SM358 - 2007

=PART 1=

Question 1
Generally, a de Broglie wave takes the form
 * $$\Psi_\mathrm{dB}(x,t)=A\mathrm{e}^{\mathrm{i}(kx-\omega t)},

$$ so in this case $$B$$ is the wavenumber $$k=1.90\times 10^{20}\ \mathrm{m}^{-1},$$ and $$C$$ is the angular frequency $$\omega=2.86\times 10^{16}\ \mathrm{s}^{-1}$$.

The energy of a de Broglie wave is given by
 * $$E=\hbar\omega=3.03\times 10^{-18}\ \mathrm J,$$

and the momentum of a de Broglie wave is given by
 * $$p=\hbar k=2.01\times 10^{-14}\ \mathrm{J\ s\ m}^{-1}.$$

Part (a)
Born's rule states that the probability of finding a particle in a small region $$\delta x$$, centred on $$x$$, is given by
 * $$p=\left| \Psi(x,t) \right|^2\delta x$$

The wave function needs to be normalized in order to ensure that the sum of the probabilities of all possible states is equal to 1. The normalization conditions is

\int_{-\infty}^{\infty}\!\!|\Psi(x,t)|^2\ \mathrm dx=1. $$

Part (b)
If the wave function is normalized, we must have
 * $$\int_{-a/2}^{a/2}\!\!|N|^2\ \mathrm dx=1,$$

since there is no chance of finding the particle outside the well. Since $$N$$ is real, we can write this as

\begin{align} \int_{-a/2}^{a/2}\!\!N^2\ \mathrm dx&=1\\ N^2\left( \tfrac{a}{2}-\left(-\tfrac{a}{2}    \right)      \right)&=1 \\ N^2a &= 1 \\ N &= \tfrac{1}{\sqrt a}. \end{align} $$

Part (a)
The coefficient rule says that the probability of measuring the state with amplitude $$C$$ is given by $$p=|C|^2$$. The sum of the probabilities for all states must be equal to 1, so in this case we must have
 * $$|C|^2+|2C|^2=1.$$

There is no requirement that the normalization constant be complex, so we may choose it to be real and write

\begin{align} C^2+4C^2 &= 1 \\ 5C^2 &= 1 \\ C &= \tfrac{1}{\sqrt 5} \end{align} $$

Part (b)
A general state is a linear superposition of stationary states. When a state is measured, the only possible values for the measurement are those corresponding to the stationary states of which it is a linear superposition. In this case, the general state is made up of stationary states $$\psi_1$$ and $$\psi_2$$, so the possible energy measurements are $$E_1$$ and $$E_2$$. The probability of measuring each is given by the square of the modulus of that state's amplitude:

\mathrm P(E_1)=\left( \tfrac{1}{\sqrt 5} \right)^2=\tfrac{1}{5}, \ \ \ \ \ \ \ \ \mathrm P(E_2)=\left( -\tfrac{2}{\sqrt 5} \right)^2=\tfrac{4}{5}. $$

Part (a)
If the ground state is given by $$\psi_0(x)$$:

\begin{align} \langle x \rangle&=\int_{-\infty}^{\infty}\!\!\psi_0^*(x)\ \widehat{\mathrm{x}}\ \psi_0(x)\ \mathrm dx \\ &=\tfrac{a}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!\psi_0^*(x)\left( \widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger \right)\psi_0(x)\ \mathrm dx \\ &= \tfrac{a}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!\psi_0^*(x)\left( \widehat{\mathrm{A}}\psi_0(x)+\widehat{\mathrm{A}}^\dagger\psi_0(x)   \right)\ \mathrm dx. \end{align} $$ Lowering the ground state - $$\widehat{\mathrm{A}}\psi_0(x)$$ - returns a value of zero, and raising the ground state - $$\widehat{\mathrm{A}}^\dagger\psi_0(x)$$ - returns $$\psi_1(x)$$.

