2003 Spring Quantum Mechanics
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1. [55 points total] Electron near a surface

The potential energy of an electron of charge -e, mass m, in the neighborhood of a horizontal surface of a dielectric liquid with dielectric constant $\epsilon>1$ can be taken to be

(1)
\begin{align} V(x,y,z) = \begin{cases} -\frac{e^2(\epsilon-1)}{z(\epsilon+1)} & z > 0 \\ \infty & z < 0 \end{cases} \end{align}

The electron is confined by an infinite potential barrier inside the region,

(2)
\begin{equation} 0<x<L_1,0<y<L_2. \end{equation}

A.

Problem:

Show that the time-independent Schrodinger equation, with a potential that is a function only of z, has stationary solutions, separable in Cartesian coordinates, of the form,

(3)
\begin{align} \psi(x,y,z) = A f(x)g(y)h(z), \end{align}

and find the possible forms of $f(x),g(y)$, when the electron is confined in the region $0<x<L_1,0<y<L_2$.

Greg’s Solution:

We set $\hbar^2/2m=1$ for convenience. (We can put it back in at the end using dimensional analysis.) Then the Schrodinger equation for this potential in the confining region takes the form,

(4)
\begin{align} -\partial_x^2 \psi -\partial_y^2 \psi - \partial_z^2 \psi - \frac{e^2}{z}\frac{\epsilon-1}{\epsilon+1} \psi = E\psi \end{align}

We make the Ansatz that $\psi$ is separable in Cartesian coordinates,

(5)
\begin{align} \psi = f(x)g(y)h(z). \end{align}

Plugging 5 into [[preschro]], we obtain

(6)
\begin{align} -f’‘gh - fg’‘h - fgh’‘ - \frac{e^2}{z}\frac{\epsilon-1}{\epsilon+1}fgh = Efgh \end{align}

We then divide by $fgh$ to obtain,

(7)
\begin{align} \underbrace{-\frac{f’‘}{f}}_{\text{function of} \,\,x} +\underbrace{-\frac{g’‘}{g}}_{\text{function of}\,\, y} + \underbrace{-\frac{h’‘}{h} -\frac{e^2}{z}\frac{\epsilon-1}{\epsilon+1}}_{\text{function of} \,\,z} = E. \end{align}

Each grouped set of terms above is a function of only a single variable; thus, we conclude that each group is in fact a constant. Intuitively, one can see this from the fact that if one group were not, then that one would be able to change it independently of all the other terms in such a way as to violate the inequality.

Alternatively, we can also see this formerly by taking partial derivatives. For example, taking the partial derivative with respect to x obtains,

(8)
\begin{align} \frac{\partial}{\partial x}\left[-\frac{f’‘}{f}\right] = 0 \Rightarrow f’‘=- k_x^2 f, \end{align}

where $k_x$ is a constant. Applying the boundary condition that $0<x<L_1$, one obtains the solution

(9)
\begin{align} f(x) \propto \sin(k_x x),\quad k_x = \frac{n\pi}{L_1}, \quad n=1,2,\dots \end{align}

The constant in front is determined from the normalization condition.

Doing the same for the y-dependent term, we find that

(10)
\begin{align} g(y) \propto \sin(k_y y),\quad k_y = \frac{m\pi}{L_2}, \quad m=1,2,\dots \end{align}

B. [18 points]

Problem:

For the potential of Eq. 1, show that one solution of the equation for the z-dependent part is

(11)
\begin{align} h(z)=z^{\alpha} e^{-z/l_0}. \end{align}

Find expressions for the index $\alpha$, the length $l_0$, and the energy E of the solution $\psi(x,y,z)$.

