SM358 - 2009

=PART 1=

Part (a)
The normalization condition requires that



\begin{align} 3^2|C|^2+(-4)^2|C|^2&=1 \\ 25|C|^2&=1 \\ \end{align} $$
 * C|^2&=\tfrac{1}{25}

The probability of measuring each energy is the square of the coefficient of the associated state.



p_1=3^2|C|^2=\tfrac{9}{25},\ \ \ p_2=0,\ \ \ p_3=(-4)^2|C|^2=\tfrac{16}{25} $$

Part (b)
The second measurement must return the same value as the first, i.e. $$E_1$$. This is because the first measurement collapses the state onto the stationary states $$\psi_1$$, and if another measurement is taken immediately the system does not have time to evolve away from this state.

Part (a)
The probability of finding the particle described by $$\Psi(x,t)$$ in a small interval $$\delta x$$ centred on position $$x$$ is given by


 * $$p=\left|\Psi(x,t)\right|^2\delta x$$

Part (b)
The probability of finding the particle in the region described is



\begin{align} p = \tfrac{2}{L}\!\!\int_{L/3}^{2L/3}\!\!\sin^2{\left(\tfrac{3\pi x}{L}\right)}\ \mathrm dx. \end{align} $$

Using the substitutions



u=\tfrac{3\pi x}{L},\ \ \mathrm dx=\tfrac{L}{3\pi}\mathrm du,\ \ \mathrm{upper}\ u=2\pi,\ \ \mathrm{lower}\ u=\pi, $$

we can rewrite this integral as



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

Using the provided integral, we find this to be equal to



p=\tfrac{2}{3\pi}\times \tfrac{1}{2}(2-1)\pi=\tfrac{1}{3} $$

Part (a)
The state at a later time is given by



\Psi(x,t)=\tfrac{1}{\sqrt 2} \left( \psi_1(x)e^{-iE_1t/\hbar}+\psi_2(x)e^{-iE_2t/\hbar}  \right), $$

where $$E_1$$ and $$E_2$$ are the eigenvalues of energy associated with the stationary states $$\psi_1$$ and $$\psi_2$$. Given that $$E_2=4E_1$$, we can write



\Psi(x,t)=\tfrac{1}{\sqrt 2} \left( \psi_1(x)e^{-iE_1t/\hbar}+\psi_2(x)e^{-4iE_2t/\hbar}  \right). $$

The wave function at the time $$t=\pi\hbar/E_1$$ is described by



\begin{align} \Psi(x,t_0)&=\tfrac{1}{\sqrt 2} \left( \psi_1(x)e^{-iE_1\pi \hbar/E_1\hbar} + \psi_2(x)e^{-4iE_1\pi\hbar/E_1\hbar} \right) \\ &=\tfrac{1}{\sqrt 2}\left( \psi_1(x)e^{-i\pi} + \psi_2(x)e^{-4i\pi} \right) \\ &=\tfrac{1}{\sqrt 2}\left( \psi_2(x)-\psi_1(x) \right). \end{align} $$

Part (b)
If the states $$\Psi(x,0)$$ and $$\Psi(x,t_0)$$ are orthogonal, then their overlap integral should be equal to 0.



\begin{align} I &=\tfrac{1}{2}\!\!\int_{-\infty}^{\infty}\!\!(\psi_1^*+\psi_2^*)(\psi_2-\psi_1)\ \mathrm dx \\ &=\tfrac{1}{2}\left( \int_{-\infty}^{\infty}\!\! \psi_1^*\psi_2\ \mathrm dx- \int_{-\infty}^{\infty}\!\!  \psi_1^*\psi_1\ \mathrm dx+ \int_{-\infty}^{\infty}\!\!  \psi_2^*\psi_2\ \mathrm dx- \int_{-\infty}^{\infty}\!\!  \psi_2^*\psi_1\ \mathrm dx \right) \\ &=\tfrac{1}{2}\left( 0-1+1-0\right)\\ &=0. \end{align} $$

The overlap integral evaluates to zero, so these states are orthogonal.

Part (a)
Since we have perfect transmission, we have no reflected wave function in the region $$x<0$$. This requires that $$B=0$$, since $$\mathrm e^{-\mathrm ik_1x}$$ implies a wave function which is travelling in the negative $$x$$-direction.

$$|F|=A$$, since the $$A$$ term is the entire incident wave, and if we have perfect transmission the amplitude of the wave function beyond the barrier should be identical.

Part (b)
The wave function and its first derivative should both be continuous at $$x=0$$, so



\begin{align} A\mathrm e^{\mathrm ik_1x}&=C\mathrm e^{\mathrm ik_2x}+D\mathrm e^{-\mathrm ik_2x},\\ \mathrm ik_1A\mathrm e^{\mathrm ik_1x}&=\mathrm ik_2C\mathrm e^{\mathrm ik_2x}-\mathrm ik_2D\mathrm e^{-\mathrm ik_2x}. \end{align} $$

Various cancellations, and the substitutions $$k_1=2k_2$$ and $$x=0$$, give



\begin{align} A&=C+D,\\ 2A&=C-D. \end{align} $$

This system is easily solved to give


 * $$C=\tfrac{3A}{2},\ \ \ D=-\tfrac{A}{2}.$$

Part (a)
The expectation value of A:



\langle A \rangle=\langle \psi_n | a_n | \psi_n \rangle=a_n\langle\psi_n|\psi_n\rangle=a_n. $$

The expectation value of A^2:



\langle A \rangle=\langle \psi_n | a_n^2 | \psi_n \rangle=a_n^2\langle\psi_n|\psi_n\rangle=a_n^2. $$

So we have



\Delta A=\sqrt{\langle A^2\rangle-\langle A \rangle^2} =\sqrt{a_n^2-(a_n)^2}=0 $$

as required.

Part (b)
The generalized uncertainty principle is



\Delta A\Delta B\geq \tfrac{1}{2}\left| \left\langle \left[\widehat{\mathrm{A}},\widehat{\mathrm{B}}   \right]  \right\rangle  \right|. $$

The uncertainty product is therefore given by



\begin{align} \Delta L_y\Delta L_z &\geq \tfrac{1}{2}\left|\left\langle\left[ \widehat{\mathrm{L}}_y,\widehat{\mathrm{L}}_z \right]\right\rangle\right|\\ &\geq \tfrac{1}{2}\left|\mathrm i\hbar\left\langle L_x\right\rangle\right| \\ &\geq \tfrac{\hbar}{2}\left\langle L_x\right\rangle. \end{align} $$

In the state with a definite value of $$L_z$$, we have $$\Delta L_z=0$$ and therefore $$\langle L_x\rangle\leq 0=0$$.