\langle x \rangle=\tfrac{a}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!\psi_0^*(x)\psi_1(x)\ \mathrm dx. $$ Since the states $$\psi_0$$ and $$\psi_1$$ are orthogonal, this integral must evaluate to zero, and so we have
 * $$\langle x \rangle = 0.$$

Part (b)
Consider the integral
 * $$\langle x \rangle=\int_{-\infty}^{\infty}\!\!\psi_0^*(x)\ \widehat{\mathrm{x}} \ \psi_0(x)\ \mathrm dx.$$

Knowing that $$\psi_0(x)$$ is an even function, we can see that the integral argument is a composite function with the form
 * $$\mathrm{even}\times \mathrm{odd}\times \mathrm{even}.$$

This composite is an odd function. Integrating any odd function over a symmetrical region - $$(-\infty,\infty)$$ - gives a value of zero. Therefore,
 * $$\langle x \rangle = 0.$$

Part (a)
If an operator is Hermitian, then for a states $$\psi$$ and $$\phi$$ and operator $$\widehat{\mathrm{O}}$$,

\langle\psi | \widehat{\mathrm O} \phi\rangle=\langle \widehat{\mathrm O}\psi|\phi \rangle. $$

Part (b)
The eigenvalues of a Hermitian operator are real-valued.

The eigenfunctions of a Hermitian operator are all mutually orthogonal.

Part (c)
Since $$\widehat{\mathrm{B}}$$ is Hermitian, we can say that


 * $$\langle\phi|\widehat{\mathrm{B}}^2|\phi\rangle=\langle\widehat{\mathrm{B}}\phi|\widehat{\mathrm{B}}\phi\rangle.$$

Using operator $$\widehat{\mathrm{B}}$$ on state $$\phi$$ simply transforms it into a new state, $$\psi$$:


 * $$\langle\widehat{\mathrm{B}}\phi|\widehat{\mathrm{B}}\phi\rangle=\langle\psi|\psi\rangle.$$

The inner product of any state with itself is necessarily greater than or equal to zero, so


 * $$\langle\psi|\psi\rangle\geq 0\ \ \ \therefore\ \ \ \langle\phi|\widehat{\mathrm{B}}^2|\phi\rangle\geq 0.$$

Part (a)
If $$\mathrm e^{\mathrm im\phi}$$ is an eigenfunction of $$\widehat{\mathrm{L}}_z$$, then it must be true that


 * $$\widehat{\mathrm{L}}_z \mathrm e^{\mathrm im\phi}=\lambda \mathrm e^{\mathrm im\phi}.$$

The left hand side evaluates to


 * $$-\mathrm i\hbar\tfrac{\partial}{\partial \phi}\mathrm e^{\mathrm im\phi}=-\mathrm i^2m\hbar \mathrm e^{\mathrm im\phi}=m\hbar \mathrm e^{\mathrm im\phi},$$

which is indeed a real multiple of $$\mathrm e^{\mathrm im\phi}$$, where the multiplier is $$m\hbar$$, as required.

Part (b)
The possible values of $$m$$ are all integers (though this is restricted in practice by the value of $$l$$):


 * $$m=0,\pm 1,\pm 2, ...$$

Part (c)
$$m$$ is restricted to the integers since the wave function must be single-valued and therefore periodic on $$2\pi$$:


 * $$\mathrm e^{\mathrm im\phi}=\mathrm e^{\mathrm im(\phi+2\pi)}.$$

This can be written as


 * $$\mathrm e^{\mathrm im(\phi+2\pi)}=\mathrm e^{\mathrm im\phi}\mathrm e^{2\pi\mathrm im},$$

where the expression $$\mathrm e^{2\pi\mathrm im}$$ must be equal to 1. This expression is only equal to 1 where $$\cos{(2\pi m)}+\mathrm i\sin{(2\pi m)}=1$$, which occurs whenever $$m$$ is an integer. The cosine part is then equal to 1, while the sine part is equal to zero.

Part (a)
The sandwich rule gives the expectation value of $$X$$ as



\begin{align} \langle X \rangle &=\int^{\infty}_{-\infty}\!\!\Psi^*(x,t)\ \widehat{\mathrm{X}}\ \Psi(x,t)\ \mathrm{dx} \\ &=\int^{\infty}_{-\infty}\!\!\psi^*(x)\mathrm{e}^{\mathrm{iEt/\hbar}}\ \widehat{\mathrm{X}}\ \psi(x)\mathrm{e}^{-\mathrm{iEt/\hbar}}\ \mathrm{dx} \\ &=\int^{\infty}_{-\infty}\!\!\psi^*(x)\ \widehat{\mathrm{X}}\ \psi(x)\ \mathrm{dx}. \end{align} $$

This expression has no time-dependence, so $$\langle X \rangle$$ is constant in time.

