SM358 - 2014

=PART 1=

Part (a)


\begin{align} E & =\hbar \omega \\ & =(1.06\times {{10}^{-34}}\mathrm{J}\,\mathrm{s)(2}\mathrm{.5}\times \mathrm{1}{{\mathrm{0}}^{16}}{{\mathrm{s}}^{-1}}) \\ & =2.7\times {{10}^{-18}}\mathrm{ J} \end{align} $$



\begin{align} p_x & =\hbar k \\ & =(1.06\times {{10}^{-34}}\mathrm{J}\,\mathrm{s)(1}\mathrm{.6}\times \mathrm{1}{{\mathrm{0}}^{10}}{{\mathrm{m}}^{-1}}) \\ & =1.7\times {{10}^{-24}}\mathrm{kg}\,\mathrm{m}\,{{\mathrm{s}}^{-1}} \end{align}$$

Part (b)


\begin{align} {{\lambda }_{\mathrm{dB}}} & =\frac{2\mathrm{ }\pi\text{ }}{k} \\ & =\frac{2\mathrm{ }\pi\text{ }}{\mathrm{(1}\mathrm{.6}\times \mathrm{1}{{\mathrm{0}}^{10}}{{\mathrm{m}}^{-1}})} \\ & =3.9\times {{10}^{-10}}\mathrm{m} \\ & =0.39\,\mathrm{nm} \end{align} $$

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


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

Part (b)


\begin{align} \mathrm{P} & =\int\limits_{0}^{{L}/{3}\;}{{{\left| \Psi \left( x,t \right) \right|}^{2}}\,\mathrm{d}x} \\ & =\frac{3}\int\limits_{0}^{{L}/{3}\;}{{{x}^{2}}\,\mathrm{d}x} \\ & =\frac{3}\left[ \frac{3} \right]_{0}^{{L}/{3}\;} \\ & =\frac{1}{{\left( \frac{L}{3} \right)}^{3}} \\ & =\frac{1}{27} \end{align} $$

Part (a)
There is no particle beam travelling in the negative $$x$$-direction for $$x>0$$ so $$D=0$$.

Part (b)
$$\psi(x)$$ must be continuous for all $$x$$ and $$\mathrm{d} \psi /\mathrm{d}x$$ must be continuous where the potential is finite so at $$x=0$$



\begin{align} &A \exp\left(ik_1(0)\right) + B \exp\left(-ik_1(0)\right) = C \exp\left(ik_1(0)\right) \\ \Rightarrow &A + B = C \qquad\qquad \mathrm{(4.1)}\\ \end{align} $$



\begin{align} &ik_1 A \exp\left(ik_1(0)\right) - ik_1 B \exp\left(-ik_1(0)\right) = ik_2 C \exp\left(ik_1(0)\right) \\ \Rightarrow &k_1 A + k_1 B = k_2 C \\ \Rightarrow & A - B = \tfrac{k_2}{k_1} C \qquad\qquad \mathrm{(4.2)}\\ \end{align} $$

Adding equations 4,1 and 4.2 we have



\begin{align} &2A = \left(1 + \tfrac{k_2}{k_1} \right) C \\ \Rightarrow &A = \tfrac{k_1 + k_2}{2k_1} C \\ \Rightarrow &C = \tfrac{2k_1}{k_1 + k_2} A \qquad\qquad \mathrm{(4.3)}\\ \end{align} $$

and substituting equation 4.3 into 4.1 we get



\begin{align} &A + B = \tfrac{2k_1}{k_1 + k_2} A \\ \Rightarrow &B = \left( \tfrac{2k_1}{k_1 + k_2} -1 \right) A \\ \Rightarrow &B = \tfrac{2k_1 - k_1 - k_2}{k_1 + k_2} A \\ \Rightarrow &B = \tfrac{k_1 - k_2}{k_1 + k_2} A \\ \end{align} $$

So
 * $$B = \tfrac{k_1 - k_2}{k_1 + k_2} A$$
 * $$C = \tfrac{2k_1}{k_1 + k_2} A$$

