SM358 - 2010

Part (a)
The ground state is characterized by the quantum numbers


 * $$(n_x,n_y,n_z)=(1,1,1).$$

The first excited state is characterized by


 * $$(n_x,n_y,n_z)=(1,1,2)\ \mathrm{or}\ (1,2,1)\ \mathrm{or}\ (2,1,1).$$

Part (b)
The energy of the photon emitted upon transition from the first excited state to the ground state is



\begin{align} E_\mathrm{ph} &=\left| \tfrac{(1^2+1^2+2^2)\hbar^2\pi^2}{2mL^2}-\tfrac{(1^2+1^2+1^2)\hbar^2\pi^2}{2mL^2} \right| \\ &=\left| 6\times 9.90-3\times 9.90 \right|\ \mathrm{eV} \\ &=29.7\ \mathrm{eV} \\ &=4.752\times 10^{-18}\ \mathrm{J}. \end{align} $$

The frequency of a photon with this energy is


 * $$f=\tfrac{E_\mathrm{ph}}{2\pi\hbar}=7.135\times 10^{15}\ \mathrm{Hz}.$$

Part (a)
The name of the rule is Born's rule, and it states that the probability of finding a particle in a region of size $$\delta x$$ centred on $$x$$ is given by


 * $$\mathrm{probability}=|\Psi(x,t)|^2\delta x.$$

Part (b)
The probability P of finding the particle in the stated range is



P = \tfrac{2}{L}\int_{-L/3}^{L/3}\!\!\cos^2{\left(\tfrac{3\pi x}{L}\right)}\ \mathrm dx $$

Making the substitutions


 * $$u=\tfrac{3\pi x}{L},\ \ \mathrm dx=\tfrac{L}{3\pi}\mathrm du,\ \ \mathrm{upper}\ u=\pi,\ \ \mathrm{lower}\ u=-\pi,$$

enables us to rewrite the integral as


 * $$P=\tfrac{2}{3\pi}\int_{-\pi}^{\pi}\!\!\cos^2u\ \mathrm du.$$

The provided integral gives


 * $$P=\tfrac{2}{3\pi}\times \pi=\tfrac{2}{3}.$$

Part (a)
If the state is normalised, then



\begin{align} 9|C|^2&=1 \\ \end{align} $$
 * C|^2+\left|-2\mathrm i\right|^2|C|^2+2^2|C|^2&=1 \\
 * C|^2&=\tfrac{1}{9}.

The probability of measuring each energy is then



\begin{align} P(E_1)&= |C|^2&&=\tfrac{1}{9}, \\ P(E_2)&= \left| -2\mathrm i \right|^2|C|^2&&=\tfrac{4}{9}, \\ P(E_3)&= 2^2|C|^2&&=\tfrac{4}{9}. \end{align} $$

Part (b)
The expectation value for energy is



\begin{align} \langle E \rangle &= \sum_i p_iE_i \\ &= \tfrac{1}{9}E_1+\tfrac{4}{9}E_2+\tfrac{4}{9}E_3 \\ &= \tfrac{1}{9}\cdot \varepsilon+\tfrac{4}{9}\cdot 2\varepsilon+\tfrac{4}{9}\cdot 3\varepsilon \\ &= \tfrac{7}{3}\varepsilon. \end{align} $$

Part (a)
To answer this question you need to know the position operator in terms of raising and lowering operators. This is given in the equation booklet, or you could reverse engineer it from part (b).


 * $$\widehat{\mathrm{x}}=\tfrac{a}{\sqrt 2}(\widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger).$$

The expectation value of $$x$$ is then



\begin{align} \langle x \rangle &= \tfrac{a}{\sqrt 2}\langle \psi_3|\widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger |\psi_3\rangle \\ &= \tfrac{a}{\sqrt 2}\langle \psi_3|\widehat{\mathrm{A}}\psi_3 \rangle + \tfrac{a}{\sqrt 2}\langle \psi_3|\widehat{\mathrm{A}}^\dagger \psi_3  \rangle \\ &= \tfrac{\sqrt 3 a}{\sqrt 2}\langle \psi_3 | \psi_2 \rangle + \tfrac{2a}{\sqrt 2}\langle \psi_3|\psi_4  \rangle. \end{align} $$

Since $$\psi_2$$, $$\psi_3$$ and $$\psi_4$$ are all mutually orthogonal, we have


 * $$\langle x \rangle = 0,$$

as expected.

Part (b)
We don't need to worry about the parts of the operator which are twice-lowering or twice-raising, since they will necessarily produce an orthogonal pair and therefore will make no contribution to the result. The expectation value of $$x^2$$ is then

$$ \begin{align} \langle x^2 \rangle &= \tfrac{a^2}{2}\langle \psi_3|\widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger+\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}|\psi_3   \rangle \\ &= \tfrac{a^2}{2}\left( \langle \psi_3|\widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger\psi_3 \rangle+\langle \psi_3|\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}\psi_3  \rangle     \right) \\ &= \tfrac{a^2}{2}\left( 2\times 2\times \langle\psi_3|\psi_3\rangle+\sqrt 3\times \sqrt 3 \times \langle\psi_3|\psi_3\rangle         \right) \\ &=\tfrac{a^2}{2}(4+3) \\ &=\tfrac{7a^2}{2}. \end{align} $$

