2002 Spring Statistical Mechanics
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Two level systems and beyond.

(The convention used in this page is that $\tau:=kT$, so that $T$ is the temperature expressed in Kelvin and $\tau$ is the temperature expressed in units of energy.)

A. [20 points] Consider a two-level system (TLS) with energy states 0 and $\epsilon$.

i. [10 points] In the canonical ensemble, derive an expression for the heat capacity of the TLS.

Greg's Solution:

To obtain the specific heat, we need to have the energy expressed as a function of temperature. We will obtain this by first writing down the partition function, which immediately gives us the free energy, and then using the fact that $E=F+\tau S$ and $S=-\frac{dF}{d\tau}$ to get the energy from the free energy.

The partition function for the two-level system is given by,

\begin{align} \Omega_0(\mu_0+\mu_B H)-\Omega_0(\mu_0-\mu_B H)\right] Z = \underbrace{1}_{\text{ground state}}+\underbrace{e^{-\epsilon/\tau}}_{\text{excited state}}. \end{align}

Thus, the free energy is given by,

\begin{align} F = -\tau\log Z = -\tau\log\left[1+e^{-\epsilon/\tau}\right] \end{align}

We want to find the total energy, given by $E=F+\tau S$, so first we find the entropy,

\begin{aligned} S &= -\frac{dF}{d\tau} \newline &= \log Z + \frac{-\frac{\epsilon}{\tau} e^{-\epsilon/\tau}}{1+e^{-\epsilon/\tau}}, \newline \end{aligned}

and so

\begin{align} E = F+\tau S = -\frac{\epsilon e^{-\epsilon/\tau}}{1+e^{-\epsilon/\tau}} \end{align}

(Note that as $\tau\to 0$, $E\to 0$, and as $\tau\to\infty$, $E\to\frac{\epsilon}{2}$.)

Finally, we obtain the heat capacity from the energy,

\begin{align} C_V = \frac{dE}{dT} = k\frac{dE}{d\tau}|_{\tau=kT} = k\frac{\epsilon^2}{k^2T^2}\frac{e^{-\epsilon/kT}}{\left(1+e^{-\epsilon/kT}\right)^2} \end{align}

ii. [10 points] In the 1970's, it was discovered that the low-temperature heat capacity of insulating glasses is linear in T. Approximate the internal degrees of freedom for a glass as a superposition of independent TLS's with a broad distribution of energy differences $n(\epsilon)$. Show tha the heat capacity of a glass is $C_V(T)\sim k^2 Tn(0)A$, where A is a constant of order unity.

Greg's Solution:

Based on Eq. (5), we see that the specific heat for a TLS as a function of the excited energy $\epsilon$ is given by,

\begin{align} \frac{dE_\epsilon}{d\tau} = \frac{\epsilon^2}{\tau^2}\frac{e^{-\epsilon/\tau}}{\left(1+e^{-\epsilon/\tau}\right)^2} \end{align}

Thus, for the glass we have that

\begin{align} \frac{dE}{d\tau} = \sum_\epsilon n(\epsilon) \frac{dE_\epsilon}{d\tau} \end{align}
\begin{align} \approx \int_0^\infty \,d\epsilon\, n(\epsilon) \frac{dE_\epsilon}{d\tau} \end{align}
\begin{align} \approx \int_0^\infty \,d\epsilon\, n(\epsilon) \frac{\epsilon^2}{\tau^2}\frac{e^{-\epsilon/\tau}}{\left(1+e^{-\epsilon/\tau}\right)^2} \end{align}

This integral will be easiest to deal with when we make it dimensionless, so we set $x:=\epsilon/\tau$ so that

\begin{align} \frac{dE}{d\tau} = \tau \int_0^\infty dx\, n(\tau x) \frac{e^{-x}}{\left(1+e^{-x}\right)^2} \end{align}

