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# Posts Tagged ‘free energy’

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## probability forms of entropy

Posted by peeterjoot on March 16, 2013

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## Question: Entropy as probability

[1] points out that entropy can be written as

\begin{aligned}S = - k_{\mathrm{B}} \sum_i P_i \ln P_i\end{aligned} \hspace{\stretch{1}}(1.0.1)

where

\begin{aligned}P_i = \frac{e^{-\beta E_i}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}Z = \sum_i e^{-\beta E_i}.\end{aligned} \hspace{\stretch{1}}(1.0.2b)

Show that this follows from the free energy $F = U - T S = -k_{\mathrm{B}} \ln Z$.

In terms of the free and average energies, we have

\begin{aligned}\frac{S}{k_{\mathrm{B}}} &= \frac{U - F}{k_{\mathrm{B}} T} \\ &= \beta \left( -\frac{\partial {\ln Z}}{\partial {\beta}} \right) - \beta \left( -k_{\mathrm{B}} T \ln Z \right) \\ &= \frac{\sum_i \beta E_i e^{-\beta E_i}}{Z} +\ln Z \\ &= -\sum_i P_i \ln e^{-\beta E_i} + \sum_i P_i \ln Z \\ &= -\sum_i P_i \ln \frac{e^{-\beta E_i}}{Z} P_i \\ &= -\sum_i P_i \ln P_i.\end{aligned} \hspace{\stretch{1}}(1.0.3)

## Question: Entropy in terms of grand partition probabilites ( [2] pr 4.1)

Generalize \cref{pr:entropyProbabilityForm:1} to the grand canonical scheme, where we have

\begin{aligned}P_{r, s} = \frac{e^{-\alpha N_r - \beta E_s}}{Z_{\mathrm{G}}}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}Z_{\mathrm{G}} = \sum_{r,s} e^{-\alpha N_r - \beta E_s}\end{aligned} \hspace{\stretch{1}}(1.0.4b)

\begin{aligned}z = e^{-\alpha} = e^{\mu \beta}\end{aligned} \hspace{\stretch{1}}(1.0.4c)

\begin{aligned}q = \ln Z_{\mathrm{G}},\end{aligned} \hspace{\stretch{1}}(1.0.4d)

and show

\begin{aligned}S = - k_{\mathrm{B}} \sum_{r,s} P_{r,s} \ln P_{r,s}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

With

\begin{aligned}\beta P V = q,\end{aligned} \hspace{\stretch{1}}(1.0.6)

the free energy takes the form

\begin{aligned}F = N \mu - P V = N \mu - q/\beta,\end{aligned} \hspace{\stretch{1}}(1.0.7)

so that the entropy (scaled by $k_{\mathrm{B}}$) leads us to the desired result

\begin{aligned}\frac{S}{k_{\mathrm{B}}} &= \beta U - N \mu \beta + q/(\beta k_{\mathrm{B}} T) \\ &= -\beta \frac{\partial {q}}{\partial {\beta}} - z \mu \beta \frac{\partial {q}}{\partial {z}} + q \\ &= \frac{1}{{Z_{\mathrm{G}}}}\sum_{r, s}\left( -\beta (-E_s) - \mu \beta N_r \right) e^{-\alpha N_r - \beta E_s}+ \ln Z_{\mathrm{G}} \\ &= \sum_{r, s} \ln e^{ \alpha N_r + \beta E_s } P_{r,s} + \left( \sum_{r, s} P_{r, s} \right)\ln Z_{\mathrm{G}} \\ &= -\sum_{r, s} \ln \frac{e^{ -\alpha N_r - \beta E_s }}{Z_{\mathrm{G}}} P_{r,s} \\ &= -\sum_{r, s} P_{r, s} \ln P_{r, s}\end{aligned} \hspace{\stretch{1}}(1.0.8)

# References

[1] E.A. Jackson. Equilibrium statistical mechanics. Dover Pubns, 2000.

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## PHY452H1S Basic Statistical Mechanics. Problem Set 5: Temperature

Posted by peeterjoot on March 10, 2013

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# Disclaimer

## Question: Polymer stretching – “entropic forces” (2013 problem set 5, p1)

Consider a toy model of a polymer in one dimension which is made of $N$ steps (amino acids) of unit length, going left or right like a random walk. Let one end of this polymer be at the origin and the other end be at a point $X = \sqrt{N}$ (viz. the rms size of the polymer) , so $1 \ll X \ll N$. We have previously calculated the number of configurations corresponding to this condition (approximate the binomial distribution by a Gaussian).

### Part a

Using this, find the entropy of this polymer as $S = k_{\mathrm{B}} \ln \Omega$. The free energy of this polymer, even in the absence of any other interactions, thus has an entropic contribution, $F = -T S$. If we stretch this polymer, we expect to have fewer available configurations, and thus a smaller entropy and a higher free energy.

### Part b

Find the change in free energy of this polymer if we stretch this polymer from its end being at $X$ to a larger distance $X + \Delta X$.

### Part c

Show that the change in free energy is linear in the displacement for small $\Delta X$, and hence find the temperature dependent “entropic spring constant” of this polymer. (This entropic force is important to overcome for packing DNA into the nucleus, and in many biological processes.)

Typo correction (via email):
You need to show that the change in free energy is quadratic in the displacement $\Delta X$, not linear in $\Delta X$. The force is linear in $\Delta X$. (Exactly as for a “spring”.)

### Entropy.

In lecture 2 probabilities for the sums of fair coin tosses were considered. Assigning $\pm 1$ to the events $Y_k$ for heads and tails coin tosses respectively, a random variable $Y = \sum_k Y_k$ for the total of $N$ such events was found to have the form

\begin{aligned}P_N(Y) = \left\{\begin{array}{l l}\left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-Y}{2}\right)!\left(\frac{N+Y}{2}\right)!}& \quad \mbox{if Y and N have same parity} \\ 0& \quad \mbox{otherwise} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.1.1)

For an individual coin tosses we have averages $\left\langle{{Y_1}}\right\rangle = 0$, and $\left\langle{{Y_1^2}}\right\rangle = 1$, so the central limit theorem provides us with a large $N$ Gaussian approximation for this distribution

\begin{aligned}P_N(Y) \approx\frac{2}{\sqrt{2 \pi N}} \exp\left( -\frac{Y^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.2)

