2004 Spring Quantum Mechanics
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1. [30 points total] 1-dimensional Schrödinger Equation

A. [12 points]


A particle of mass m moves in a 1-dimensional potential given by

\begin{align} V(x) = G\delta(x),\quad G\ge 0 \end{align}

Derive the boundary conditions for the wave function and show that the transmission coefficient, $T(E)$, through the `$\delta$ barrier’ for a given energy E is,

\begin{align} T(E) = \frac{1}{1+\frac{mG^2}{2E\hbar^2}} \end{align}

The transmission coefficient is the ratio of the fluxes of probabilities for the transmitted and incident waves.

Greg’s Solution:

Let $\psi_L$ be the wave function in the region left of the δ-function, and $\psi_R$ the the wave function in the region to the right. There are two boundary conditions. First, the wave function must be continuous, so that

\begin{align} \psi_L(0) = \psi_R(0) \end{align}

Second, by integrating the Schrodinger equation from $-\epsilon\to \epsilon$ and taking the limit as $\epsilon\to 0$, we obtain the boundary condition,

\begin{align} \psi_L’(0) = \psi_R’(0) + \frac{2mG}{\hbar^2}\psi_R(0) \end{align}

We now assume that the wave function takes the general scattering form,

\begin{aligned} \psi_L = Ie^{ikx} + Re^{-ikx} \\ \psi_R = Te^{ikx} \\ \end{aligned}

Plugging these into the two boundary conditions, we obtain

\begin{aligned} Ie^{ikx} + Re^{-ikx} &= Te^{ikx} \\ ikIe^{ikx} - ikRe^{-ikx} &= T\left(ik+\frac{2mG}{\hbar^2}\right)e^{ikx} \\ \end{aligned}

Dividing the second equation by $ik$ and then adding it to the first, we obtain

\begin{align} 2I e^{ikx} = T\left(2-i\frac{2mG}{k\hbar^2}\right)e^{ikx} \quad \Rightarrow\quad \left|\frac{T}{I}\right|^2 = \frac{1}{1+\frac{m^2G^2}{k^2\hbar^4}} \equiv \frac{1}{1+\frac{2mG^2}{2E\hbar^2}}, \end{align}

where in the last equality we used the fact that $E=\frac{\hbar^2k^2}{2m}$.

B. Consider the potential

\begin{align} V(x) = \begin{cases} G\delta(x) & -a\le x \le \infty; \quad G\ge 0 \\ \infty & x < -a \end{cases} \end{align}

for which the wave function corresponding to the eigen-state with energy E is,

\begin{align} \chi_E(x) = \begin{cases} \alpha(k)\sin k(x+a) & -a\le x < 0 \\ \beta(k) e^{ikx} + \gamma(k) e^{-ikx} & x > 0 \end{cases} \end{align}

where $k=\frac{\sqrt{2mE}}{\hbar}$, and $\alpha(k),\beta(k),\gamma(k)$ are functions of k.

Consider the situation where the wave function at $t=0$ is given by,

\begin{align} \Psi(x,t=0) = \begin{cases} \sqrt{\frac{2}{a}}\sin\frac{nx }{a} & -a \le x < 0 \\ 0 & x > 0 \end{cases} \end{align}

[Note: This situation resembles particle emission from unbound nuclear states, where a particle is trapped by a barrier and can slowly tunnel through it.]

Write an expression (perhaps a complicated integral that you are not required to solve) for $\Psi_{x<0}(x,t)$, the wave function for $x<0$, $t>0$.

Greg’s Solution:

The time evolution operator is $e^{-iHt/\hbar}$. Thus, what we want to compute is

\begin{aligned} |Psi(t)\right> &= e^{-iHt/\hbar} \left|\Psi(0)\right> \\ &= \int \,dE\, e^{-iEt/\hbar}\left|\chi_E\right>\left<\chi_E\big| \Psi(0)\right> \\ &= \int \,dE\, e^{-iEt/\hbar}\left|\chi_E\right>\int\,dx\,\alpha\left(\frac{\sqrt{2mE}}{\hbar}\right) \sin\left[ \left(\frac{\sqrt{2mE}}{\hbar}\right)(x+a)\right]\sqrt{\frac{2}{a}}\sin\frac{\pi x}{a} \\ \end{aligned}

C. [7 points]


For the potential of point 1.B, consider the quantity,

\begin{align} \bar E = \frac{\int_0^\infty \,dx |\Psi(x,t)|^2 E}{\int_0^\infty |\Psi(x,t)|^2}, \end{align}

which yields the mean energy `outside the δ-potential’. Will this quantity increase, remain constant, or decrease with time? Give a qualitative expression.