Part (a)
If $$\widehat{\mathrm{B}}$$ is Hermitian, we can write


 * $$\langle \widehat{\mathrm{B}}\Psi|\widehat{\mathrm{B}}\Psi \rangle = \langle \Psi|\widehat{\mathrm{B}}^2|\Psi \rangle.$$

This expression has the form of the sandwich integral, which gives the expectation value of the observable $$B^2$$ in the state $$\Psi$$, so


 * $$\langle \Psi|\widehat{\mathrm{B}}^2|\Psi \rangle=\langle B^2\rangle.$$

Part (b)
The expectation values for $$B$$ and $$B^2$$:


 * $$\begin{align}\langle B\rangle&=\langle\Psi|\widehat{\mathrm{B}}|\Psi\rangle&\langle B^2\rangle&=\langle\Psi|\widehat{\mathrm{B}}^2|\Psi\rangle\\&=\langle\Psi|\widehat{\mathrm{B}}\Psi\rangle&&=\langle\widehat{\mathrm{B}}\Psi|\widehat{\mathrm{B}}\Psi\rangle\\&=\tfrac{1}{2}(\langle f|-\mathrm i\langle g|)(2|f\rangle+3\mathrm i|g\rangle)&&=\tfrac{1}{2}(2\langle f|-3\mathrm i\langle g|)(2|f\rangle+3\mathrm i|g\rangle)\\&=\tfrac{1}{2}(2\langle f|f\rangle-3\mathrm i^2\langle g|g\rangle)&&=\tfrac{1}{2}(4\langle f|f\rangle-9\mathrm i^2\langle g|g\rangle)\\&=\tfrac{1}{2}(2+3)&&=\tfrac{1}{2}(4+9)\\&=\tfrac{5}{2}.&&=\tfrac{13}{2}.\end{align}$$

The uncertainty in $$B$$:



\Delta B=\sqrt{\langle B^2\rangle-\langle B\rangle^2}=\sqrt{\tfrac{13}{2}-\tfrac{25}{4}}=\tfrac{1}{2}. $$

Part (a)
Acting on an arbitrary state $$f(x)$$:



\begin{align} \left[ \widehat{\mathrm{p}}_x,\widehat{\mathrm{x}}^4 \right] &=\widehat{\mathrm{p}}_x\widehat{\mathrm{x}}^4f(x)-\widehat{\mathrm{x}}^4\widehat{\mathrm{p}}_xf(x) \\ &=-\mathrm i\hbar\tfrac{\partial}{\partial x}(x^4f(x))+x^4\mathrm i\hbar\tfrac{\partial}{\partial x}f(x) \\ &=-\mathrm i\hbar\left( x^4\tfrac{\partial}{\partial x}f(x)+4x^3f(x)   \right)+x^4\mathrm i\hbar\tfrac{\partial}{\partial x}f(x) \\ &=-4\mathrm i\hbar x^3f(x) \end{align} $$

The commutator of $$\widehat{\mathrm{p}}_x$$ and $$\widehat{\mathrm{x}}^4$$ can therefore be written as


 * $$\left[\widehat{\mathrm{p}}_x,\widehat{\mathrm{x}}^4\right]=-4\mathrm i\hbar x^3$$

Part (b)
The generalized Ehrenfest theorem:


 * $$\tfrac{\mathrm d}{\mathrm dt}\langle p_x\rangle=\tfrac{1}{\mathrm i\hbar}\left\langle\left[\widehat{\mathrm{p}}_x,\widehat{\mathrm{H}}\right]\right\rangle.$$

If we call the kinetic part of the Hamiltonian $$\widehat{\mathrm{E}}$$, and the potential part $$\widehat{\mathrm{V}}$$, the commutator in the above becomes


 * $$\left[\widehat{\mathrm{p}}_x,\widehat{\mathrm{E}}+\widehat{\mathrm{V}}\right]=\widehat{\mathrm{p}}_x\widehat{\mathrm{E}}+\widehat{\mathrm{p}}_x\widehat{\mathrm{V}}-\widehat{\mathrm{E}}\widehat{\mathrm{p}}_x-\widehat{\mathrm{V}}\widehat{\mathrm{p}}_x.$$

By writing them out fully, it can be shown that the momentum-kinetic parts sum to zero, and we are left with the momentum-potential parts.



\begin{align} \tfrac{\mathrm d}{\mathrm dt}\langle p_x\rangle &=\tfrac{1}{\mathrm i\hbar}\left\langle  \widehat{\mathrm{p}}_x\widehat{\mathrm{V}}-\widehat{\mathrm{V}}\widehat{\mathrm{p}}_x    \right\rangle \\ &=\tfrac{1}{\mathrm i\hbar}\left\langle  -\mathrm i\hbar\tfrac{\partial}{\partial x}Cx^4+Cx^4\mathrm i\hbar\tfrac{\partial}{\partial x} \right\rangle \\ &=\tfrac{1}{\mathrm i\hbar}\left\langle  C[\widehat{\mathrm{p}}_x,\widehat{\mathrm{x}}^4] \right\rangle \\ &=\tfrac{1}{\mathrm i\hbar}\left\langle -4\mathrm i\hbar Cx^3 \right\rangle \\ &=-4C\langle x^3 \rangle. \end{align} $$

The rate of change of momentum is proportional to the expectation value of $$x^3$$, so is unchanging in time. This is because the expectation value of $$x^3$$ is given by


 * $$\langle x^3\rangle=\int_{-\infty}^{\infty}\!\!\Psi^*(x,t)\ \widehat{\mathrm{x}}^3\ \Psi(x,t)\ \mathrm dx.$$

This evaluates to zero, since it is the integral of an odd function over a symmetrical region. Therefore the rate of change of momentum with time is zero.