Part (a)
From the coefficient rule and the normalisation condition we have



\begin{align} &1 = \left|C\right|^2\left( |2|^2 + |-i|^2 + |1|^2\right) \\ \Rightarrow \qquad &\frac{1}{\left|C\right|^2} = 2^2 + 1^2 + 1^2 = 6 \\ \Rightarrow \qquad &\left|C\right|^2 = \frac{1}{6} \\ \end{align} $$

So the probabilities, $$P_1 \text{, } P_2 \text{ and } P_3$$, of the three Eigenvalues $$E_1 \text{, } E_2 \text{ and } E_3$$ are



\begin{align} & P_1 = \frac{4}{6} = \frac{2}{3} \\ & P_2 = \frac{1}{6} \\ & P_3 = \frac{1}{6} \\ \end{align} $$

Part (b)


\begin{align} \left\langle E \right\rangle &= \sum_{i} P_i E_i \\ &= P_1 E_1 + P_2 E_2 + P_3 E_3 \\ &= \frac{2}{3} \times \epsilon + \frac{1}{6} \times 2\epsilon + \frac{1}{6} \times 3\epsilon \\ &= \frac{3}{2}\epsilon \end{align} $$

Part (a)
For any $$f(x)$$



\begin{align} \left[ \widehat{\mathrm{p}}_x, \widehat{\mathrm{x}}^2 \right] f(x) &=\widehat{\mathrm{p}}_x \widehat{\mathrm{x}}^2 f(x)-\widehat{\mathrm{x}}^2 \widehat{\mathrm{p}}_x f(x) \\ &=-\mathrm i\hbar \left( \tfrac{\mathrm d}{\mathrm d x}(x^2f(x)) - x^2 \mathrm \tfrac{\mathrm d}{\mathrm d x}f(x) \right) \\ &=-\mathrm i\hbar \left( x^2 f^'(x) + 2x f(x) - x^2 f^'x) \right) \\ &=-2\mathrm i\hbar x f(x) \end{align} $$

so



\begin{align} \left[ \widehat{\mathrm{p}}_x, \widehat{\mathrm{x}}^2 \right] &= -2 \mathrm i\hbar x \end{align} $$

Part (b)
Using the facts that Hermitian operators are linear and that any operator commutes with any power of itself we have



\begin{align} \frac{\mathrm{d}\left\langle {{{\mathrm{\widehat{p}}}}_{x}} \right\rangle }{\mathrm{d}t} & =\frac{1}{i\hbar }\left\langle \left[ {{{\mathrm{\widehat{p}}}}_{x}},{{{\mathrm{\widehat{H}}}}_{x}} \right] \right\rangle \\ & =\frac{1}{i\hbar }\left\langle \left[ {{{\mathrm{\widehat{p}}}}_{x}},\left( \frac{\mathrm{\widehat{p}}_{x}^{2}}{2m} + \frac{1}{2}C{{{\mathrm{\widehat{x}}}}^{2}} \right) \right] \right\rangle \\ & =\frac{1}{i\hbar }\left\langle \left[ {{{\mathrm{\widehat{p}}}}_{x}},\frac{\mathrm{\widehat{p}}_{x}^{2}}{2m} \right] \right\rangle +\frac{1}{i\hbar }\left\langle \left[ {{{\mathrm{\widehat{p}}}}_{x}},\frac{1}{2}C{{{\mathrm{\widehat{x}}}}^{2}} \right] \right\rangle \\ & =0+\frac{C}{2i\hbar }\left\langle \left[ {{{\mathrm{\widehat{p}}}}_{x}},{{{\mathrm{\widehat{x}}}}^{2}} \right] \right\rangle \\ & =\frac{C}{2i\hbar }\left\langle -2i\hbar x \right\rangle \\ & =-C\left\langle x \right\rangle \end{align} $$

Part (a)