The bulk of the contribution of the integral is around $x=1$. Furthermore, as $\tau\to 0$, the peak of the integral gets shrunk until it is approximately a delta function times a constant of order unity. Thus, we have that

\begin{align} \frac{dE}{d\tau} \to \tau n(0) A \end{align}

And thus,

\begin{align} C_V = k\frac{dE}{d\tau}|_{\tau=kT} \to k^2T n(0) A. \end{align}

B. A myoglobin molecule in solution can either have exactly one adsorbed O2 molecule or else zero adsorbed O2 molecules. Let $\epsilon$ denote the energy of an adsorbed molecule of O2 relative to an O2 in solution at infinite distance from the myoglobin.

i. It will be helpful to approximate the O2 molecules in solution as an ideal gas (exclude rotational and vibrational degrees of freedom). Prove that the chemical potential of an ideal has is $\mu=kT\log\left(n/n_Q\right)$ where $n_Q:=\left(MkT/2\pi\hbar^2\right)^{3/2}$.

Greg's Solution:

The game plan is that if we can find the free energy, then we have the chemical potential since $\mu=\frac{\partial F}{\partial N}$. Thus, we seek the partition function from which we will obtain the free energy.

The partition function for a single particle of gas is given by,

\begin{align} Z_1 = \sum_\epsilon e^{-\epsilon/\tau}. \end{align}

In our problem we have that $\epsilon = p^2/2M$. Plugging this in, and converting our sum to an integral via the standard formula,

\begin{align} \sum \to \int \frac{dp\,dq}{(2\pi\hbar)^3}, \end{align}

we have that

\begin{align} Z_1 =\int \frac{dp\,dq}{(2\pi\hbar)^3} e^{-p^2/2M\tau} \end{align}

Since the integrand is independant of $q$, we integrate it out and obtain the volume,

\begin{align} \begin{aligned} Z_1 = V\int \frac{dp}{(2\pi\hbar)^3} e^{-p^2/2M\tau} \end{align}
\begin{align} = V\frac{(2\pi M\tau)^{3/2}}{(2\pi\hbar)^3} e^{-p^2/2M\tau} \end{align}
\begin{align} &= V\left(\frac{M\tau}{2\pi\hbar^2}\right)^{3/2} \end{align}

The N-particle partition function is given by,

\begin{align} Z_N = \frac{Z_1^N}{N!}, \end{align}

where the factor of $1/N!$ is needed to avoid overcounting states (as the particles are indistinguishable). Thus, the free energy is given by

\begin{align} F = -\tau\log Z_N \end{align}
\begin{align} = -\tau\log \frac{1}{N!}\left(V\left(\frac{M\tau}{2\pi\hbar^2}\right)^{3/2}\right)^N \end{align}
\begin{align} = -N\tau\log \frac{1}{N!}V\left(\frac{M\tau}{2\pi\hbar^2}\right)^{3/2} \end{align}
\begin{align} = -N\tau\log \frac{V}{N}\left(\frac{M\tau}{2\pi\hbar^2}\right)^{3/2} \end{align}
\begin{align} = N\tau\log \frac{n}{\underbrace{\left(\frac{M\tau}{2\pi\hbar^2}\right)^{3/2}}_{n_Q}} \end{align}
\begin{align} = N\tau\log \frac{n}{n_Q} \end{align}

And so we see that indeed, $\mu=\frac{\partial F}{\partial N}=\tau\log \frac{n}{n_Q}$, where $n_Q:=\left(MkT/2\pi\hbar^2\right)^{3/2}$. Q.E.D.

ii. [8 points] Prove that the fraction of myoglobin carrying an O_2 molecule is given by $f=\frac{n}{n_Q \exp{\epsilon/k T}+n}$, where n is the concentration of O_2 molecules in the surrounding solution.

Greg's Solution:

The key to solving this problem is to work in the grand canonical ensemble. That is, we consider that at each myoglobin site, there are two possible states: zero particles, or one particle. Thus, the Gibbs distribution partition function is given by,

\begin{align} \xi = 1+e^{(\mu-\epsilon)/\tau}, \end{align}

and so the thermodynamic potential (a.k.a. Gibbs free energy) is given by

\begin{align} \Omega = -\tau\log\left[1+e^{(\mu-\epsilon)/\tau}\right]. \end{align}

Now we can obtain the expected number of particules from

\begin{align} N = -\pdrv{\Omega}{\mu} = \frac{e^{(\mu-\epsilon)/\tau}}{1+e^{(\mu-\epsilon)/\tau}} = \frac{1}{e^{(-\mu+\epsilon)/\tau}+1} \end{align}

(Note that we have just derived the distribution function for a Fermi gas.)

