SM358 - 2013

=PART 1=

Part (a)
For the state with energy $$E_\mathrm{init}$$ we need


 * $$n_x^2+n_y^2+n_z^2=11,$$

so the possible quantum numbers for this state are


 * $$(n_x,n_y,n_z)=(3,1,1),\ (1,3,1),\ (1,1,3),$$

and this state has threefold degeneracy.

For the state with energy $$E_\mathrm{fin}$$ we need


 * $$n_x^2+n_y^2+n_z^2=3,$$

so the only possible quantum numbers are


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

and the state has no (or onefold) degeneracy.

Part (b)
I really dislike this question, due to the form of the provided constant. Depending on the approach you take you can end up with some very messy algebra which is an unreasonable distraction on a timed test. Nevertheless ...

Using $$E_\mathrm{ph}=hc/\lambda$$, we have


 * $$\lambda=\tfrac{hc}{E_\mathrm{ph}}$$

where $$E_\mathrm{ph}$$ is given by



\begin{align} E_\mathrm{ph} &=\left| E_\mathrm{init}-E_\mathrm{fin} \right|\\ &=\tfrac{h}{mL^2}\left| \tfrac{11\hbar\pi}{4}-\tfrac{3\hbar\pi}{4} \right| \\ &=5\times 10^{14}\times \left| 6.66\times 10^{-34} \right|\ \mathrm J \\ &=3.33\times 10^{-19}\ \mathrm J,\\ &\therefore\\ \lambda &= \tfrac{6.63\times 10^{-34}\times 3\times 10^8}{3.33\times 10^{-19}}\ \mathrm m \\ &=5.97\times 10^{-7}\ \mathrm m. \end{align} $$

Part (a)
The normalization condition requires that (assuming N is real and positive)



\begin{align} 25|N|^2&=1 \\ N&=\tfrac{1}{5}. \end{align} $$
 * 4N^2|+|3\mathrm iN|^2&=1 \\

Part (b)
The only possible energy measurements in the given state are the energy eigenvalues corresponding to the eigenfunctions, i.e. $$E_2=2\varepsilon$$ or $$E_3=3\varepsilon$$.

At $$t=0$$, the probability of measuring each energy eigenvalue is



\begin{align} P_{2\varepsilon}=|4N|^2=\left(\tfrac{4}{5}\right)^2=\tfrac{16}{25},\\ P_{3\varepsilon}=|3\mathrm iN|^2=\left(\tfrac{3}{5}\right)^2=\tfrac{9}{25}. \end{align} $$

Part (c)
The expectation value for energy is given by


 * $$\langle E\rangle=\sum_i P_iE_i=\tfrac{16}{25}\times 2\varepsilon+\tfrac{9}{25}\times 3\varepsilon=2.36\varepsilon.$$

Part (a)
Squaring the given momentum operator:



\begin{align} \widehat{\mathrm{p}}_x^2 &=\tfrac{\mathrm i^2\hbar^2}{2a^2}(\widehat{\mathrm{A}}-\widehat{\mathrm{A}}^\dagger)(\widehat{\mathrm{A}}-\widehat{\mathrm{A}}^\dagger)\\ &=-\tfrac{\hbar^2}{2a^2}(\widehat{\mathrm{A}}\widehat{\mathrm{A}}-\widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger-\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}^\dagger). \end{align} $$

Part (b)
The expectation value of the square of momentum:



\begin{align} \langle p_x^2\rangle &=-\tfrac{\hbar^2}{2a^2}\langle \psi_2 | \widehat{\mathrm{A}}\widehat{\mathrm{A}}-\widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger-\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}^\dagger | \psi_2 \rangle\\ &=-\tfrac{\hbar^2}{2a^2}\left( \langle \psi_2|\widehat{\mathrm{A}}\widehat{\mathrm{A}}|\psi_2 \rangle -\langle \psi_2| \widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger|\psi_2 \rangle-\langle \psi_2|\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}} |\psi_2 \rangle+\langle\psi_2 |\widehat{\mathrm{A}}^\dagger \widehat{\mathrm{A}}^\dagger |\psi_2 \rangle \right)\\ &=-\tfrac{\hbar^2}{2a^2}\left( \sqrt 2 \langle \psi_2|\psi_0\rangle-3\langle\psi_2|\psi_2\rangle-2\langle\psi_2|\psi_2\rangle+2\sqrt 3\langle \psi_2|\psi_4\rangle     \right) \\ &=-\tfrac{\hbar^2}{2a^2}(-3-2) \\ &=\tfrac{5}{2}\tfrac{\hbar^2}{a^2}. \end{align} $$