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

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

Part (b)
We require that $$\exp (im\phi )$$ is single valued so that



\begin{align} \exp (im\phi ) & =\exp \left( im(\phi +2 \mathrm\pi \right) \\ & =\exp (2im \mathrm\pi )\exp (im\phi ) \end{align} $$

and $$\exp (2im \mathrm\pi )=1$$ when $$m=0,\,\pm 1,\,\pm 2\,\ldots $$

so m may take the values 0, ±1, ±2, …

Part (a)


\begin{align} \Psi ({{x}_{2}},{{x}_{1}}) & =\frac{1}{\sqrt{2}}\left( {{\psi }_{\text{A}}}({{x}_{2}}){{\psi }_{\text{B}}}({{x}_{1}})-{{\psi }_{\text{B}}}({{x}_{2}}){{\psi }_{\text{A}}}({{x}_{1}}) \right) \\ & =\frac{1}{\sqrt{2}}\left( -{{\psi }_{\text{A}}}({{x}_{1}}){{\psi }_{\text{B}}}({{x}_{2}})+{{\psi }_{\text{B}}}({{x}_{1}}){{\psi }_{\text{A}}}({{x}_{2}}) \right) \\ & =-\frac{1}{\sqrt{2}}\left( {{\psi }_{\text{A}}}({{x}_{1}}){{\psi }_{\text{B}}}({{x}_{2}})-{{\psi }_{\text{B}}}({{x}_{1}}){{\psi }_{\text{A}}}({{x}_{2}}) \right) \\ & =-\Psi ({{x}_{1}},{{x}_{2}}) \end{align} $$

So the wave function $$\Psi ({{x}_{1}},{{x}_{2}})$$ is anti-symmetric with respect to particle label exchange

Part (b)
Identical fermions have anti-symmetric total wave functions so a spin-½ particle pair with the anti-symmetric spatial wave function $$\Psi ({{x}_{1}},{{x}_{2}})$$ must have a symmetric spin state.

Part (c)
In this state $$S=1$$ and $${{M}_{S}}=-1,\,0\ \text{or}\ \text{1}$$

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

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

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

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

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

Question 9
Using the variational method the ground state energy is estimated by


 * $${{E}_{\text{gs}}}\le \min \frac{\left\langle {{\phi }_{t}} \right|\left. {\mathrm{\widehat{H}}} \right|\left. {{\phi }_{t}} \right\rangle }{\left\langle  {{\phi }_{t}} \right|\left. {{\phi }_{t}} \right\rangle }$$

Let



\begin{align} f(\lambda ) &= \frac{\left\langle {{\phi }_{t}} \right|\left. {\mathrm{\widehat{H}}} \right|\left. {{\phi }_{t}} \right\rangle }{\left\langle {{\phi }_{t}} \right|\left. {{\phi }_{t}} \right\rangle } \\ & =\frac{(4{{\lambda }^{2}}+1)C}{\lambda } \\ & =\left( 4\lambda +\frac{1}{\lambda } \right)C \end{align} $$

where $$\lambda >0$$

Now $$f(\lambda )\to \infty $$ as $$\lambda \to 0$$ and $$f(\lambda )\to \infty $$ as $$\lambda \to \infty $$ and $$f(\lambda )$$ is finite on $$0<\lambda <\infty $$ so $$\min f(\lambda )$$ occurs when $${\text{d}f}/{\text{d}\lambda }\;=0$$.


 * $$\begin{align}

& \frac{\text{d}f}{\text{d}\lambda }=0 \\ & \Rightarrow \frac{\text{d}}{\text{d}\lambda }\left( 4\lambda +\frac{1}{\lambda } \right)C=0 \\ & \Rightarrow \left( 4-\frac{1} \right)C=0 \\ & \Rightarrow {{\lambda }^{2}}=\frac{1}{4} \\ & \Rightarrow \lambda =\frac{1}{2}(\lambda >0) \\ \end{align}$$
 * $$\begin{align}

& f({1}/{2}\;)=\left( 4({1}/{2}\;)+\frac{1}{({1}/{2)}\;} \right)C =4C \end{align}$$

Hence the ground state energy is less than or equal to $$4C$$.