Question 6
Acting on an arbitrary function:



\begin{align} \left[\widehat{\mathrm{p}}_x,\widehat{\mathrm{V}}(x)\right]\psi &=(\widehat{\mathrm{p}}_x\widehat{\mathrm{V}}(x)-\widehat{\mathrm{V}}(x)\widehat{\mathrm{p}}_x)\psi \\ &=-\mathrm i\hbar\tfrac{\partial}{\partial x}(V(x)\psi)+V(x)\mathrm i\hbar\tfrac{\partial}{\partial x}\psi \\ &=-\mathrm i\hbar V(x)\tfrac{\partial\psi}{\partial x}-\mathrm i\hbar\tfrac{\partial V}{\partial x}\psi+V(x)\mathrm i\hbar\tfrac{\partial\psi}{\partial x}\\ &=-\mathrm i\hbar\tfrac{\partial V}{\partial x}\psi, \end{align} $$

so


 * $$\left[\widehat{\mathrm{p}}_x,\widehat{\mathrm{V}}(x)\right]=-\mathrm i\hbar\tfrac{\partial V}{\partial x},$$

as required.

The generalized Ehrenfest theorem gives


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

We need the commutation relation between $$\widehat{\mathrm{p}}_x$$ and $$\widehat{\mathrm{H}}$$:



\begin{align} \left[ \widehat{\mathrm{p}}_x,\widehat{\mathrm{H}} \right] &=\widehat{\mathrm{p}}_x\widehat{\mathrm{H}}-\widehat{\mathrm{H}}\widehat{\mathrm{p}}_x \\ &=\widehat{\mathrm{p}}_x\left( \tfrac{\widehat{\mathrm{p}}^2_x}{2m} +\widehat{\mathrm{V}}(x)    \right)-\left(  \tfrac{\widehat{\mathrm{p}}^2_x}{2m}+\widehat{\mathrm{V}}(x)   \right)\widehat{\mathrm{p}}_x \\ &=\tfrac{\widehat{\mathrm{p}}^3_x}{2m}-\widehat{\mathrm{p}}_x\widehat{\mathrm{V}}(x)-\tfrac{\widehat{\mathrm{p}}^3_x}{2m}-\widehat{\mathrm{V}}\widehat{\mathrm{p}}_x \\ &=\widehat{\mathrm{p}}_x\widehat{\mathrm{V}}(x)-\widehat{\mathrm{V}}(x)\widehat{\mathrm{p}}_x \\ &=\left[ \widehat{\mathrm{p}}_x,\widehat{\mathrm{V}}(x) \right]\\ &=-\mathrm i\hbar\tfrac{\partial V}{\partial x}. \end{align} $$

At this point I'm not sure how the question wishes us to phrase the answer, since we can give a general solution (which actually appears on the formula sheet), but this doesn't really use the $$Cx^2/2$$ we're given. We have either


 * $$\tfrac{\mathrm d}{\mathrm dt}\langle p_x\rangle=\tfrac{1}{\mathrm i\hbar}\left\langle \left[ \widehat{\mathrm{p}}_x,\widehat{\mathrm{H}}  \right]  \right\rangle=-\left\langle \tfrac{\partial V}{\partial x} \right\rangle,$$

or, since $$V(x)=Cx^2/2$$, we have



\tfrac{\mathrm d}{\mathrm dt}\langle p_x\rangle =-C\langle x \rangle

$$

Preamable: spatial wave function symmetries
$$f(x_1,x_2)$$ is symmetric with respect to label exchange, since $$f(x_1,x_2)=f(x_2,x_1)$$.

$$g(x_1,x_2)$$ is symmetric with respect to label exchange, since $$g(x_1,x_2)=g(x_2,x_1)$$.

$$h(x_1,x_2)$$ is antisymmetric with respect to label exchange, since $$h(x_1,x_2)=-h(x_2,x_1)$$.

Part (a)
The total wave function for a system of identical fermions must be antisymmetric. The spin state given us is antisymmetric, so the spatial state must be symmetric for a total wave function which is antisymmetric. The spatial states which are symmetric are $$f(x_1,x_2)$$ and $$g(x_1,x_2)$$, so the total wave function is


 * $$f(x_1,x_2)|0,0\rangle\ \ \ \mathrm{or}\ \ \ g(x_1,x_2)|0,0\rangle.$$

Part (b)
Again we have a system of identical fermions, so the total wave function must be antisymmetric. This time, though, the given spin state is symmetric, so the spatial state must be antisymmetric, which means h(x_1,x_2) is the only option. The total wave function is


 * $$h(x_1,x_2)|1,1\rangle.$$

Part (c)
A system of identical bosons must have a symmetric total wave function. Since they are spinless, the spin state is automatically symmetric, so the spatial state must also be symmetric to give a symmetric total. The symmetric spatial functions are f(x_1,x_2) and g(x_1,x_2). The total wave function is therefore


 * $$f(x_1,x_2)\ \ \ \mathrm{or}\ \ \ g(x_1,x_2).$$

There is no spin ket since the particles are spinless.

Question 9
The first order approximation to the ground-state energy of a system is described by


 * $$E_n=E_n^{(0)}+\langle \psi_n^{(0)}|\delta \widehat{\mathrm{H}} |\psi_n^{(0)} \rangle,$$

where $$\delta\widehat{\mathrm{H}}$$ is given (in the region of interest) by


 * $$\begin{align}\delta\widehat{\mathrm{H}}&=\widehat{\mathrm{H}}-\widehat{\mathrm{H}}^{(0)} \\

&=\tfrac{1}{2}Cb|x|-\tfrac{1}{2}Cx^2 \\ &=\tfrac{1}{2}C(b|x|-x^2).\end{align}$$

The first-order correction is given by



\begin{align} E_n^{(1)} &=\tfrac{1}{2}C\langle \psi_0^{(0)}| (b|x|-x^2)  |\psi_0^{(0)}  \rangle \\ &=\tfrac{1}{2}C\!\!\int_{-b}^{b}\!\!\psi_0^{(0)*}\!(x)(b|x|-x^2)\psi_0^{(0)}\!(x)\ \mathrm dx. \end{align} $$

The zeroth-order energy for a harmonic oscillator is known to be $$\hbar\omega/2$$, so in total, the first-order approximation to the ground-state energy is



E_n=\tfrac{1}{2}\hbar\omega_0+\tfrac{1}{2}C\!\!\int_{-b}^{b}\!\!\psi_0^{(0)*}\!(x)(b|x|-x^2)\psi_0^{(0)}\!(x)\ \mathrm dx. $$

Part (a)
The scaled Bohr radius of a hydrogen-like system is given in the equation booklet as


 * $$a_0^\mathrm{scaled}=\tfrac{1}{Z}\tfrac{\mu_\mathrm H}{\mu}a_0$$,

where $$Z=1$$, $$\mu$$ is the reduced mass of the system, and $$\mu_\mathrm H$$ is the reduced mass of hydrogen.