Greg’s Solution:

We make the Ansatz,

(12)
\begin{align} h(z) = z^\alpha e^{-z/l_0} \end{align}

and then take derivatives,

(13)
\begin{aligned} h’(z) &= \alpha z^{\alpha-1} e^{-z/l_0} - \frac{z^{\alpha}}{l_0}e^{-z/l_0} \\ h’‘(z) &=\alpha(\alpha-1) z^{\alpha-2} e^{-z/l_0} -\frac{2\alpha z^{\alpha-1}}{l_0} e^{-z/l_0} + \frac{z^{\alpha}}{l_0^2}e^{-z/l_0} \end{aligned}

From the z-terms in Eq. $postschro$, we obtain the equation

(14)
\begin{align} h’‘ + \frac{e^2}{z}\frac{\epsilon-1}{\epsilon+1} h = -E_z h \end{align}

We plug in 12 and 13 to obtain,

(15)
\begin{align} \frac{z^\alpha}{l_0^2}e^{-z/l_0} -\frac{2\alpha z^{\alpha-1}}{l_0}e^{-z/l_0} + \alpha(\alpha-1) z^{\alpha-2}e^{-z/l_0} + e^2\frac{\epsilon-1}{\epsilon+1}z^{\alpha-1}e^{-z/l_0} + E_z z^\alpha e^{-z/l_0} = 0 \end{align}

Dividing $\epsilon^{-z/l_0}$ from both sides and grouping powers of z, we rewrite the above to see that

(16)
\begin{align} \alpha(\alpha-1) z^{\alpha-2}e^{-z/l_0} + \left(e^2\frac{\epsilon-1}{\epsilon+1} - \frac{2\alpha}{l_0}\right)z^{\alpha-1}+ \left(\frac{1}{l_0^2}+ E_z \right)z^\alpha = 0 \end{align}

Since each coefficient must independently vanish, we have that $\alpha=0,1$, $E_z=-\frac{1}{l_0^2}$, and $l_0=\frac{2\alpha(\epsilon+1)}{e^2(\epsilon-1)}$. Now, $\alpha=0\Rightarrow l_0=0\Rightarrow C=\infty$, so we conclude that $\alpha=1$, and therefore $l_0=\frac{2(\epsilon+1)}{e^2(\epsilon-1)}$ and $E_z=-\frac{e^4(\epsilon-1)^2}{4(\epsilon+1)^2}$.

Thus, the energy of the solution is given by,

(17)
\begin{align} E = \left(\frac{n\pi}{L_1}\right)^2 + \left(\frac{m\pi}{L_2}\right) - \frac{e^4(\epsilon-1)^2}{4(\epsilon+1)^2} \end{align}

Observe that the first two terms have dimensions of $1/\text{length}^2$, and the last term has units of $\text{energy}^2\text{length}^2$ (since dividing that term by $z^2/2$ gives $V^2$ which has dimensions of $\text{energy}^2$. Thus, to express all terms in SI units of energy, we need to multiply the first two terms by $\hbar^2/2m$ (which has units of $\text{energy}\cdot\text{\length}^2$) and divide the last term by the same, obtaining

(18)
\begin{align} E_{\text{SI}} = \frac{\hbar^2}{2m}\left(\frac{n\pi}{L_1}\right)^2 + \frac{\hbar^2}{2m}\left(\frac{m\pi}{L_2}\right) - \frac{2m}{\hbar^2} \frac{e^4(\epsilon-1)^2}{4(\epsilon+1)^2} \end{align}

C. [10 points] What is the mean distance of an electron from the surface $z=0$ in terms of the length $l_0$?

Since our wavefunction is unnormalized, the mean distance is given by,

(19)
\begin{align} \left<z\right> = \frac{\left<\psi\bigg|z\bigg|\psi\right>}{\left<\psi\bigg|\psi\right>} \end{align}

Since $\psi=f(x)g(y)h(z)$ with $h(z)=z e^{-z/l_0}$, we have that

(20)
\begin{align} \left<z\right> = \frac{\int_0^\infty z^3 e^{-2z/l_0}}{\int_0^\infty z^2 e^{-2z/l_0}} \end{align}

To compute these integrals, we recall that

(21)
\begin{align} \int_0^\infty z^n e^{-\lambda z} = \frac{(-1)^n}{n!}\frac{\partial^n}{\partial \lambda^n} \int_0^\infty e^{-\lambda z} = \frac{(-1)^n}{n!}\frac{\partial^n}{\partial \lambda^n} \frac{1}{\lambda} = \frac{1}{\lambda^n}, \end{align}

and so

(22)
\begin{align} \left<z\right> = \frac{l_0^3/8}{l_0^2/4} = \frac{l_0}{2}. \end{align}

D. [[12 points]] The dielectric constant of liquid helium is 1.057. Assuming that $L_1,L_2$ are large, and using the facts that the binding energy of the electron in a hydrogen atom is $E_H=\frac{me^4}{2\hbar^2}=13.6\,\text{eV}$, and the Bohr radius is $a_0=\frac{\hbar^2}{me^2}=0.053\text{nm}$, deduce the binding energy of an electron to the surface of liquid helium and its mean distance from the surface. You can assume that the solution, Eq. 11, gives the lowest possible energy.

