SM358 - 2011

=PART 1=

Part (a)
Apply the momentum operator to state $$\Psi(x,t)$$:



\begin{align} \widehat{\mathrm{p}}_x\Psi(x,t) &=-\mathrm i\hbar\tfrac{\partial}{\partial x}\left(A\mathrm e^{-\mathrm ikx}\mathrm e^{-\mathrm i\omega t}\right) \\ &=-\mathrm i\hbar A\mathrm e^{-\mathrm i\omega t}\tfrac{\partial}{\partial x}\mathrm e^{-\mathrm ikx} \\ &=\mathrm i^2\hbar kA\mathrm e^{-\mathrm i\omega t}\mathrm e^{-\mathrm ikx} \\ &=-\hbar k\Psi(x,t). \end{align} $$

The eigenvalue of momentum in state $$\Psi(x,t)$$ is $$-\hbar k$$.

Part (b)
The energy of the particle is


 * $$E=\hbar\omega=(1.06\times 10^{-34}\times 2.50\times 10^{14})\ \mathrm J=2.65\times 10^{-20}\ \mathrm J.$$

Part (c)
Since there is no potential energy, the particle's energy is entirely kinetic:


 * $$E=\tfrac{p^2}{2m}.$$

We rearrange for the mass (and use the fact that $$p=\hbar k$$):


 * $$m=\tfrac{p^2}{2E}=\tfrac{\hbar^2k^2}{2E}=2.75\times 10^{-28}\ \mathrm{kg}.$$

Part (a)
We use the normalization condition for the state to derive a value for $$N$$:



\begin{align} \int_0^L\!\!N^2x^2\ \mathrm dx &=  1 \\ N^2 \left[\tfrac{x^3}{3}\right]^{x=L}_{x=0}   &=  1 \\ N^2\tfrac{L^3}{3} &=  1 \\ N &=  \sqrt{\tfrac{3}{L^3}}. \end{align} $$

Part (b)
The probability of finding the particle in the region $$0\leq x\leq L/2$$ is



\begin{align} P &= \int_0^{L/2}\!\!\Psi^*(x,0)\Psi(x,0)\ \mathrm dx \\ &=\tfrac{3}{L^3}\!\!\int_0^{L/2}\!\!x^2\ \mathrm dx \\ &=\tfrac{3}{L^3}\tfrac{L^3}{24} \\ &=\tfrac{1}{8}. \end{align} $$

Part (a)
The probabilities of measuring $$E_0$$ and $$E_1$$ in state $$\Psi(x,0)$$ are



P(E_0)=\left( \tfrac{3}{5} \right)^2=\tfrac{9}{25}, $$



P(E_1)=\left( \tfrac{4}{5} \right)^2=\tfrac{16}{25}. $$

Part (b)
The state at a later time is



\begin{align} \Psi(x,t)&=\tfrac{3}{5}\psi_0(x)\mathrm e^{-\mathrm iE_ut/\hbar}  +\tfrac{4}{5}\psi_1(x)\mathrm e^{-\mathrm iE_dt/\hbar} \\ &=\tfrac{3}{5}\psi_0(x)\mathrm e^{-\mathrm i\omega_0t/2}  +\tfrac{4}{5}\psi_1(x)\mathrm e^{-3\mathrm i\omega_0t/2}. \end{align} $$

Part (c)
The state at time $$t=2\pi/\omega_0$$ is



\begin{align} \Psi(x,t)&=\tfrac{3}{5}\psi_0(x)\mathrm e^{-\mathrm i\pi}  +\tfrac{4}{5}\psi_1(x)\mathrm e^{-3\mathrm i\pi} \\ &=-\tfrac{3}{5}\psi_0(x)  -\tfrac{4}{5}\psi_1(x). \end{align} $$

The probabilities of measuring $$E_0$$ and $$E_1$$ in this state are



P(E_0)=\left( -\tfrac{3}{5} \right)^2=\tfrac{9}{25}, $$



P(E_1)=\left( -\tfrac{4}{5} \right)^2=\tfrac{16}{25}, $$

which is exactly the same result as part (a).

