**1.** [30 points total] 1-dimensional Schrödinger Equation

**A.** [12 points]

Problem:

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

(1)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,

(2)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

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

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

(5)Plugging these into the two boundary conditions, we obtain

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

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

**B.** Consider the potential

(8)
for which the wave function corresponding to the eigen-state with energy E is,

(9)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,

(10)[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

(11)**C.** [7 points]

Problem:

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

(12)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,

(13)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

(14)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,

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

(16)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]

Problem:

Define lowering and raising operators,

(17)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

(18)The expectation in terms of the ladder operators is,

(19)**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:

(20)**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

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

**A.** [15 points]

Problem:

An operator $\hat R$ is defined such that

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

Greg’s Solution:

In matrix form,

(23)By simple matrix-multiplication, we see that

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

(25)**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

(26)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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