Greg’s Solution:

Since the lengths $L_1,L_2$ are large, we may neglect the parts of the energy which came from the x and y constraints. Thus,

(23)
\begin{align} E \approx - \frac{2m}{\hbar^2} \frac{e^4(\epsilon-1)^2}{4(\epsilon+1)^2}. \end{align}

Since we are given that $\frac{2m e^4}{\hbar^2}=13.6\,\text{eV}$, and $\left(\frac{.057}{2.057}\right)^2 \approx \frac{1}{41^2} \approx .03$, we have that the binding energy is given by,

(24)
\begin{align} E \approx 4\times 13.6\,\text{eV} \times .03 \approx 1.4 \,\text{eV} \end{align}

Recall that the mean distance from the surface is given by,

(25)
\begin{align} \left<\psi\bigg| z\bigg|\right> = \frac{1}{2}\frac{2(\epsilon+1)}{e^2(\epsilon-1)} = \frac{(\epsilon+1)}{e^2(\epsilon-1)} \end{align}

To convert to SI units, we note that the above has dimensions of $\frac{1}{\text{energy}\cdot\text{length}}$ and so we need to multiply by $\hbar^2/2m$ which has dimensions of $\text{energy}\cdot\text{\length}^2$, to obtain a length; thus,

(26)
\begin{align} \left<\psi\bigg| z\bigg|\right>_{\text{SI}} = \frac{\hbar^2(\epsilon+1)}{2me^2(\epsilon-1)} = a_0 \frac{41}{2} = 0.053\,\text{nm} \times 20.5 \approx 1\,\text{nm}. \end{align}

2. [[45 points total]] Oscillating potential

The raising and lowering operators $\hat b_j^\dagger, \hat b_j$ for the three modes of oscillation of a particle of mass M in a potential of the form

(27)
\begin{align} V(\hat x_j) = V_0 + \frac{1}{2}\sum_{j=1}^3 K_j \hat x_j^2 \end{align}

satisfy the commutation relations

(28)
\begin{align} [\hat b_j,\hat b_{j’}] = [\hat b_j,\hat b_{j’}]=0,[\hat b_j,\hat b_{j’}^\dagger] =\delta_{jj’}. \end{align}

The position and momentum operators are related to the raising and lowering operators by

[[math position-and-momentum-operators]]
\hat x_j = \sqrt{\frac{\hbar}{2\sqrt{K_j M}}}(\hat b_j^\dagger+\hat b_j),\quad \hat p_j=i\sqrt{\frac{\hbar\sqrt{K_j M}}{2}}(\hat b_j^\dagger -\hat b_j)
[[/math]]

A. [5 points] Show that the possible eigenvalues of $\hat n_j = \hat b_j^\dagger \hat b_j$ are non-negative.

Greg’s Solution:

Suppose that we have some eigenket $\left|\psi\right>$ of $\hat n_j$ with eigenvalue $n$. The requirement that the norm of all kets (including in particular the ket $b_j\ket{\psi}$) must be nonnegative implies that,

(29)
\begin{aligned} \left|\hat b_j\left|\psi\right>\right|^2 \ge 0 \\ \left<\psi\bigg|\hat b_j^\dagger \hat b_j\bigg|\psi\right> \ge 0 \\ n \ge 0, \end{aligned}

and so we conclude that eigenvalues of $\hat n_j$ must be non-negative.

B. [5 points] Show that $\hat b_j\hat b_j^\dagger=\hat n_j+1$.

Greg’s Solution:

(30)
\begin{aligned} \hat b_j\hat b_j^\dagger &= \hat b_j^\dagger \hat b_j + [\hat b_j,\hat b_j^\dagger] \\ &= \hat b_j^\dagger \hat b_j + 1 \\ &= \hat n_j + 1 \\ \end{aligned}

C. [10 points] What are the commutators $[\hat n_j,\hat b_{j’}]$ and $[\hat n_j,\hat b_{j’}^\dagger]$, and why are operators $b_j^\dagger,b_j$ called raising and lowering operators?