Part (b)
The generalized Ehrenfest theorem states that


 * $$\tfrac{\mathrm{d}}{\mathrm{dt}}\langle X \rangle=\tfrac{1}{\mathrm{i\hbar}}\left\langle \left[\widehat{\mathrm{X}},\widehat{\mathrm{H}} \right]   \right\rangle.$$

The fact that $$\widehat{\mathrm{X}}$$ commutes with $$\widehat{\mathrm{H}}$$ means that $$[\widehat{\mathrm{X}},\widehat{\mathrm{H}}]=0$$. The expectation value of zero is of course zero, so the right-hand side is equal to zero. This shows that the rate of change of $$\langle X \rangle$$ with time is equal to zero, as expected.

Part (a)
The two bases have an angle of 45 degrees between them. The bases are complementary so the probability of Bob finding the photon to be vertically polarised using the other basis from Alice is 0.5.

Part (b)
The no cloning principle prevents Eve from sending an exact copy of the photon to Bob and keeping a copy.

Part (c)
In the case where Alice and Bob use the same basis Eve will use the wrong basis to measure them on 50% of occasions. When she sends a replacement photon to Bob using the incorrect polarisation he will disagree with Alice 50% of the time so, for the photons measured by Alice and Bob in the same basis, 25% of the intercepted photons would be expected to be unmatched revealing the eavesdropper's presence.

Question 9
The energy levels in the Coulomb model of a hydrogen atom are given by


 * $$E_n=-\tfrac{E_\mathrm{R}}{n^2},$$

so in this case we have the principal quantum number $$n=\sqrt 9=3$$. The possible values of orbital angular momentum are $$l=0,1,...,n-1$$, so in this case there are three possible $$l$$-values. For each $$l$$, the magnetic quantum number $$m$$ can take values in the range $$m=-l,...,l$$, so there is 1 state with $$l=0$$, 3 states with $$l=1$$ and 5 states with $$l=2$$, for a total of $$1+3+5=9$$ states. For each value of $$m$$, there are two possible spin states, so the total number of states doubles to $$18$$.

(As a check, the degeneracy of a Coulomb-model hydrogen atom with principal quantum number $$n$$ is given by $$2n^2=18$$, as required.)

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


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

but we are advised to ignore reduced mass effects, so


 * $$E_\mathrm R^{\mathrm{scaled}}=Z^2E_\mathrm R=4\times 13.6\ \mathrm{eV}=217.6\ \mathrm{eV}.$$

The energy levels of this hydrogen-like system are then given by $$E_n=-E_\mathrm R/n^2$$, so


 * $$E_1=-\tfrac{E_\mathrm R^{\mathrm{scaled}}}{1^2}=-217.6\ \mathrm{eV},\ \ \ \mathrm{and}\ \ \ E_2=\tfrac{E_\mathrm R^{\mathrm{scaled}}}{2^2}=-54.4\ \mathrm{eV}.$$

The energy of the photon emitted when the system transitions from state 2 to state 1 is given by


 * $$E_\mathrm{ph}=\left| E_2-E_1 \right|=163.2\ \mathrm{eV}.$$

Part (b)
We will need a new scaled Rydberg energy, but this time the reduced mass effect will certainly be too large to ignore. The question does not specify the value we should use for the reduced mass of hydrogen, so I have calculated it explicitly, but the answer will be almost identical if you simply take it as equal to 1.



\begin{align} E_\mathrm R^\mathrm{scaled} &=Z^2\tfrac{\mu}{\mu_\mathrm H}E_\mathrm R \\ &=4^2\times \tfrac{\left(\tfrac{207\times 16560}{207+16560}\right)}{\left(\tfrac{1840\times 1}{1840+1}\right)}\times 13.6\ \mathrm{eV}\\ &=44511.28...\ \mathrm{eV}. \end{align}$$

Notice that I have omitted all occurrences of $$m_e$$; they would all have cancelled out anyway.

