2002 Autumn Quantum Mechanics
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Problem 1 [35 points total]

A. [5 points]


Derive, from the one-dimensional Time-Independent Schroedinger Equation, the expression for the discontinuity

\begin{align} \lim_{\epsilon\to0} \left[\frac{d\psi}{dx}_{x_0+\epsilon}-\frac{d\psi}{dx}_{x_0-\epsilon}\right] = \frac{2mV_0}{\hbar^2}\psi(x_0} \end{align}

at a singularity of potential

\begin{align} V(x) = W(x) + V_0\delta(x-x_0) \end{align}

where $W(x)$ is any finite function.

Greg's Solution:

The Time-Independent Schroedinger Equation is given by,

\begin{align} \frac{-\hbar^2}{2m}\psi''(x) + V\psi(x) = E\psi(x) \end{align}

Integrate both sides from $x_0-\epsilon$ to $x_0+\epsilon$,

\begin{align} \int_{x_0+\epsilon}^{x_0-\epsilon}\left[\frac{-\hbar^2}{2m}\psi''(x) + (W(x)+V_0\delta(x-x_0)\psi(x)\right]\,dx = \int_{x_0+\epsilon}^{x_0-\epsilon} E\psi(x)\,dx \end{align}

Now, as we take the limit $\epsilon\to 0$, all the continuous terms get killed off, leaving only the delta function and the discontinuity in the derivative,

\begin{align} \frac{-\hbar^2}{2m}\lim_{\epsilon\to 0} \left[\frac{d\psi}{dx}_{x_0+\epsilon}-\frac{d\psi}{dx}_{x_0-\epsilon}\right] +V_0 \psi(x_0)=0 \end{align}

Rearranging terms,

\begin{align} \lim_{\epsilon\to 0} \left[\frac{d\psi}{dx}_{x_0+\epsilon}-\frac{d\psi}{dx}_{x_0-\epsilon}\right] = \frac{2m}{\hbar^2}V_0 \psi(x_0), \end{align}


B. [8 points]


Use this result to obtain an equation which determines the energies of the even parity energy eigenstates for a particle in the potential,

\begin{align} V(x) = \begin{cases} \lambda\frac{\hbar^2}{2m}\delta(x) & |x| < a \\\\ \infty & |x| > a \end{cases}, \end{align}

where $\lambda\ge 0$ is a real constant.

Greg's Solution:

The general even solution to this problem (up to a normalization) is,

\begin{align} \psi(x) = \sin(k|x|+\delta_0), \end{align}

where $k:=\frac{\sqrt{2mE}}{\hbar}$ and $\delta_0$ are constants that will be determined from the two boundary conditions. The first boundary condition is that $\psi(\pm)=0$. For $x>0$, this translates into requiring that

\begin{align} ka+\delta_0=n\pi \Rightarrow \delta_0=n\pi-ka \end{align}

Thus, we see that

\begin{align} \psi(x) = \sin(k|x-a|+n\pi) \equiv \sin(k|x-a|), \end{align}

Next, we invoke Eq. 6 to obtain a condition on k,

\begin{aligned} 2k\cos(ka)&=\frac{2m}{\hbar^2}\frac{\lambda\hbar^2}{2m}\sin(ka) \\\\ \frac{k}{\lambda}&= \tan(ka) \end{aligned}

C. [7 points] Draw the wavefunctions for the lowest energy even and lowest energy odd state. Show the wavefunctions in the limits of $\lambda\to0$ and $\lambda\to\infty$. Compare with the result of part B.

D. [15 points]


In the limit of large but finite $\lambda$ determine the period of oscillations if at time $t=0$ the particle is in a mixture of the first even state and first odd state,

\begin{align} \psi = \frac{\psi_{\text{even}}+\psi_{\text{odd}}}{\sqrt{2}} \end{align}

(determine the period explicitly — i.e., in terms of m, a, and $\lambda$)

Greg's Solution:

They key here is to recognize that since $\lambda$ is large, the even and odd parts have nearly the same frequency and so the period of oscillation is given by the period of the "beat" of interference between the two.