Part (b)
The ground state electronic configuration of 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 (a)
By Bloch’s theorem $${{u}_{\mathbf{k}}}(\mathbf{r})$$ has the periodicity of the lattice so for any lattice vector $$\mathbf{R}$$



{{u}_{\mathbf{k}}}(\mathbf{r}+\mathbf{R})={{u}_{\mathbf{k}}}(\mathbf{r}) $$

Part (b)


\begin{align} {{\psi }_{\mathbf{k}}}(\mathbf{r}+\mathbf{R}) & = {{u}_{\mathbf{k}}}(\mathbf{r}+\mathbf{R})\exp \left( i\mathbf{k}\cdot (\mathbf{r}+\mathbf{R}) \right) \\ & ={{u}_{\mathbf{k}}}(\mathbf{r})\exp \left( i\mathbf{k}\cdot (\mathbf{r}+\mathbf{R}) \right) \qquad\qquad (\text{by}\ \text{part}\ a) \\ & ={{u}_{\mathbf{k}}}(\mathbf{r})\exp \left( i\mathbf{k}\cdot \mathbf{r} \right)\exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \\ & ={{\psi }_{\mathbf{k}}}(\mathbf{r})\exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \end{align} $$

Part (c)
The electron probability density function is given by $${{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}) \right|}^{2}}$$

Now

\begin{align} {{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}+\mathbf{R}) \right|}^{2}} & ={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}+\mathbf{R}) \right|}^{2}} \\ & ={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r})\exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \right|}^{2}} \qquad\qquad (\text{by}\ \text{part}\ b) \\ & ={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}) \right|}^{2}}\exp \left( -i\mathbf{k}\cdot \mathbf{R} \right)\exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \\ & ={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}) \right|}^{2}} \end{align} $$

so

{{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}+\mathbf{R}) \right|}^{2}}={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r}) \right|}^{2}} $$

and the electron probability density has the periodicity of the lattice.

Question 12
According to first-order time dependent perturbation theory a radiative transition is forbidden if the matrix element for the transition is zero.

For the transition to from state  to state   in response to the perturbation   the matrix element is

$$ \begin{align} \left\langle u \right|\left. {\mathrm{\widehat{V}}} \right|\left. f \right\rangle & =\int{{{u}^{*}}(x,y,z)V(t)f(x,y,z)\ \text{d}V} \\ & =BC\cos (\omega t) \\ & \quad \int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{x\exp (-{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/\;)z\exp (-{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ & =BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{xz\exp (-{2({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ & =BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{x\exp (-{2{{x}^{2}}}/\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{\exp (-{2{{y}^{2}}}/\;)\ \text{d}y} \\ & \qquad \,\int\limits_{-\infty }^{\infty }{z\exp (-{2{{z}^{2}}}/\;)\ \text{d}z} \\ & =0 \qquad (\text{since the }x\text{-integrand is an odd (}x\text{) }\times \text{ even (Guassian)} \\ & \qquad\qquad\qquad \text{= odd function}) \end{align} $$

similarly

$$ \begin{align} \left\langle v \right|\left. {\mathrm{\widehat{V}}} \right|\left. f \right\rangle & =\int{{{v}^{*}}(x,y,z)V(t)f(x,y,z)\ \text{d}V} \\ & = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{yz\exp (-{2({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ & = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\exp (-{2{{x}^{2}}}/\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{y\exp (-{2{{y}^{2}}}/\;)\ \text{d}y} \\ & \qquad \,\int\limits_{-\infty }^{\infty }{z\exp (-{2{{z}^{2}}}/\;)\ \text{d}z} \\ & = 0 \qquad (\text{since the }y\text{-integrand is an odd }\times \text{ even = odd function) } \end{align} $$

and

$$ \begin{align} \left\langle w \right|\left. {\mathrm{\widehat{V}}} \right|\left. f \right\rangle & = \int{{{w}^{*}}(x,y,z)V(t)f(x,y,z)\ \text{d}V} \\ & = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{z}^{2}}\exp (-{2({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ & = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\exp (-{2{{x}^{2}}}/\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{\exp (-{2{{y}^{2}}}/\;)\ \text{d}y} \\ & \qquad \,\int\limits_{-\infty }^{\infty }{{{z}^{2}}\exp (-{2{{z}^{2}}}/\;)\ \text{d}z} \\ & \ne 0 \qquad (\text{since all the integrands are even functions)} \end{align} $$

Hence, according to first-order time dependent perturbation theory, the only excited state that can be reached is state $$w$$.