Greg’s Solution:

This quantity will increase over time as the particle tunnels through the barrier. This can be seen from the expression above in that although at time $t=0$ the particle may not have any components outside the barrier, its wave function consists of a sum of eigenstates which do have components outside the barrier, and although at one particular instant in time a sum of them was contrived to make everything cancel for $x>0$, at later times the phase of each eigenstate will change independently so it is unlikely that they will continue to cancel in such a fashon.

2. [35 points total] Atom in harmonic oscillator

Consider an atom of mass M and electronic states with energies $E_0$ (ground state) and $E_1=E_0+\hbar\omega_0$ (first excited state). This atom is in a potential $V(x,y,z)=\frac{1}{2}M\Omega^2(x^2+y^2+z^2)$. Assume that the motion of the atom as a whole in the potential and the electronic degrees of freedom are not coupled and that $\omega_0\gg\Omega$.

At time $t=0$ the atom is in the ground state of both electronic and atomic center-of-mass motion. At that time the system starts interacting with an electromagnetic plane wave with wave number k propagating in the z direction. The interaction is of the form,

\begin{align} \hat H_{\text{int}} = \epsilon \hat F(k)(e^{i(k\hat Z-\omega t)}+e^{-i(k\hat Z-\omega t)}) \end{align}

where $\epsilon \ll \hbar\Omega$ has units of energy, $\hat F(k)$ is a hermitian operator that acts only on the internal electronic coordinates, and $\hat Z$ is the operator corresponding to the z-coordinate of the atomic mass.

A. [10 points] Show that, to lowest order in $\epsilon$, the probabilitiy for finding the system in state of oscillation $\left|0,0,n\right>$ with energy $(n+3/2)\hbar\Omega$, and excited electronic state with energy $E_1$ at time t is maximal when $\omega = \pm(\omega_0+\Omega)$.

Greg’s Solution:

To first order in $\epsilon$ we have that

\begin{align} P_{i\to f} = \int \,dt\,\left<0,0,n,E_1\big|\epsilon \hat F(k)(e^{i(k\hat Z-\omega t)}+e^{-i(k\hat Z-\omega t)})\big|0,0,0,E_0\right> e^{it[E_0+\hbar \omega_0-E_0 + (n+3/2)\hbar\Omega - (3/2)\hbar\Omega)]/\hbar} \end{align}

However, observe that we can factor our $e^{-i\omega t} + e^{+i\omega t}$, and then pull it out of the expectation since it is a scalar! Thus,

\begin{align} P_{i\to f} = \int \,dt\,\left<0,0,n,E_1\big|\epsilon \hat F(k)e^{ik\hat Z}\big|0,0,0,E_0\right> \left[e^{-i\omega t} + e^{+i\omega t}\right]e^{it[\omega_0+n\hbar\Omega]} \end{align}

Furthermore, since the expectation does not depend on time, we can pull it outside of the integral,

\begin{align} P_{i\to f} = \left<0,0,n,E_1\big|\epsilon \hat F(k)e^{ik\hat Z}\big|0,0,0,E_0\right> \int \,dt\,\left[e^{-i\omega t} + e^{+i\omega t}\right] e^{it[\omega_0+n\hbar\Omega]} \end{align}

This integral is just a Fourier transform which picks out the component with frequency $\omega_0+n\hbar\Omega$; thus, the integral will have its maximum value when $\om=\pm \omega_0+n\hbar\Omega$.

B. [10 points]


Define lowering and raising operators,

\begin{align} a_Z = \hat Z \sqrt{\frac{M\Omega}{2\hbar}} + i\hat p_Z \sqrt{\frac{1}{2M\hbar\Omega}};\quad \a_Z^\dagger = \hat Z\sqrt{\frac{M\Omega}{2\hbar}} - i\hat p_Z\sqrt{\frac{1}{2M\hbar\Omega}} \end{align}

where $\hat p_Z$ is the momentum operator conjugate to $\hat Z$. Calculate the commutator $[a_Z,a_Z^\dagger]$ and express the matrix element $\left<0,0,n\big|e^{ik\hat Z}\big |0,0,0\right>$ in terms of $a_Z,a_Z^\dagger$.

Greg’s Solution:

When computing the commutator, only the cross terms matter so that

\begin{aligned} [a_Z,a_Z^dagger] = -\frac{1}{2\hbar}[\hat Z,i\hat p_z] + \frac{1}{2\hbar}[i\hat p_z,\hat Z] = 1 \end{aligned}

The expectation in terms of the ladder operators is,

\begin{align} \left<0,0,n|e^{ik\hat Z}|0,0,0\right> = \left<0,0,n|e^{ik(\hat a + \hat a^\dagger)/2\sqrt{M\Omega/2\hbar}}|0,0,0\right> \end{align}

C. [15 points] Calculate $\left<0,0,0|e^{ik\hat Z}|0,0,0\right>$. What is the relationship between $\Omega,M$ and $\omega_0$ such that $|\left<0,0,0|e^{ik\hat Z}|\right>$ is approximately unity? Does this contradict the assumption that $\omega_0\gg \Omega$ above?