Part (a)
The boundary conditions require that $$\psi(x)$$ and its first deriviatve are both continuous across the boundary, which in this case is at $$x=0$$.


 * $$\begin{align}A+B&=C,\\k_1A-k_1B&=k_2C.\end{align}$$

Part (b)
We are to eliminate $$C$$ and rearrange the result to make $$B/A$$ the subject.



\begin{align} A+B &=  \tfrac{k_1}{k_2}A-\tfrac{k_1}{k_2}B \\ A\left( 1-\tfrac{k_1}{k_2} \right) &=  B\left(  -1-\tfrac{k_1}{k_2} \right) \\ \tfrac{B}{A} &=  \tfrac{\left(\tfrac{k_1}{k_2}-1\right)}{\left(1+\tfrac{k_1}{k_2}\right)} \\ \tfrac{B}{A} &= \tfrac{k_2-k_1}{k_2+k_1}. \end{align} $$

If $$k_1=2k_2$$, the reflection coefficient is given by


 * $$R=\left| \tfrac{B}{A} \right|^2=\left| -\tfrac{1}{3} \right|^2=\tfrac{1}{9}.$$

Part (a)

 * $$\langle f|\widehat{\mathrm{B}}g\rangle=\langle \widehat{\mathrm{B}}f|g\rangle.$$

Part (b)


\int_{-\infty}^{\infty}\!\!f^*(x)\left( \widehat{\mathrm{B}}g(x) \right)\ \mathrm dx=\int_{-\infty}^{\infty}\!\!\left(\widehat{\mathrm{B}}f(x)\right)^*g(x)\ \mathrm dx. $$

Part (c)
From (a) we have


 * $$\langle f|\widehat{\mathrm{B}}g\rangle=\langle \widehat{\mathrm{B}}f|g\rangle.$$

If $$B$$ is the eigenvalue of $$\widehat{\mathrm{B}}$$, then it can replace the operator in the bracket:


 * $$\langle f|Bg\rangle=\langle Bf|g\rangle.$$

A general property of Dirac brackets allows us to bring the constant $$B$$ out of the bracket (not forgetting that the $$B$$ on the R.H.S. must be complex conjugated):


 * $$B\langle f|g\rangle=B^*\langle f|g\rangle.$$

The brackets cancel, since we have the same term on both sides, and we are left with


 * $$B=B^*.$$

A number which is its own complex conjugate must be real-valued, so $$B$$ is real.

Part (a)


\begin{align} \left[\widehat{\mathrm{S}}_x, \widehat{\mathrm{S}}_y \right] &= \widehat{\mathrm{S}}_x \widehat{\mathrm{S}}_y - \widehat{\mathrm{S}}_y \widehat{\mathrm{S}}_x \\ &= \frac{\hbar^2}{4} \left[\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right] \left[\begin{matrix} 0 & -i \\ i & 0 \\ \end{matrix}\right] - \frac{\hbar^2}{4} \left[\begin{matrix} 0 & -i \\ i & 0 \\ \end{matrix}\right] \left[\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right] \\ &= \frac{\hbar^2}{4} \left[\begin{matrix} i & 0 \\ 0 & -i \\ \end{matrix}\right] - \frac{\hbar^2}{4} \left[\begin{matrix} -i & 0 \\ 0 & i \\ \end{matrix}\right] \\ &= \frac{\hbar^2}{4} \left[\begin{matrix} 2i & 0 \\ 0 & -2i \\ \end{matrix}\right] \\ &= \frac{i}{\hbar}\frac{\hbar}{2} \left[\begin{matrix} 1 & 0 \\ 0 & -1 \\ \end{matrix}\right] \\ &= i\hbar \widehat{\mathrm{S}}_z \\ \end{align} $$

Part (b)
From the generalized uncertainty relation



\begin{align} \Delta \widehat{\mathrm{S}}_x \Delta \widehat{\mathrm{S}}_y &\geq \frac{1}{2} \left|\left\langle\left[ \widehat{\mathrm{S}}_x, \widehat{\mathrm{S}}_y \right]\right\rangle\right| \\ &\geq \frac{\hbar}{2} \left|\left\langle \widehat{\mathrm{S}}_z \right\rangle\right| \\ \end{align} $$