This fair coin toss problem can also be thought of as describing the coordinate of the end point of a one dimensional polymer with the beginning point of the polymer is fixed at the origin. Writing $\Omega(N, Y)$ for the total number of configurations that have an end point at coordinate $Y$ we have

\begin{aligned}P_N(Y) = \frac{\Omega(N, Y)}{2^N},\end{aligned} \hspace{\stretch{1}}(1.1.3)

From this, the total number of configurations that have, say, length $X = \left\lvert {Y} \right\rvert$, in the large $N$ Gaussian approximation, is

\begin{aligned}\Omega(N, X) &= 2^N \left( P_N(+X) +P_N(-X) \right) \\ &= \frac{2^{N + 2}}{\sqrt{2 \pi N}} \exp\left( -\frac{X^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.4)

The entropy associated with a one dimensional polymer of length $X$ is therefore

\begin{aligned}S_N(X) &= - k_{\mathrm{B}} \frac{X^2}{2N} + k_{\mathrm{B}} \ln \frac{2^{N + 2}}{\sqrt{2 \pi N}} \\ &= - k_{\mathrm{B}} \frac{X^2}{2N} + \text{constant}.\end{aligned} \hspace{\stretch{1}}(1.1.5)

Writing $S_0$ for this constant the free energy is

\begin{aligned}\boxed{F = U - T S = U + k_{\mathrm{B}} T \frac{X^2}{2N} + S_0 T.}\end{aligned} \hspace{\stretch{1}}(1.1.6)

### Change in free energy.

At constant temperature, stretching the polymer from its end being at $X$ to a larger distance $X + \Delta X$, results in a free energy change of

\begin{aligned}\Delta F &= F( X + \Delta X ) - F(X) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( (X + \Delta X)^2 - X^2 \right) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( 2 X \Delta X + (\Delta X)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.1.7)

If $\Delta X$ is assumed small, our constant temperature change in free energy $\Delta F \approx (\partial F/\partial X)_T \Delta X$ is

\begin{aligned}\boxed{\Delta F = \frac{k_{\mathrm{B}} T}{N} X \Delta X.}\end{aligned} \hspace{\stretch{1}}(1.1.8)

### Temperature dependent spring constant.

I found the statement and subsequent correction of the problem statement somewhat confusing. To figure this all out, I thought it was reasonable to step back and relate free energy to the entropic force explicitly.

Consider temporarily a general thermodynamic system, for which we have by definition free energy and thermodynamic identity respectively

\begin{aligned}F = U - T S,\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}dU = T dS - P dV.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

The differential of the free energy is

\begin{aligned}dF &= dU - T dS - S dT \\ &= -P dV - S dT \\ &= \left( \frac{\partial {F}}{\partial {T}} \right)_V dT+\left( \frac{\partial {F}}{\partial {V}} \right)_T dV.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Forming the wedge product with $dT$, we arrive at the two form

\begin{aligned}0 &= \left( \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV + \left( S + \left( \frac{\partial {F}}{\partial {T}} \right)_V \right) dT \right)\wedge dT \\ &= \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV \wedge dT,\end{aligned} \hspace{\stretch{1}}(1.0.11)

This provides the relation between free energy and the “pressure” for the system

\begin{aligned}P = - \left( \frac{\partial {F}}{\partial {V}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.12)

For a system with a constant cross section $\Delta A$, $dV = \Delta A dX$, so the force associated with the system is

\begin{aligned}f &= P \Delta A \\ &= - \frac{1}{{\Delta A}} \left( \frac{\partial {F}}{\partial {X}} \right)_T \Delta A,\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}f = - \left( \frac{\partial {F}}{\partial {X}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Okay, now we have a relation between the force and the rate of change of the free energy

\begin{aligned}f(X) = -\frac{k_{\mathrm{B}} T}{N} X.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Our temperature dependent “entropic spring constant” in analogy with $f = -k X$, is therefore

\begin{aligned}\boxed{k = \frac{k_{\mathrm{B}} T}{N}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

## Question: Independent one-dimensional harmonic oscillators (2013 problem set 5, p2)

Consider a set of $N$ independent classical harmonic oscillators, each having a frequency $\omega$.

### Part a

Find the canonical partition at a temperature $T$ for this system of oscillators keeping track of correction factors of Planck constant. (Note that the oscillators are distinguishable, and we do not need $1/N!$ correction factor.)

### Part b

Using this, derive the mean energy and the specific heat at temperature $T$.

### Part c

For quantum oscillators, the partition function of each oscillator is simply $\sum_n e^{-\beta E_n}$ where $E_n$ are the (discrete) energy levels given by $(n + 1/2)\hbar \omega$, with $n = 0,1,2,\cdots$. Hence, find the canonical partition function for $N$ independent distinguishable quantum oscillators, and find the mean energy and specific heat at temperature $T$.

### Part d

Show that the quantum results go over into the classical results at high temperature $k_{\mathrm{B}} T \gg \hbar \omega$, and comment on why this makes sense.

### Part e

Also find the low temperature behavior of the specific heat in both classical and quantum cases when $k_{\mathrm{B}} T \ll \hbar \omega$.

### Classical partition function

For a single particle in one dimension our partition function is

\begin{aligned}Z_1 = \frac{1}{{h}} \int dp dq e^{-\beta \left( \frac{1}{{2 m}} p^2 + \frac{1}{{2}} m \omega^2 q^2 \right)},\end{aligned} \hspace{\stretch{1}}(1.0.17)

with

\begin{aligned}a = \sqrt{\frac{\beta}{2 m}} p\end{aligned} \hspace{\stretch{1}}(1.0.18a)

\begin{aligned}b = \sqrt{\frac{\beta m}{2}} \omega q,\end{aligned} \hspace{\stretch{1}}(1.0.18b)

we have

\begin{aligned}Z_1 &= \frac{1}{{h \omega}} \sqrt{\frac{2 m}{\beta}} \sqrt{\frac{2}{\beta m}} \int da db e^{-a^2 - b^2} \\ &= \frac{2}{\beta h \omega}2 \pi \int_0^\infty r e^{-r^2} \\ &= \frac{2 \pi}{\beta h \omega} \\ &= \frac{1}{\beta \hbar \omega}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

So for $N$ distinguishable classical one dimensional harmonic oscillators we have

\begin{aligned}\boxed{Z_N(T) = Z_1^N = \left( \frac{k_{\mathrm{B}} T}{\hbar \omega} \right)^N.}\end{aligned} \hspace{\stretch{1}}(1.0.20)