Part (c)
The expectation value of kinetic energy:



\langle E_\mathrm{kin}\rangle=\tfrac{\langle p_x^2\rangle}{2m}=\tfrac{5}{2}\tfrac{\hbar^2}{a^2}\tfrac{1}{2m}=\tfrac{5}{4}\tfrac{\hbar^2}{ma^2}. $$

Part (a)
For a particle in one dimension with wave function $$\Psi \left( x, t \right)$$, the probability of finding the particle in a small interval $$\delta x$$ centred on $$x$$ at time $$t$$ is given by


 * $$\mathrm P = \left| {\Psi \left( x, t \right)} \right|^2 \delta x$$

At time $$t=0$$



\begin{align} & x=a \\ & \delta x=\frac{a}{50} \\ & \left| {\Psi \left( x, 0 \right)} \right|^2 = \frac{1}{a\sqrt{\pi}} \exp\left(-{x^2}/{a^2}\right) \\ \end{align} $$

So

\begin{align} P &= \frac{1}{a\sqrt{\pi}} \exp\left(-{a^2}/{a^2}\right) \frac{a}{50} \\ &= \frac{1}{50\sqrt{\pi}} \exp{\left(- 1 \right)} \\ &= \frac{1}{50\mathrm{e}\sqrt{\pi}}\\ &\approx 0.00415 \end{align} $$

Part (b)
The probability of measuring the momentum of a particle with wave function $$\Psi \left( x, t \right)$$ to be in the small range $$\hbar \delta k$$ centred on $$\hbar k$$ is


 * $$\mathrm P = \left| {A(k)} \right|^2 \delta k$$

At time $$t=0$$



\begin{align} & k=\frac{1}{a} \\ & \delta k=\frac{1}{50a} \\ & \left| A(k) \right|^2 = \frac{a}{\sqrt{\pi}} \exp\left(- a^2 k^2 \right) \\ \end{align} $$

So

\begin{align} P &= \frac{a}{\sqrt{\pi}} \exp\left(- a^2 / a^2 \right) \frac{1}{50a} \\ &= \frac{1}{50\sqrt{\pi}} \exp\left(- 1\right)\\ &= \frac{1}{50\mathrm{e}\sqrt{\pi}}\\ &\approx 0.00415 \end{align} $$

Part (a)

 * $$\left\langle \mathrm\Psi \right|=\tfrac{3}{5}\left\langle  f \right|-i\tfrac{4}{5}\left\langle  g \right|$$

so



\begin{align} \left\langle \mathrm\Psi | \mathrm{\widehat B } \mathrm\Psi \right\rangle & =\left( \tfrac{3}{5}\left\langle f \right|-i\tfrac{4}{5}\left\langle  g \right| \right)\left( 6\left| f \right\rangle -i2\left| g \right\rangle  \right) \\ & =\tfrac{18}{5}\left\langle f | f \right\rangle -\tfrac{8}{5}\left\langle g | g \right\rangle -i\tfrac{6}{5}\left\langle f | g \right\rangle -\tfrac{20}{5}\left\langle g | f \right\rangle \\ & =\tfrac{18}{5}-\tfrac{8}{5} \\ & =2 \end{align} $$

Part (b)