Plugging in our expression for the chemical potential derived earlier, we obtain

\begin{align} N = \frac{1}{\frac{n_Q}{n}e^{\epsilon/\tau}+1} \equiv \frac{n}{n_Q e^{\epsilon/kT} +n} \end{align}

as required. Q.E.D.

C. Consider a gas of free electrons at T=0. An electron in a magnetic field has an energy of $\pm \mu_B H$, according to whether the spin is parallel or antiparallel to the field $\bar H$.

i. [5 points] Sketch the spin-dependent density of states and indicate a typical occupancy.

i. [15 points] Show that the spin-paramagnetic susceptiblity is $\frac{3}{2}n\frac{\mu_B^2}{\mu_0}$, where $\mu_0$ is the chemical potential at T=0.

Greg's Solution:

Recall that if you know that,

\begin{align} \delta E = X\delta\lambda, \end{align}


\begin{align} \bar X = \frac{dE}{d\lambda}. \end{align}

In this case, we know that

\begin{align} \delta E = M\delta H, \end{align}

since for a small perturbation of the external field, the magnetization will remain essentially constant and so the energy will increase by an amount proportional to the change in the field. As a result, we can apply [[eqref findavalue]] to obtain the magnetization,

\begin{align} \bar M = \frac{dE}{dH}, \end{align}

and from that the magnetic susceptibility,

\begin{align} \chi = \frac{\bar M}{V}. \end{align}

So it remains to find the energy as a function of the external field. This need not be the total energy — any of the other energies shall do; for this problem it is most convenient to work with the Gibbs free energy, $\Omega_0$, in the grand canonical ensemble.

We assume that in the absense of the field the chemical potential is given by $\mu_0$ and the Gibbs free energy by a known function $\Omega_0(\mu_0)$. The effect of the field is to split this system into two smaller systems with new chemical potentials $\mu_0\pm \mu_B H$, so that the Gibbs free energy is given by

\begin{align} \Omega = \frac{1}{2}\left[\Omega_0(\mu_0+\mu_B H)-\Omega_0(\mu_0-\mu_B H)\right] \end{align}

The factor of 1/2 is required as in each system the electron is only allowed to have one spin, thus halving the number of possible states.

Now that we have an energy for the system, we can find the magnetization by taking the derivative with respect to H,

\begin{align} \bar M = \frac{d\Omega}{dH} = \frac{\mu_B}{2}\left[\frac{d\Omega_0}{d\mu}(\mu_0+\mu_B H)-\frac{d\Omega_0}{d\mu}(\mu_0-\mu_B H)\right] \end{align}

If we assume that the perturbing magnetic field is sufficiently small, then we can approximate the above by

\begin{align} \bar M \approx \mu_B^2 H\frac{d^2\Omega_0}{d\mu^2} = \mu_B^2 H\frac{dN}{d\mu} \end{align}

Since T=0, the system is completely degenerate and so

\begin{align} N \propto \mu^{3/2}. \end{align}

We know this because for a one dimensional square well we have that $\epsilon_F\propto k_F^2 \propto N^2\Rightarrow N\propto \epsilon_F^{1/2}$ — that is, the number of states that can "fit" into a system is proportional to the square root of the Fermi energy. In three dimensions, the number of states that can fit is cubed, and so $N\propto \epsilon_F^{3/2}$, and when T=0 we have that $\mu=\epsilon_F$, resulting in Eq. [[eqref easyNfromMU]] above.

Given Eq. 38, we see that

\begin{align} \frac{\partial N}{\partial \mu} = \frac{3}{2}\frac{N}{\mu}, \end{align}

and so

\begin{align} \bar M = \frac{3N\mu_B^2}{2\mu_0} \Rightarrow \chi = \frac{\bar M}{V} = \frac{3n\mu_B^2}{2\mu_0}, \end{align}


2. [40 points] Photons and radiation pressure

A. [20 points] Consider a 3-dimensional photon gas with energy spectrum $E=\hbar cq$, where $q=|\vec{q}|$ and $\vec{q}$ is the wave vector. Solve the questions below using quantum statistical mechanics.

i. Discuss why the chemical potential of the photon gas is zero.