 * $$\tfrac{\mu_\mathrm H}{\mu}=\tfrac{\left(\tfrac{1840\times 1}{1840+1}\right)}{\left(\tfrac{1\times 1}{1+1}\right)}=1.9989.$$

I have omitted all occurences of the electron mass, since they will ultimately cancel each other. The scaled Bohr radius is therefore


 * $$a_0^\mathrm{scaled}=1.9989\times 5.29\times 10^{-11}\ \mathrm m=1.0574\times 10^{10}\ \mathrm m.$$

The scaled Rydberg energy is also given by the equation booklet as


 * $$E_\mathrm{R}^\mathrm{scaled}=Z^2\tfrac{\mu}{\mu_\mathrm H}E_\mathrm R=\tfrac{13.6}{1.9989}=6.8037\ \mathrm{eV}.$$

Part (b)
The energy of a photon emitted upon transition from state 2 to state 1 is given by


 * $$\begin{align}

E_\mathrm{ph} &=\left| \tfrac{E_\mathrm R}{2^2}-\tfrac{E_\mathrm R}{1^2} \right| \\ &=\left| \tfrac{6.8037}{4}- 6.8037 \right|\ \mathrm{eV} \\ &=5.1028\ \mathrm{eV} \\ &=8.1747\times 10^{-19}\ \mathrm J. \end{align}$$

The frequency of a photon with this energy is given by


 * $$\begin{align}

f &=\tfrac{E_\mathrm{ph}}{h} \\ &=\tfrac{8.1747\times 10^{-19}}{2\times \pi\times 1.06\times 10^{-34}}\ \mathrm s^{-1} \\ &=1.2274\times 10^{15}\ \mathrm{Hz}. \end{align}$$

Part (a)
The symbol $$\mathrm\sigma$$ indicates a molecular orbital created from the linear combination of atomic orbitals with $$m=0$$; the subscript u stands for Ungerade, the German for odd indicating the orbital is anti-symmetric and the factor 2 denotes that this is the second lowest energy $$\mathrm{\sigma_u}$$ orbital.

Part (b)
The ground state electronic configuration of diatomic Carbon (Z=6) 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} $$

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

Bonding = (2 + 2 + 4) = 8

Anti-bonding = (2 + 2) = 4

Bond order = 1/2(8 - 4) = 2

Part (a)
The position vector $$\mathbf{r}$$ has real valued components and we require $$\mathbf{k}\cdot \mathbf{R}$$ to be real so that $$\exp \left( i\mathbf{k}\cdot \mathbf{r} \right)$$ is a phase factor hence $$\mathbf{k}$$ must also have real valued components.

Part (b)
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 (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| {u_{\mathbf{k}}}(\mathbf{r}) \exp \left( i\mathbf{k}\cdot (\mathbf{r}+\mathbf{R}) \right) \right|}^{2}} \\ & ={{\left| {u_{\mathbf{k}}}(\mathbf{r}) \exp \left( i\mathbf{k}\cdot \mathbf{r} \right) \exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \right|}^{2}} \\ & ={{\left| {{\psi }_{\mathbf{k}}}(\mathbf{r})\exp \left( i\mathbf{k}\cdot \mathbf{R} \right) \right|}^{2}} \\ & ={{\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.

=PART 2=

Part (a)
The time-dependent Schrödinger equation is


 * $$\mathrm i\hbar\tfrac{\partial}{\partial t}\Psi=\widehat{\mathrm{H}}\Psi=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\Psi.$$

where


 * $$\Psi=\psi(x)\mathrm e^{-\mathrm iEt/\hbar}.$$

Left-hand side:


 * $$\mathrm i\hbar\tfrac{\partial}{\partial t}\Psi=-\tfrac{\mathrm i^2\hbar E}{\hbar}\psi(x)\mathrm e^{-\mathrm iEt/\hbar}=E\psi(x)\mathrm e^{-\mathrm iEt/\hbar}.$$

Right-hand side:


 * $$-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\Psi=-\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\psi(x)\mathrm e^{-\mathrm iEt/\hbar}.$$

When we equate these:



\begin{align} -\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\psi(x)\mathrm e^{-\mathrm iEt/\hbar}&=E\psi(x)\mathrm e^{-\mathrm iEt/\hbar}\\ -\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\psi(x)&=E\psi(x). \end{align} $$

This result is the time-independent Schroedinger equation inside the well.

Part (b)
To show that this function satisfies the TISE, and finding $$E$$ in terms of $$k$$:



\begin{align} -\sqrt{\tfrac{2}{L}}\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\sin{(kx)}&=E\sqrt{\tfrac{2}{L}}\sin{(kx)} \\ -\tfrac{\hbar^2}{2m}\tfrac{\partial^2}{\partial x^2}\sin{(kx)}&=E\sin{(kx)} \\ \tfrac{k^2\hbar^2}{2m}\sin{(kx)}&=E\sin{(kx)} \\ E&=\tfrac{k^2\hbar^2}{2m}. \end{align} $$

We can use the fact the the wave function must be continuous at the boundary to find the possible values of $$k$$. If $$x=L$$, the wave function in both regions must have a value of zero to be continuous:



\begin{align} \sqrt{\tfrac{2}{L}}\sin{(kL)}&=0 \\ \sin{(kL)}&=0 \end{align} $$

This equality can only hold where $$kL$$ is an integer multiple of $$\pi$$, so



\begin{align} kL&=n\pi \\ k&=\tfrac{n\pi}{L},\ \ \ \mathrm{where}\ \ n=0,\pm 1,\pm 2,... \end{align} $$

We can disregard those cases where $$n=0$$, since that corresponds to a non-existent wave function, and where $$n$$ is negative, since they will only reproduce the positive cases. Substituting this $$k$$ into the expression for $$E$$, obtained earlier, gives the allowed values for $$E$$ as


