PHYS 481 - Quantum Mechanics

Assignment 03

Problem 1

In a harmonic oscillator a coherent state \(\psi_{\alpha}(x)\) is defined as follows: When acted on by the lowering operator \(\hat{a}_-\), we get the wave function back times a constant:

\[\hat{a}_- \psi_\alpha(x) = \alpha \psi_{\alpha}(x)\]

or in linear algebra language, \(\psi_{\alpha}(x)\) is an eigenvector of \(\hat{a}_-\) with eigenvalue \(\alpha\). Different coherent states have different values of \(\alpha\). Do not in general assume that the constant \(\alpha\) is real. (Coherent states have many applications in atomic, molecular, and optical physics. For instance, lasers and Bose-Einstein condensates are examples of coherent states.)

1. Show that for any square-integrable functions \(f(x)\) and \(g(x)\) \[\int_{-\infty}^{\infty}f^*(x)(\hat{a}_\pm g(x))dx = \int_{\infty}^{\infty}(\hat{a}_\mp f(x))^*g(x)dx\]
2. Use Eq. (2) along with the eigenvector equation (1) to evaluation \(\left<x\right>\) and \(\left<p\right>\) for the coherent state wave function \(\psi_\alpha\) in terms of \(\alpha\) and constants. You may assume \(\psi_{\alpha}(x)\) is normalized. (Hint: how can \(\hat{x}\) and \(\hat{p}\) be written in terms of \(\hat{a}_+\) and \(\hat{a}_-\)?)
3. Is the ground state \(\chi_0(x)\) of the harmonic oscillator a coherent state? What is the value of \(\alpha\)?
4. Any wave function of the harmonic oscillator can be expressed as a linear combination of stationary states \(\chi_n(x)\) of the harmonic oscillator. Assume therefore that \[\displaystyle\psi_\alpha(x) = \sum_{n=0}^{\infty}c_n\chi_n(x),\] and show that the \(c_n\) are given by \[c_n = \frac{\alpha^{n}}{\sqrt{n!}}c_0.\] (Hint: Operate with \(\hat{a}_-\) on Eq. (3).)
5. Another interesting property of coherent states is how their expectation values evolve in time. Recall that stationary states have time-independent expectation values; coherent states are different. Assume that \(\Psi_\alpha(x, t=0) = \psi_\alpha(x)\) and show that \(\Psi_\alpha(x, t)\) is still a coherent state - that is, show it satisfies \[\hat{a}_- \Psi_\alpha(x,t) = \alpha(t) \Psi_{\alpha}(x,t)\]. What is \(\alpha(t)\) in terms of \(\alpha\) and other quantities?
6. In this part, for simplicity assume \(\alpha\) is real (but \(\alpha(t)\) might not be real). Take the results for \(\left<x\right>\) and \(\left<p\right>\) from part b) and put the value of \(\alpha(t)\) into them to find \(\left<x\right>(t)\) and \(\left<p\right>(t)\) for \(\Psi_\alpha(x,t)\)$. How does the result compare to the classical motion of a particle in a harmonic oscillator?

Response: Before we start, rewriting the following terms and simplications will make our calculations easier throughout the problem.

\[\begin{align} \hat{p} &= -i\hbar \frac{d}{dx} \\ \\ \hat{a}_\pm &= \frac{1}{\sqrt{2\hbar m \omega}}\left(\mp i\hat{p} + m\omega\hat{x}\right) \\ \hat{a}_+ &= -\sqrt{\frac{\hbar}{2m\omega}}\frac{d}{dx} + \sqrt{\frac{m\omega}{2\hbar}}x \\ &= -A\frac{d}{dx} + Bx,\quad A=\sqrt{\frac{\hbar}{2m\omega}},\; B=\sqrt{\frac{m\omega}{2\hbar}}\\ \hat{a}_- &= A\frac{d}{dx} + Bx\end{align}\]

For part a) (just using \(\hat{a}_+\))