### Classical mean energy and heat capacity

From the free energy

\begin{aligned}F = -k_{\mathrm{B}} T \ln Z_N = N k_{\mathrm{B}} T \ln (\beta \hbar \omega),\end{aligned} \hspace{\stretch{1}}(1.0.21)

we can compute the mean energy

\begin{aligned}U &= \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {}}{\partial {\beta}} \left( \frac{F}{T} \right) \\ &= N \frac{\partial {}}{\partial {\beta}} \ln (\beta \hbar \omega) \\ &= \frac{N }{\beta},\end{aligned} \hspace{\stretch{1}}(1.0.22)

or

\begin{aligned}\boxed{U = N k_{\mathrm{B}} T.}\end{aligned} \hspace{\stretch{1}}(1.0.23)

The specific heat follows immediately

\begin{aligned}\boxed{C_{\mathrm{V}} = \frac{\partial {U}}{\partial {T}} = N k_{\mathrm{B}}.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

### Quantum partition function, mean energy and heat capacity

For a single one dimensional quantum oscillator, our partition function is

\begin{aligned}Z_1 &= \sum_{n = 0}^\infty e^{-\beta \hbar \omega \left( n + \frac{1}{{2}} \right)} \\ &= e^{-\beta \hbar \omega/2}\sum_{n = 0}^\infty e^{-\beta \hbar \omega n} \\ &= \frac{e^{-\beta \hbar \omega/2}}{1 - e^{-\beta \hbar \omega}} \\ &= \frac{1}{e^{\beta \hbar \omega/2} - e^{-\beta \hbar \omega/2}} \\ &= \frac{1}{{\sinh(\beta \hbar \omega/2)}}.\end{aligned} \hspace{\stretch{1}}(1.0.25)

Assuming distinguishable quantum oscillators, our $N$ particle partition function is

\begin{aligned}\boxed{Z_N(\beta) = \frac{1}{{\sinh^N(\beta \hbar \omega/2)}}.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

This time we don’t add the $1/\hbar$ correction factor, nor the $N!$ indistinguishability correction factor.

Our free energy is

\begin{aligned}F = N k_{\mathrm{B}} T \ln \sinh(\beta \hbar \omega/2),\end{aligned} \hspace{\stretch{1}}(1.0.27)

our mean energy is

\begin{aligned}U &= \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {}}{\partial {\beta}} \frac{F}{T} \\ &= N \frac{\partial {}}{\partial {\beta}}\ln \sinh(\beta \hbar \omega/2) \\ &= N \frac{\cosh( \beta \hbar \omega/2 )}{\sinh(\beta \hbar \omega/2)} \frac{\hbar \omega}{2},\end{aligned} \hspace{\stretch{1}}(1.0.28)

or

\begin{aligned}\boxed{U(T)= \frac{N \hbar \omega}{2} \coth \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right).}\end{aligned} \hspace{\stretch{1}}(1.0.29)

This is plotted in fig. 1.1.

Fig 1.1: Mean energy for N one dimensional quantum harmonic oscillators

With $\coth'(x) = -1/\sinh^2(x)$, our specific heat is

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {U}}{\partial {T}} \\ &= \frac{N \hbar \omega}{2} \frac{-1}{\sinh^2 \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)} \frac{\hbar \omega}{2 k_{\mathrm{B}}} \left( \frac{-1}{T^2} \right),\end{aligned} \hspace{\stretch{1}}(1.0.30)

or

\begin{aligned}\boxed{C_{\mathrm{V}} = N k_{\mathrm{B}}\left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T \sinh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right) } \right)^2.}\end{aligned} \hspace{\stretch{1}}(1.0.31)

### Classical limits

In the high temperature limit $1 \gg \hbar \omega/k_{\mathrm{B}} T$, we have

\begin{aligned}\cosh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)\approx 1\end{aligned} \hspace{\stretch{1}}(1.0.32)

\begin{aligned}\sinh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)\approx \frac{\hbar \omega}{2 k_{\mathrm{B}} T},\end{aligned} \hspace{\stretch{1}}(1.0.33)

so

\begin{aligned}U \approx N \frac{\not{{\hbar \omega}}}{\not{{2}}} \frac{\not{{2}} k_{\mathrm{B}} T}{\not{{\hbar \omega}}},\end{aligned} \hspace{\stretch{1}}(1.0.34)

or

\begin{aligned}U(T) \approx N k_{\mathrm{B}} T,\end{aligned} \hspace{\stretch{1}}(1.0.35)

matching the classical result of eq. 1.0.23. Similarly from the quantum specific heat result of eq. 1.0.31, we have

\begin{aligned}C_{\mathrm{V}}(T) \approx N k_{\mathrm{B}}\left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right) } \right)^2= N k_{\mathrm{B}}.\end{aligned} \hspace{\stretch{1}}(1.0.36)

This matches our classical result from eq. 1.0.24. We expect this equivalence at high temperatures since our quantum harmonic partition function eq. 1.0.26 is approximately

\begin{aligned}Z_N \approx \frac{2}{\beta \hbar \omega},\end{aligned} \hspace{\stretch{1}}(1.0.37)

This differs from the classical partition function only by this factor of $2$. While this alters the free energy by $k_{\mathrm{B}} T \ln 2$, it doesn’t change the mean energy since ${\partial {(k_{\mathrm{B}} \ln 2)}}/{\partial {\beta}} = 0$. At high temperatures the mean energy are large enough that the quantum nature of the system has no significant effect.

### Low temperature limits

For the classical case the heat capacity was constant ($C_{\mathrm{V}} = N k_{\mathrm{B}}$), all the way down to zero. For the quantum case the heat capacity drops to zero for low temperatures. We can see that via L’hopitals rule. With $x = \hbar \omega \beta/2$ the low temperature limit is

\begin{aligned}\lim_{T \rightarrow 0} C_{\mathrm{V}} &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{x^2}{\sinh^2 x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{2x }{2 \sinh x \cosh x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{1 }{\cosh^2 x + \sinh^2 x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{1 }{\cosh (2 x) } \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.38)

We also see this in the plot of fig. 1.2.