 * $$E=\tfrac{n^2\hbar^2\pi^2}{2mL^2},\ \ \ \mathrm{where}\ \ n=1,2,3,...$$

Part (c)
If $$\psi(x)$$ is indeed normalized, then we expect the overlap integral $$I$$ of $$\psi$$ with itself to be equal to 1 in the region of the infinite square well. Recalling that we must have $$k=n\pi/L$$:



\begin{align} I &= \int_0^L\!\!\psi^*(x)\psi(x)\ \mathrm dx\\ &=\tfrac{2}{L}\!\!\int_0^L\!\!\sin{\left(\tfrac{n\pi x}{L}\right)}\sin{\left(\tfrac{n\pi x}{L}\right)}\ \mathrm dx \\ &=\tfrac{2}{L}\!\!\int_0^L\!\!\sin^2{\left(\tfrac{n\pi x}{L}\right)}\ \mathrm dx. \end{align} $$

Making the following substitution:


 * $$u=\tfrac{n\pi x}{L},\ \ \mathrm dx=\tfrac{L}{2\pi}\mathrm du,\ \ \mathrm{upper}\ u=n\pi,\ \ \mathrm{lower}\ u=0,$$

enables us to recast the integral as


 * $$I=\tfrac{2}{L}\tfrac{L}{n\pi}\!\!\int_0^{n\pi}\!\!\!\sin^2{u}\ \mathrm du.$$

We can use one of the provided integrals to solve this and we have


 * $$I=\tfrac{2}{L}\tfrac{L}{n\pi}\tfrac{n\pi}{2}=1,$$

as required.

Part (d)
The stationary-state wave function for the ground state (where n=1) is


 * $$\Psi_1(x,t)=\sqrt{\tfrac{2}{L}}\sin{\left(\tfrac{\pi x}{L}\right)}\mathrm e^{-\mathrm iEt/\hbar}$$

The expectation value of x^4 is



\langle x^4 \rangle = \int_0^L\!\! \left( \sqrt{ \tfrac{2}{L} }\sin{\left(\tfrac{\pi x}{L}\right)}\mathrm e^{-\mathrm iEt/\hbar} \right)^*\widehat{\mathrm{x}}^4\left( \sqrt{ \tfrac{2}{L} }\sin{\left(\tfrac{\pi x}{L}\right)}\mathrm e^{-\mathrm iEt/\hbar} \right) \mathrm dx. $$

This expression has no time-dependence, since


 * $$\left(\mathrm e^{-\mathrm iEt/\hbar}\right)^*\mathrm e^{-\mathrm iEt/\hbar}=\mathrm e^{\mathrm iEt/\hbar}\mathrm e^{-\mathrm iEt/\hbar}=\mathrm e^0=1.$$

This integral simplifies to


 * $$\langle x^4 \rangle

= \tfrac{2}{L}\!\!\int_0^L\!\! x^4\sin^2{\left(\tfrac{\pi x}{L}\right)}\ \mathrm dx. $$

The substitutions



u=\tfrac{\pi x}{L},\ \ \mathrm dx=\tfrac{L}{\pi}\mathrm du,\ \ x^4=\tfrac{L^4u^4}{\pi^4},\ \ \mathrm{upper}\ u=\pi,\ \ \mathrm{lower}\ u=0, $$

enables us to recast the integral as



\langle x^4\rangle = \tfrac{2L^4}{\pi^5}\!\!\int_0^{\pi}\!\! u^4\sin^2u\ \mathrm du. $$

A standard integral is provided to enable us to find a solution, and we get



\langle x^4 \rangle = 17.443\times\tfrac{2L^4}{\pi^5}\simeq 0.114L^4. $$

Part (a)
The expectation value of $$x$$ is given by



\begin{align} \langle x \rangle &=\int_{-\infty}^{\infty}\!\! \psi_n^*(x)\ \widehat{\mathrm{x}}\ \psi_n(x)\ \mathrm dx \\ &=\tfrac{a}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\! \psi_n^*(x)\ \left(\widehat{\mathrm{A}}\psi_n(x)+\widehat{\mathrm{A}}^\dagger\psi_n(x)\right)\ \mathrm dx \\ &=\tfrac{a\sqrt n}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!\psi_n^*(x)\psi_{n-1}(x)\ \mathrm dx+\tfrac{a\sqrt{n+1}}{\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!\psi_n^*(x)\psi_{n+1}(x)\ \mathrm dx. \end{align} $$

Both integrals represent orthogonal sets of functions, so


 * $$\langle x \rangle=0.$$

Part (b)
The square of operator $$\widehat{\mathrm{x}}$$ is


 * $$\widehat{\mathrm{x}}^2=\tfrac{a^2}{2}\left( \widehat{\mathrm{A}}\widehat{\mathrm{A}}+\widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger+\widehat{\mathrm{A}}^\dagger\widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger\widehat{\mathrm{A}}^\dagger \right).$$

Applying each of these to the state $$\psi_n$$ gives



\begin{align} \widehat{\mathrm{A}}\widehat{\mathrm{A}}\psi_n &=\widehat{\mathrm{A}}\sqrt n\psi_{n-1}&&=\sqrt n\sqrt{n-1}\psi_{n-2}&&\\ \widehat{\mathrm{A}}\widehat{\mathrm{A}}^\dagger\psi_n &=\widehat{\mathrm{A}}\sqrt{n+1}\psi_{n+1}&&=\sqrt{n+1}\sqrt{n+1}\psi_n&&=(n+1)\psi_n\\ \widehat{\mathrm{A}}^\dagger\widehat{\mathrm{A}}\psi_n &=\widehat{\mathrm{A}}\sqrt n\psi_{n-1}&&=\sqrt n\sqrt n\psi_n&&=n\psi_n\\ \widehat{\mathrm{A}}^\dagger\widehat{\mathrm{A}}^\dagger\psi_n &=\widehat{\mathrm{A}}^\dagger\sqrt{n+1}\psi_{n+1}&&=\sqrt{n+2}\sqrt{n+1}\psi_{n+2}.&& \end{align} $$

The first and last will result in orthogonal sets inside the integral, so we know they will evaluate to zero. What we have left is


 * $$\widehat{\mathrm{x}}^2=\tfrac{a^2}{2}(n+1+n)=a^2\!\left(n+\tfrac{1}{2}\right)$$

So the expectation value of $$x^2$$ is



\begin{align} \langle x^2\rangle &=\int_{-\infty}^{\infty}\!\!\psi_n^*(x)\ a^2\!\left(n+\tfrac{1}{2}\right)\psi_n(x) \\ &=a^2\!\left(n+\tfrac{1}{2}\right)\!\!\int_{-\infty}^{\infty}\!\!\psi_n^*(x)\psi_n(x)\ \mathrm dx \\ &=a^2\!\left(n+\tfrac{1}{2}\right) \end{align} $$

The uncertainty in $$x$$ is given by


 * $$\Delta x=\left(\langle x^2\rangle-\langle x \rangle^2\right)^{1/2}=a\sqrt{n+\tfrac{1}{2}}.$$

The product of the uncertainties in momentum and position is


 * $$\Delta x\Delta p_x=\left(n+\tfrac{1}{2}\right)\tfrac{\hbar}{a}\ a\sqrt{n+\tfrac{1}{2}}=\left(n+\tfrac{1}{2}\right)^{3/2}\hbar.$$

$$(n+1/2)^{3/2}$$ is greater than $$1/2$$ for $$n\geq 1$$, so the Heisenberg uncertainty principle is satisfied.