In any state where $$S_z = \pm {\hbar}/{2}$$ we have $$\Delta \widehat{\mathrm{S}}_x \Delta \widehat{\mathrm{S}}_y \geq {\hbar^2}/{4}$$. In such a state if either $$S_x$$ or $$S_y$$ were certain we would have $$\Delta \widehat{\mathrm{S}}_x \Delta \widehat{\mathrm{S}}_y = 0 $$ which would violate the uncertainty principle and hence neither $$S_x$$ or $$S_y$$ may be in a definite state when $$S_z$$ is.

Part (a)
Neutrons are fermions and obey the Pauli exclusion principle.

Part (b)
Identical fermions have an anti-symmetric total wave function so if a pair of neutrons have an anti-symmetric spin state then their spatial wave function must be symmetric. A pair of particles with a symmetric spatial wave function tend to huddle together.

Part (c)
If the particles were in a symmetric spin state they would have an anti-symmetric spatial wave function and would tend to segregate.

Part (a)
FALSE: Entanglement is a fundamental property and is basis-independant.

Part (b)
TRUE: Quantum mechanics predicts that Bell's inequality can be violated.

Part (c)
TRUE: Rather the point of the experiment!

Part (d)
TRUE: The BB84 does not use entangled photons, however the Eckert protocol does.

Part (e)
FALSE: The no-cloning theorem requires that the state to be teleported remains unknown.

Part (a)


\begin{align} \left\langle r \right\rangle &= \left\langle R \left| r \right| R \right\rangle \\ &= \int\limits_{r} r\left| R \right|^2 r^2 \mathrm{d}r \\ &= \frac{1}{24 a_0^5} \int\limits_{0}^{\infty} r^5 \exp(-{r}{a_0}) \mathrm{d}r \\ &= \frac{1}{24 a_0^5} \times 5! \times a_0^6 \\ &= 5 a_0 \\ \end{align} $$

Part (b)
In part (a) $$ E_n = {E_{\mathrm{R}}}/{n^2} = {E_{\mathrm{R}}}/{4}$$ so $$n=2$$

For the hydrogen like system $$\mathrm{Li}^{2+}$$ we have $$Z=3$$ and, ignoring reduced mass effects, $$\mu = \mu_{\mathrm{H}}$$ so

\begin{align} a_0^{\mathrm{scaled}} &= \frac{1}{Z}\frac{\mu_{\mathrm H}}{\mu}a_0 \\ &= \frac{a_0}{3} \\ \end{align} $$

hence

\left\langle r \right\rangle = \frac{5}{3} a_0 $$

and

\begin{align} E_{\mathrm{R}}^{\mathrm{scaled}} &= Z^2 \frac{\mu}{\mu_{\mathrm H}} E_{\mathrm{R}} \\ &= 9 E_{\mathrm{R}} \\ \end{align} $$

so

\begin{align} E_2 &= - \frac{E_{\mathrm{R}}^{\mathrm{scaled}}}{4} \\ &= - \frac{9}{4} E_{\mathrm{R}} \\ \end{align} $$

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


 * $${{L}_{\min }}=\left| {{l}_{1}}-{{l}_{2}} \right| =\left| 3-1 \right| = 2$$
 * $${{L}_{\max }}=\left| {{l}_{1}}+{{l}_{2}} \right| =\left| 3+1 \right| = 4$$
 * $${{S}_{\min }}=\left| {{s}_{1}}-{{s}_{2}} \right| =\left| \tfrac{1}{2}-\tfrac{1}{2} \right| = 0$$
 * $${{S}_{\max }}=\left| {{s}_{1}}+{{s}_{2}} \right| =\left| \tfrac{1}{2}+\tfrac{1}{2} \right|= 1$$
 * $$L=L_{\min}, L_{\min} +1, \ldots, L_{\max} -1, L_{\max}$$
 * $$S=S_{\min}, S_{\min} +1, \ldots, S_{\max} -1, S_{\max}$$

So $$L=2,3 \text{ or } 4$$ and $$S=0 \text{ or } 1$$ and the allowed combinations in $$\left(L,S\right)$$ notation are

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

Part (b)
The atomic terms split into levels as a result of perturbations due to spin-orbit interactions. In the LS-coupling scheme 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 $$L=2, S=1$$ we have the levels with $$J=1,2 \text{ or } 3$$.