Part (a)
Any particle with half-integer spin is a fermion by definition. Here we have an odd number of fermions; any odd multiple of a half-integer value must be another half-integer value, so the system as a whole has half-integer spin. Since the system has half-integer spin, it is a fermion.

Part (b)
A system of two identical fermions (the $$^+\mathrm{He}$$ atoms) must have a total wave function which is antisymmetric. The given spin ket is symmetric, so the spatial wave function must be antisymmetric to ensure an antisymmetric total. Since the spatial state is antisymmetric, the particles will tend to separate.

Part (c)
No, the answer will not change. A system of identical bosons must have a total wave function which is symmetric. The spin ket is antisymmetric, so the spatial state must also be antisymmetric to ensure a symmetric total. An antisymmetric spatial state indicates that the bosons will tend to separate.

Part (a)
An entangled state cannot be factorized into a product of two terms, each representing the independent state of a single particle. In an entangled system, the individual particles do not have independent states.

Part (b)
State $$|A\rangle$$ and $$|C\rangle$$ can be factorized;



\begin{align} \end{align} $$
 * A\rangle &= \tfrac{1}{\sqrt 2}(|\!\uparrow\uparrow\rangle-|\!\downarrow\downarrow\rangle) &&=\tfrac{1}{\sqrt 2}|\!\uparrow\rangle(|\!\downarrow\rangle-|\!\uparrow\rangle), \\
 * C\rangle &= \tfrac{1}{\sqrt{10}}(|\!\uparrow\uparrow\rangle+2|\!\downarrow\uparrow\rangle+|\!\downarrow\uparrow\rangle+2|\!\downarrow\downarrow\rangle) &&=\tfrac{1}{\sqrt{10}}(|\!\uparrow\rangle+|\!\downarrow\rangle)(2|\!\downarrow\rangle+|\!\uparrow\rangle).

State $$|B\rangle$$ cannot be factorized and so it is an entangled state.

Part (c)
The Eckert protocol relies on entanglement, since it involves a pair of entangled photons - one sent to each end of a communication channel.

Quantum teleportation relies on entanglement, since it makes similar use of a pair of entangled qubits, called an EPR pair.

Part (a)
The eigenvalues of $$\widehat{\mathrm{L}}^2$$ are $$l(l+1)\hbar^2$$. Applying the operator to the state 'picks out' the values of $$l$$, of which there is only one in this case.



\begin{align} \widehat{\mathrm{L}}^2|A\rangle &=l(l+1)\hbar^2|A\rangle \\ &=3(3+1)\hbar^2|A\rangle \\ &=12\hbar^2|A\rangle. \end{align} $$

The value $$L^2=12\hbar^2$$ will be measured with certainty.

Part (b)
The eigenvalues of $$\widehat{\mathrm{L}}_z$$ are $$m\hbar$$, where $$m$$ is an integer. This state supplies two possible values of $$m$$, so the possible measurements of $$L_z$$ are $$\hbar$$ and $$2\hbar$$.

Part (c)


\begin{align} \widehat{\mathrm{J}}_z|A\rangle &=(\widehat{\mathrm{L}}_z+\widehat{\mathrm{S}}_z)|A\rangle \\ &=(\widehat{\mathrm{L}}_z+\widehat{\mathrm{S}}_z)(f(r)Y_{3,1}(\theta,\phi)|\!\uparrow\rangle+g(r)Y_{3,2}(\theta,\phi)|\!\downarrow\rangle) \\ &= \widehat{\mathrm{L}}_zf(r)Y_{3,1}(\theta,\phi)|\!\uparrow\rangle+ \widehat{\mathrm{S}}_zf(r)Y_{3,1}(\theta,\phi)|\!\uparrow\rangle+ \widehat{\mathrm{L}}_zg(r)Y_{3,2}(\theta,\phi)|\!\downarrow\rangle+ \widehat{\mathrm{S}}_zg(r)Y_{3,2}(\theta,\phi)|\!\downarrow\rangle \\ &= \hbar f(r)Y_{3,1}(\theta,\phi)|\!\uparrow\rangle+ \tfrac{\hbar}{2}f(r)Y_{3,1}(\theta,\phi)|\!\uparrow\rangle+ 2\hbar g(r)Y_{3,2}(\theta,\phi)|\!\downarrow\rangle- \tfrac{\hbar}{2}g(r)Y_{3,2}(\theta,\phi)|\!\downarrow\rangle \\ &=\tfrac{3\hbar}{2}|A\rangle. \end{align} $$

The value $$J_z=3\hbar/2$$ will be measured with certainty.

