Consider a two-state quantum system with a Hamiltonian \begin{equation} \hat{H}\doteq \begin{pmatrix} E_1&0\\ 0&E_2 \end{pmatrix} \end{equation} Another physical observable \(M\) is described by the operator \begin{equation} \hat{M}\doteq \begin{pmatrix} 0&c\\ c&0 \end{pmatrix} \end{equation} where \(c\) is real and positive. Let the initial state of the system be \(\left|{\psi(0)}\right\rangle =\left|{m_1}\right\rangle \), where \(\left|{m_1}\right\rangle \) is the eigenstate corresponding to the larger of the two possible eigenvalues of \(\hat{M}\). What is the frequency of oscillation of the expectation value of \(M\)? This frequency is the Bohr frequency.

1. Find the entropy of a set of \(N\) oscillators of frequency \(\omega\) as a function of the total quantum number \(n\). Use the multiplicity function: \begin{equation} g(N,n) = \frac{(N+n-1)!}{n!(N-1)!} \end{equation} and assume that \(N\gg 1\). This means you can make the Sitrling approximation that \(\log N! \approx N\log N - N\). It also means that \(N-1 \approx N\).

2. Let \(U\) denote the total energy \(n\hbar\omega\) of the oscillators. Express the entropy as \(S(U,N)\). Show that the total energy at temperature \(T\) is \begin{equation} U = \frac{N\hbar\omega}{e^{\frac{\hbar\omega}{kT}}-1} \end{equation} This is the Planck result found the hard way. We will get to the easy way soon, and you will never again need to work with a multiplicity function like this.

Calculate the temperature of the surface of the Earth on the assumption that as a black body in thermal equilibrium it reradiates as much thermal radiation as it receives from the Sun. Assume also that the surface of the Earth is a constant temperature over the day-night cycle. Use the sun's surface temperature \(T_{\odot}=5800\text{K}\); and the sun's radius \(R_{\odot}=7\times 10^{10}\text{cm}\); and the Earth-Sun distance of \(1.5\times 10^{13}\text{cm}\).

(modified from K&K 4.6) We discussed in class that \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \end{align} Use this relationship to show that

1. \begin{align} p &= -\sum_j \langle n_j\rangle\hbar \left(\frac{d\omega_j}{dV}\right), \end{align} where \(\langle n_j\rangle\) is the number of photons in the mode \(j\);

2. Solve for the relationship between pressure and internal energy.

A one-dimensional harmonic oscillator has an infinite series of equally spaced energy states, with \(\varepsilon_n = n\hbar\omega\), where \(n\) is an integer \(\ge 0\), and \(\omega\) is the classical frequency of the oscillator. We have chosen the zero of energy at the state \(n=0\) which we can get away with here, but is not actually the zero of energy! To find the true energy we would have to add a \(\frac12\hbar\omega\) for each oscillator.

1. Show that for a harmonic oscillator the free energy is \begin{equation} F = k_BT\log\left(1 - e^{-\frac{\hbar\omega}{k_BT}}\right) \end{equation} Note that at high temperatures such that \(k_BT\gg\hbar\omega\) we may expand the argument of the logarithm to obtain \(F\approx k_BT\log\left(\frac{\hbar\omega}{kT}\right)\).

2. From the free energy above, show that the entropy is \begin{equation} \frac{S}{k_B} = \frac{\frac{\hbar\omega}{kT}}{e^{\frac{\hbar\omega}{kT}}-1} - \log\left(1-e^{-\frac{\hbar\omega}{kT}}\right) \end{equation}

   

   

   This entropy is shown in the nearby figure, as well as the heat capacity.

As discussed in class, we can consider a black body as a large box with a small hole in it. If we treat the large box a metal cube with side length \(L\) and metal walls, the frequency of each normal mode will be given by: \begin{align} \omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2} \end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have positive integer values. This simply comes from the fact that a half wavelength must fit in the box. There is an additional quantum number for polarization, which has two possible values, but does not affect the frequency. Note that in this problem I'm using different boundary conditions from what I use in class. It is worth learning to work with either set of quantum numbers. Each normal mode is a harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\) where we will not include the zero-point energy \(\frac12\hbar\omega\), since that energy cannot be extracted from the box. (See the Casimir effect for an example where the zero point energy of photon modes does have an effect.)

Note

This is a slight approximation, as the boundary conditions for light are a bit more complicated. However, for large \(n\) values this gives the correct result.

1. Show that the free energy is given by \begin{align} F &= 8\pi \frac{V(kT)^4}{h^3c^3} \int_0^\infty \ln\left(1-e^{-\xi}\right)\xi^2d\xi \\ &= -\frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3} \\ &= -\frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3} \end{align} provided the box is big enough that \(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a slightly different dimensionless integral that numerically evaluates to the same result, which would be fine. I also do not expect you to solve this definite integral analytically, a numerical confirmation is fine. However, you must manipulate your integral until it is dimensionless and has all the dimensionful quantities removed from it!

2. Show that the entropy of this box full of photons at temperature \(T\) is \begin{align} S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3 \\ &= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3 \end{align}

3. Show that the internal energy of this box full of photons at temperature \(T\) is \begin{align} \frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3} \\ &= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3} \end{align}