The $$\left(L=2,S=1\right)$$ term has $$\left(2L+1\right)\left(2S+1\right) = 5\times3 = 15$$ degenerate states.

Part (a)
In the LCAO approximation $$\mathrm\sigma$$ molecular orbitals are created by the combination of atomic orbitals with $$m=0$$ so $$\mathrm\sigma$$ orbitals will be produced by the atomic orbital pairs


 * $$ (1{{\mathrm{s}}_{0}},1{{\mathrm{s}}_{0}}), \ \ (2{{\mathrm{p}}_{0}},2{{\mathrm{p}}_{0}})$$

Part (b)
The ground state electronic configuration of diatomic Fluorine (Z=9) is



1\mathrm\sigma _{g}^{2} \color{red} 1\mathrm\sigma _{u}^{2} \color{black} 2\mathrm\sigma _{g}^{2} \color{red} 2\mathrm\sigma _{u}^{2} \color{black} 2\mathrm\sigma _{g}^{2} \color{black} 1\mathrm\pi _{u}^{4} \color{red} 1\mathrm\pi _{g}^{4} $$

Formal bond order = 1/2(bonding - anti-bonding)

Bonding = (2 + 2 + 2 + 4) = 10

Anti-bonding = (2 + 2 + 4) = 8

Bond order = 1/2(10 - 8) = 1

Part (c)
=PART 2=

Part (d)
=PART 3=

Part (a)
The Hamiltonian for the system is

\widehat{\mathrm{H}} = -\omega \widehat{\mathrm{S}}_z = \frac{\hbar}{2} \left[\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right] $$

so the generalised Ehrenfest theorem gives



\begin{align} \frac{\mathrm{d}\left\langle S_y \right\rangle}{\mathrm{d}t} &= \frac{1}{i\hbar}\left\langle\left[ \widehat{\mathrm{S}}_y, \widehat{\mathrm{H}} \right]\right\rangle \\ &= \frac{1}{i\hbar}\left\langle\left[ \widehat{\mathrm{S}}_y, - \omega \widehat{\mathrm{S}}_x \right]\right\rangle \\ &= \frac{-\omega}{i\hbar}\left\langle\left[ \widehat{\mathrm{S}}_y, \widehat{\mathrm{S}}_x \right]\right\rangle \\ &= \frac{\omega}{i\hbar}\left\langle\left[ \widehat{\mathrm{S}}_x, \widehat{\mathrm{S}}_y \right]\right\rangle \\ &= \frac{\omega}{i\hbar}\left\langle i\hbar \widehat{\mathrm{S}}_z \right\rangle \\ &= \omega \left\langle \widehat{\mathrm{S}}_z \right\rangle \\ \end{align} $$

Part (b)


\begin{align} \widehat{\mathrm{H}} \left|\uparrow_x\right\rangle &= - \frac{\hbar\omega}{2}\frac{1}{\sqrt{2}} \left[\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right] \left[\begin{matrix} 1 \\ 1 \\ \end{matrix}\right] \\ &= - \frac{\hbar\omega}{2}\frac{1}{\sqrt{2}} \left[\begin{matrix} 1 \\ 1 \\ \end{matrix}\right] \\ &= - \frac{\hbar\omega}{2} \widehat{\mathrm{H}} \left|\uparrow_x\right\rangle \end{align} $$