Part (a)
The scaled Bohr radius is



\begin{align} a_0^\mathrm{scaled} &=\tfrac{1}{Z}\tfrac{\mu_\mathrm H}{\mu}a_0 \\ &=\tfrac{1}{5}\times\tfrac{\left(\tfrac{1840\times 1}{1841 }\right)}{\left(\tfrac{11\times 1840\times 207}{11\times 1840+207 }\right)}\times 5.29\times 10^{-11}\ \mathrm m \\ &=5.16\times 10^{-14}\ \mathrm m. \end{align} $$

Part (b)
We need the scaled Rydberg energy, which is



\begin{align} E_\mathrm{R}^\mathrm{scaled} &=Z^2\tfrac{\mu}{\mu_\mathrm H}E_\mathrm R \\ &=69705.35\ \mathrm{eV}. \end{align} $$

The energy of the emitted photon is



\begin{align} E_\mathrm{ph}&=\left| \tfrac{E_\mathrm{R}^\mathrm{scaled}}{3^2} - \tfrac{E_\mathrm{R}^\mathrm{scaled}}{1^2}   \right| \\ &=61,960\ \mathrm{eV}. \end{align} $$

Part (a)
In the central-field approximation the atomic orbitals split into atomic terms as a result of the perturbation due to residual electron-electron interactions (repulsive).

For a configuration with the non-equivalent open shell electron states $$({{l}_{1}}=2,{{s}_{1}}=\tfrac{1}{2})$$and$$({{l}_{2}}=3,{{s}_{2}}=\tfrac{1}{2})$$we have


 * $${{L}_{\min }}=\left| {{l}_{1}}-{{l}_{2}} \right| =\left| 2-3 \right| = 1$$
 * $${{L}_{\max }}=\left| {{l}_{1}}+{{l}_{2}} \right| =\left| 2+3 \right| = 5$$
 * $${{S}_{\min }}=\left| {{s}_{1}}-{{s}_{2}} \right| =\left| \tfrac{1}{2}-\tfrac{1}{2} \right| = 0$$
 * $${{S}_{\max }}=\left| {{s}_{1}}+{{s}_{2}} \right| =\left| \tfrac{1}{2}+\tfrac{1}{2} \right|= 1$$
 * $$L=L_{\min}, L_{\min} +1, \ldots, L_{\max} -1, L_{\max}$$
 * $$S=S_{\min}, S_{\min} +1, \ldots, S_{\max} -1, S_{\max}$$

So in $$(L,S)$$ notation the atomic terms are



\begin{matrix} (1,0) & (1,1) \\   (2,0) & (2,1)  \\   (3,0) & (3,1)  \\   (4,0) & (4,1)  \\   (5,0) & (5,1)  \\ \end{matrix} $$

Part (b)
The atomic terms split into levels as a result of perturbations due to spin-orbit interactions. In the $$LS$$ regime the good quantum numbers are $$J, L, S$$ where


 * $${{J}_{\min }}=\left| L-S \right|$$
 * $${{J}_{\max }}=\left| L+S \right|$$
 * $$J=J_{\min}, J_{\min} +1, \ldots, J_{\max} -1, J_{\max}$$

So for the term $$(5,1)$$ the atomic levels in $$(J,L,S)$$ notation the levels are



\begin{matrix} (4,5,1) \\  (5,5,1) \\   (6,5,1) \\ \end{matrix} $$

Part (a)
The classical Hamiltonian function for a harmonic oscillator is


 * $$H=\tfrac{p^2_x}{2m}+\tfrac{1}{2}m\omega_0^2x^2.$$

We get the Hamiltonian operator by replacing momentum and position by their corresponding quantum-mechanical operators:


 * $$\widehat{\mathrm{p}}_x=-\mathrm i\hbar\tfrac{\partial}{\partial x},\ \ \ \ \widehat{\mathrm{x}}=x.$$

The Hamiltonian operator is therefore


 * $$\widehat{\mathrm{H}}=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}+\tfrac{1}{2}m\omega_0^2x^2,$$

and the TISE for a harmonic oscillator is


 * $$\widehat{\mathrm{H}}\psi=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\psi+\tfrac{1}{2}m\omega_0^2x^2\psi=E\psi.$$

Part (b)
We apply the Hamiltonian operator to the given state (using the provided derivative):



\begin{align} \widehat{\mathrm{H}}\phi(x) &=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\left( A\mathrm e^{-\alpha x^2}\right)+\tfrac{1}{2}m\omega_0^2x^2\cdot A\mathrm e^{-\alpha x^2} \\ &=-\tfrac{\hbar^2}{2m}\left( 4\alpha^2x^2-2\alpha \right)A\mathrm e^{-\alpha x^2}+\tfrac{1}{2}m\omega_0^2x^2\cdot A\mathrm e^{-\alpha x^2}. \end{align} $$

So, this function satisfies the TISE, provided that the energy of the system has the value



\begin{align} E &=-\tfrac{\hbar^2}{2m}\left( 4\alpha^2x^2-2\alpha \right)+\tfrac{1}{2}m\omega_0^2x^2 \\ &=-\tfrac{\hbar^2}{2m}4\alpha^2x^2+\tfrac{\hbar^2}{2m}2\alpha+\tfrac{1}{2}m\omega_0^2x^2 \\ &=-\tfrac{\hbar^2}{2m}\tfrac{4x^2m^2\omega_0^2}{4\hbar^2}+\tfrac{\hbar^2}{2m}\tfrac{2m\omega_0}{2\hbar}+\tfrac{1}{2}m\omega_0^2x^2 \\ &=-\tfrac{1}{2}m\omega_0^2x^2+\tfrac{1}{2}\hbar\omega_0+\tfrac{1}{2}m\omega_0^2x^2 \\ &=\tfrac{1}{2}\hbar\omega_0. \end{align} $$

Part (c)
To find the normalization constant $$A$$ of the state $$\phi$$, we form the overlap integral of $$\phi$$ with itself.