Now, the energy of a photon that would be released is again given by


 * $$E_\mathrm{ph}=\left| \tfrac{E_\mathrm{R}^\mathrm{scaled}}{2^2}-\tfrac{E_\mathrm{R}^\mathrm{scaled}}{1^2} \right|=33383.46...\ \mathrm{eV}\simeq 3.34\times 10^{4}\ \mathrm{eV}.$$

(Note that due to the precision of the data we are given, only 3 significant figures are justified in our answer, and taking $$\mu=1$$ will change the fourth significant figure only.)

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 Oxygen (Z=8) 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} 3\mathrm\sigma _{g}^{2} \color{black} 1\mathrm\pi _{u}^{4} \color{red} 1\mathrm\pi _{g}^{2} $$

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

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

Anti-bonding = (2 + 2 + 2) = 6

Bond order = 1/2(10 - 6) = 2

Part (c)
$$\mathrm\pi$$orbitals are doubly degenerate so the two highest energy electrons can occupy separate $$\mathrm\pi$$orbitals and the electrons can be in a spin-symmetric triplet state. The triplet state is of lower energy than the singlet state hence the ground state of diatomic oxygen is in this state.

Part (a)
Spontaneous absorption: an incoming photon is absorbed and the energy level of the system is raised by an amount equal to the energy of the photon.



Spontaneous emission: a system spontaneously lowers its energy level, and in doing so emits a photon with energy equal to the change in energy of the system.



Stimulated emission: an incoming photon causes an excited system to lower its energy level and emit a photon. The incoming photon is not absorbed. The incoming photon and both outgoing photons all have energy equal to the change in energy of the system.



Part (b)
=PART 2=

Part (a)
Substituting the function $$\psi_2$$ into the TISE gives



\begin{align} -\tfrac{\hbar^2}{2m} \tfrac{\partial^2}{\partial x^2} \left( \sqrt{\tfrac{2}{L}}\sin{\left(\tfrac{2\pi x}{L}\right)} \right) &= E_2 \sqrt{\tfrac{2}{L}}\sin{\left(\tfrac{2\pi x}{L}\right)} \\ \tfrac{\hbar^2}{2m}\sqrt{\tfrac{2}{L}}\tfrac{4\pi^2}{L^2}\sin{\left(\tfrac{2\pi x}{L}\right)} &= E_2\sqrt{\tfrac{2}{L}}\sin{\left(\tfrac{2\pi x}{L}\right)} \\ \tfrac{2\hbar^2\pi^2}{mL^2}&=E_2 \end{align} $$

Part (b)
The expectation value of position is given by the integral


 * $$\langle x \rangle=\tfrac{2}{L}\!\!\int_{-L/2}^{L/2}\!\!x\sin^2{\left(\tfrac{2\pi x}{L}\right)}\ \mathrm dx$$

The integrand is the product of an even function (sine squared) and an odd function ($$x$$), so is itself an odd function. The integral of an odd function over a symmetric region is equal to zero, so the expectation value of $$x$$ is equal to zero.

The expectation value of momentum is given by the integral



\begin{align} \langle p_x \rangle &= \tfrac{2}{L}\!\!\int_{-L/2}^{L/2}\!\!\sin{\left(\tfrac{2\pi x}{L}\right)}\left(-\mathrm i\hbar\tfrac{\partial}{\partial x}\sin{\left(\tfrac{2\pi x}{L}\right)} \right)\ \mathrm dx \\ &=-\mathrm i\hbar\tfrac{4\pi}{L^2}\!\!\int_{-L/2}^{L/2}\!\!\sin{\left(\tfrac{2\pi x}{L}\right)}\cos{\left(\tfrac{2\pi x}{L}\right)}\ \mathrm dx \end{align} $$

The integrand is again a product of one odd function (sine) and one even function (cosine), so the integral is equal to zero and the expectation value of momentum is equal to zero.