So $$\left|\uparrow_x\right\rangle$$ is an eigenvector $$\widehat{\mathrm{H}}$$ with eigenvalue $$E_{\mathrm u} = -{\hbar\omega}/{2}$$



\begin{align} \widehat{\mathrm{H}} \left|\downarrow_x\right\rangle &= - \frac{\hbar\omega}{2}\frac{1}{\sqrt{2}} \left[\begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix}\right] \left[\begin{matrix} -1 \\ 1 \\ \end{matrix}\right] \\ &= - \frac{\hbar\omega}{2}\frac{1}{\sqrt{2}} \left[\begin{matrix} 1 \\ -1 \\ \end{matrix}\right] \\ &= \frac{\hbar\omega}{2} \widehat{\mathrm{H}} \left|\downarrow_x\right\rangle \end{align} $$

So $$\left|\downarrow_x\right\rangle$$ is an eigenvector $$\widehat{\mathrm{H}}$$ with eigenvalue $$E_{\mathrm d} = {\hbar\omega}/{2}$$

Part (c)
$$\left|\uparrow_x\right\rangle$$ and $$\left|\downarrow_x\right\rangle$$ form an orthonormal., spanning basis for spinor space so, at $$t=0$$



\begin{align} \left|A\right\rangle_{\mathrm{initial}} &=\left\langle\uparrow_x | A\right\rangle_{\mathrm{initial}}\left|\uparrow_x\right\rangle + \left\langle\downarrow_x | A\right\rangle_{\mathrm{initial}}\left|\downarrow_x\right\rangle \\ &= \frac{1}{\sqrt{2}} \left[\begin{matrix} 1 & 1\\ \end{matrix}\right] \left[\begin{matrix} 1 \\ 0 \\ \end{matrix}\right] \left|\uparrow_x\right\rangle + \frac{1}{\sqrt{2}} \left[\begin{matrix} -1 & 1\\ \end{matrix}\right] \left[\begin{matrix} 1 \\ 0 \\ \end{matrix}\right] \left|\downarrow_x\right\rangle \\ &= \frac{1}{\sqrt{2}} \left|\uparrow_x\right\rangle - \frac{1}{\sqrt{2}} \left|\downarrow_x\right\rangle \\ \end{align} $$

so, at time $$t \geq 0 $$

\begin{align} \left|A\right\rangle &= \tfrac{1}{\sqrt{2}} \left|\uparrow_x\right\rangle \exp \left( {-i E_{\mathrm{u}}t}/{\hbar} \right) - \tfrac{1}{\sqrt{2}} \left|\downarrow_x\right\rangle \exp \left( {-i E_{\mathrm{d}}t}/{\hbar} \right) \\ &= \tfrac{1}{\sqrt{2}} \left|\uparrow_x\right\rangle \exp \left( {i \omega t}/{2} \right) - \tfrac{1}{\sqrt{2}} \left|\downarrow_x\right\rangle \exp \left( {-i \omega t}/{2} \right) \\ &= \frac{1}{2} \left[\begin{matrix} 1 \\ 1 \\ \end{matrix}\right] \exp \left( {i \omega t}/{2} \right) - \frac{1}{2} \left[\begin{matrix} -1 \\ 1 \\ \end{matrix}\right] \exp \left( {-i \omega t}/{2} \right) \\ &= \frac{1}{2} \left[\begin{matrix} \exp \left( {i \omega t}/{2} \right) + \exp \left( {-i \omega t}/{2} \right) \\ \exp \left( {i \omega t}/{2} \right) - \exp \left( {-i \omega t}/{2} \right) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} \tfrac{1}{2} \left( \exp \left( {i \omega t}/{2} \right) + \exp \left( {-i \omega t}/{2} \right) \right) \\ \tfrac{i}{2i} \left( \exp \left( {i \omega t}/{2} \right) - \exp \left( {-i \omega t}/{2} \right) \right) \\ \end{matrix}\right] \\ &= \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) \\ i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \\ \end{align} $$