\[\begin{align}\int_{-\infty}^{\infty}f^*(x)(\hat{a}_+ g(x))dx &= \int_{-\infty}^{\infty}f^*(x)\left(\left(-A\frac{d}{dx} + Bx\right) g(x)\right)dx \\ &= -A\int_{-\infty}^{\infty}f^*(x)g'(x)dx + B\int_{-\infty}^{\infty}f^*(x)xg(x) dx \\ &= -A\left[-\int_{-\infty}^{\infty}(f^*(x))'g(x)dx + f^*(x)g(x)\Big|_{-\infty}^{\infty}\right] + B\int_{-\infty}^{\infty}f^*(x)xg(x) dx \\ &= A\int_{-\infty}^{\infty}(f^*(x))'g(x)dx + B\int_{-\infty}^{\infty}f^*(x)xg(x) dx \\ &= \int_{-\infty}^{\infty}\left[A(f^*(x))' + Bf^*(x)x\right]gdx \\ &= \int_{-\infty}^{\infty}\left[A\frac{d}{dx}f^*(x) + Bxf^*(x)\right]gdx \\ &= \int_{-\infty}^{\infty}\left(A\frac{d}{dx}+ Bx\right)f^*(x)g(x) dx \\ &= \int_{-\infty}^{\infty}\hat{a}_-f^*(x)g(x) dx \\ &= \int_{\infty}^{\infty}(\hat{a}_- f(x))^*g(x)dx \end{align}\]

Note that we can do the same process starting with \(\hat{a}_-\), the only difference is that every \(A\) will have a \(-\) in front of it, but the end result will still satisfy the relationship.

For part b), note that we can rewrite \(\hat{x}\) and \(\hat{p}\) as \[\hat{x} =\sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}_+ + \hat{a}_-\right) ,\quad \hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}\left(\hat{a}_+ - \hat{a}_-\right)\]

It follows that

\[\begin{align} \left<x\right> &= \int_{-\infty}^{\infty}\psi_\alpha^*\left[\sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a}_+ + \hat{a}_-\right)\right]\psi_\alpha dx \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(\int_{-\infty}^{\infty}\psi_\alpha^*\hat{a}_+\psi_\alpha dx + \int_{-\infty}^{\infty}\psi_\alpha^*\hat{a}_-\psi_\alpha dx\right) \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(\int_{-\infty}^{\infty}(\hat{a}_-\psi_\alpha)^*\psi_\alpha dx + \alpha\int_{-\infty}^{\infty}\psi_\alpha^*\psi_\alpha dx\right) \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(\alpha^* + \alpha\right) \\ \\ \left<p\right> &= \int_{-\infty}^{\infty}\psi_\alpha^*\left[i\sqrt{\frac{\hbar m \omega}{2}}\left(\hat{a}_+ - \hat{a}_-\right)\right]\psi_\alpha dx \\ &= i\sqrt{\frac{\hbar m \omega}{2}}\left(\int_{-\infty}^{\infty}\psi_\alpha^*\hat{a}_+\psi_\alpha dx - \int_{-\infty}^{\infty}\psi_\alpha^*\hat{a}_-\psi_\alpha dx\right) \\ &= i\sqrt{\frac{\hbar m \omega}{2}}\left(\int_{-\infty}^{\infty}(\hat{a}_-\psi_\alpha)^*\psi_\alpha dx - \alpha\int_{-\infty}^{\infty}\psi_\alpha^*\psi_\alpha dx\right) \\ &= i\sqrt{\frac{\hbar m \omega}{2}}(\alpha^* - \alpha)\end{align}\]

For part c), if we apply \(\hat{a}_-\) to \(\psi_0\) and get a constant and \(\psi_0\) back out, then the ground state is a coherent state.

\[\begin{align}\psi_0 &= \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^2} \\ &= \left(\frac{2}{\pi}\right)^{1/4}B^{1/2}e^{-B^2x^2} \\ \\ \hat{a}_-\psi_0 &= \left(A\frac{d}{dx}+ Bx\right)\left(\frac{2}{\pi}\right)^{1/4}B^{1/2}e^{-B^2x^2} \\ &= A\frac{d}{dx}\left(\frac{2}{\pi}\right)^{1/4}B^{1/2}e^{-B^2x^2} + Bx\left(\frac{2}{\pi}\right)^{1/4}B^{1/2}e^{-B^2x^2} \\ &= \underbrace{\left(-2AB^2x + Bx\right)}_{\alpha}\underbrace{\left(\frac{2}{\pi}\right)^{1/4}B^{1/2}e^{-B^2x^2}}_{\psi_0} \\ \\ \alpha &= B(1-2AB)x,\quad A = \frac{1}{2B} \\ \alpha &= 0\end{align}\]

Noting that \(A = \frac{1}{2B}\), we can see that in order \(\alpha\) to be constant for all \(x\), \(\alpha\) must be 0. Thus, the ground state is a coherent state if \(\alpha = 0\).

