SM358 - 2011

=PART 1=
==Question 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}.
==Question 2==
===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}
==Question 3==
===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).
==Question 4==
===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}.
==Question 5==
===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.
==Question 6==
===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.
==Question 7==
===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. 
==Question 8==
===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.
==Question 9==
===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}
==Question 10==
===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.
==Question 11==
===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
 
==Question 12==
===Part (a)===
===Part (b)===
===Part (c)===
=PART 2=
==Question 13==
===Part (a)===
===Part (b)===
===Part (c)===
===Part (d)===
===Part (e)===
==Question 14==
===Part (a)===
===Part (b)===
===Part (c)===
===Part (d)===
=PART 3=
==Question 15==
===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).
==Question 16==
===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 (c)===
===Part (d)===
===Part (e)===
=PART 4=
==Question 17==
===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}.
==Question 18==
===Part (a)===
===Part (b)===
===Part (c)===