Part (d)


\begin{align} \left\langle S_y \right\rangle &= \left\langle A \left| \widehat{\mathrm{S}}_y \right| A \right\rangle \\ &= \frac{\hbar}{2} \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) & -i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \left[\begin{matrix} 0 & -i \\ i & 0 \\ \end{matrix}\right] \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) \\ i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \\ &= \frac{\hbar}{2} \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) & -i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \left[\begin{matrix} \sin \left( {\omega t}/{2} \right) \\ i\cos \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \\ &= \frac{\hbar}{2} \left( \sin\left( {\omega t}/{2} \right)\cos\left( {\omega t}/{2} \right)  + \sin\left( {\omega t}/{2} \right)\cos\left( {\omega t}/{2} \right) \right) \\ &= \frac{\hbar}{2} \sin\left( \omega t \right) \\ \end{align} $$



\begin{align} \left\langle S_z \right\rangle &= \left\langle A \left| \widehat{\mathrm{S}}_z \right| A \right\rangle \\ &= \frac{\hbar}{2} \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) & -i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \left[\begin{matrix} 1 & 0 \\ 0 & -1  \\ \end{matrix}\right] \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) \\ i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \\ &= \frac{\hbar}{2} \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) & -i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \left[\begin{matrix} \cos \left( {\omega t}/{2} \right) \\ -i\sin \left( {\omega t}/{2} \right) \\ \end{matrix}\right] \\ &= \frac{\hbar}{2} \left( \cos^2\left( {\omega t}/{2} \right) - \sin^2\left( {\omega t}/{2} \right) \right) \\ &= \frac{\hbar}{2} \cos\left( \omega t \right) \\ \end{align} $$

Now

\begin{align} \frac{\mathrm{d}\left\langle S_y \right\rangle}{\mathrm{d}t} &= \frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\hbar}{2} \sin\left(\omega t\right)\right) \\ &= \omega \frac{\hbar}{2} \cos\left(\omega t\right) \\ &= \omega \left\langle \widehat{\mathrm{S}}_z \right\rangle \\ \end{align} $$

in agreement with part (a).

Part (a)
A two particle state is entangled if it cannot be factorised into the product of two single particle states, hence state $$\left|A\right\rangle$$ is entangled. If observer 1 measures the polarisation to be verticle then, using the same basis, entanglement will ensure that observer 2 measures the polarisation to be horizontal. Similarly if observer 2 measures the polarisation to be vertical then observer one must measure the polarisation to be horizontal so $$\mathrm{P}\left(V_1V_2\right)=0$$.

Part (b)


\tfrac{1}{\sqrt{2}} \left( \left|\mathrm V_{\theta} \right\rangle_1 \left|\mathrm H_{\theta} \right\rangle_2 -        \left|\mathrm V_{\theta} \right\rangle_1 \left|\mathrm H_{\theta} \right\rangle_2 \right) $$

\begin{align} &= \tfrac{1}{\sqrt{2}} \left( \cos\theta \left| \mathrm V \right\rangle_1 + \sin\theta \left| \mathrm H \right\rangle_1 \right) \left(-\sin\theta \left| \mathrm V \right\rangle_2 + \cos\theta \left| \mathrm H \right\rangle_2 \right) \\ &\qquad - \left(-\sin\theta \left| \mathrm V \right\rangle_1 + \cos\theta \left| \mathrm H \right\rangle_1 \right) \left( \cos\theta \left| \mathrm V \right\rangle_2 + \sin\theta \left| \mathrm H \right\rangle_2 \right) \\ &= \tfrac{1}{\sqrt{2}} (- \sin\theta\cos\theta \left| \mathrm V \right\rangle_1 \left| \mathrm V \right\rangle_2       + \cos^2\theta \left| \mathrm V \right\rangle_1 \left| \mathrm H \right\rangle_2 \\ &\qquad - \sin^2\theta \left| \mathrm H \right\rangle_1 \left| \mathrm V \right\rangle_2           + \sin\theta\cos\theta \left| \mathrm H \right\rangle_1 \left| \mathrm H \right\rangle_2 \\ &\qquad + \sin\theta\cos\theta \left| \mathrm V \right\rangle_1 \left| \mathrm V \right\rangle_2        + \sin^2\theta \left| \mathrm V \right\rangle_1 \left| \mathrm H \right\rangle_2 \\ &\qquad - \cos^2\theta \left| \mathrm H \right\rangle_1 \left| \mathrm V \right\rangle_2           - \sin\theta\cos\theta \left| \mathrm H \right\rangle_1 \left| \mathrm H \right\rangle_2 ) \\ &= \tfrac{1}{\sqrt{2}} ( \left( \cos^\theta + \sin^2\theta \right) \left| \mathrm V \right\rangle_1 \left| \mathrm H \right\rangle_2 \\ &\qquad - \left( \cos^\theta + \sin^2\theta \right) \left| \mathrm H \right\rangle_1 \left| \mathrm V \right\rangle_2 ) \\ &= \tfrac{1}{\sqrt{2}} \left( \left| \mathrm V \right\rangle_1 \left| \mathrm H \right\rangle_2       - \left| \mathrm H \right\rangle_1 \left| \mathrm V \right\rangle_2 \right) \\ &= \left| A \right\rangle \\ \end{align} $$