Greg’s Solution:

\begin{aligned} \left|\left<0,0,0|e^{ik\hat Z}|0,0,0\right>\right|^2 &= \left\left<0,0,0|e^{ik(\hat a + \hat a^\dagger)/2\sqrt{M\Omega/2\hbar}}|0,0,0\right>\right|^2 \\ &= \left\left<0,0,0|e^{ik\hat a/2\sqrt{M\Omega/2\hbar}}e^{ik\hat a^\dagger/2\sqrt{M\Omega/2\hbar}}e^{ik/4\sqrt{M\Omega/2\hbar}}|0,0,0\right>\right|^2 \\ &= \left\left<0,0,0|e^{ik\hat a/2\sqrt{M\Omega/2\hbar}}e^{ik\hat a^\dagger/2\sqrt{M\Omega/2\hbar}}|0,0,0\right>\right|^2 \\ &= \left\left<0,0,0|\left[\sum_{n=0}^\infty \left(2ik\sqrt{\frac{2\hbar}{M\Omega}}\right)^n \frac{\overbrace{\hat a^n}^{\prod_{m=1^n} \sqrt{m}}}{n!}\right]\left[\sum_{n=0}^\infty \left(2ik\sqrt{\frac{2\hbar}{M\Omega}}\right)^n \frac{\overbrace{\hat a^{\dagger\,\,n}}^{\prod_{m=1^n} \sqrt{m}}}{n!}\right] |0,0,0\right>\right|^2 \\ &= \end{aligned}

3. Molecular Hamiltonian

A system of atoms is arranged in a molecule in an equilateral triangle array as shown in the figure.

The eigenstates of this molecule can be written as linear combinations of states $\left|\alpha_i\right>$ with $i=1,2,3$ such that

\begin{align} \left<\alpha_i|\hat H|\alpha_j\right> = \begin{cases} |\epsilon| & \text{if}\.\. i=j \\ -|t| & \text{if}\,\, i\ne j \end{cases} \end{align}

where $\epsilon$ and $t$ are non-zero constants.

A. [15 points]


An operator $\hat R$ is defined such that

\begin{aligned} \hat R \left|\alpha_1\right> &= \left|\alpha_2\right> \\ \hat R \left|\alpha_2\right> &= \left|\alpha_3\right> \\ \hat R \left|\alpha_3\right> &= \left|\alpha_1\right> \\ \end{aligned}

Prove that $\hat R$ commutes with $\hat H$ and find the eigenvalues and eigenvectors of $\hat R$.

Greg’s Solution:

In matrix form,

\begin{align} H = \begin{bmatrix} \epsilon & -t & -t \\ -t & \epsilon & -t \\ -t & -t & \epsilon \\ \end{bmatrix},\quad R = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} \end{align}

By simple matrix-multiplication, we see that

\begin{align} \hat R \hat H = \hat H \hat R = \begin{bmatrix} -t & -t & \epsilon \\ \epsilon & -t & -t \\ -t & \epsilon & -t \\ \end{bmatrix} \end{align}

By inspection, we see that the eigenvalues and corresponding eigenvectors are

\begin{aligned} 1, &\left|\alpha_1\right>+\left|\alpha_2\right>+\left|\alpha_3\right> \\ e^{-i2\pi/3}, &\left|\alpha_1\right>+e^{+i2\pi/3}\left|\alpha_2\right>+e^{-i2\pi/3}\left|\alpha_3\right> \\ e^{+i2\pi/3}, &\left|\alpha_1\right>+e^{-i2\pi/3}\left|\alpha_2\right>+e^{+i2\pi/3}\left|\alpha_3\right> \\ \end{aligned}

B. [20 points] Calculate the eigenvalues and eigenvectors of H.

Greg’s Solution:

First, we observe that finding the eigenvalues $\lambda$ of H is equivalent to finding the eigenvalues $\lambda’$ of the matrix

\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{bmatrix}

where $\lambda = -t\lambda’ +\epsilon$. After solving the eigenvalue equation, we see that $\lambda’=2,-1,-1$, and so $\lambda=-2t+\epsilon,t+\epsilon,t+\epsilon$. The eigenvector corresponding to $\lambda’=2$ is $\left|\alpha_1\right>+\left|\alpha_2\right>+\left|\alpha_3\right>$; since this is also an eigenvector of $\hat R$, we conclude that the eigenvectors of $\hat H$ corresponding to $\lambda’=-1$ are any linear combination of the other two eigenvectors of $\hat R$.

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