Part (c)
The uncertainty in $$x$$ is given by


 * $$\Delta x=\sqrt{\langle x^2 \rangle-\langle x \rangle^2},$$

but we have already confirmed that $$\langle x \rangle=0$$, so


 * $$\Delta x=\sqrt{\langle x^2 \rangle}.$$

So we need the expectation value of $$x^2$$:



\langle x^2 \rangle =\tfrac{2}{L}\!\!\int_{-L/2}^{L/2}\!\!x^2\sin^2{\left(\tfrac{2\pi x}{L}\right)}\ \mathrm dx. $$

Using the substitutions


 * $$u=\tfrac{2\pi x}{L},\ \ \mathrm dx=\tfrac{L}{2\pi}\mathrm du,\ \ x^2=\tfrac{L^2}{4\pi^2}u^2,\ \ \mathrm{upper}\ u=\pi,\ \ \mathrm{lower}\ u=-\pi,$$

we can recast this integral as


 * $$\langle x^2 \rangle

=\tfrac{L^2}{4\pi^3}\!\!\int_{-\pi}^{\pi}\!\!u^2\sin^2{u}\ \mathrm du.$$

A standard integral is given to enable us to solves this, and the solution is


 * $$\langle x^2\rangle=\tfrac{L^2}{4\pi^3}\left( \tfrac{\pi^3}{3}-\tfrac{\pi}{2} \right).$$

The uncertainty in $$x$$ is therefore



\begin{align} \Delta x &= \sqrt{\tfrac{L^2}{4\pi^3}\left( \tfrac{\pi^3}{3}-\tfrac{\pi}{2} \right)} \\ &\simeq 0.2658L. \end{align} $$

Part (d)
For the same reason as in part (c), we will need the expectation value of $$p_x^2$$.



\begin{align} \langle p_x^2 \rangle &= \tfrac{2}{L}\!\!\int_{-L/2}^{L/2}\!\!\sin{\left(\tfrac{2\pi x}{L}\right)}\left( -\hbar^2\tfrac{\partial^2}{\partial x^2}\sin{\left(\tfrac{2\pi x}{L}\right)}\right)\ \mathrm dx \\ &= \tfrac{8\pi^2\hbar^2}{L^3}\!\!\int_{-L/2}^{L/2}\!\!\sin^2{\left(\tfrac{2\pi x}{L}\right)}\ \mathrm dx \end{align} $$

Using the same substitutions as in part (c) we can rewrite as



\langle p_x^2 \rangle=\tfrac{4\pi\hbar^2}{L^2}\!\!\int_{-\pi}^{\pi}\!\!\sin^2u\ \mathrm du. $$

We can use the remaining provided integral to give


 * $$\langle p_x^2 \rangle=\tfrac{4\pi^2\hbar^2}{L^2}.$$

The uncertainty in $$p_x$$ is then given by


 * $$\Delta p_x=\sqrt{\langle p_x^2 \rangle}=\sqrt{\tfrac{4\pi^2\hbar^2}{L^2}}=\tfrac{2\pi\hbar}{L}=\tfrac{h}{L}.$$

Note
For interest you might like to check that $$\Delta x\Delta p_x\geq \hbar/2$$, as is required if our answer is to be consistent with the uncertainty principle.

For x < 0:


\begin{align} -\tfrac{\hbar^2}{2m}\tfrac{\mathrm d^2}{\mathrm dx^2}\left( A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x} \right)&=E\left( A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x}  \right) \\ -\tfrac{\hbar^2}{2m}\left( A\mathrm i^2k_1^2\mathrm e^{\mathrm ik_1x}+B(-\mathrm i)^2k_1^2\mathrm e^{-\mathrm ik_1x} \right) &= E\left( A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x}  \right) \\ \tfrac{k_1^2\hbar^2}{2m}\left( A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x} \right)&=E\left( A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x}  \right). \end{align} $$ So we have $$E=k_1^2\hbar^2/2m$$, and therefore


 * $$k_1=\tfrac{\sqrt{2Em}}{\hbar}.$$

For x > 0:


\begin{align} -\tfrac{\hbar^2}{2m}\tfrac{\mathrm d^2}{\mathrm dx^2}\left( C\mathrm e^{\mathrm ik_2x}+D\mathrm e^{-\mathrm ik_2x} \right)+\tfrac{3E}{4}\left(...\right)&=E(...)\\ \tfrac{k_2^2\hbar^2}{2m}\left( C\mathrm e^{\mathrm ik_2x}+D\mathrm e^{-\mathrm ik_2x}  \right)+\tfrac{3E}{4}\left(...\right)&=E(...). \end{align} $$ So we have $$E/4=k^2_2\hbar^2/2m$$, and therefore


 * $$k_2=\tfrac{\sqrt{2Em}}{2\hbar}=\tfrac{k_1}{2}.$$

Part (b)
$$D=0$$, because the term $$\mathrm e^{-\mathrm ik_2x}$$ implies a beam moving in the wrong direction in the region $$x>0$$. Since we know no such beam exists, we can require that this term be equal to zero on physical grounds.