For part d), we can derive the relationship by exploring what happens when \(\hat{a}_-\) acts of the sum.

\[\begin{align} \psi_\alpha &= \sum_{n=0}^{\infty} c_n \chi_n \\ \hat{a}_-\psi_\alpha &= \sum_{n=0}^{\infty} c_n \hat{a}_-\chi_n \\ \alpha\psi_n&= \sum_{n=0}^{\infty} c_n\sqrt{n}\chi_{n-1} \\ \alpha\psi_n &= \sum_{n=0}^{\infty}\alpha c_n\chi_n \\ \sum_{n=0}^{\infty}c_n\sqrt{n}\chi_{n-1} &= \sum_{n=0}^{\infty}\alpha c_n\chi_n \\ (n=1)&\quad \sum_{n=1}^{\infty}c_{n+1}\sqrt{n+1}\chi_n \\ \\ c_{n+1} \sqrt{n+1} &= \alpha c_n \\ c_{n+1} &= \frac{\alpha}{\sqrt{n+1}}c_n \\ c_n &= \frac{\alpha^n}{\sqrt{n!}}c_0 \end{align}\]

For part e), the calulcation below derives \(\alpha(t) = \alpha e^{-i\omega t}\)

\[\begin{align} \hat{a}_-\Psi_\alpha &= \sum_{n=0}^{\infty}c_n \hat{a}_-\chi_n e^{-iE_nt/\hbar} \\ &= \sum_{n=0}^{\infty}c_n \sqrt{n}\chi_{n-1} e^{-iE_nt/\hbar} \\ (\text{shift to }n+1) &= \sum_{n=0}^{\infty}c_{n+1} \sqrt{n+1}\chi_{n} e^{-iE_{n+1}t/\hbar} \\ &= \sum_{n=0}^{\infty}\alpha c_{n}\chi_{n} e^{-i(E_n + \hbar\omega)t/\hbar} \\ &= \underbrace{\alpha e^{-i\omega t}}_{\alpha(t)}\underbrace{\sum_{n=0}^{\infty}c_n\chi_n e^{-i E_n t/\hbar}}_{\Psi_\alpha}\end{align}\]

For part f), combining the results from part b) with \(\alpha(t)\) from e) (and only considering the real part), we get the following for \(\left<x\right>\) and \(\left<p\right>\)

\[\begin{align}\left<x\right> &= \sqrt{\frac{\hbar}{2m \omega}}\left(\alpha e^{i\omega t} + \alpha e^{-i\omega t}\right) \\ &= \sqrt{\frac{\hbar}{2m \omega}}\alpha\left(\cos\omega t + i \sin\omega t + \cos(-\omega t) + i\sin(-\omega t)\right) \\ &= 2\alpha\sqrt{\frac{\hbar}{2m \omega}}\cos\omega t \\ &= \alpha\sqrt{\frac{2\hbar}{m \omega}}\cos\omega t \\ \\ \left<p\right> &= i\sqrt{\frac{\hbar m \omega}{2}}\left(\alpha e^{i\omega t} - \alpha e^{-i\omega t}\right) \\ &= i\alpha\sqrt{\frac{\hbar m \omega}{2}}\left(\cos\omega t + i\sin\omega t - \cos(-\omega t) - i \sin(-\omega t)\right) \\ &= i\alpha\sqrt{\frac{\hbar m \omega}{2}}\left(\cos\omega t + i\sin\omega t - \cos\omega t + i \sin\omega t\right) \\ &= -\alpha\sqrt{2\hbar m \omega}\sin\omega t \end{align}\]

For comparison, classically, we get

\[\begin{align} x &= A \cos \omega t \\ p &= m \frac{dx}{dt} &= -m\omega A\sin\omega t\end{align}\]

Look’s pretty similar!

Problem 2

A system is initially in the state \[\Psi(x, t=0) = \frac{1}{\sqrt{7}}\left(\sqrt{2}\chi_1(x) + \sqrt{3}\chi_2(x) - i\chi_3(x) + \chi_4(x)\right)\] where \(\chi_n(x)\) are eigenstates of the systems’s Hamiltonian such that \(\hat{H}\chi_n(x) = n^2E_0\chi_n(x)\)

1. Is \(\Psi(x,t)\) normalized for \(t>0\)?
2. If the energy of state is measured, what values will be obtained and with what probabilities?
3. Calculate the expectation value of the total energy of the system.
4. Suppose that a measurement yields \(4E_0\), write down the wave function immediately after the measurement.