Part (c)
(Not overly sure about this one.)

The expectation value of $$x$$ is the state $$\Psi(x,0)$$ is



\begin{align} \langle x \rangle &=\tfrac{1}{2}\!\!\int_{-\infty}^{\infty}\!\!(\psi_2^*+\psi_3^*)\ \widehat{\mathrm{x}}\ (\psi_2+\psi_3)\ \mathrm dx \\ &=\tfrac{a}{2\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!(\psi_2^*+\psi_3^*)(\widehat{\mathrm{A}}+\widehat{\mathrm{A}}^\dagger)(\psi_2+\psi_3)\ \mathrm dx \\ &=\tfrac{a}{2\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!(\psi_2^*+\psi_3^*)(\widehat{\mathrm{A}}\psi_2+\widehat{\mathrm{A}}\psi_3+\widehat{\mathrm{A}}^\dagger\psi_2+\widehat{\mathrm{A}}^\dagger\psi_3)\ \mathrm dx \\ &=\tfrac{a}{2\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!(\psi_2^*+\psi_3^*)(\sqrt 2\psi_1+\sqrt 3\psi_2+\sqrt 3\psi_3+2\psi_4)\ \mathrm dx \\ &=\tfrac{a}{2\sqrt 2}\!\!\int_{-\infty}^{\infty}\!\!(\sqrt 2\psi_2^*\psi_1+\sqrt 3\psi_2^*\psi_2+\sqrt 3\psi_2^*\psi_3+2\psi_2^*\psi_4+\sqrt 2\psi_3^*\psi_1+\sqrt 3\psi_3^*\psi_2+\sqrt 3\psi_3^*\psi_3+2\psi_3^*\psi_4)\ \mathrm dx \\ &=\tfrac{a}{2\sqrt 2}(0+\sqrt 3+0+0+0+0+\sqrt 3+0) \\ &=\tfrac{a\sqrt 6}{2}. \end{align} $$

=PART 3=

Part (a)
For the up spinor:



\widehat{\mathrm{H}} |\!\uparrow_\mathbf{n}\rangle =-\tfrac{\hbar\omega}{2}\begin{bmatrix}\tfrac{1}{2}&\tfrac{\sqrt 3}{2}\\\tfrac{\sqrt 3}{2}&-\tfrac{1}{2}\end{bmatrix}\begin{bmatrix}\tfrac{\sqrt 3}{2}\\\tfrac{1}{2}\end{bmatrix} =-\tfrac{\hbar\omega}{2}\begin{bmatrix}\tfrac{\sqrt 3}{4}+\tfrac{\sqrt 3}{4}\\\tfrac{3}{4}-\tfrac{1}{4}\end{bmatrix}=-\tfrac{\hbar\omega}{2}\begin{bmatrix}\tfrac{\sqrt 3}{2}\\\tfrac{1}{2}\end{bmatrix}=-\tfrac{\hbar\omega}{2}|\!\uparrow_\mathbf{n}\rangle. $$

For the down spinor:



\widehat{\mathrm{H}} |\!\downarrow_\mathbf{n}\rangle =-\tfrac{ \hbar\omega }{2 }\begin{bmatrix} \tfrac{1}{2}&\tfrac{\sqrt 3}{2}\\\tfrac{\sqrt 3}{2}&-\tfrac{1}{2} \end{bmatrix}\begin{bmatrix} -\tfrac{1}{2}\\\tfrac{\sqrt 3}{2} \end{bmatrix} =-\tfrac{\hbar\omega}{2} \begin{bmatrix} -\tfrac{1}{4}+\tfrac{3}{4}  \\  -\tfrac{\sqrt 3}{4}-\tfrac{\sqrt 3}{4} \end{bmatrix} =-\tfrac{\hbar\omega}{2}  \begin{bmatrix} \tfrac{1}{2}\\-\tfrac{\sqrt 3}{2} \end{bmatrix} =\tfrac{\hbar\omega}{2} |\!\downarrow_\mathbf{n}\rangle. $$

To verify that the two spinors are orthogonal, the up-down inner product must be equal to zero (for orthogonality), and the up-up and down-down inner products must both be equal to 1 (for normalization).



\langle\uparrow_\mathbf{n}|\downarrow_\mathbf{n}\rangle =\begin{bmatrix} \tfrac{\sqrt 3}{2} & \tfrac{1}{2}\end{bmatrix}\begin{bmatrix} -\tfrac{1}{2} \\ \tfrac{\sqrt 3}{2}\end{bmatrix} =-\tfrac{\sqrt 3}{4}+\tfrac{\sqrt 3}{4}=0. $$



\langle\uparrow_\mathbf{n}|\uparrow_\mathbf{n}\rangle =\begin{bmatrix} \tfrac{\sqrt 3}{2} & \tfrac{1}{2}\end{bmatrix}\begin{bmatrix} \tfrac{\sqrt 3}{2} \\ \tfrac{1}{2}\end{bmatrix} =\tfrac{3}{4}+\tfrac{1}{4}=1. $$



\langle\downarrow_\mathbf{n}|\downarrow_\mathbf{n}\rangle =\begin{bmatrix} -\tfrac{1}{2} & \tfrac{\sqrt 3}{2}\end{bmatrix}\begin{bmatrix} -\tfrac{1}{2} \\ \tfrac{\sqrt 3}{2}\end{bmatrix} =\tfrac{1}{4}+\tfrac{3}{4}=1. $$

Part (b)
The initial state is given as


 * $$\begin{bmatrix}1\\0\end{bmatrix}=1|\!\uparrow_z\rangle+0|\!\downarrow_z\rangle=c_1|\!\uparrow_\mathbf n\rangle+c_2|\!\downarrow_\mathbf n\rangle,$$

where



\begin{align} c_1&=\langle \uparrow_\mathbf n | \uparrow_z  \rangle&&=\begin{bmatrix} \tfrac{\sqrt 3}{2}&\tfrac{1}{2}  \end{bmatrix}\begin{bmatrix} 1\\0  \end{bmatrix}&&=\tfrac{\sqrt 3}{2}, \\ c_2&=\langle \downarrow_\mathbf n | \uparrow_z  \rangle&&=\begin{bmatrix} -\tfrac{1}{2}&\tfrac{\sqrt 3}{2}  \end{bmatrix}\begin{bmatrix} 1\\0  \end{bmatrix}&&=-\tfrac{1}{2}, \end{align} $$

so we have


 * $$|A\rangle_\mathrm{initial}=\tfrac{\sqrt 3}{2}|\!\uparrow_\mathbf n\rangle-\tfrac{1}{2}|\!\downarrow_\mathbf n\rangle.$$