Greg's Solution:

The chemical potential is zero because it costs nothing to add another photon to the system, as the same amount of total energy can be divided among an arbitrary number of photons.

ii. Show that the radiation pressure of the photon gas is $p=\frac{4\sigma}{3c}T^4$, where $\sigma:=\frac{\pi^2k^4}{60\hbar^3c^3}$.

Greg's Solution:

The game plan is to find the free energy, since once we have that we can find the pressure via $P=-\left(\frac{\partial F}{\partial V}\right)_\tau$. (Note that finding the total energy will not help us, since $P\ne -\left(\frac{\partial E}{\partial V}\right)_\tau$.) To find the free energy, we compute the partition function for the system.

For a given mode $\vec{q}$, the partition function is given by

\begin{align} Z_{\vec{q}} = \sum_{n=0}^\infty e^{n(\overbrace{\mu}^{=0}-\epsilon_q)/\tau} = \frac{1}{1-e^{-\epsilon_q/\tau}}, \end{align}

where observe that we have set the chemical potential $\mu$ to zero for the reasons given in the previous part.

Since each mode is independent, the partition function for the whole system is a product of the partition functions for the individual modes,

\begin{align} Z = \prod_{\vec{q}} Z_{\vec{q}}, \end{align}

and so the free energy (note that $F=\Omega$ — i.e. the Helmholtz and Gibbs free energies are equal — since $\mu=0$) is given by

\begin{align} F = -\tau\log Z = -\tau \log\left[\prod_{\vec{q}} Z_{\vec{q}}\right] \equiv +\tau \sum_{\vec{q}} \log\left[1-e^{-\epsilon_q/\tau}\right] \end{align}

Now, we have that $q=nq_0$ where $n=|\vec{n}|$ and the components of $\vec{n}$ are integers, and $q_0=\frac{2\pi}{L}$ is the smallest possible wave number. Thus,

\begin{align} F = 2\tau \sum_{\vec{n}} \log\left[1-e^{-\hbar c q_0 n/\tau}\right] \end{align}

(The extra factor of 2 comes from the fact that for every mode there are two independent photon polarizations.)

We convert this to an integral over $\vec{n}$$]; since the integrand only depends on [[$|\vec{n}$, we write the integral in spherical coordinate form,

\begin{align} F = 2\tau \int_0^\infty\,4\pi n^2 \,dn\,\log\left[1-e^{-\hbar c q_0 n/\tau}\right] \end{align}

Our next goal is to make this integral dimensionless by setting $x=-\hbar c q_0n/\tau$ so that

\begin{align} F = \frac{8\pi \tau^4}{(\hbar c q_0)^3}\int_0^\infty \,dx\,x^2\log\left[1-e^{-x}\right] \end{align}

Integrate by parts to see that,

\begin{align} F = -\frac{8\pi \tau^4}{3(\hbar c q_0)^3}\int_0^\infty \frac{x^3\,dx}{e^{x}-1}\right], \end{align}

and it turns out that $\int_0^\infty \frac{x^3\,dx}{e^{x}-1}\right]=\frac{\pi^4}{15}$ so therefore

\begin{align} F = -\frac{8\pi^5 \tau^4}{45(\hbar c q_0)^3}, \end{align}

(Note that even if you did not know the value of that integral, you could reverse-engineer it since you are given the answer as part of the question. :-) )

Now, the largest possible wavelength is just the length of the side of a "box" containing our system, and so the smallest possible wave number is given by $q_0=\frac{2\pi}{L}$. Thus,

\begin{align} F = -\frac{8\pi^5 \tau^4 L^3}{45\hbar^3 c^3 8\pi^3} \equiv \frac{\pi^2 \tau^4 V}{45\hbar^3 c^3} \equiv \frac{4\sigma VT^4}{3c} \end{align}

where $\sigma=\frac{\pi^2k^4}{60 \hbar^3c^2}$$ is the Stefan-Boltzmann constant.