Fig 1.2: Specific heat for N quantum oscillators

## Question: Quantum electric dipole (2013 problem set 5, p3)

A quantum electric dipole at a fixed space point has its energy determined by two parts – a part which comes from its angular motion and a part coming from its interaction with an applied electric field $\mathcal{E}$. This leads to a quantum Hamiltonian

\begin{aligned}H = \frac{\mathbf{L} \cdot \mathbf{L}}{2 I} - \mu \mathcal{E} L_z,\end{aligned} \hspace{\stretch{1}}(1.0.39)

where $I$ is the moment of inertia, and we have assumed an electric field $\mathcal{E} = \mathcal{E} \hat{\mathbf{z}}$. This Hamiltonian has eigenstates described by spherical harmonics $Y_{l, m}(\theta, \phi)$, with $m$ taking on $2l+1$ possible integral values, $m = -l, -l + 1, \cdots, l -1, l$. The corresponding eigenvalues are

\begin{aligned}\lambda_{l, m} = \frac{l(l+1) \hbar^2}{2I} - \mu \mathcal{E} m \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.40)

(Recall that $l$ is the total angular momentum eigenvalue, while $m$ is the eigenvalue corresponding to $L_z$.)

### Part a

Schematically sketch these eigenvalues as a function of $\mathcal{E}$ for $l = 0,1,2$.

### Part b

Find the quantum partition function, assuming only $l = 0$ and $l = 1$ contribute to the sum.

### Part c

Using this partition function, find the average dipole moment $\mu \left\langle{{L_z}}\right\rangle$ as a function of the electric field and temperature for small electric fields, commenting on its behavior at very high temperature and very low temperature.

### Part d

Estimate the temperature above which discarding higher angular momentum states, with $l \ge 2$, is not a good approximation.

### Sketch the energy eigenvalues

Let’s summarize the values of the energy eigenvalues $\lambda_{l,m}$ for $l = 0, 1, 2$ before attempting to plot them.

$l = 0$

For $l = 0$, the azimuthal quantum number can only take the value $m = 0$, so we have

\begin{aligned}\lambda_{0,0} = 0.\end{aligned} \hspace{\stretch{1}}(1.0.41)

$l = 1$

For $l = 1$ we have

\begin{aligned}\frac{l(l+1)}{2} = 1(2)/2 = 1,\end{aligned} \hspace{\stretch{1}}(1.0.42)

so we have

\begin{aligned}\lambda_{1,0} = \frac{\hbar^2}{I} \end{aligned} \hspace{\stretch{1}}(1.0.43a)

\begin{aligned}\lambda_{1,\pm 1} = \frac{\hbar^2}{I} \mp \mu \mathcal{E} \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.43b)

$l = 2$

For $l = 2$ we have

\begin{aligned}\frac{l(l+1)}{2} = 2(3)/2 = 3,\end{aligned} \hspace{\stretch{1}}(1.0.44)

so we have

\begin{aligned}\lambda_{2,0} = \frac{3 \hbar^2}{I} \end{aligned} \hspace{\stretch{1}}(1.0.45a)

\begin{aligned}\lambda_{2,\pm 1} = \frac{3 \hbar^2}{I} \mp \mu \mathcal{E} \hbar\end{aligned} \hspace{\stretch{1}}(1.0.45b)

\begin{aligned}\lambda_{2,\pm 2} = \frac{3 \hbar^2}{I} \mp 2 \mu \mathcal{E} \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.45c)

These are sketched as a function of $\mathcal{E}$ in fig. 1.3.

Fig 1.3: Energy eigenvalues for l = 0,1, 2

### Partition function

Our partition function, in general, is

\begin{aligned}Z &= \sum_{l = 0}^\infty \sum_{m = -l}^l e^{-\lambda_{l,m} \beta} \\ &= \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = -l}^l e^{ m \mu \hbar \mathcal{E} \beta}.\end{aligned} \hspace{\stretch{1}}(1.0.46)

Dropping all but $l = 0, 1$ terms this is

\begin{aligned}Z \approx 1 + e^{-\hbar^2 \beta/I} \left( 1 + e^{- \mu \hbar \mathcal{E} \beta } + e^{ \mu \hbar \mathcal{E} \beta} \right),\end{aligned} \hspace{\stretch{1}}(1.0.47)

or

\begin{aligned}\boxed{Z \approx 1 + e^{-\hbar^2 \beta/I} (1 + 2 \cosh\left( \mu \hbar \mathcal{E} \beta \right)).}\end{aligned} \hspace{\stretch{1}}(1.0.48)

### Average dipole moment

For the average dipole moment, averaging over both the states and the partitions, we have

\begin{aligned}Z \left\langle{{ \mu L_z }}\right\rangle &= \sum_{l = 0}^\infty \sum_{m = -l}^l {\left\langle {l m} \right\rvert} \mu L_z {\left\lvert {l m} \right\rangle} e^{-\beta \lambda_{l, m}} \\ &= \sum_{l = 0}^\infty \sum_{m = -l}^l \mu {\left\langle {l m} \right\rvert} m \hbar {\left\lvert {l m} \right\rangle} e^{-\beta \lambda_{l, m}} \\ &= \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = -l}^l m e^{ \mu m \hbar \mathcal{E} \beta} \\ &= \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = 1}^l m \left( e^{ \mu m \hbar \mathcal{E} \beta} -e^{-\mu m \hbar \mathcal{E} \beta} \right) \\ &= 2 \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = 1}^l m \sinh (\mu m \hbar \mathcal{E} \beta).\end{aligned} \hspace{\stretch{1}}(1.0.49)

For the cap of $l = 1$ we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle \approx\frac{2 \mu \hbar }{Z}\left( 1 (0) + e^{-\hbar^2 \beta/ I} \sinh (\mu \hbar \mathcal{E} \beta) \right)\approx2 \mu \hbar \frac{e^{-\hbar^2 \beta/ I} \sinh (\mu \hbar \mathcal{E} \beta) }{1 + e^{-\hbar^2 \beta/I} \left( 1 + 2 \cosh( \mu \hbar \mathcal{E} \beta) \right)},\end{aligned} \hspace{\stretch{1}}(1.0.50)

or

\begin{aligned}\boxed{\left\langle{{ \mu L_z }}\right\rangle \approx\frac{2 \mu \hbar \sinh (\mu \hbar \mathcal{E} \beta) }{e^{\hbar^2 \beta/I} + 1 + 2 \cosh( \mu \hbar \mathcal{E} \beta)}.}\end{aligned} \hspace{\stretch{1}}(1.0.51)

This is plotted in fig. 1.4.