The lowest energy odd state is a sine wave with one node, so we have that

\begin{align} k_{\text{odd}}=\frac{\pi}{a} \Rightarrow f_{\text{odd}} = \frac{E_{\text{odd}}}{h} = \frac{h}{2ma^2}. \end{align}

To find $k_{\text{even}}$ for the lowest energy even state, we need to solve $\frac{k}{\lambda}= \tan(ka)$. Since $\lambda$ is large, we assume that the left-hand side is nearly zero, which allows us to expand $\tan(ka)$ about $ka\approx \pi$. (This comes from the fact that the left-hand side is nearly a flat line, so the first point at which it intersects the right-hand side is near the second zero of the tangent function.) Since the tangent is approximately linear around its zeros, we have that

\begin{align} \frac{k}{\lambda} = ka-\pi \Rightarrow k = \frac{\pi }{a-\lambda^{-1}} \end{align}


\begin{align} f_{\text{even}} = \frac{h}{2m(a-\lambda^{-1})^2} \end{align}

Since $\lambda$ is large — and in particular $\lambda^{-1}\ll a$ — we can expand the frequency about $1/\lambda a$ using the formula $(1-\epsilon)^{-2}\approx 1+2\epsilon$,

\begin{align} f_{\text{even}} = \frac{h}{2ma^2}\left(1+\frac{2}{a\lambda}\right) \end{align}

Thus, the beat frequency is

\begin{align} \Delta f = f_{\text{even}}-f_{\text{odd}} = \frac{h}{ma^3\lambda} \end{align}

and so the period of a full oscillation is

\begin{align} T = \frac{1}{\Delta f} = \frac{ma^3\lambda}{h} \end{align}

Note that this has the correct limits; if the well is infinitely large, or if the barrier between the two halves of the well is infinitely large, or in the classical limit $h\to 0$, the particle will take infinitely long to switch sides.

Problem 2. [35 points total]

A. [20 points]


Determine, from first principles, the Clebsh-Gordan coefficients for $j_1=1$ and $j_2=1$ forming states with $m=0$.

Hints: Don't try to remember some complicated recustion formulas, just use simple reasoning using elementary concepts (normalization, orthogonality, …) plus

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} J_{\pm}\ket{jm}=(J_x+iJ_y)\ket{jm}=\hbar\sqrt{j(j+1)-m(m\pm1 )}\ket{j,m\pm1} \end{align}

Greg's Solution:

We start with the observation that the state with highest total and z-component angular momentum must be given by

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \ket{2,2}=\ket{1,1}\ket{1,1} \end{align}

After hitting each side with the lowering operator, we obtain

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \ket{2,1}=\frac{1}{\sqrt{2}}\left[\ket{1,0}\ket{1,1} + \ket{1,1}\ket{1,0}\right] \end{align}

(Note that we didn't need to use the formula above to find the coefficients, since we know from symmetry that each of the two terms should have the same coefficient, and thus the only constant should be an overall normalization.)

We hit each side with the lowering operator again to obtain,

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \begin{aligned} \sqrt{6}\ket{2,0}&=2\ket{1,0}\ket{1,0} + \ket{1,1}\ket{1,-1}+ \ket{1,1}\ket{1,-1} \\ \ket{2,0} &= \sqrt{\frac{2}{3}} \ket{1,0}\ket{1,0} + \frac{1}{\sqrt{2}}\left( \ket{1,1}\ket{1,-1}+ \ket{1,1}\ket{1,-1}\right) \end{aligned} \end{align}

Now, we know that the state with total angular momentum 1 must be anti-symmetric, so immediately we can write

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \ket{1,1} = \frac{1}{\sqrt{2}}\left(\ket{1,1}\ket{1,0}-\ket{1,0}\ket{1,1}\right) \end{align}

Now, when we hit both sides with the lowering operator, we will get a symmetric term and an anti-symmetric term. Without doing any computations, we know that the symmetric term must vanish since the right-hand side must be overall anti-symmetric, so we are left with

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \ket{1,0} = \frac{1}{\sqrt{2}}\left(\ket{1,1}\ket{1,-1}-\ket{1,-1}\ket{1,1}\right) \end{align}

B. [15 points]


Express the operator of total angular momentum $J^2$ in terms of $J_i^2,J_{iz}^2,\,\,\text{and}\, J_{i\pm}$ of the two $j=1$ ingrediants (example: $J_z=J_{1z}+J_{2z}$, and use this to show that the state

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \ket{\psi}=\frac{\ket{1,-1}\ket{1,1}+2\ket{1,0}\ket{1,0}+\ket{1,1}\ket{1,-1}}{\sqrt{6}} \end{align}

is an eigenstate of $J^2$.

Greg's Solution:

First, by expanding $J^2=(\vec{J}_1+\vec{J}_2)^2$ we see that

\begin{equation} J^2 = J_1^2 + J_2^2 + 2J_{1z}J_{2z} + J_{1+}J_{2-} + J_{1-}J_{2+} \end{equation}

So now we will apply each of these operators (ignoring the normalization constant) to our state $\left|\psi\right>$ and then add them up to verify that we get an eigenstate.