 * $$\left\langle \mathrm{\widehat B } \mathrm\Psi \right|=6\left\langle f \right| + 2i\left\langle  g \right|$$

so



\begin{align} \left\langle \mathrm{\widehat B } \mathrm\Psi | \mathrm{\widehat B } \mathrm\Psi \right\rangle & =\left( 6\left\langle f \right|+ 2i \left\langle  g \right| \right)\left( 6\left| f \right\rangle -2i\left| g \right\rangle  \right) \\ & =36\left\langle f | f \right\rangle + 4\left\langle g | g \right\rangle - 12i\left\langle f | g \right\rangle + 12i\left\langle g | f \right\rangle \\ & =36+4 \\ & =40 \end{align} $$

Now

\begin{align} \left\langle \mathrm\Psi \left| \mathrm{\widehat B }_2 \right| \mathrm\Psi \right\rangle & = \left\langle \mathrm\Psi \left| \mathrm{\widehat B } \right| \mathrm{\widehat B } \mathrm\Psi \right\rangle \\ & = \left\langle \mathrm{\widehat B } \mathrm\Psi \left| \mathrm{\widehat B } \right| \mathrm\Psi \right\rangle \qquad\qquad \text{since } \mathrm{\widehat B } \text{ is Hermitian} \\ & = \left\langle \mathrm{\widehat B } \mathrm\Psi | \mathrm{\widehat B } \mathrm\Psi \right\rangle \\ \end{align} $$

so
 * $$ \left\langle \mathrm\Psi \left| \mathrm{\widehat B }_2 \right| \mathrm\Psi \right\rangle = \left\langle B^2 \right\rangle = 40 $$

Part (c)


\begin{align} \mathrm\Delta B & = \left( {\left\langle B^2 \right\rangle} - {\left\langle B \right\rangle}^2 \right)^\tfrac{1}{2} \\ & = \left( 40 - 2^2 \right)^\tfrac{1}{2} \\ & = 6 \\ \end{align} $$

so
 * $$ \mathrm\Delta B = 6 $$

Part (a)

 * $$ {\mathrm{\widehat{L}}}_{z} = -i\hbar \frac{\partial }{\partial \phi } $$

so

\begin{align} {{\mathrm{\widehat{L}}}_{z}}\exp (-5i\phi ) & =-i\hbar \frac{\partial }{\partial \phi }\exp (-5i\phi ) \\ & =(-i\hbar )(-5i)\exp (-5i\phi ) \\ & =-5\hbar \exp (-5i\phi ) \end{align} $$

so $$\exp (-5i\phi )$$ is an eigenfunction of $${{\mathrm{\widehat{L}}}_{z}}$$ with eigenvalue $$-5\hbar $$

Part (b)


\begin{align} \Delta L_x \Delta L_y & \geq \tfrac{1}{2} \left| \left\langle \left[ {\mathrm{\widehat{L}}}_{x}, {\mathrm{\widehat{L}}}_{y} \right] \right\rangle \right| \\ & \geq \tfrac{1}{2} \left| \left\langle i\hbar {\mathrm{\widehat{L}}}_{z} \right\rangle \right| \\ & \geq \tfrac{\hbar}{2} \left| \left\langle {\mathrm{\widehat{L}}}_{z} \right\rangle \right| \\ \end{align} $$

so, for any state with the $$\phi$$ dependence of $$f(\phi)$$,



\begin{align} \Delta L_x \Delta L_y & \geq \tfrac{\hbar}{2} \left| \left\langle {\mathrm{\widehat{L}}}_{z} \right\rangle \right| \\ & \geq \tfrac{\hbar}{2} \left| {\left\langle f \left| {\mathrm{\widehat{L}}}_{z} \right| f \right\rangle} \right| \\ & \geq \tfrac{\hbar}{2} \left| -5\hbar \right| \\ & \geq \frac{5\hbar^2}{2} \\ \end{align} $$

If either $$L_x$$ or $$L_y$$ were to have a definite value then $$\Delta L_x \Delta L_y = 0$$ and the uncertainty principle would be violated.