Now that we have the free energy expressed in this form, finding the pressure is a piece of cake!

\begin{align} p = -\frac{\partial F}{\partial V} = \frac{4\sigma}{3c}T^4, \end{align}


iii. Show that the energy density of the gas, u, can be expressed as $u=3p$.

Note that,

\begin{align} E = F + TS = F-T\frac{\partial F}{\partial T} = F-4F = -3F \end{align}

Thus, since $u=E/V$ and $p=-F/V$, from Eq. 51 we have that $u=3p$, Q.E.D.

B. [20 points] You are given an evacuated container of volume V whose walls are perfectly reflective for EM radiation.

i. [6 points] By considering the energy density of states of the 3-dimensional photon gas in the container, show that the energy U of the gas is a linear function of V.

Greg's Solution:

(This result was essentially already derived in the previous part, however there is an alternative method of obtaining the same conclusion.)

For a photon gas, the occupancy of a mode with energy $\epsilon_k$ is

\begin{align} n_k = \frac{2}{e^{\epsilon_k/\tau}-1} \end{align}

In particular, we have that $\epsilon_k = \hbar \om_0 n$, where $\om_0=\frac{2\pi c}{L}$ and $n=|\vec{n}|$ is the magnitude of a 3-vector with integer components. Thus, the expected value of the energy is given by

\begin{align} E = \sum \epsilon_k n_k = \sum_{\vec{n}} \frac{2\hbar \om_0 n}{e^{\hbar \om_0 n/\tau}-1} \end{align}

We convert this sum to a radial integral in spherical coordinates,

\begin{align} E = \int_0^\infty \,4\pi n^2\,dn\,\frac{2\hbar \om_0 n}{e^{\hbar \om_0 n/\tau}-1} \end{align}

We set $x=\hbar\om_0 n/\tau$ in order to make the integral dimensionless,

\begin{align} E = \frac{4\pi \tau^4}{\hbar^3 \om_0^3}\int_0^\infty \frac{ x^3\,dx}{e^{x}-1} \end{align}

Since $E\propto \om_0^{-3} \propto L^3 \propto V$, we conclude that the energy of the gas is a linear function of the volume.

ii. [7 points] Starting from the second law of thermodynamics and using Maxwell's relations, derive the general relation $\left(\frac{\partial U}{\partial V}\right_)_T=\left(\frac{\partial P}{\partial T}\right_)_V-P$.

Greg's Solution:

First, observe that

\begin{align} \frac{\partial F}{\partial V} = -P \end{align}


\begin{align} \frac{\partial F}{\partial T} = -S \end{align}

Thus, by taking second derivatives, we obtain the Maxwell relation,

\begin{align} \left(\frac{\partial P}{\partial T}\right)_V = -\frac{\partial^2 F}{\partial T\partial V}= -\frac{\partial^2 F}{\partial T\partial V}=\left(\frac{\partial S}{\partial V}\right)_T. \end{align}

Now, the second law of thermodynamics gives us the thermodynamic relation,

\begin{align} dU = T\,dS - P\,dV, \end{align}

from which we obtain

\begin{align} \left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial S}{\partial V}\right)_T-P = T\left(\frac{\partial P}{\partial T}\right)_V-P, \end{align}


iii.* [7 points] Using the results of parts B.i. and B.ii. above, prove that the radiation pressure is given by $p=aT^4$ where a is an undetermined constant. The result $u=3p$, which was derived statistically mechanically, may be used here.

Greg's Solution

In part B.i. we showed that the energy was linear with respect to the volume, so we expect that $u=\frac{\partial U}{\partial V}\equiv \frac{U}{V}$ should be a constant with respect to volume, and therefore so should $p=u/3$. Furthermore, upon examination of the right-hand side of B.ii., we see that P should be proportional to some power of T — that is, $P\propto aT^m$. Plugging this into B.ii., we obtain

\begin{equation} u = amT^m - aT^m = (m-1)aT^m = 3aT^m, \end{equation}

where the last equality came from $u=3p$. By inspection, we see that $m=4$ and so $p\propto T^4$, Q.E.D.

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