Part (c)
The boundary conditions require that $$\psi$$ and its first derivative are both continuous across the boundary, where $$x = 0$$:



\begin{align} A\mathrm e^{\mathrm ik_1x}+B\mathrm e^{-\mathrm ik_1x} =C\mathrm e^{\mathrm ik_2x}\ \ &\Longrightarrow\ \ A+B=C, \\ k_1A-k_1B=k_2C \ \ &\Longrightarrow\ \ 2(A-B)=C. \end{align} $$

Some simple algebra should quickly yield


 * $$B=\tfrac{A}{3}\ \ \ \mathrm{and}\ \ \ C=\tfrac{4A}{3}.$$

Part (d)
The reflection coefficient, $$R$$ is given by $$R=1-T$$. The formula for $$T$$ is given in the equation booklet, and so we have


 * $$R=1-\tfrac{4k_1k_2}{(k_1+k_2)^2}.$$

Recalling that $$k_1=2k_2$$, we can write


 * $$R=1-\tfrac{8k_2^2}{(3k_2)^2}=1-\tfrac{8}{9}=\tfrac{1}{9}.$$

=PART 3=

Part (a)


\widehat{\mathrm{S}}_z|A\rangle_\mathrm{initial}= \tfrac{\hbar}{2}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\tfrac{\hbar}{2}\begin{bmatrix}1\\0\end{bmatrix}=\tfrac{\hbar}{2}|A\rangle_\mathrm{initial}.$$

so $$|A\rangle_{\mathrm{initial}}$$ is an eigenvector of $$\widehat{\mathrm{S}}_z$$ with eigenvalue $$\hbar/2$$. Measuring $$S_z$$ in the state $$|A\rangle_{\mathrm{initial}}$$ will return the value $$\hbar/2$$ with certainty.

Part (b)


\widehat{\mathrm{H}}|\!\uparrow_y\rangle=-\tfrac{\gamma_sB\hbar}{2}\begin{bmatrix}0&-\mathrm i\\\mathrm i&0\end{bmatrix}\begin{bmatrix}1\\\mathrm i\end{bmatrix}=-\tfrac{\gamma_sB\hbar}{2}\begin{bmatrix}1\\\mathrm i\end{bmatrix} $$

If $$\omega=-\gamma_sB$$, the eigenvalue associated with $$|\!\uparrow_y\rangle$$ is $$\hbar\omega/2$$.



\widehat{\mathrm{H}}|\!\downarrow_y\rangle=-\tfrac{\gamma_sB\hbar}{2}\begin{bmatrix}0&-\mathrm i\\\mathrm i&0\end{bmatrix}\begin{bmatrix}\mathrm i\\1\end{bmatrix}=\tfrac{\gamma_sB\hbar}{2}\begin{bmatrix}\mathrm i\\1\end{bmatrix} $$

If $$-\omega=\gamma_sB$$, the eigenvalue associated with $$|\!\downarrow_y\rangle$$ is $$-\hbar\omega/2$$.

Part (c)
The initial state can be written as


 * $$|A\rangle_\mathrm{initial}=|\!\uparrow_z \rangle=a_1|\!\uparrow_y\rangle+a_2|\!\downarrow_y\rangle.$$

The amplitudes $$a_1$$ and $$a_2$$ are given by


 * $$a_1=\langle\uparrow_y|\uparrow_z\rangle=\tfrac{1}{2}\begin{bmatrix}1&-\mathrm i\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\tfrac{1}{2},$$


 * $$a_2=\langle\downarrow_y|\uparrow_z\rangle=\tfrac{1}{2}\begin{bmatrix}-\mathrm i&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=-\tfrac{\mathrm i}{2}.$$

So we have the initial state


 * $$|A\rangle_\mathrm{initial}=\tfrac{1}{2}|\!\uparrow_y\rangle-\tfrac{\mathrm i}{2}|\!\downarrow_y\rangle.$$