I=|A|^2\!\!\int_{-\infty}^{\infty}\!\!\mathrm e^{-2\alpha x^2}\ \mathrm dx. $$

A solution is provided for this integral.



I=\tfrac{|A|^2\sqrt\pi}{\sqrt{2\alpha}}. $$

If the state is indeed normalized, then we require that its self-overlap integral is $$I=1$$, and we can therefore solve for $$A$$ (assuming it to be real and positive);



\begin{align} \tfrac{|A|^2\sqrt\pi}{\sqrt{2\alpha}}&=1 \\ A&=\left( \tfrac{2\alpha}{\pi} \right)^{1/4}. \end{align} $$
 * A|^2&=\tfrac{\sqrt{2\alpha}}{\sqrt\pi} \\

Part (d)
The expectation value of $$Cx^2/2$$ in the state $$\phi$$ is



\begin{align} \left\langle \tfrac{1}{2}Cx^2 \right\rangle &= \int_{-\infty}^{\infty}\!\!\phi^*(x)\left(\tfrac{1}{2}Cx^2\right)\phi(x)\ \mathrm dx \\ &=\tfrac{|A|^2C}{2}\!\!\int_{-\infty}^{\infty}\!\!x^2\mathrm e^{-2\alpha x^2}\ \mathrm dx. \end{align} $$

A provided standard integral gives



\left\langle \tfrac{1}{2}Cx^2 \right\rangle =\tfrac{|A|^2C}{2}\tfrac{\sqrt{\pi}}{\sqrt{8\alpha^3}}. $$

The force constant $$C$$ for a harmonic oscillator is $$m\omega_0^2$$, and the values of $$\alpha$$ and $$|A|^2$$ are given elsewhere. Making these substitutions and simplifying gives



\begin{align} \left\langle \tfrac{1}{2}Cx^2 \right\rangle=\tfrac{1}{4}\hbar\omega_0. \end{align} $$

Part (e)
The probability of measuring state $$\phi$$ is the square of the overlap integral of $$\phi$$ with $$\Psi$$.



\begin{align} P_\phi &=\left|  \int_{-\infty}^{\infty}\!\!\phi^*(x)\Psi(x,0)\ \mathrm dx  \right|^2 \\ &=\left|A\left(\tfrac{8\alpha}{\pi}\right)^{1/4}\!\!\int^{-\infty}_{\infty}\!\!\mathrm e^{-5\alpha x^2}\ \mathrm dx\right|^2 \\ &=\left| \tfrac{\sqrt{4\alpha}}{\sqrt\pi}\!\!\int_{-\infty}^{\infty}\!\!\mathrm e^{-5\alpha x^2}\ \mathrm dx\right|^2. \end{align} $$

The provided standard integral of the appropriate form gives



P_\phi =\left|  \tfrac{\sqrt{4\alpha}}{\sqrt\pi}\tfrac{\sqrt\pi}{\sqrt{5\alpha}}\right|^2=\tfrac{4}{5}. $$

Part (a)
To solve for $$k$$ we apply the TISE to the state $$\psi$$ in the region $$x\leq 0$$.



\begin{align} \widehat{\mathrm{H}}\psi=E\psi &=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\left( A\mathrm e^{\mathrm ikx}+B\mathrm e^{-\mathrm ikx} \right) \\ E\psi &=\tfrac{\hbar^2k^2}{2m}\left( A\mathrm e^{\mathrm ikx}+B\mathrm e^{-\mathrm ikx} \right) \\ E\psi &=\tfrac{\hbar^2k^2}{2m}\psi \\ E&=\tfrac{\hbar^2k^2}{2m}. \end{align} $$

We can replace $$E$$ with $$V_0/2$$, and rearrange for $$k$$:



k=\tfrac{\sqrt{mV_0}}{\hbar}. $$

We can take an exactly similar approach to find $$\alpha$$, by applying the TISE in the region $$x>0$$ (not forgetting the potential energy term).



\begin{align} \widehat{\mathrm{H}}\psi=E\psi &=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\left( C\mathrm e^{\alpha x}+D\mathrm e^{-\alpha x} \right)+V_0\left( C\mathrm e^{\alpha x}+D\mathrm e^{-\alpha x} \right) \\ E\psi &=-\tfrac{\hbar^2\alpha^2}{2m}\psi+V_0\psi\\ E &= -\tfrac{\hbar^2\alpha^2}{2m}+V_0. \end{align} $$

Again replacing $$E$$ with $$V_0/2$$ and rearranging for $$\alpha$$, we find


 * $$\alpha=\tfrac{\sqrt{mV_0}}{\hbar}=k,$$

as expected.