The state at a later time is then given by


 * $$|A\rangle=\tfrac{\sqrt 3}{2}|\!\uparrow_\mathbf n\rangle \mathrm e^{-\mathrm iE_ut/\hbar}-\tfrac{1}{2}|\!\downarrow_\mathbf n\rangle \mathrm e^{-\mathrm iE_dt/\hbar}.$$

If we make the appropriate substitutions for energy (from part (a)) and time (given in this part), we have



\begin{align} &=\tfrac{\sqrt 3}{2}|\!\uparrow_\mathbf n\rangle \mathrm e^{\mathrm i\pi/2}-\tfrac{1}{2}|\!\downarrow_\mathbf n\rangle \mathrm e^{-\mathrm i\pi/2}\\ &=\mathrm i\left(\tfrac{\sqrt 3}{2}|\!\uparrow_\mathbf n\rangle+\tfrac{1}{2}|\!\downarrow_\mathbf n\rangle\right). \end{align} $$
 * A\rangle

Substituting in the n-direction spinors gives



\begin{align} &=\mathrm i\left( \tfrac{\sqrt 3}{2}\begin{bmatrix}\tfrac{\sqrt 3}{2}\\\tfrac{1}{2} \end{bmatrix} +\tfrac{1}{2}\begin{bmatrix}-\tfrac{1}{2}\\\tfrac{\sqrt 3}{2}\end{bmatrix} \right)\\ &=\mathrm i\left( \begin{bmatrix} \tfrac{3}{4} \\ \tfrac{\sqrt 3}{4} \end{bmatrix} +\begin{bmatrix} -\tfrac{1}{4} \\ \tfrac{\sqrt 3}{4} \end{bmatrix} \right) \\ &=\mathrm i\begin{bmatrix} \tfrac{1}{2}\\\tfrac{\sqrt 3}{2} \end{bmatrix}, \end{align} $$
 * A\rangle

as required.

Part (c)
At time $$t=0$$, the spin state is exclusively up in the $$z$$-direction, so



p_u=1,\ \ p_d=0,\ \ \langle S_z\rangle=\tfrac{\hbar}{2}. $$

The spin expectation value is what it is since that is the eigenvalue corresponding to the spin-up eigenfunction.

At time $$t=\pi/\omega$$:



\begin{align} p_u &=\left| \langle \uparrow_z | A  \rangle   \right|^2 &&=\left| \mathrm i\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix} \tfrac{1}{2} \\ \tfrac{\sqrt 3}{2} \end{bmatrix} \right|^2 &&=\tfrac{\mathrm i}{2}\tfrac{-\mathrm i}{2} &&=\tfrac{1}{4} \\ p_d &=\left| \langle \downarrow_z | A  \rangle   \right|^2 &&=\left| \mathrm i\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix} \tfrac{1}{2} \\ \tfrac{\sqrt 3}{2} \end{bmatrix} \right|^2 &&=\tfrac{\mathrm i\sqrt 3}{2}\tfrac{-\mathrm i\sqrt 3}{2} &&=\tfrac{3}{4} \end{align} $$

The expectation value of the spin at time $$t=\pi/\omega$$ takes a little more work:



\begin{align} \langle S_z \rangle &=\langle A  | \widehat{\mathrm{S}}_z  | A  \rangle\\ &=\tfrac{\mathrm i\hbar}{2}\begin{bmatrix} -\tfrac{\mathrm i}{2} & -\tfrac{\mathrm i\sqrt 3}{2} \end{bmatrix}\begin{bmatrix} 1&0\\0&-1  \end{bmatrix} \begin{bmatrix}  \tfrac{1}{2}  \\  \tfrac{\sqrt 3}{2} \end{bmatrix}\\ &=\tfrac{\mathrm i\hbar}{2}\begin{bmatrix} -\tfrac{\mathrm i}{2} & -\tfrac{\mathrm i\sqrt 3}{2} \end{bmatrix}\begin{bmatrix} \tfrac{1}{2} \\ -\tfrac{\sqrt 3}{2} \end{bmatrix}\\ &=\tfrac{\mathrm i\hbar}{2}\tfrac{\mathrm i}{2}\\ &=-\tfrac{\hbar}{4} \end{align} $$

Part (a)
An entangled state cannot be factorised; the system as a whole has a state function, but the individual particles do not. States $$|A\rangle$$ and $$|C\rangle$$ can both be factorised, so they are not entangled states:



\begin{align} \end{align} $$
 * A\rangle&=|\mathrm V\mathrm V\rangle=|\mathrm V\rangle|\mathrm V\rangle, \\
 * C\rangle&=\tfrac{1}{2}(|\mathrm V\mathrm V\rangle-|\mathrm V\mathrm H\rangle-|\mathrm H\mathrm V\rangle+|\mathrm H\mathrm H\rangle)=\tfrac{1}{2}(|\mathrm V\rangle-|\mathrm H\rangle)(|\mathrm V\rangle-|\mathrm H\rangle).

State $$|B\rangle$$ cannot be factorised, so it must represent an entangled state.

Part (b)
If $$\mathrm V_{\theta_1}\mathrm V_{\theta_2}$$ is the state in which photon 1 is vertically polarized with respect to an axis at $$\theta_1$$ to the $$|\mathrm V\rangle$$-axis, and similar for photon 2, then for state $$|A\rangle$$:



\begin{align} P_{\mathrm V\mathrm V}(\theta_1,\theta_2) &=\left| \langle \mathrm V_{\theta_1}\mathrm V_{\theta_2}|A\rangle   \right|^2\\ &=\left| (\cos\theta_1\langle \mathrm V|+\sin\theta_1\langle \mathrm H|)(\cos\theta_2\langle \mathrm V|+\sin\theta_2\langle \mathrm H|)(|\mathrm V\mathrm V\rangle) \right|^2\\ &=\left|\cos\theta_1\cos\theta_2\langle \mathrm V\mathrm V|\mathrm V\mathrm V\rangle\right|^2 \\ &= \left| \cos\theta_1\cos\theta_2\right|^2\\ &=\cos^2\theta_1\cos^2\theta_2. \end{align} $$