Greg’s Solution:

(31)
\begin{aligned} \left[\hat n_j,\hat b_{j’}\right] &= \left[\hat b_j^\dagger \hat b_j,\hat b_{j’}\right] \\ &= \hat b_j^\dagger \left[\hat b_j,\hat b_{j’}\right] + \left[\hat b_j^\dagger,\hat b_{j’}\right] \hat b_j\\ &= -\hat b_j \delta_{jj’}\\ \end{aligned}

Putting daggers on the $\hat b_{j’}$ operators, it is straightforward to see that

(32)
\begin{align} \left[\hat n_j,\hat b_{j’}\right] = +\hat b_j^\dagger \delta_{jj’} \end{align}

$\hat b_j$ is called a lowering operator because if $\left|n\right>$ is a eigenket of $\hat n_j$ with eigenvalue n, then we have that

(33)
\begin{align} \newcommand{\ket}[1]{\left|#1\right>} \begin{aligned} \hat n_j \hat b_j \ket{n} &= \left(\hat b_j\hat n_j +[\hat n_j,\hat b_j]\right)\ket {n} = (n-1)\ket{n} \\ \hat n_j \hat b_j^\dagger \ket{n} &= \left(\hat b_j^\dagger\hat n_j +[\hat n_j,\hat b_j^\dagger]\right)\ket {n} = (n+1)\ket{n} \end{aligned} \end{align}

and so we see that $b_j\left|n\right>$ and $b_j^\dagger\left|n\right>$ are eigenkets of $\hat n_j$ with respective eigenvalues $n-1$ and $n+1$, and thus we call $b_j$ and $b_j^\dagger$ respsectively lowering and raising operators.

D. [15 points] Use Eq. to deduce the expectation values of $\hat x_j^2$ and $\hat p_j^2$ in an eigenstate of $\hat n_j$ with eigenvalue $N_j$.

Greg’s Solution:

Observe that

(34)
\begin{align} \hat x_j^2 = \frac{\hbar}{2\sqrt{K_jM}}\left(\hat b_j^{\dagger\,2} + \underbrace{\hat b_j^\dagger \hat b_j}_{\hat n_j} + \underbrace{\hat b_j \hat b_j^\dagger}_{\hat n_j+1} + \hat b_j^2\right) \end{align}

and so

(35)
\begin{align} \left<N_j\bigg| \hat x_j^2 \bigg|N_j\right> = \frac{\hbar}{2\sqrt{K_jM}}\left(2N_j+1\right). \end{align}

Likewise,

(36)
\begin{align} \hat p_j^2 = \frac{-2\sqrt{K_jM}}{\hbar}\left(\hat b_j^{\dagger\,2} - \underbrace{\hat b_j^\dagger \hat b_j}_{-\hat n_j} - \underbrace{\hat b_j \hat b_j^\dagger}_{-\hat n_j-1} + \hat b_j^2\right) \end{align}

and so

(37)
\begin{align} \left<N_j\bigg| \hat p_j^2 \bigg|N_j\right> = -\frac{2\sqrt{K_jM}}{\hbar}\left(-2N_j-1\right) = \frac{2\sqrt{K_jM}}{\hbar}\left(2N_j+1\right). \end{align}

E. [10 points] A set of noninteracting bosons is in the ground sate of such a harmonic trap. The trap is suddenly switched off, so that the potential is now uniform. In which direction will the particles spread out most rapidly, and why? Take $K_1<K_2<K_3$.

Greg’s Solution:

Intuitively, one would guess that they’d spread out in the direction in which they were most “bunched up” — that is, the direction in which the strongest force was applies, direction 3. We can see this directly from the expectation of the root mean square of the magnitude,

(38)
\begin{align} \sqrt{\left<\hat p_j^2 \right>} \propto 4 K_j M, \end{align}

which is proportional to the magnitude of the constraining force, $K_j$. Thus, since $K_3>K_2>K_1$, we expect that the particles will spread out most rapidly in direction 3 since that is where they have the most momentum.

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