Part (b)
Since $$\alpha$$ is real and positive, the term for which $$C$$ is a coefficient implies that the number of particles in the region increases exponentially the further we are beyond the barrier; on physical grounds we know that this is nonsensical, and therefore require that $$C$$ must be equal to zero.

Part (c)
The continuity boundary conditions require that, at $$x=0$$, the wave function and its first derivative are continuous:



A\mathrm e^{\mathrm i kx}+B\mathrm e^{-\mathrm ikx}=D\mathrm e^{-\alpha x}\ \ \ \ \ \mathrm{and}\ \ \ \ \ \tfrac{\mathrm d}{\mathrm dx}\left( A\mathrm e^{\mathrm i kx}+B\mathrm e^{-\mathrm ikx} \right)=\tfrac{\mathrm d}{\mathrm dx}\left(  D\mathrm e^{-\alpha x} \right). $$

Using $$x=0$$, $$\alpha=k$$, we can simplify these expressions to


 * $$A+B=D\ \ \ \ \ \mathrm{and}\ \ \ \ \ A-B=\mathrm iD.$$

Solving this system should be straight-forward, and we can find coefficients $$B$$ and $$D$$ in terms of $$A$$ to be


 * $$D=\tfrac{2A}{1+\mathrm i}\ \ \ \ \ \mathrm{and}\ \ \ \ \ B=\tfrac{A(1+\mathrm i)}{(1-\mathrm i)}.$$

Note that due to the nature of the imaginary number $$\mathrm i$$ it is possible to get several equivalent answers to this problem which at first glance may look very different. All I can advise is to either make sure you understand the complex numbers well or just trust your working.

Part (d)
The reflection coefficient is given by the probability current of the reflected part of the wave, over the probability current of the incident part of the wave.


 * $$R=\tfrac{|B\mathrm e^{-\mathrm ikx}|^2}{|A\mathrm e^{\mathrm ikx}|^2}=\tfrac{|B|^2}{|A|^2}.$$

Using the expression we found for B in part (c), this should be fairly straight-forward.



\begin{align} R &=\tfrac{|B|^2}{|A|^2} \\ &=\left|\tfrac{A(1+\mathrm i)}{A(1-\mathrm i)}\right|^2 \\ &=\tfrac{1+\mathrm i}{1-\mathrm i}\cdot\tfrac{1-\mathrm i}{1+\mathrm i} \\ &=1. \end{align} $$

Part (e)
The total number of particles in a region is the integral of the number density of said particles over the region of interest.



\int_0^{\infty}\!\!|\psi(x)|^2\ \mathrm dx=|D|^2\!\!\int_0^\infty \!\!\mathrm e^{-2\alpha x}\ \mathrm dx. $$

Part (a)
If the given spinors are indeed eigenvectors of the Hamiltonian, then applying the Hamiltonian to them should give a scalar multiple of those same vectors.



\widehat{\mathrm{H}}|\!\uparrow_x\rangle=\tfrac{\hbar\omega}{2}\tfrac{1}{\sqrt 2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\tfrac{\hbar\omega}{2}\tfrac{1}{\sqrt 2}\begin{bmatrix}1\\1\end{bmatrix}=\tfrac{\hbar\omega}{2}|\!\uparrow_x\rangle. $$



\widehat{\mathrm{H}}|\!\downarrow_x\rangle=\tfrac{\hbar\omega}{2}\tfrac{1}{\sqrt 2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\tfrac{\hbar\omega}{2}\tfrac{1}{\sqrt 2}\begin{bmatrix}-1\\1\end{bmatrix}=-\tfrac{\hbar\omega}{2}\tfrac{1}{\sqrt 2}\begin{bmatrix}1\\-1\end{bmatrix}=-\tfrac{\hbar\omega}{2}|\!\downarrow_x\rangle. $$

Part (b)
The state $$|A_0\rangle$$ can be written in the form


 * $$|A_0\rangle=c_1|\!\uparrow_x\rangle+c_2|\!\downarrow_x\rangle$$

where



c_1=\langle\uparrow_x|A_0\rangle=\tfrac{1}{\sqrt 2}\tfrac{1}{5}\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}3\\4\mathrm i\end{bmatrix}=\tfrac{1}{5\sqrt 2}(3+4\mathrm i), $$



c_1=\langle \downarrow_x|A_0\rangle=\tfrac{1}{\sqrt 2}\tfrac{1}{5}\begin{bmatrix}1&-1\end{bmatrix}\begin{bmatrix}3\\4\mathrm i\end{bmatrix}=\tfrac{1}{5\sqrt 2}(3-4\mathrm i). $$

Therefore,



\begin{align} \end{align} $$
 * A_0\rangle&=\tfrac{1}{5\sqrt 2}\left( (3+4\mathrm i)|\!\uparrow_x\rangle+(3-4\mathrm i)|\!\downarrow_x\rangle \right) \\
 * A_0\rangle&=\tfrac{1}{10}\left( (3+4\mathrm i)\begin{bmatrix}1\\1\end{bmatrix}+(3-4\mathrm i)\begin{bmatrix}1\\-1\end{bmatrix} \right).