For state $$|B\rangle$$



\begin{align} P_{\mathrm V\mathrm V}(\theta_1,\theta_2) &=\left| \langle \mathrm V_{\theta_1}\mathrm V_{\theta_2}| B \rangle  \right|^2\\ &=\tfrac{1}{2}\left| (\cos\theta_1\langle \mathrm V|+\sin\theta_1\langle \mathrm H|)(\cos\theta_2\langle \mathrm V|+\sin\theta_2\langle \mathrm H|)(|\mathrm V\mathrm V\rangle-|\mathrm H\mathrm H\rangle) \right|^2\\ &=\tfrac{1}{2}\left| (\cos\theta_1\cos\theta_2\langle \mathrm V\mathrm V|+\sin\theta_1\sin\theta_2\langle \mathrm H\mathrm H |) (|\mathrm V\mathrm V\rangle-|\mathrm H\mathrm H\rangle) \right|^2\\ &=\tfrac{1}{2}\left| \cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2 \right|^2. \end{align} $$

For state $$|A\rangle$$, which is not entangled, the probability that the measurement in question is made can be factorised into independent expressions for particle 1 and particle 2. This is typical for probability calculations for multiple independent events: to find the probability of outcome X and outcome Y both occurring, we multiply their independent probabilities. No such factorisation is possible for the entangled state $$|B\rangle$$, which suggests that the probability of measuring one outcome is intrinsically linked to the outcome of the other; they are not independent measurements.

Part (c)
We can rewrite the described state by subtstituing in the provided expressions for V_45 and H_45:



\begin{align} &=\tfrac{1}{\sqrt 2}\left[ \tfrac{1}{2}\left(|\mathrm H\rangle+|\mathrm V\rangle\right) \left(|\mathrm H\rangle-|\mathrm V\rangle\right)+ \tfrac{1}{2}\left(|\mathrm H\rangle-|\mathrm V\rangle\right) \left(|\mathrm H\rangle+|\mathrm V\rangle\right)   \right]\\ &=\tfrac{1}{\sqrt 2}\left[ \tfrac{1}{2}\left(|\mathrm H\mathrm H\rangle-|\mathrm H\mathrm V\rangle+|\mathrm V\mathrm H\rangle-|\mathrm V\mathrm V\rangle+|\mathrm H\mathrm H\rangle+|\mathrm H\mathrm V\rangle-|\mathrm V\mathrm H\rangle-|\mathrm V\mathrm V\rangle\right)  \right]\\ &=\tfrac{1}{\sqrt 2}\left[|\mathrm H\mathrm H\rangle-|\mathrm V\mathrm V\rangle\right]\\ &=-|B\rangle. \end{align} $$

If we measure the first particle on the $$(\mathrm V,\mathrm H)$$ axes, then the state must collapse onto either $$|\mathrm V\mathrm V\rangle$$ or $$|\mathrm H\mathrm H\rangle$$. This means that if particle 1 is meaured as $$|\mathrm V\rangle$$, the other must be measured as $$|\mathrm V\rangle$$ as well, and similar for $$|\mathrm H\rangle$$. The measurements must be the same.

If we measure the first particle on the $$(\mathrm V_{45},\mathrm H_{45})$$ axes, then the state must collapse onto either $$|\mathrm V_{45}\mathrm H_{45}\rangle$$ or $$|\mathrm H_{45}\mathrm V_{45}\rangle$$. This means that if particle 1 is measured as $$|\mathrm V_{45}\rangle$$, the total state must be $$|\mathrm V_{45}\mathrm H_{45}\rangle$$, and so the second particle must be measured as $$|\mathrm H_{45}\rangle$$. If particle 1 is meaured as $$|\mathrm H_{45}\rangle$$, the total state must be $$|\mathrm H_{45}\mathrm V_{45}\rangle$$, so the second particle must be $$|\mathrm V_{45}\rangle$$. The measurements will be opposite.

=PART 4=

Part (a)
Applying the operator $$\widehat{\mathrm{L}}_z$$ to $$f(\theta,\phi)$$:



\begin{align} \widehat{\mathrm{L}}_zf(\theta,\phi) &=-\mathrm i\hbar\tfrac{\partial}{\partial \phi}\left(\sin^2\theta\ \mathrm e^{2\mathrm i\phi}\right)\\ &=-\mathrm i\hbar\sin^2\theta\tfrac{\partial}{\partial\phi}\ \mathrm e^{2\mathrm i\phi}\\ &=-2\mathrm i^2\hbar\sin^2\theta\ \mathrm e^{2\mathrm i\phi}\\ &=2\hbar f(\theta,\phi). \end{align} $$

The eigenvalue of $$\widehat{\mathrm{L}}_z$$ corresponding to this eigenfunction is therefore $$2\hbar$$, which means the magnetic quantum number is $$m=2$$.

Applying the operator $$\widehat{\mathrm{L}}^2$$ to $$f(\theta, \phi)$$:



\begin{align} \widehat{\mathrm{L}}^2f(\theta,\phi) &=-\hbar^2\left[ \tfrac{1}{\sin\theta}\tfrac{\partial}{\partial\theta}\left( \sin\theta\tfrac{\partial}{\partial\theta}(\sin^2\theta\ \mathrm e^{2\mathrm i\phi})  \right) + \tfrac{1}{\sin^2\theta}\tfrac{\partial^2}{\partial\phi^2}(\sin^2\theta\ \mathrm e^{2\mathrm i\phi})   \right]\\ &=-\hbar^2\left[ \mathrm e^{2\mathrm i\phi}\tfrac{1}{\sin\theta}\tfrac{\partial}{\partial\theta}\left( \sin\theta\tfrac{\partial}{\partial\theta}\sin^2\theta \right) +\tfrac{\partial^2}{\partial\phi^2}\mathrm e^{2\mathrm i\phi}          \right]. \end{align} $$

The first term can be simplified using the provided identity, and the second should be fairly trivial.



\widehat{\mathrm{L}}^2f(\theta,\phi) =-\hbar^2\left[  \mathrm e^{2\mathrm i\phi}(4-6\sin^2\theta)-4\mathrm e^{2\mathrm i\phi}      \right]. $$

We can factor out the function $$f(\theta,\phi)$$ to give



\begin{align} \widehat{\mathrm{L}}^2f(\theta,\phi) &=-\hbar^2\left[  \tfrac{4-6\sin^2\theta}{\sin^2\theta}-\tfrac{4}{\sin^2\theta}        \right]\sin^2\theta\ \mathrm e^{2\mathrm i\phi}\\ &=-\hbar^2\left[  \tfrac{4-6\sin^2\theta-4}{\sin^2\theta}      \right]\sin^2\theta\ \mathrm e^{2\mathrm i\phi}\\ &=6\hbar^2f(\theta,\phi). \end{align} $$

The eigenvalue of $$\widehat{\mathrm{L}}^2$$ corresponding to this eigenfunction is therefore $$6\hbar^2$$, which means the orbital angular momentum quantum number is $$l=2$$.