Response: For part a), we know \(\Psi(x,t)\) is normalized if \(\int_{-\infty}^{\infty}|\Psi(x,t)|^2dx = 1\). If we assume

\[\Psi(x,t) = \frac{1}{\sqrt{7}}\left(\sqrt{2}\chi_1(x)e^{-iE_1t/\hbar} + \sqrt{3}\chi_2(x)e^{-iE_2t/\hbar} - i\chi_3(x)e^{-iE_3t/\hbar} + \chi_4(x)e^{-iE_4t/\hbar}\right)\]

Then \(\Psi(x,t)^*\Psi(x,t)\) will set all exponential terms to 1, and all we will have to worry about will be the coefficients, since the eigenstates form an orthonormal basis. Thus, we can prove \(\Psi(x,t)\) is normalized by checking:

\[\begin{align}\int_{-\infty}^{\infty}\Psi(x,t)^*\Psi(x,t) dx &= \frac{1}{7}\left(2 + 3 + 1 + 1\right) \\ &= 1\end{align}\]

For part b), a measurment of energy gives one of the eigenvalues \(E_n = n^2E_0\) with a probability \(P = |c_n|^2\). Thus, the associated probabilities for each possible energy level is:

\[\begin{align} n=1,\quad E_1 = E_0,\quad P = \frac{2}{7} \\ n=2,\quad E_2 = 4E_0,\quad P = \frac{3}{7} \\ n=3,\quad E_3 = 9E_0,\quad P = \frac{1}{7} \\ n=4,\quad E_4 = 16E_0, \quad P = \frac{1}{7} \end{align}\]

For part c) the expectation value is given by:

\[\begin{align}\left<E\right> &= \sum_n P(E_n)E_n \\ &= \frac{E_0}{7}\left(2\cdot1 + 3\cdot4 + 1\cdot9 + 1\cdot16\right) \\ &= \frac{39}{7}E_0\end{align}\]

For part d), once a measurement is made, the wavefunction collapses and the only eigenstate that can exist is \(\chi_2\), indicating a wave function of \[\psi(x) = \chi_2(x)\]

Problem 3

The wave function for a particle of mass \(m\) in a potential energy well for which \(V=0\) for \(0<x<L\) and \(V=\infty\) elsewhere is given by \[\Psi(x) = \cases{\sqrt{\frac{105}{L^7}}x^2(L-x), \hspace{1cm} 0<x<L \\ 0, \hspace{3.3cm} \text{elsewhere.}}\]

What is the probability that a measurement of the energy of the particle yields the ground-state energy?

Response: This problem can be solved easily with Mathematica. See Problem 3 in the Mathematica file.

The probability that a measurement of the energy would return the value \(E_n\) is given by

\[P(E_n) = |c_n|^2 = \left|\int_{-\infty}^{\infty}\psi_n(x)^*f(x)dx\right|^2\]

In our case, \(f(x)\) is the ground-state energy of the particle in an infinite square well, specifically

\[f(x) = \sqrt{\frac{2}{L}}\sin\frac{pi n}{L}x\]

where \(n=0\). If we plug \(f(x)\) and \(\psi_n(x)\)$ into our equation for \(P(E_n)\), we get a probability of \[P=0.873736\]

Problem 4

Find \(\left<x\right>\), \(\left<p\right>\), \(\left<x^2\right>\), \(\left<p^2\right>\), \(\left<V\right>\) and \(\left<T\right>\), for the nth stationary state of the harmonic oscillator, [use the method of Example 2.5 of textbook]. Check that the uncertainty principle is satisfied.

Response: Using the method of Example 2.5 in the textbook, we can define \(\hat{x}\) and \(\hat{p}\) in terms of \(\hat{a}_+\) and \(\hat{a}_-\) to simplify integration.

\[\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}_+ + \hat{a}_-),\quad \hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a}_+ - \hat{a}_-)\]

Also, a useful reminder:

\[\hat{a}_+ \hat{a}_- \psi_n = n \psi_n,\quad \hat{a}_- \hat{a}_+ \psi_n = (n+1)\psi_n\]

Beginning with \(\left<x\right>\),

\[\begin{align}\left<x\right> &= \int_{-\infty}^{\infty}\psi_n^*\hat{x}\psi_n dx \\ &= \sqrt{\frac{\hbar}{2m\omega}}\int_{-\infty}^{\infty}\psi_n^*(\hat{a}_+ + \hat{a}_-)\psi_n dx \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(\int_{-\infty}^{\infty}\psi_n^*\hat{a}_+\psi_n dx + \int_{-\infty}^{\infty}\psi_n^*\hat{a}_-\psi_n dx\right) \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n+1}\int_{-\infty}^{\infty}\psi_n^*\psi_{n+1} dx + \sqrt{n}\int_{-\infty}^{\infty}\psi_n^*\psi_{n-1} dx\right) \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left(0 - 0\right) \\ &= 0\end{align}\]

Similarly, for \(\left<p\right>\),

\[\left<p\right> = m\frac{d\left<x\right>}{dt} = 0\]