Fig 1.4: Dipole moment

For high temperatures $\mu \hbar \mathcal{E} \beta \ll 1$ or $k_{\mathrm{B}} T \gg \mu \hbar \mathcal{E}$, expanding the hyperbolic sine and cosines to first and second order respectively and the exponential to first order we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle &\approx 2 \mu \hbar \frac{ \frac{\mu \hbar \mathcal{E}}{k_{\mathrm{B}} T}}{ 4 + \frac{h^2}{I k_{\mathrm{B}} T} + \left( \frac{\mu \hbar \mathcal{E}}{k_{\mathrm{B}} T} \right)^2}=\frac{2 (\mu \hbar)^2 \mathcal{E} k_{\mathrm{B}} T}{4 (k_{\mathrm{B}} T)^2 + \hbar^2 k_{\mathrm{B}} T/I + (\mu \hbar \mathcal{E})^2 } \\ &\approx\frac{(\mu \hbar)^2 \mathcal{E}}{4 k_{\mathrm{B}} T}.\end{aligned} \hspace{\stretch{1}}(1.0.52)

Our dipole moment tends to zero approximately inversely proportional to temperature. These last two respective approximations are plotted along with the all temperature range result in fig. 1.5.

Fig 1.5: High temperature approximations to dipole moments

For low temperatures $k_{\mathrm{B}} T \ll \mu \hbar \mathcal{E}$, where $\mu \hbar \mathcal{E} \beta \gg 1$ we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle \approx\frac{ 2 \mu \hbar e^{\mu \hbar \mathcal{E} \beta} }{ e^{\hbar^2 \beta/I} + e^{\mu \hbar \mathcal{E} \beta} }=\frac{ 2 \mu \hbar }{ 1 + e^{ (\hbar^2 \beta/I - \mu \hbar \mathcal{E})/{k_{\mathrm{B}} T} } }.\end{aligned} \hspace{\stretch{1}}(1.0.53)

Provided the electric field is small enough (which means here that $\mathcal{E} < \hbar/\mu I$) this will look something like fig. 1.6.

Fig 1.6: Low temperature dipole moment behavior

### Approximation validation

In order to validate the approximation, let’s first put the partition function and the numerator of the dipole moment into a tidier closed form, evaluating the sums over the radial indices $l$. First let’s sum the exponentials for the partition function, making an $n = m + l$

\begin{aligned}\sum_{m = -l}^l a^m &= a^{-l} \sum_{n=0}^{2l} a^n \\ &= a^{-l} \frac{a^{2l + 1} - 1}{a - 1} \\ &= \frac{a^{l + 1} - a^{-l}}{a - 1} \\ &= \frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}}.\end{aligned} \hspace{\stretch{1}}(1.0.54)

With a substitution of $a = e^b$, we have

\begin{aligned}\boxed{\sum_{m = -l}^l e^{b m}=\frac{\sinh(b(l + 1/2))}{\sinh(b/2)}.}\end{aligned} \hspace{\stretch{1}}(1.0.55)

Now we can sum the azimuthal exponentials for the dipole moment. This sum is of the form

\begin{aligned}\sum_{m = -l}^l m a^m &= a \left( \sum_{m = 1}^l + \sum_{m = -l}^{-1} \right)m a^{m-1} \\ &= a \frac{d}{da}\sum_{m = 1}^l\left( a^{m} + a^{-m} \right) \\ &= a \frac{d}{da}\left( \sum_{m = -l}^l a^m - \not{{1}} \right) \\ &= a \frac{d}{da}\left( \frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.56)

With $a = e^{b}$, and $1 = a db/da$, we have

\begin{aligned}a \frac{d}{da} = a \frac{db}{da} \frac{d}{db} = \frac{d}{db},\end{aligned} \hspace{\stretch{1}}(1.0.57)

we have

\begin{aligned}\sum_{m = -l}^l m e^{b m}= \frac{d}{db}\left( \frac{ \sinh(b(l + 1/2)) }{ \sinh(b/2) } \right).\end{aligned} \hspace{\stretch{1}}(1.0.58)

With a little help from Mathematica to simplify that result we have

\begin{aligned}\boxed{\sum_{m = -l}^l m e^{b m}=\frac{l \sinh(b (l+1)) - (l+1) \sinh(b l) }{2 \sinh^2(b/2)}.}\end{aligned} \hspace{\stretch{1}}(1.0.59)

We can now express the average dipole moment with only sums over radial indices $l$

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle &= \mu \hbar \frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sum_{m = -l}^l m e^{ \mu m \hbar \mathcal{E} \beta}}{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sum_{m = -l}^l e^{ m \mu \hbar \mathcal{E} \beta}} \\ &= \mu \hbar\frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \frac { l \sinh(\mu \hbar \mathcal{E} \beta (l+1)) - (l+1) \sinh(\mu \hbar \mathcal{E} \beta l) } { 2 \sinh^2(\mu \hbar \mathcal{E} \beta/2) }}{\sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \frac { \sinh(\mu \hbar \mathcal{E} \beta(l + 1/2)) } { \sinh(\mu \hbar \mathcal{E} \beta/2) }}.\end{aligned} \hspace{\stretch{1}}(1.0.60)

So our average dipole moment is

\begin{aligned}\boxed{\left\langle{{ \mu L_z }}\right\rangle = \frac{\mu \hbar }{2 \sinh(\mu \hbar \mathcal{E} \beta/2)}\frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\left( l \sinh(\mu \hbar \mathcal{E} \beta (l+1)) - (l+1) \sinh(\mu \hbar \mathcal{E} \beta l) \right)}{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sinh(\mu \hbar \mathcal{E} \beta(l + 1/2))}.}\end{aligned} \hspace{\stretch{1}}(1.0.61)

The hyperbolic sine in the denominator from the partition function and the difference of hyperbolic sines in the numerator both grow fast. This is illustrated in fig. 1.7.

