}=\frac{\widehat{P}}{\sqrt{\hbar m\omega}}\), so
\begin{equation*}
\hat{a}^{\dagger}-\hat{a}=-i\sqrt{2}\hat{\mathcal{P}}=-i\sqrt{2}\frac{\widehat{P}}{\sqrt{\hbar m\omega}}
\end{equation*}

which means
\begin{equation*}
\widehat{P}=i\sqrt{\frac{\hbar m\omega}{2}}\left(\hat{a}^{\dagger}-\hat{a}\right).
\end{equation*}

To show that the position operator \(\widehat{X}\) can be written in terms of the ladder operators \(\hat{a}^{\dagger}\) and \(\hat{a}\) as
\begin{equation*}
\widehat{X}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}^{\dagger}+\hat{a}),
\end{equation*}

begin by adding the equations for the ladder operators \(\hat{a}^{\dagger}\) (Eq. 5.102) and \(\hat{a}\) (5.103):
\begin{equation*}
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})
\end{equation*}

and

\begin{equation*}
\hat{a}=\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}}),
\end{equation*}

which gives

\begin{align*}
\hat{a}^{\dagger}+\hat{a}&=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})+\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\\
&=\frac{2}{\sqrt{2}}\hat{\xi}=\sqrt{2}\hat{\xi}.
\end{align*}

Recall that \(\hat{\xi}=\widehat{X}/x_{ref}=\sqrt{\frac{m\omega}{\hbar}}\widehat{X}\), so
\begin{equation*}
\hat{a}^{\dagger}+\hat{a}=\sqrt{2}\hat{\xi}=\sqrt{2}\sqrt{\frac{m\omega}{\hbar}}\widehat{X}
\end{equation*}

which means
\begin{equation*}
\widehat{X}=\sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}^{\dagger}+\hat{a}\right).
\end{equation*}

To show that the momentum operator \(\widehat{P}\) can be written in terms of the ladder operators \(\hat{a}^{\dagger}\) and \(\hat{a}\) as
\begin{equation*}
\widehat{P}=i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a}^{\dagger}-\hat{a}),
\end{equation*}

begin by subtracting the equation for \(\hat{a}\) (5.103) from the equation for \(\hat{a}^{\dagger}\) (Eq. 5.102):
\begin{align*}
\hat{a}^{\dagger}-\hat{a}&=\frac{1}{\sqrt{2}}(\hat{\xi}-i\hat{\mathcal{P}})-\frac{1}{\sqrt{2}}(\hat{\xi}+i\hat{\mathcal{P}})\\
&=\frac{-2i}{\sqrt{2}}\hat{\mathcal{P}}=-i\sqrt{2}\hat{\mathcal{P}}.
\end{align*}

Recall that \(\hat{\mathcal{P}}=\widehat{P}/p_{ref}=\frac{\widehat{P}}{\sqrt{\hbar m\omega}}\), so
\begin{equation*}
\hat{a}^{\dagger}-\hat{a}=-i\sqrt{2}\hat{\mathcal{P}}=-i\sqrt{2}\frac{\widehat{P}}{\sqrt{\hbar m\omega}}
\end{equation*}

which means
\begin{equation*}
\widehat{P}=i\sqrt{\frac{\hbar m\omega}{2}}\left(\hat{a}^{\dagger}-\hat{a}\right).
\end{equation*}