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \begin{aligned} (J_1^2+J_2^2) \ket\psi &= 8\ket\psi \\ 2J_1J_2 \ket\psi &= - 2\ket\psi + 4\ket{1,0}\ket{1,0} \\ J_{1+}J_{2-} \ket\psi &= 2\ket{1,0}\ket{1,0}+4\ket{1,1}\ket{1,-1} \\ J_{1-}J_{2+} \ket\psi &= 2\ket{1,0}\ket{1,0}+4\ket{1,-1}\ket{1,1} \\ \end{aligned} \end{align}

Thus, we see that

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} J^2\ket\psi = 4\ket\psi, \end{align}


Problem 3. [30 points total]

Consider the two-dimensional harmonics oscillator:

\begin{align} V(x) = \frac{1}{2}\left(k_1x^2+k_2y^2\right) \end{align}

Recall that for a one-dimensional Simple Harmonic Oscillator,

\begin{align} \newcommand{\ket}[1]{\left|#1\right>} x=\sqrt{\frac{\hbar}{2m\om}}(a+a^\dagger),\quad a^\dagger \ket{n_x}=\sqrt{n_x+1}\ket{n_x+1},\quad a\ket{n_x}=\sqrt{n_x}\ket{n_x-1} \end{align}

Determine, at first order in $\lambda$, the effect of a perturbation $H'=\lambda xy$ on the eigen-energies $E_{00},E_{10},E_{01}$ (where $E_{n_xn_y}$ is the energy of the unperturbed state $\left|n_x,n_y\right>$.

A. [5 points] for the non-degenerate case $k_1\ne k_2$

We express the perturbation in terms of raising and lowering operators,

\begin{align} H' = \lambda xy \equiv \frac{\hbar}{2m\sqrt{\om_x\om_y}} (a_x+a_x^\dagger)(a_y+a_y^\dagger) = \frac{\hbar}{2m\sqrt{\om_x\om_y}} (a_xa_y + a_x^\dagger a_y + a_xa_y^\dagger + a_x^\dagger a_y^\dagger) \end{align}

As you can see, each term raises and/or lowers the quantum numbers of a state, so as a result any expectation of the form $\left<n_xn_y\big|H'\big|n_xn_y\right>$ is zero, and so we conclude that to first order the energy levels are unmodified.

B. [10 points] for the case with a degeneracy $k_1=k_2$

Again we have that $\left<n_xn_y\big|H'\big|n_xn_y\right>=0\quad\forall\,\,n_x,n_y$, so naively one might think that again the energies are the same to first order. This is true for the $E_{00}$ energy level, however it does not hold for $E_{10}$ and $E_{01}$ since they are degenerate and so one must us an entirely different machinery with them.

In particular, we need to find a new basis which spans this space, and we do this by diagonalizing the operator

\begin{align} V_{ij} := \left<i\big|H'\big|j\right> \end{align}

If we label the $E_{10}$ state by 1 and the $E_{01}$ by 2, then our matrix becomes

\begin{align} V=\frac{\lambda\hbar}{m\om}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \end{align}

which has eigenvalues $E_\pm := \pm \frac{\lambda\hbar}{m\om}$ with respective corresponding eigenvectors (expressed in the original basis) $\left|\psi_\pm\right>:=\left|1,0\right>\pm\left|0,1\right>$. Thus we have obtained both the first-order perturbation and the "good" set of zero-th order eigenstates.

C. [10 points] Describe the transition between cases A and B — i.e., determine the eigen-energies for the "quasi-degenerate case".

Greg's Solution:

(In the following analysis, we only consider the excited states since the ground state is the same in both the degenerate and non-degenerate cases.)

In this transition regime, the constants $k_1$ and $k_2$ are nearly but not quite equal. We may consider this difference as being a perturbation from the degenerate case — that is,

\begin{align} H = k_1(x^2+y^2) + H' + H'',\quad H'' := \frac{1}{2}(k_2-k_1)y_2^2. \end{align}

To first order, the perturbation that $H''$ introduces to the energies of the excited states $\left|\psi_\pm\right>$ (as found in the previous part) is given by,

\begin{aligned} \delta E_\pm &= \left<\psi_\pm\big|H''\big|\psi_\pm\right>\\ &= \left(\left<1,0\right|\pm\left<0,1\right|\right)H''\left(\left|1,0\right>\pm\left|0,1\right>\right)\\ \end{aligned}

D. [5 points] What is the criterion to apply the approximation in part A and when would you apply the approximation in part B?

When $|k_2-k_1|\gg \lambda$, apply the approximation in part A. Otherwise, apply the approximation in part B.

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