Fig 1.7: Hyperbolic sine plots for dipole moment

Let’s look at the order of these hyperbolic sines for large arguments. For the numerator we have a difference of the form

\begin{aligned}x \sinh( x + 1 ) - (x + 1) \sinh ( x ) &= \frac{1}{{2}} \left( x \left( e^{x + 1} - e^{-x - 1} \right) -(x +1 ) \left( e^{x } - e^{-x } \right) \right)\approx\frac{1}{{2}} \left( x e^{x + 1} -(x +1 ) e^{x } \right) \\ &= \frac{1}{{2}} \left( x e^{x} ( e - 1 ) - e^x \right) \\ &= O(x e^x).\end{aligned} \hspace{\stretch{1}}(1.0.62)

For the hyperbolic sine from the partition function we have for large $x$

\begin{aligned}\sinh( x + 1/2) = \frac{1}{{2}} \left( e^{x + 1/2} - e^{-x - 1/2} \right)\approx \frac{\sqrt{e}}{2} e^{x}= O(e^x).\end{aligned} \hspace{\stretch{1}}(1.0.63)

While these hyperbolic sines increase without bound as $l$ increases, we have a negative quadratic dependence on $l$ in the $\mathbf{L}^2$ contribution to these sums, provided that is small enough we can neglect the linear growth of the hyperbolic sines. We wish for that factor to be large enough that it dominates for all $l$. That is

\begin{aligned}\frac{l(l+1) \hbar^2}{2 I k_{\mathrm{B}} T} \gg 1,\end{aligned} \hspace{\stretch{1}}(1.0.64)

or

\begin{aligned}T \ll \frac{l(l+1) \hbar^2}{2 I k_{\mathrm{B}} T}.\end{aligned} \hspace{\stretch{1}}(1.0.65)

Observe that the RHS of this inequality, for $l = 1, 2, 3, 4, \cdots$ satisfies

\begin{aligned}\frac{\hbar^2 }{I k_{\mathrm{B}}}<\frac{3 \hbar^2 }{I k_{\mathrm{B}}}<\frac{6 \hbar^2 }{I k_{\mathrm{B}}}<\frac{10 \hbar^2 }{I k_{\mathrm{B}}}< \cdots\end{aligned} \hspace{\stretch{1}}(1.0.66)

So, for small electric fields, our approximation should be valid provided our temperature is constrained by

\begin{aligned}\boxed{T \ll \frac{\hbar^2 }{I k_{\mathrm{B}}}.}\end{aligned} \hspace{\stretch{1}}(1.0.67)

## PHY452H1S Basic Statistical Mechanics. Lecture 13: Interacting spin. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 5, 2013

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# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Interacting spin

For these notes

\begin{aligned}\boxed{\hbar = k_{\mathrm{B}} = 1}\end{aligned}

This lecture requires concepts from phy456 [1].

We’ll look at pairs of spins as a toy model of interacting spins as depicted in fig. 1.1.

Fig 1.1: Pairs of interacting spins

Example:

Simple atomic system, with the nucleus and the electron can interact with each other (hyper-fine interaction).

Consider two interacting spin $1/2$ operators $\mathbf{S}$ each with components $\hat{S}^x$, $\hat{S}^y$, $\hat{S}^z$

\begin{aligned}H = J \mathbf{S}_1 \cdot \mathbf{S}_2 - B (\hat{S}_1^z + \hat{S}_2^z)\end{aligned} \hspace{\stretch{1}}(1.2.1)

\begin{aligned}\hat{S}_1^z + \hat{S}_2^z \propto \mbox{magnetization along z}\end{aligned} \hspace{\stretch{1}}(1.2.2)

We rewrite the dot product term of the Hamiltonian in terms of just the squares of the spin operators

\begin{aligned}H = J \frac{(\mathbf{S}_1 + \mathbf{S}_2)^2 - \mathbf{S}_1^2 - \mathbf{S}_2^2}{2}- B (\hat{S}_1^z + \hat{S}_2^z)\end{aligned} \hspace{\stretch{1}}(1.2.3)

The squares $\mathbf{S}_1^2$, $\mathbf{S}_2^2$, $(\mathbf{S}_1 + \mathbf{S}_2)^2$ can be thought of as “length”s of the respective angular momentum vectors.

We write

\begin{aligned}\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2,\end{aligned} \hspace{\stretch{1}}(1.2.4)

for the total angular momentum. We recall that we have

\begin{aligned}\hat{S}^z_2 = \hat{S}^z_1 = S(S + 1),\end{aligned} \hspace{\stretch{1}}(1.2.5)

where $S = 1/2$, and $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$ implies that $S_{\mathrm{total}} \in \{0, 2\}$.

$S_{\mathrm{total}} = 0$ (singlet).
$S_{\mathrm{total}} = 1$. Triplet: $(-1, 0, +1)$.

$S_{\mathrm{total}} = 0$ state.

For $m = 0$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \downarrow - \downarrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

energies

\begin{aligned}J \frac{-3/4 -3/4}{2} = -\frac{3}{4} J\end{aligned} \hspace{\stretch{1}}(1.2.7)

For $m = 1$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.8)

energies

\begin{aligned}J \left( 1 - \frac{3}{4} \right) - B \rightarrow \frac{J}{4} - B\end{aligned} \hspace{\stretch{1}}(1.2.9)

$S_{\mathrm{total}} = 1$ state

For $m = 0$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \uparrow \downarrow + \downarrow \uparrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.10)

energies

\begin{aligned}\frac{J}{4} \end{aligned} \hspace{\stretch{1}}(1.2.11)

For $m = 1$

\begin{aligned}\frac{1}{{\sqrt{2}}} \left( \downarrow \downarrow \right)\end{aligned} \hspace{\stretch{1}}(1.2.12)

energies

\begin{aligned}\frac{J}{4} + B.\end{aligned} \hspace{\stretch{1}}(1.2.13)

These are illustrated schematically in fig. 1.2.

Fig 1.2: Energy levels for two interacting spins as a function of magnetic field

Our single pair partition function is

\begin{aligned}Z_1 = e^{ +\beta 3 J/4}+e^{ -\beta (J/4 - B)}e^{ -\beta 3 J/4}+e^{ -\beta (J/4 + B)}\end{aligned} \hspace{\stretch{1}}(1.2.14)

So for $N$ pairs our partition function is

\begin{aligned}Z = Z_1^N = \left( e^{ +\beta 3 J/4} +e^{ -\beta (J/4 - B)} e^{ -\beta 3 J/4} +e^{ -\beta (J/4 + B)} \right)^N.\end{aligned} \hspace{\stretch{1}}(1.2.15)

Our free energy

\begin{aligned}F = - T \ln Z = - T N \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.2.16)

\begin{aligned}-\frac{\partial {F}}{\partial {\beta}} = T N \frac{\partial {}}{\partial {\beta}} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.2.17)