The state $$|A\rangle$$ at a later time is given by



\begin{align} &=\tfrac{1}{2}\mathrm e^{-\mathrm i\omega t/2}\begin{bmatrix}1\\\mathrm i\end{bmatrix}-\tfrac{\mathrm i}{2}\mathrm e^{i\omega t/2}\begin{bmatrix}\mathrm i\\1\end{bmatrix} \\ &=\tfrac{1}{2}\left(\begin{bmatrix}\mathrm e^{-\mathrm i\omega t/2}\\ \mathrm i\mathrm e^{-\mathrm i\omega t/2}\end{bmatrix}-\begin{bmatrix}\mathrm i^2\mathrm e^{\mathrm i\omega t/2}\\ \mathrm i\mathrm e^{\mathrm i\omega t/2}\end{bmatrix}\right) \\ &=\tfrac{1}{2}\left( \begin{bmatrix}\cos{(\omega t/2)}-\mathrm i\sin{(\omega t/2)} \\ \mathrm i\cos{(\omega t/2)}+\sin{(\omega t/2)}\end{bmatrix}-\begin{bmatrix}-\cos{(\omega t/2)}-\mathrm i\sin{(\omega t/2)} \\ \mathrm i\cos{(\omega t/2)}-\sin{(\omega t/2)}\end{bmatrix} \right)\\ &=\tfrac{1}{2}\begin{bmatrix}2\cos{(\omega t/2)} \\ 2\sin{(\omega t/2)}\end{bmatrix} \\ &=\begin{bmatrix}\cos{(\omega t/2)} \\ \sin{(\omega t/2)}\end{bmatrix}, \end{align} $$
 * A\rangle

as required.

Part (d)
The state $$|A\rangle$$ is a superposition of an up-spin state and down-spin state. The probability of measuring the spin to be $$+\hbar/2$$ is equal to the square of the amplitude of the eigenvector corresponding to upward spin, and the probability of measuring $$-\hbar/2$$ is the square of the amplitude of the eigenvector corresponding to downward spin.


 * $$P(\uparrow_z)=\cos^2{(\omega t/2)},$$
 * $$P(\downarrow_z)=\sin^2{(\omega t/2)}.$$

Part (a)
The state cannot be factorized, which means that the individual photons do not have independent states, although the system as a whole does. This is the definition of entanglement.

Part (b)
We are looking for the probability of measuring the state $$|\mathrm V\mathrm V_\theta\rangle$$, which is given by



\begin{align} p_{\mathrm V\mathrm V_\theta} &=\left| \langle \mathrm V\mathrm V_\theta| A \rangle \right|^2 \\ &=\left|\tfrac{1}{\sqrt 2}\langle \mathrm V|(\cos\theta\langle \mathrm V|+\sin\theta\langle \mathrm H|) (|\mathrm V\mathrm V\rangle+|\mathrm H\mathrm H\rangle)\right|^2\\ &=\left|\tfrac{1}{\sqrt 2} \left( \cos\theta\langle \mathrm V\mathrm V|+\sin\theta\langle \mathrm V\mathrm H| \right) \left( \right)\right|^2\\ &=\left|\tfrac{1}{\sqrt 2} \left( \cos\theta\langle \mathrm V\mathrm V|\mathrm V\mathrm V\rangle+\cos\theta\langle \mathrm V\mathrm V|\mathrm H\mathrm H\rangle+\sin\theta\langle \mathrm V\mathrm H|\mathrm V\mathrm V\rangle+\sin\theta\langle \mathrm V\mathrm H|\mathrm H\mathrm H\rangle \right)\right|^2 \end{align} $$
 * \mathrm V\mathrm V\rangle+|\mathrm H\mathrm H\rangle

The $$\langle \mathrm V\mathrm V|\mathrm V\mathrm V\rangle$$ term is the only one that survives, since all of the others represent orthogonal sets. The probability of measureing $$|\mathrm V\mathrm V_\theta\rangle$$ is therefore


 * $$p_{VV_\theta}=\left|\tfrac{1}{\sqrt 2}\cos\theta\langle \mathrm V\mathrm V|\mathrm V\mathrm V\rangle\right|^2=\tfrac{\cos^2\theta}{2}$$

To calculate the probability of measuring the state $$|\mathrm V\mathrm H_\theta\rangle$$, we follow an very similar procedure to find


 * $$p_{\mathrm V\mathrm H_\theta}=\left| \langle \mathrm V\mathrm H_\theta|A\rangle \right|^2=\tfrac{\sin^2\theta}{2}$$

Part (c)
The probability that both photons are measured to be vertically polarized is $$\cos^2\theta/2$$, and both horizontally polarized is $$\cos^2\theta/2$$ (we are told we may assume them to be the same. Similarly, the probability that the photon pair is vertical--horizontal is $$\sin^2\theta/2$$ and the probability that the pair is horizontal--vertical is also $$\sin^2\theta/2$$.