Part (a)
The two particle spin state $$\left|B\right\rangle$$ is entangled if it cannot be factorised into the product of two single particle spin states. For the general spin state
 * $$\left|B\right\rangle = a\left| \uparrow_z\uparrow_z \right\rangle + b\left| \downarrow_z\uparrow_z \right\rangle + c\left| \uparrow_z\downarrow_z \right\rangle + d\left| \downarrow_z\downarrow_z \right\rangle $$

can only be factorised if $$ad - bc = 0$$

Now, for the state $$\left|A\right\rangle$$,
 * $$ad - bc = \tfrac{4}{5}\times\tfrac{2}{5} - \tfrac{2}{5}\times\tfrac{1}{5} = \tfrac{6}{25} \neq 0$$

so the state is entangled.

Part (b)


\begin{align} \mathrm{P}(opp) & = \mathrm{P}(\uparrow_z\downarrow_z) + \mathrm{P}(\downarrow_z\uparrow_z) \\ & = \left|\tfrac{2}{5}\right|^2 + \left|\tfrac{1}{5}\right|^2 \\ & = \tfrac{4}{25} + \tfrac{1}{25} \\ & = \tfrac{1}{5} \\ \end{align} $$

Part (c)


\begin{align} \mathrm{P}(\uparrow_x\uparrow_x) & = \left|\left\langle \uparrow_x\uparrow_x | A \right\rangle\right|^2 \\ & = \tfrac{1}{25} \left|\left\langle \uparrow_x\uparrow_x \right| \left(	4\left| \uparrow_z\uparrow_z \right\rangle +	2\left| \downarrow_z\uparrow_z \right\rangle +	 \left| \uparrow_z\downarrow_z \right\rangle +	2\left| \uparrow_z\uparrow_z \right\rangle	\right)\right|^2 \\ & = \tfrac{1}{25} \left| 4\left\langle \uparrow_x\uparrow_x | \uparrow_z\uparrow_z \right\rangle + 2\left\langle \uparrow_x\uparrow_x | \downarrow_z\uparrow_z \right\rangle + \left\langle \uparrow_x\uparrow_x | \uparrow_z\downarrow_z \right\rangle + 2\left\langle \uparrow_x\uparrow_x | \downarrow_z\downarrow_z \right\rangle \right|^2 \\ & = \tfrac{1}{25} \left| 4\left\langle \uparrow_x | \uparrow_z \right\rangle_1 \left\langle \uparrow_x | \uparrow_z \right\rangle_2 + 2\left\langle \uparrow_x | \downarrow_z \right\rangle_1 \left\langle \uparrow_x | \uparrow_z \right\rangle_2 + \right. \\ &\left.\qquad\qquad \left\langle \uparrow_x | \uparrow_z \right\rangle_1 \left\langle \uparrow_x | \downarrow_z \right\rangle_2 + 2\left\langle \uparrow_x | \downarrow_z\ \right\rangle_1 \left\langle \uparrow_x | \downarrow_z \right\rangle_2 \right|^2 \\ & = \tfrac{1}{25} \left| 4 \times \tfrac{1}{\sqrt{2}} \times \tfrac{1}{\sqrt{2}} + 2 \times \tfrac{1}{\sqrt{2}} \times \tfrac{1}{\sqrt{2}} + \times \tfrac{1}{\sqrt{2}} \times \tfrac{1}{\sqrt{2}} + 2 \times \tfrac{1}{\sqrt{2}} \times \tfrac{1}{\sqrt{2}} \right|^2 \\ & = \tfrac{1}{100} \left| 4 + 2 + 1 + 2 \right|^2 \\ & = \tfrac{81}{100} \\ \end{align} $$

so the probability of both particles having $$S_x = +\tfrac{\hbar}{2} \text{ is } \tfrac{81}{100} $$

Part (a)
FALSE: Violation of the Bell inequality in the Aspect experiment rules out any theory relying on local realism.

Part (b)
TRUE:

Part (c)
TRUE: The BB84 protocol does not rely on entanglement, Eckert's protocol does.

Part (d)
TRUE: Only entangled photons can violate the CHSH inequality, eavesdropping destroys the entanglement used in the Exckert protocol.

Part (e)
FALSE: The no cloning theorem prohibits knowledge of the state to be teleported.