Our magnetization $\mu$ is

\begin{aligned}\mu = \frac{T N}{Z_1} \left( \beta e^{-\beta(J/4 - B)} -\beta e^{-\beta(J/4 + B)} \right)\end{aligned} \hspace{\stretch{1}}(1.2.18)

The moment per particle, after $T \beta$ cancellation, is

\begin{aligned}m = \frac{\mu}{N} = \frac{1}{Z_1} \left( e^{-\beta(J/4 - B)} -e^{-\beta(J/4 + B)} \right)=2 \frac{e^{-\beta J/4}}{Z_1} \sinh\left( \frac{B}{T} \right).\end{aligned} \hspace{\stretch{1}}(1.2.19)

Low temperatures, small $B$ ($T \ll J, B \ll J$)

The $e^{3 \beta J/4}$ term will dominate.

\begin{aligned}Z_1 \approx e^{3 J \beta/4}\end{aligned} \hspace{\stretch{1}}(1.2.20)

\begin{aligned}m \approx 2 e^{-\beta J} \sinh\left( \frac{B}{T} \right).\end{aligned} \hspace{\stretch{1}}(1.2.21)

Fig 1.3: magnetic moment

The specific heat has a similar behavior

\begin{aligned}C_V \sim e^{-\beta J}.\end{aligned} \hspace{\stretch{1}}(1.2.22)

Considering a single spin $1/2$ system, we have energies as illustrated in fig. 1.4.

Fig 1.4: Single particle spin energies as a function of magnetic field

At zero temperatures we have a finite non-zero magnetization as illustrated in fig. 1.5, but as we heat the system up, the state of the system will randomly switch between the 1, and 2 states. The partition function democratically averages over all such possible states.

Fig 1.6: Single spin magnetization

Once the system heats up, the spins are democratically populated within the entire set of possible states.

We contrast this to this interacting spins problem which has a magnetization of the form fig. 1.6.

Fig 1.6: Interacting spin magnetization

For the single particle specific heat we have specific heat of the form fig. 1.7.

Fig 1.7: Single particle specific heat

We’ll see the same kind of specific heat distribution with temperature for the interacting spins problem, but the peak will be found at an energy that’s given by the difference in energies of the two states as illustrated in fig. 1.8.

\begin{aligned}\Delta E = \frac{J}{4} - \frac{-3J}{4} = J\end{aligned} \hspace{\stretch{1}}(1.2.23)

# References

[1] Peeter Joot. Quantum Mechanics II., chapter: Two spin systems, angular momentum, and Clebsch-Gordon convention. URL http://sites.google.com/site/peeterjoot2/math2011/phy456.pdf.

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## Some problems from Kittel chapter 3

Posted by peeterjoot on February 10, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

## Question: Classical gas partition function

[1] expresses the classical gas partition function (3.77) as

\begin{aligned}Z_1 \propto \int \exp\left( - \frac{p_x^2 + p_y^2 + p_z^2 }{2 M \tau}\right) dp_x dp_y dp_z\end{aligned} \hspace{\stretch{1}}(1.0.1)

Show that this leads to the expected $3 \tau/2$ result for the thermal average energy.

Let’s use the adjustment technique from the text for the $N$ partition case and write

\begin{aligned}Z_N = \frac{1}{{N!}} Z_1^N,\end{aligned} \hspace{\stretch{1}}(1.0.2)

with $Z_1$ as above. This gives us

\begin{aligned}U &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z_N \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}} \left(N \ln Z_1 - \ln N!\right) \\ &= N \tau^2 \frac{\partial {\ln Z_1 }}{\partial {\tau}} \\ &= N \tau^2 \frac{\partial {}}{\partial {\tau}} \sum_{k = 1}^{3} \ln\int \exp\left( - \frac{p_k^2 }{2 M \tau}\right) dp_k \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k}{\int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k} \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \sqrt{ 2 \pi M \tau }}{\sqrt{ 2 \pi M \tau}} \\ &= 3 N \tau^2\frac{\frac{1}{{2}} \tau^{-1/2}}{\sqrt{ \tau}} \\ &= \frac{3}{2} N \tau \\ &= \frac{3}{2} N k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.3)

## Question: Two state system

[1] problem 3.1.

Find an expression for the free energy as a function of $\tau$ of a system with two states, one at energy $0$ and one at energy $\epsilon$. From the free energy, find expressions for the energy and entropy of the system.

Our partition function is

\begin{aligned}Z = 1 + e^{-\epsilon /\tau}\end{aligned} \hspace{\stretch{1}}(1.0.4)

The free energy is just

\begin{aligned}F = -\tau \ln Z = -\tau \ln (1 + e^{-\epsilon/\tau})\end{aligned} \hspace{\stretch{1}}(1.0.5)

The entropy follows immediately

\begin{aligned}\sigma \\ &= -\frac{\partial {F}}{\partial {\tau}} \\ &= \frac{\partial {}}{\partial {\tau}}\left( \tau \ln \left( 1 + e^{-\epsilon/\tau} \right) \right) \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)-\tau \epsilon \frac{-1}{\tau^2} \frac{1}{{1 + e^{-\epsilon/\tau}}} \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)+\frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}\end{aligned} \hspace{\stretch{1}}(1.0.6)

The energy is

\begin{aligned}U \\ &= F + \tau \sigma \\ &= -\tau \ln (1 + e^{-\epsilon/\tau}) + \tau \sigma \\ &= \tau\left( \not{{\ln \left( 1 + e^{-\epsilon/\tau} \right)}} + \frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}} -\not{{\ln (1 + e^{-\epsilon/\tau}) }} \right)\end{aligned} \hspace{\stretch{1}}(1.0.7)

This is

\begin{aligned}U=\frac{\epsilon e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}=\frac{\epsilon}{1 + e^{\epsilon/\tau}}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

These are all plotted in (Fig 1).

Fig1: Plots for two state system

\imageFigure{kittelCh3Problem1PlotsFig1}{Plots for two state system}{fig:kittelCh3Problem1Plots:kittelCh3Problem1PlotsFig1}{0.2}

## Question: Magnetic susceptibility

[1] problem 3.2.

Use the partition function to find an exact expression for the magnetization $M$ and the susceptibility $\chi = dM/dB$ as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization, found by other means, was $M = n m \tanh( m B/\tau)$, where $n$ is the particle concentration. Find the free energy and express the result as a function only of $\tau$ and the parameter $x = M/nm$. Show that the susceptibility is $\chi = n m^2/\tau$ in the limit $m B \ll \tau$.