At a later time $$t$$, the state $$|A\rangle$$ is given by



$$
 * A\rangle=\tfrac{1}{10}\left( (3+4\mathrm i)\begin{bmatrix}1\\1\end{bmatrix}\mathrm e^{-\mathrm iE_ut/\hbar}+(3-4\mathrm i)\begin{bmatrix}1\\-1\end{bmatrix}\mathrm e^{-\mathrm iE_dt/\hbar} \right).

We substitute in values for $$E_u$$ and $$E_d$$ (found earlier), and also $$t=\pi/\omega$$, and we have



$$
 * A\rangle=\tfrac{1}{10}\left( (3+4\mathrm i)\begin{bmatrix}1\\1\end{bmatrix}\mathrm e^{-\mathrm i\pi/2}+(3-4\mathrm i)\begin{bmatrix}1\\-1\end{bmatrix}\mathrm e^{\mathrm i\pi/2} \right).

The exponential terms associated with the up-spinor and down-spinor evaluate to $$-\mathrm i$$ and $$\mathrm i$$ respectively, so



\begin{align} &=\tfrac{1}{10}\left( -\mathrm i(3+4\mathrm i)\begin{bmatrix}1\\1\end{bmatrix}+\mathrm i(3-4\mathrm i)\begin{bmatrix}1\\-1\end{bmatrix} \right)\\ &=\tfrac{1}{10}\left( \begin{bmatrix}-3\mathrm i+4\\-3\mathrm i+4\end{bmatrix}+\begin{bmatrix}3\mathrm i+4\\-3\mathrm i-4\end{bmatrix} \right) \\ &=\tfrac{1}{10}\begin{bmatrix}8\\-6\mathrm i\end{bmatrix} \\ &=\tfrac{1}{5}\begin{bmatrix}4\\-3\mathrm i\end{bmatrix}. \end{align} $$
 * A\rangle

Part (c)
The expectation value of $$S_y$$ in the state $$|A_0\rangle$$:



\begin{align} \langle S_y\rangle &=\langle A_0|\widehat{\mathrm{S}}_y|A_0\rangle \\ &=\tfrac{\hbar}{2}\tfrac{1}{5}\tfrac{1}{5}\begin{bmatrix} 3&-4\mathrm i \end{bmatrix}\begin{bmatrix} 0&-\mathrm i\\\mathrm i & 0  \end{bmatrix}\begin{bmatrix} 3\\4\mathrm i  \end{bmatrix}\\ &=\tfrac{\hbar}{50}\begin{bmatrix} 3&-4\mathrm i \end{bmatrix}\begin{bmatrix} 4\\3\mathrm i  \end{bmatrix} \\ &=\tfrac{\hbar}{50}(12+12)\\ &=\tfrac{12}{25}\hbar. \end{align} $$

The expectation value of $$S_y$$ in the state $$|A\rangle$$:



\begin{align} \langle S_y\rangle &=\langle A|\widehat{\mathrm{S}}_y|A\rangle \\ &=\tfrac{\hbar}{2}\tfrac{1}{5}\tfrac{1}{5}\begin{bmatrix} 4&3\mathrm i \end{bmatrix}\begin{bmatrix} 0&-\mathrm i\\\mathrm i & 0  \end{bmatrix}\begin{bmatrix} 4\\-3\mathrm i  \end{bmatrix}\\ &=\tfrac{\hbar}{50}\begin{bmatrix} 4&3\mathrm i \end{bmatrix}\begin{bmatrix} -3\\4\mathrm i  \end{bmatrix} \\ &=\tfrac{\hbar}{50}(-12-12)\\ &=-\tfrac{12}{25}\hbar. \end{align} $$

Part (d)
(Tip: sometimes these questions seem to expect you to remember the spin matrices, but actually $$S_x$$ was given implicitly at the beginning of this question, embedded in the Hamiltonian.)

The generalized Ehrenfest theorem:



\begin{align} \tfrac{\mathrm d}{\mathrm dt}\langle S_x\rangle &=\tfrac{1}{\mathrm i\hbar}\left\langle \left[\widehat{\mathrm{S}}_x,\widehat{\mathrm{H}} \right]  \right\rangle \\ &=\tfrac{1}{\mathrm i\hbar}\left\langle  \tfrac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\tfrac{\hbar\omega}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}-\tfrac{\hbar\omega}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix}\tfrac{\hbar}{2}\begin{bmatrix}0&1\\1&0\end{bmatrix} \right\rangle \\ &=\tfrac{1}{\mathrm i \hbar}\tfrac{\hbar^2\omega}{4}\left\langle \begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}-\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix} \right\rangle \\ &=\tfrac{\hbar\omega}{2}\left\langle 0 \right\rangle \\ &=0. \end{align} $$

So the rate of change of $$\langle S_x\rangle$$ with time is zero, as expected.

Question 17
Hint: the questions concerning the radial function and spherical harmonics invariably involve a lot of integration. It's important to remember that we're performing these integrations in a spherical coordinate system, and so you should take care not to forget the extra factor or $$r^2$$ or $$\sin\theta$$ (or both) that should appear in the integrand. If you forget it, it should be easy to spot (for $$r^2$$, at least), since your answers will not be dimensionally consistent.