=PART 2=

Part (a)

 * $$\widehat\mathrm p_x = - \frac{\mathrm i \hbar}{a\sqrt{2}}\left( \widehat\mathrm A - \widehat\mathrm A^{\dagger} \right)$$

so



\begin{align} \widehat\mathrm p_x^2 &= - \frac{\hbar^2}{2a^2}\left( \widehat\mathrm A - \widehat\mathrm A^{\dagger} \right)^2 \\ &= - \frac{\hbar^2}{2a^2}\left( \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} \widehat\mathrm A \right) \\ &= \frac{\hbar^2}{2a^2}\left( \widehat\mathrm A \widehat\mathrm A^{\dagger} + \widehat\mathrm A^{\dagger} \widehat\mathrm A			- \widehat\mathrm A \widehat\mathrm A - \widehat\mathrm A^{\dagger} \widehat\mathrm A^{\dagger} \right) \\ \end{align} $$

hence



\begin{align} \left\langle p_x^2 \right\rangle &= \left\langle \psi_n \right| \widehat\mathrm p_x^2 \left| \psi_n \right\rangle \\ &= \frac{\hbar^2}{2a^2} \left\langle \psi_n \right| \widehat\mathrm A \widehat\mathrm A^{\dagger} + \widehat\mathrm A^{\dagger} \widehat\mathrm A - \widehat\mathrm A \widehat\mathrm A 			- \widehat\mathrm A^{\dagger} \widehat\mathrm A^{\dagger} \left| \psi_n \right\rangle \\ \end{align} $$

Now



\left\langle \psi_n \right| \widehat\mathrm A^{\dagger} \widehat\mathrm A^{\dagger} \left| \psi_n \right\rangle = \left\langle \psi_n \right| \widehat\mathrm A \widehat\mathrm A \left| \psi_n \right\rangle = 0 $$

so, for the the state $$n$$,



\begin{align} \left\langle p_x^2 \right\rangle_n &= \frac{\hbar^2}{2a^2} \left\langle \psi_n \right| \widehat\mathrm A \widehat\mathrm A^{\dagger} + \widehat\mathrm A^{\dagger} \widehat\mathrm A \left| \psi_n \right\rangle \\ &= \frac{\hbar^2}{2a^2} \left\langle \psi_n \right| 1 + 2\widehat\mathrm A^{\dagger} \widehat\mathrm A \left| \psi_n \right\rangle \\ &= \frac{\hbar^2}{2a^2} \left\langle \psi_n \right| 1 + 2n \left| \psi_n \right\rangle \\ &= \frac{\hbar^2}{2a^2} \left(2n+1\right) \left\langle \psi_n \left. \right| \psi_n \right\rangle \\ &= \frac{\hbar^2}{2a^2} \left(2n+1\right) \\ \end{align} $$

so



\left\langle p_x^2 \right\rangle_1 = \frac{3\hbar^2}{2a^2} $$ and

\left\langle p_x^2 \right\rangle_2 = \frac{5\hbar^2}{2a^2} $$

Part (b)
For a system which is a combination of discreet states the expectation value is



\left\langle E \right\rangle = \sum_{i}{} \mathrm P_i E_i $$

Now $$\mathrm P_1 = \mathrm P_2 = \left|{1}/{\sqrt{2}}\right|^2={1}/{2}$$ so

$$ \begin{align} \left\langle E \right\rangle_{\phi} &= \frac{1}{2} E_1 + \frac{1}{2} E_2\\ &= \frac{1}{2} \times \frac{3}{2}\hbar \omega_0 + \frac{1}{2} \times \frac{5}{2}\hbar \omega_0 \\ &= \left( \frac{3}{5} + \frac{5}{4} \right) \hbar \omega_0 \\ &= 2 \hbar \omega_0\\ \end{align} $$

Part (d)
=PART 3=

(iii)
=PART 4=