Part (e)
=PART 4=

Part (a)


\begin{align} \widehat{\mathrm{L}}_z f\left( \theta, \phi \right) &= - \mathrm i \hbar \frac{\partial}{\partial\theta} \left( \sin\theta \exp\left(-\mathrm i \phi \right) \right) \\ &= - \hbar \sin\theta \exp\left(-\mathrm i \phi \right) \\ &= - \hbar f\left( \theta, \phi \right) \\ \end{align} $$

so $$f\left( \theta, \phi \right)$$ is an eigenfunction of $$\widehat{\mathrm{L}}_z$$ with eigenvalue $$-\hbar$$.



\begin{align} \widehat{\mathrm{L}}^2 f\left( \theta, \phi \right) &= -\hbar^2 \left[\frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2} \right] f\left( \theta, \phi \right) \\ &= -\hbar^2 \left[\frac{1}{\sin\theta} \frac{\partial}{\partial\theta} \left( \sin\theta\cos\theta \exp\left(-\mathrm i \phi\right) \right) + \frac{1}{\sin\theta} \exp\left(-\mathrm i \phi \right) \left(-1\right) \right] \\ &= -\hbar^2 \left[ \left( \frac{1}{\sin\theta} \left(1 - 2\sin^2\theta\right) - \frac{1}{\sin\theta} \right) \exp\left(-\mathrm i \phi \right) \right] \\ &= 2\hbar^2 \sin\theta \exp\left(-\mathrm i \phi \right) \\ &= 2\hbar^2 f\left( \theta, \phi \right) \\ \end{align} $$

so $$f\left( \theta, \phi \right)$$ is an eigenfunction of $$\widehat{\mathrm{L}}^2$$ with eigenvalue $$2\hbar^2$$.

Part (b)
In the Coulomb model


 * $$\psi\left( \theta, \phi \right) = R_{n,l}\left( r \right) Y_{l,m}\left( \theta, \phi \right)$$

from part (a) $$Y_{l,m}\left( \theta, \phi \right) = f\left( \theta, \phi \right)$$ so
 * $$\widehat{\mathrm{L}}_z f = m\hbar f = -\hbar f \Rightarrow m=-1$$
 * $$\widehat{\mathrm{L}}^2 f = l(l+1)\hbar^2 f = 2\hbar^2 f \Rightarrow l=1$$

and we know that
 * $$E_n = -\frac{E_{\mathrm R}}{n^2} = -\frac{E_{\mathrm R}}{4} \Rightarrow n=2 $$

so the quantum number of this eigenfunction are $$n=2,\; l=1, \; m=-1$$

Part (c)
$$\psi\left(r,\theta,\phi\right)$$ is normalised if $$\left\langle \psi \right.\left| \psi \right\rangle = 1 $$ so



\begin{align} 1 &= \int\limits_V \left| \psi \right|^2 \mathrm{d}V \\ &= \left|A\right|^2 \int\limits_{V} \left(\frac{r}{a_0}\right)^2 \exp\left( -{r}/{a_0} \right) \sin^2\theta \, \mathrm{d}V\\ \end{align} $$