However, \(\left<x^2\right>\) is non-zero:

\[\begin{align}\left<x^2\right> &= \int_{-\infty}^{\infty}\psi_n^*\hat{x}^2\psi_n dx \\ &= \frac{\hbar}{2m\omega}\int_{-\infty}^{\infty}\psi_n^*(\hat{a}_+^2 + \hat{a}_+\hat{a}_- + \hat{a}_-\hat{a}_+ + \hat{a}_-^2)\psi_n dx \\ &= \frac{\hbar}{2m\omega}\left(\int_{-\infty}^{\infty}\psi_n^*\hat{a}_+\hat{a}_-\psi_n dx + \int_{-\infty}^{\infty}\psi_n^*\hat{a}_-\hat{a}_+\psi_n dx\right) \\ &= \frac{\hbar}{2m\omega}\left(n + (n+1)\right) \\ &= \frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)\end{align}\]

Same goes for \(\left<p^2\right>\): \[\begin{align}\left<p^2\right> &= \int_{-\infty}^{\infty}\psi_n^*\hat{p}^2\psi_n dx \\ &= -\frac{\hbar m \omega}{2}\int_{-\infty}^{\infty}\psi_n^*(\hat{a}_+^2 - \hat{a}_+\hat{a}_- - \hat{a}_-\hat{a}_+ + \hat{a}_-^2)\psi_n dx \\ &= -\frac{\hbar m \omega}{2}\left(-\int_{-\infty}^{\infty}\psi_n^*\hat{a}_+\hat{a}_-\psi_n dx - \int_{-\infty}^{\infty}\psi_n^*\hat{a}_-\hat{a}_+\psi_n dx\right) \\ &= -\frac{\hbar m \omega}{2}\left(-n - (n+1)\right) \\ &= \frac{1}{2}\hbar m \omega\left(n+\frac{1}{2}\right)\end{align}\]

\(\left<V\right>\) and \(\left<T\right>\) are calulated as follows:

\[\begin{align}\left<V\right> &= \left<\frac{1}{2}m\omega^2x^2\right> \\ &= \frac{1}{2}m\omega^2\left<x^2\right> \\ &= \frac{1}{2}m\omega^2\left(\frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)\right) \\ &= \frac{1}{2}\hbar\omega\left(n + \frac{1}{2}\right) \\ \\ \left<T\right> &= \left<\frac{p^2}{2m}\right> \\ &= \frac{\left<p^2\right>}{2m} \\ &= \frac{1}{2}\hbar\omega\left(n+ \frac{1}{2}\right)\end{align}\]

Checking the uncertainty principle, we can see our results satisfy the condition.

\[\begin{align} \sigma_x &= \sqrt{\left<x^2\right> - \left<x\right>^2} \\ &= \sqrt{\frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)} \\ \\ \sigma_p &= \sqrt{\left<p^2\right> - \left<p\right>^2} \\ &= \sqrt{\hbar m \omega\left(n+\frac{1}{2}\right)} \\ \\ \sigma_x\sigma_p &= \sqrt{\frac{\hbar}{m\omega}\left(n+\frac{1}{2}\right)} \sqrt{\hbar m \omega\left(n+\frac{1}{2}\right)} \\&= \hbar\left(n + \frac{1}{2}\right) \\ &\geq \frac{\hbar}{2}\end{align}\]

Problem 5

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials.

1. The Rodrigues formula says that \[H_n(\xi) = (-1)^n e^{\xi^2}\left(\frac{d}{d\xi}\right)^ne^{-\xi^2}\] Use it to derive \(H_3\) and \(H_4\).
2. The following recursion relation gives you \(H_{n+1}\) in terms of the two preceding Hermite polynomials: \[H_{n+1}(\xi)=2\xi H_n(\xi)-2nH_{n-1}(\xi).\] Use it, together with your answer in (a), to obtain \(H_5\) and \(H_6\).
3. If you differentiate an \(n\)-th order polynomial, you get a polynomial of order \((n-1)\). For Hermite polynomials, in fact, \[\frac{dH_n}{d\xi}=2nH_{n-1}(\xi).\] Check this, by differentiating \(H_5\) and \(H_6\).
4. \(H_n(\xi)\) is the \(n\)-th \(z\) derivative, at \(z=0\), of the generating function \(e^{-z^2 + 2z\xi}\); or to put it another way, it is the coefficient of \(z^n/n!\) in the Taylor series expansion for this function: \[e^{-z^2 + 2z\xi} = \sum_{n=0}^{\infty}\frac{z^n}{n!}H_n(\xi)\] Use this to obtain \(H_1\), \(H_2\) and \(H_3\).

Response: This problem was also solved with Mathematica. See Problem 5 in the Mathematica file.