Part (a)
The differential $$\widehat{\mathrm{L}}_z$$ operator is


 * $$\widehat{\mathrm{L}}_z=-\mathrm i\hbar\tfrac{\partial}{\partial\phi}.$$

Applying this operator to $$Y_{10}(\theta,\phi)$$:



\widehat{\mathrm{L}}_zY_{10}(\theta,\phi)=-\mathrm i\hbar\tfrac{\partial}{\partial\phi}\left( \left(\tfrac{3}{4\pi}\right)^{1/2}\cos\theta \right). $$

Since $$Y$$ has no $$\phi$$-dependence, the derivative evaluates to zero and we see that the eigenvalue of $$\widehat{\mathrm{L}}_z$$ corresponding to eigenfunction $$Y$$ is zero. Since the eigenvalues of $$\widehat{\mathrm{L}}_z$$ are $$m\hbar$$, we have $$m=0$$.

The differential $$\widehat{\mathrm{L}}^2$$ operator is


 * $$\widehat{\mathrm{L}}^2=-\hbar^2\left[

\tfrac{1}{\sin\theta} \tfrac{\partial}{\partial\theta} \left( \sin\theta\tfrac{\partial}{\partial\theta} \right) + \tfrac{1}{\sin^2\theta} \tfrac{\partial^2}{\partial\phi^2} \right].$$

Since $$Y$$ has no $$\phi$$-dependence the second term will disappear, and we have



\begin{align} \widehat{\mathrm{L}}^2Y_{10}(\theta,\phi) &=-\hbar^2\tfrac{\sqrt 3}{\sqrt{4\pi}} \tfrac{1}{\sin\theta}\tfrac{\partial}{\partial\theta}\!\left( \sin\theta\tfrac{\partial}{\partial\theta}\cos\theta \right) \\ &=\hbar^2\tfrac{\sqrt 3}{\sqrt{4\pi}}\tfrac{1}{\sin\theta}\tfrac{\partial}{\partial\theta}\sin^2\theta \\ &=2\hbar^2\tfrac{\sqrt 3}{\sqrt{4\pi}}\cos\theta \\ &=2\hbar^2\ Y_{10}(\theta,\phi). \end{align} $$

The eigenvalue of $$\widehat{\mathrm{L}}^2$$ corresponding to eigenfunction $$Y$$ is $$2\hbar^2$$. Since the eigenvalues of $$\widehat{\mathrm{L}}^2$$ are $$l(l+1)\hbar^2$$, we have $$l=1$$.

Part (b)
The probability of finding a separation in the range between $$r$$ and $$r+\delta r$$ is given by


 * $$P=R^2_{nl}(r)r^2\ \delta r,$$

where $$R_{nl}(r$$) is the radial wave function. The most likely electron--proton separation occurs at the maximum of the probability function. We take the first derivative with respect to $$r$$ of the radial probability density function, set it equal to zero, and solve for $$r$$.



\begin{align} \tfrac{\mathrm d}{\mathrm dr}\left(R^2_{nl}(r)r^2\right)  &=   0 \\ \tfrac{1}{24a_0^5}\tfrac{\mathrm d}{\mathrm dr}\left(r^4\mathrm e^{-r/a_0}\right)  &=   0 \\ \tfrac{1}{24a_0^5}\left( 4r^3\mathrm e^{-r/a_0}-\tfrac{r^4}{a_0}\mathrm e^{-r/a_0}  \right)   &=   0 \\ 4r^3-\tfrac{r^4}{a_0} &=  0 \\ r^3\left( 4-\tfrac{r}{a_0} \right) &=  0 \\ r&=0,\ \ \ \ r=4a_0. \end{align} $$

We can reject the solution where $$r=0$$ on physical grounds, so the most likely value of the electron--proton separation is $$r=4a_0$$.

Part (c)
To find the expectation value of $$r$$ we place it in a sandwich integral with the radial function. We don't need to consider the spherical harmonic since it has no $$r$$-dependence and is normalized, and so well make a contribution to the answer which amounts to multiplying by 1. Not really worth the extra work!



\begin{align} \langle r \rangle &=\int_{0}^{\infty}\!\!R^*_{nl}(r)\ r\ R_{nl}(r)\ r^2\ \mathrm dr \\ &=\tfrac{1}{24a_0}\!\!\int_{0}^{\infty}\!\!r^5\mathrm e^{-r/a_0}\ \mathrm dr. \end{align} $$

This is solved using a provided integral.



\langle r \rangle =\tfrac{1}{24a_0^5}\cdot 5!\cdot a_0^6=5a_0. $$

The most likely electron--proton separation is $$5a_0$$.

Part (d)
The first-order correction to the energy is found by placing the perturbation to the Hamiltonian in a sandwich integral with the radial function. Again we do not need to consider the spherical harmonic, since the new Hamiltonian has no $$\theta$$- or $$\phi$$-dependence, and so the angular part of the wave function is irrelevant.



\begin{align} E_1 &=\int_{0}^{\infty}\!\!R^*_{nl}(r)\ \delta\widehat{\mathrm{H}}\ R_{nl}(r)\ r^2\ \mathrm dr \\ &=\tfrac{A}{24a_0}\!\!\int_{0}^{\infty}\!\!r\mathrm e^{-r/a_0}\ \mathrm dr \\ &=\tfrac{A}{24a_0^5}\cdot 1!\cdot a_0^2 \\ &=\tfrac{A}{24a_0^3}. \end{align} $$