so

\begin{align} \frac{1}{\left|A\right|^2} &= \int\limits_{-\pi}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty} \left(\frac{r}{a_0}\right)^2 \exp\left( -{r}/{a_0} \right) \sin^2\theta \,\mathrm{d}r \, r \mathrm{d}\theta \, r\sin\theta \mathrm{d}\phi \\ &= \frac{1}{a_0^2} \int\limits_{0}^{\infty} r^5 \exp\left( -{r}/{a_0} \right) \,\mathrm{d}r \int\limits_{0}^{\pi} \sin^3\theta \,\mathrm{d}\theta \int\limits_{-\pi}^{\pi} \,\mathrm{d}\phi \\ &= \frac{1}{a_0^2} \times 4!\,a_0^5 \times \frac{4}{3} \times 2\pi \\ &= 64\pi a_0^3 \\ \end{align} $$

so, choosing $$A$$ to be positive and real,



\begin{align} A &= \left( \frac{1}{64\pi a_0^3} \right)^{{1}/{2}} \\ &= \frac{1}{8\sqrt{\pi} a_0^{{3}/{2}}} \end{align} $$

Part (d)


\begin{align} \widehat{\mathrm{J}}_z \left|A\right\rangle &= \left( \widehat{\mathrm{L}}_z + \widehat{\mathrm{S}}_z \right) \left|A\right\rangle \\ &= \widehat{\mathrm{L}}_z \left|A\right\rangle + \widehat{\mathrm{S}}_z \left|A\right\rangle \\ &= a \widehat{\mathrm{L}}_z Y_{1,0} \left|\uparrow\right\rangle + b \widehat{\mathrm{L}}_z Y_{1,1} \left|\downarrow\right\rangle + a Y_{1,0} \widehat{\mathrm{S}}_z \left|\uparrow\right\rangle + b Y_{1,1} \widehat{\mathrm{S}}_z \left|\downarrow\right\rangle \\ &= 0\hbar a Y_{1,0} \left|\uparrow\right\rangle + \hbar b Y_{1,1} \left|\downarrow\right\rangle + \frac{\hbar}{2} a Y_{1,0} \left|\uparrow\right\rangle - \frac{\hbar}{2} b Y_{1,1} \left|\downarrow\right\rangle \\ &= \frac{\hbar}{2} a Y_{1,0} \left|\uparrow\right\rangle + \frac{\hbar}{2} b Y_{1,1} \left|\downarrow\right\rangle \\ &= \frac{\hbar}{2} \left|A\right\rangle\\ \end{align} $$

so $$\left|A\right\rangle$$ is an eigenfunction of $$\widehat{\mathrm{J}}_z$$ with eigenvalue $$\frac{\hbar}{2}$$.

Part (e)


\begin{align} &\widehat{\mathrm{J}}^2 \left|A\right\rangle = j(j+1)\hbar^2 = \left(\frac{3}{2}\times\frac{5}{2}\right)\hbar^2 = \frac{15}{4}\hbar^2 \\ &\widehat{\mathrm{L}}^2 \left|A\right\rangle = l(l+1)\hbar^2 = \left(1 \times 2 \right)\hbar^2 = 2\hbar^2 \\ &\widehat{\mathrm{S}}^2 \left|A\right\rangle = s(s+1)\hbar^2 = \left(\frac{1}{2}\times\frac{3}{2}\right)\hbar^2 = \frac{3}{4}\hbar^2 \\ \end{align} $$

Now



\begin{align} \widehat{\mathbf{L}} \mathbf\cdot \widehat{\mathbf{S}} \left|A\right\rangle &= \frac{1}{2} \left( \widehat{\mathrm{J}}^2 - \widehat{\mathrm{L}}^2 - \widehat{\mathrm{S}}^2 \right) \left|A\right\rangle \\ &= \frac{1}{2} \left( \frac{15}{4} - 2 - \frac{3}{4} \right) \hbar^2 \left|A\right\rangle \\ &= \frac{\hbar^2}{2} \left|A\right\rangle \\ \end{align} $$

so $$\left|A\right\rangle$$ is an eigenfunction of $$\widehat{\mathbf{L}}\mathbf\cdot \widehat{\mathbf{S}}$$ with eigenvalue $$\frac{\hbar^2}{2}$$.