SM358 - 2014

=PART 1=
==Question 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}
==Question 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
:\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}{{{L}^{3}}}\int\limits_{0}^{{L}/{3}\;}{{{x}^{2}}\,\mathrm{d}x} \\ 
& =\frac{3}{{{L}^{3}}}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{{L}/{3}\;} \\ 
& =\frac{1}{{{L}^{3}}}{{\left( \frac{L}{3} \right)}^{3}} \\ 
& =\frac{1}{27} 
\end{align}
==Question 3==
===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
==Question 4==
===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}
==Question 5==
===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}
==Question 6==
===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, … 
==Question 7==
===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}
==Question 8==
===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}{{{\lambda }^{2}}} \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.
==Question 10==
===Part (a)===
===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
==Question 11==
===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}})}/{{{a}^{2}}}\;)z\exp (-{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}/{{{a}^{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}})}/{{{a}^{2}}}\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ 
& =BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{x\exp (-{2{{x}^{2}}}/{{{a}^{2}}}\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{\exp (-{2{{y}^{2}}}/{{{a}^{2}}}\;)\ \text{d}y} \\
& \qquad \,\int\limits_{-\infty }^{\infty }{z\exp (-{2{{z}^{2}}}/{{{a}^{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}})}/{{{a}^{2}}}\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ 
& = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\exp (-{2{{x}^{2}}}/{{{a}^{2}}}\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{y\exp (-{2{{y}^{2}}}/{{{a}^{2}}}\;)\ \text{d}y} \\
& \qquad \,\int\limits_{-\infty }^{\infty }{z\exp (-{2{{z}^{2}}}/{{{a}^{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}})}/{{{a}^{2}}}\;)}\ \text{d}x\,\text{d}y\,\text{d}z}} \\ 
& = BC\cos (\omega t)\int\limits_{-\infty }^{\infty }{\exp (-{2{{x}^{2}}}/{{{a}^{2}}}\;)\ \text{d}x}\,\int\limits_{-\infty }^{\infty }{\exp (-{2{{y}^{2}}}/{{{a}^{2}}}\;)\ \text{d}y} \\
& \qquad \,\int\limits_{-\infty }^{\infty }{{{z}^{2}}\exp (-{2{{z}^{2}}}/{{{a}^{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=
==Question 13==
===Part (a)===
===Part (b)===
===Part (c)===
==Question 14==
===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 (c)===
===Part (d)===
=PART 3=
==Question 15==
===Part (a)===
===Part (b)===
===Part (c)===
==Question ==
===Part (a)===
===Part (b)===
===Part (c)===
===Part (d)===
====(i)====
====(ii)====
====(iii)====
=PART 4=
==Question 17==
===Part (a)===
===Part (b)===
===Part (c)===
===Part (d)===
==Question 18==
===Part (a)===
===Part (b)===
===Part (c)===
===Part (d)===