 * $$C=P_{VV}+P_{HH}-P_{VH}-P_{HV}=\cos^2\theta-\sin^2\theta.$$

Part (d)
This inequality places a limit on the value of $$\Sigma$$ for any local hidden-variable theory. Quantum mechanical models allow the value of $$\Sigma$$ to exceed this limit, so experiments which support quantum mechanics by measuring this value would provide evidence against local hidden-variable theories.

Aspect's experiment found that certain quantum measurements do not obey this inequality. Since any local-hidden variable theory must obey the inequality, this is strong evidence that local hidden-variables theories cannot reproduce all of the predictions of quantum mechanics. This is Bell's theorem.

=PART 4=

Part (a)
The normalization condition for spherical harmonics requires that


 * $$\int_{0}^{2\pi}\!\!\!\!\int_{0}^{\pi}\!\!Y_{lm}^*(\theta,\phi)Y_{lm}(\theta,\phi)\ \sin\theta \ \mathrm d\theta\ \mathrm d\phi=1.$$

The right-hand side evaluates to

\begin{align} \mathrm{R.H.S.}&=\tfrac{3}{4\pi}\!\!\int_{0}^{2\pi}\!\!\!\!\int_{0}^{\pi}\!\!\cos^2\theta\ \sin\theta \ \mathrm d\theta\ \mathrm d\phi\\ &=\tfrac{3\times 2\pi}{4\pi}\!\!\int_{0}^{\pi}\!\!\cos^2\theta\ \sin\theta\ \mathrm d\theta. \end{align} $$

A provided integral enables us to find the solution as


 * $$\mathrm{R.H.S.}=\tfrac{3\times 2\pi}{4\pi}\times \tfrac{2}{3}=1,$$

as required.

Part (b)
We have just demonstrated that the spherical harmonic is normalized, but for the whole wave function to be normalized we must select an appropriate normalization constant $$A$$ for the radial part of the wave function.



\begin{align} \int_0^\infty\!\!R^*_{nl}(r)R_{nl}(r)r^2\ \mathrm dr &=1 \\ \end{align} $$
 * A|^2\!\!\int_0^\infty\!\!r^4\mathrm e^{-r/a_0}\ \mathrm dr &= 1 \\

Another provided integral enables us to solve this as



\begin{align} A&=\tfrac{1}{\sqrt{24a_0^5}}. \end{align} $$
 * A|^2\times 4!\times a_0^5&=1 \\

Part (c)
The expectation value of $$r$$:



\begin{align} \langle r \rangle &= \int_0^\infty\!\!R^*_{nl}(r)rR_{nl}(r)r^2\ \mathrm dr \\ &= \tfrac{1}{24a_0^5}\!\!\int_0^\infty\!\!r^5\mathrm e^{-r/a_0}\ \mathrm dr. \end{align} $$

Again, this can be solved using one of the provided integrals, and evaluates to



\langle r \rangle = \tfrac{1}{24a_0^5}\times 5!\times a_0^6=5a_0 $$

Part (d)
We are looking for a maximum in the radial probability density function, i.e. a value for $$r$$ at which the rate of change of the radial probability density function is 0. The



\begin{align} \mathrm{R.P.D.F.} &=|R_{nl}(r)|^2r^2\\ &=|A|^2r^4\mathrm e^{-r/a_0}. \end{align} $$

Setting the derivative of this equal to zero:



\begin{align} \tfrac{\mathrm d}{\mathrm dr}\mathrm{R.P.D.F.} =|A|^2\left( 4r^3\mathrm e^{-r/a_0}-\tfrac{1}{a_0}r^4\mathrm e^{-r/a_0} \right)&=0\\ r^3\left( 4a_0-r\right)&=0. \end{align} $$

We can reject the solution where $$r = 0$$ on physical grounds, so the most probable electron--proton separation is $$4a_0$$.