Part (b)
The normalization condition insists that integrating the wave function over all space must return a value of 1, so we set this equality and solve for A.



\begin{align} 1 &=\int_0^{2\pi}\!\!\!\int_0^\pi\!\!\int_0^\infty\!\! \psi^*(r,\theta,\phi)\psi(r,\theta,\phi)\ r^2\sin\theta\ \mathrm dr\ \mathrm d\theta\ \mathrm d\phi\\ &=\tfrac{A^2}{a_0^4}\!\!\int_0^\infty\!\!r^6\mathrm e^{-2r/3a_0}\ \mathrm dr\times \int_0^{2\pi}\!\!\mathrm d\phi\times \int_0^\pi\!\!\sin^5\theta\ \mathrm d\theta. \end{align} $$

The middle integral is simply 2\pi, and the other two can be solved using the provided integrals.



\begin{align} 1 &=\tfrac{A^2}{a_0^4}\times 6!\times \left(\tfrac{3a_0}{2}\right)^7\times 2\pi\times\tfrac{16}{15}\\ &=26244\pi A^2a_0^3\\ A^2&=\tfrac{1}{26244\pi a_0^3}\\ A&=\tfrac{1}{162\sqrt{\pi} a_0^{3/2}}, \end{align} $$

as expected. This solution is not necessarily unique; my working implicitly assumed A to be real, but in fact it can be any complex number provided that


 * $$AA^*=\tfrac{1}{26244\pi a_0^3}$$

Part (c)
The expectation value of $$r^2\sin^2\theta$$:



\begin{align} \langle r^2\sin^2\theta \rangle &=\tfrac{A^2}{a_0^4}\!\!\int_0^{2\pi}\!\!\!\int_0^\pi\!\!\int_0^\infty\!\! \left(r^4\mathrm e^{-2r/3a_0}\sin^4\theta\right)\left(r^2\sin^2\theta\right) r^2\sin\theta\ \mathrm dr\ \mathrm d\theta\ \mathrm d\phi\\ &=\tfrac{2\pi A^2}{a_0^4}\!\!\int_0^\infty\!\! r^8\mathrm e^{-2r/3a_0}\ \mathrm dr\times \int_0^\pi\!\!\sin^6\theta\ \mathrm d\theta\\ &=\tfrac{2\pi A^2}{a_0^4}\times 8!\times \left(\tfrac{3a_0}{2}\right)^9\times\tfrac{32}{35}\\ &=\tfrac{2\pi}{a_0^4}\times \tfrac{1}{26244\pi a_0^3}\times 8!\times \left(\tfrac{3}{2}\right)^9\times a_0^9\times \tfrac{32}{35}\\ &=108a_0^2 \end{align} $$

Part (a)
The symbol $$\nabla^2$$ is an operator called the Laplacian and is given in Cartesian coordinates by



\nabla^2=\tfrac{\partial^2}{\partial x^2}+\tfrac{\partial^2}{\partial y^2}+\tfrac{\partial^2}{\partial z^2} $$

The subscript $$i$$ indicates that this operator should be applied to the wave function of the $$i$$th electron.

$$r_i$$ is the separation between the nucleus and the $$i$$th electron.


 * $$r_i=|\mathbf r_i|$$

$$r_{ij}$$ is the separation between the $$i$$th and $$j$$th electrons


 * $$r_{ij}=|\mathbf r_i-\mathbf r_j|$$

The energy contributions to the Hamiltonian are
 * 1) the first term, representing the kinetic energy of the electrons,
 * 2) the second term, representing the electrical potential energy due to the separation of positive charge (the nucleus) and negative charge (the electrons),
 * 3) the third term, representing the electrical potential energy due to the interactions between the negatively charged electrons.

Part (b)
The independent-particle model simplifies the Hamiltonian by ignoring energy due to the electron-electron interactions (the third term in the above). The resulting Hamiltonian is exactly solvable, but gives a relatively poor approximation to the energy which is measured experimentally.

Part (c)
The central-field approximation invokes the laws of classical electromagnetism to introduce an effect called screening. An electron's potential energy is approximated by treating the nucleus and all charges nearer the nucleus than the electron in question as a single charge and assigning each electron an effective potential energy based on that charge. Electrons further from the nucleus have no effect on the potential of those nearer the nucleus. This approximation assumes that the electron distribution can be considered to be spherically symmetrical. Making this approximation gives an expression for the energy which depends only on the radial coordinate of the electron, rather than the separations between various electrons.

The degeneracy of states with same $$n$$ and different $$l$$ is lost in the central-field approximation, since, classically, we would expect electrons with lower orbital angular momentum to be nearer the nucleus. Since the energy given by the central-field approximation depends explicitly on the radial coordinate of the electron, states with different $$l$$, and therefore different radial coordinate, must have different energy.

Part (d)
For a configuration with the non-equivalent open shell electron states $$({{l}_{1}}=1,{{s}_{1}}=\tfrac{1}{2})$$and$$({{l}_{2}}=3,{{s}_{2}}=\tfrac{1}{2})$$we have


 * $${{L}_{\min }}=\left| {{l}_{1}}-{{l}_{2}} \right| =\left| 1-3 \right| = 2$$
 * $${{L}_{\max }}=\left| {{l}_{1}}+{{l}_{2}} \right| =\left| 1+3 \right| = 4$$
 * $${{S}_{\min }}=\left| {{s}_{1}}-{{s}_{2}} \right| =\left| \tfrac{1}{2}-\tfrac{1}{2} \right| = 0$$
 * $${{S}_{\max }}=\left| {{s}_{1}}+{{s}_{2}} \right| =\left| \tfrac{1}{2}+\tfrac{1}{2} \right|= 1$$
 * $$L=L_{\min}, L_{\min} +1, \ldots, L_{\max} -1, L_{\max}$$
 * $$S=S_{\min}, S_{\min} +1, \ldots, S_{\max} -1, S_{\max}$$

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 (e)
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 $$J=1,2 \text{ or } 3$$ each with degeneracy $$\left(2J + 1\right)$$ hence



\begin{matrix} J & \text{Degeneracy} \\ 1 & 3 \\  2 & 5 \\   3 & 7 \\ \end{matrix} $$

Thus the $$\left(L=2,S=1\right)$$ term corresponds to $$3+5+7=15$$ distinct states.