Question 9
For the radial probability function $$f(r)$$ we have $$f(0)=0$$, $$f\to 0 \text{ as } r \to \infty $$ and $$ f>0 \text{ for } 0<r<\infty$$ so the maximum probability occurs when $${\mathrm{d}f}/{\mathrm{d}r}=0$$



\begin{align} & \frac{\mathrm{d}f}{\mathrm{d}r}=0 \\ \Rightarrow \qquad &2r \frac{4}{a_{0}^{3}} \exp\left(\frac{-2r}{a_0}\right) + \left(\frac{-2}{a_0}\right) \frac{4r^2}{a_{0}^{3}} \exp\left(\frac{-2r}{a_0}\right) = 0 \\ \Rightarrow \qquad &2r = \frac{2r^2}{a_0} \\ \Rightarrow \qquad &r = {a_0} \\ \end{align} $$

so the most probable separation is $$r=a_0$$

Part (b)


\begin{align} \left\langle r \right\rangle &= \left\langle R_{10} \right| r \left| R_{10} \right\rangle \\ &= \int_{0}^{\infty } r \left| R_{10} \right|^2 r^2 {\mathrm{d}r} \\ &= \int_{0}^{\infty } r f(r) {\mathrm{d}r}  \\ &= \frac{4}{a_{0}^{3}} \int_{0}^{\infty } r^3 \exp\left(\frac{-2r}{a_0}\right) {\mathrm{d}r} \\ &= \frac{4}{a_{0}^{3}} 3! \left( \frac{a_0}{2} \right)^4 \\ &= \frac{2^3 \times 3 a_{0}^{4}}{2^4 a_{0}^{3}} \\ &= \frac{3}{2} a_{0} \\ \end{align} $$

So the expectation value of $$r$$ is $${3 a_0}/{2}$$

Part (a)
The scaled Rydberg energy of a hydrogen like system is given by


 * $$E_{\mathrm{R}}^{\mathrm{scaled}} = Z^2\frac{\mu}{\mu_{\mathrm{H}}}E_{\mathrm{R}}$$

so, for a $$\mathrm{He}^+$$ ion with $$Z=2$$ and ignoring reduced mass effects, we have



\begin{align} E_{\mathrm{R}}^{\mathrm{scaled}} &= Z^2E_{\mathrm{R}} \\ &= 4 E_{\mathrm{R}} \\ \end{align} $$

In the ground state we have $$n=1$$ so

\begin{align} E_1 &= - \frac{E_{mathrm{R}}^{\mathrm{scaled}}}{n^2} \\ &= - 4 E_{\mathrm{R}} \\ &= - 4 \times 13.6 \,\mathrm{eV} \\ &= - 54.4 \,\mathrm{eV} \\ \end{align} $$

Part (b)
In the independent particle model of the Helium atom both electrons are in the ground state and there are no electron-electron interactions so

\begin{align} E_{\mathrm{gs}} &= 2 \times E_1 \\ &= - 109 \,\mathrm{eV} \\ \end{align} $$

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


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

So $$L=1,2 \text{ or } 3$$ and $$S=0 \text{ or } 1$$

Part (b)
The degeneracy of an atomic term is given by $$\left(2L + 1\right) \left(2S+1\right)$$ so the degeneracy of the $$L=1, S=1$$ is $$\left(2\times1 + 1\right) \left(2\times1+1\right)=9$$

Part (c)
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 $$L=1, S=1$$ we have the levels with $$J=0,1 \text{ or } 2$$with degeneracies $$\left(2J + 1\right)$$ hence



\begin{matrix} J & \text{Degeneracy} \\ 0 & 1 \\  1 & 3 \\   2 & 5 \\ \end{matrix} $$

Part (a)
The ground state electronic configuration of a diatomic molecule (Nitrogen, Z=7) 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} 1\mathrm\pi _{u}^{4}3\mathrm\sigma _{g}^{2} $$

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

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

Anti-bonding = (2 + 2) = 4

Bond order = 1/2(10 - 4) = 3

Part (b)
=PART 2=

Question 14
=PART 3=

Question 16
=PART 4=