Our partition function for a unit volume containing $n$ spins is

\begin{aligned}Z=\frac{\left( e^{-m B/\tau} +e^{m B/\tau} \right)^n}{n!}=2 \frac{\left( \cosh\left( m B/\tau \right) \right)^n}{n!},\end{aligned} \hspace{\stretch{1}}(1.0.9)

so that the Free energy is

\begin{aligned}F = -\tau\left( \ln 2 - \ln n! + n \ln \cosh\left( m B/\tau \right) \right).\end{aligned} \hspace{\stretch{1}}(1.0.15)

The energy, magnetization and magnetic field were interrelated by

\begin{aligned}- M B &= U \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}}\left( -\frac{F}{\tau} \right) \\ &= \tau^2 n\frac{\partial {}}{\partial {\tau}}\ln \cosh\left( m B/\tau \right) \\ &= \tau^2 n \frac{ -m B/\tau^2\sinh\left( m B/\tau \right)}{\cosh\left( m B/\tau \right)} \\ &= - m B n \tanh \left( m B/\tau \right).\end{aligned} \hspace{\stretch{1}}(1.0.11)

This gives us

\begin{aligned}M = m n \tanh \left( m B/\tau \right),\end{aligned} \hspace{\stretch{1}}(1.0.12)

so that

\begin{aligned}\chi = \frac{dM}{dB}= \frac{m^2 n}{\tau \cosh^2 \left( m B/\tau \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

For $m B/\tau \ll 1$, the cosh term goes to unity, so we have

\begin{aligned}\chi \approx= \frac{m^2 n}{\tau},\end{aligned} \hspace{\stretch{1}}(1.0.14)

as desired.

With $x = M/nm$, or $m = M/nx$, the free energy is

\begin{aligned}F =-\tau\left( \ln 2/n! + n \ln \cosh\left( \frac{M B}{n x \tau} \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.15)

That last expression isn’t particularly illuminating. What was the point of that substitution?

## Question: Free energy of a harmonic oscillator

[1] problem 3.3.

A one dimensional harmonic oscillator has an infinite series of equally spaced energy states, with $\epsilon_s = s \hbar \omega$, where $s$ is a positive integer or zero, and $\omega$ is the classical frequency of the oscillator. We have chosen the zero of energy at the state $s = 0$. Show that for a harmonic oscillator the free energy is

\begin{aligned}F = \tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Note that at high temperatures such that $\tau \gg \hbar \omega$ we may expand the argument of the logarithm to obtain $F \approx \tau \ln (\hbar \omega/\tau)$. From 1.0.16 show that the entropy is

\begin{aligned}\sigma = \frac{\hbar\omega/\tau}{e^{\hbar \omega/\tau} - 1} -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

I found it curious that this problem dropped the factor of $\hbar\omega/2$ from the energy. Including it we have

\begin{aligned}\epsilon_s = \left( s + \frac{1}{{2}} \right) \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.0.18)

So that the partition function is

\begin{aligned}Z= \sum_{s = 0}^\infty e^{-\left( s + \frac{1}{{2}} \right) \hbar \omega/\tau }=e^{-\hbar \omega/2\tau}\sum_{s = 0}^\infty e^{-s \hbar \omega/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

The free energy is

\begin{aligned}F &= -\tau \ln Z \\ &= -\tau\left( -\frac{\hbar \omega}{2\tau} + \ln \left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right) \right) \\ &= \frac{\hbar \omega}{2} +\ln\left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

We see that the contribution of the $\hbar \omega/2$ in the energy of each state just adds a constant factor to the free energy. This will drop out when we compute the entropy. Dropping that factor now that we know why it doesn’t contribute, we can complete the summation, so have, by inspection

\begin{aligned}F = -\tau \ln Z=\tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

Taking derivatives for the entropy we have

\begin{aligned}\sigma &= -\frac{\partial {F}}{\partial {\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\tau\frac{\hbar \omega}{\tau^2} \frac{e^{-\hbar \omega/\tau}}{1 - e^{-\hbar \omega/\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\frac{\frac{\hbar \omega}{\tau}}{e^{\hbar \omega/\tau} - 1}\end{aligned} \hspace{\stretch{1}}(1.0.22)

## Question: Energy fluctuation

[1] problem 3.4.

Consider a system of fixed volume in thermal contact with a reservoir. Show that the mean square fluctuation in the energy of the system is

\begin{aligned}\left\langle{{ (\epsilon - \left\langle{\epsilon}\right\rangle)^2 }}\right\rangle = \tau^2\left( \frac{\partial {U}}{\partial {\tau}} \right)_V\end{aligned} \hspace{\stretch{1}}(1.0.23)

Here $U$ is the conventional symbol for $\left\langle{{\epsilon}}\right\rangle$. Hint: Use the partition function $Z$ to relate ${\partial {U}}/{\partial {t}}$ to the mean square fluctuation. Also, multiply out the term $(\cdots)^2$.

With a probability of finding the system in state $s$ of

\begin{aligned}P_s = \frac{e^{-\epsilon_s/\tau}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.24)

the average energy is

\begin{aligned}U &= \left\langle{{\epsilon}}\right\rangle \\ &= \sum_s P_s \epsilon_s \\ &= \sum_s \epsilon_s \frac{e^{-\epsilon_s/\tau}}{Z} \\ &= \frac{1}{{Z}} \sum_s \epsilon_s e^{-\epsilon_s/\tau}\end{aligned} \hspace{\stretch{1}}(1.0.25)

So we have

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{\tau^2}{Z}\sum_s \frac{\epsilon_s^2}{\tau^2} e^{-\epsilon_s/\tau} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

But

\begin{aligned}\frac{dZ}{d\tau}=\frac{d}{d\tau} \sum_s e^{-\epsilon_s/\tau}=\sum_s \frac{\epsilon_s}{\tau^2} e^{-\epsilon_s/\tau},\end{aligned} \hspace{\stretch{1}}(1.0.27)

giving

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} &= \frac{1}{Z^2}\sum_s \epsilon_s e^{-\epsilon_s/\tau} \sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau} \\ &= -\left\langle{{ \epsilon }}\right\rangle^2 + \left\langle{{\epsilon^2}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(1.0.28)

which shows 1.0.23 as desired.

# References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.