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Posts Tagged ‘Partition function’

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.

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Kittel Zipper problem

Posted by peeterjoot on March 20, 2013

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Question: Zipper problem ([1] pr 3.7)

A zipper has N links; each link has a state in which it is closed with energy 0 and a state in which it is open with energy \epsilon. we require, however, that the zipper can only unzip from the left end, and that the link number s can only open if all links to the left (1, 2, \cdots, s - 1) are already open. Find (and sum) the partition function. In the low temperature limit k_{\mathrm{B}} T \ll \epsilon, find the average number of open links. The model is a very simplified model of the unwinding of two-stranded DNA molecules.

Answer

The system is depicted in fig. 1.1, in the E = 0 and E = \epsilon states.

Fig 1.1: Zipper molecule model in first two states

 

The left opening only constraint simplifies the combinatorics, since this restricts the available energies for the complete molecule to 0, \epsilon, 2 \epsilon, \cdots, N \epsilon.

The probability of finding the molecule with s links open is then

\begin{aligned}P_s =\frac{e^{- \beta s \epsilon}}{Z},\end{aligned} \hspace{\stretch{1}}(1.0.1)

with

\begin{aligned}Z = \sum_{s = 0}^N \frac{e^{- \beta s \epsilon}}{Z}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

We can sum this geometric series immediately

\begin{aligned}\boxed{Z =\frac{e^{-\beta (N+1) \epsilon} - 1}{e^{-\beta \epsilon } - 1}.}\end{aligned} \hspace{\stretch{1}}(1.0.3)

The expectation value for the number of links is

\begin{aligned}\left\langle{{s}}\right\rangle &= \sum_{s = 0}^N s P_s \\ &= \frac{1}{{Z}} \sum_{s = 1}^N s e^{- \beta s \epsilon} \\ &= -\frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta \epsilon)}} \sum_{s = 1}^N e^{- \beta s \epsilon}.\end{aligned} \hspace{\stretch{1}}(1.0.4)

Let’s write

\begin{aligned}a = e^{-\beta \epsilon},\end{aligned} \hspace{\stretch{1}}(1.0.5)

and make a change of variables

\begin{aligned}-\frac{\partial {}}{\partial {(\beta \epsilon)}} &= \frac{\partial {}}{\partial {\ln a}} \\ &= \frac{\partial {a}}{\partial {\ln a}}\frac{\partial {}}{\partial {a}} \\ &= \frac{\partial {e^{-\beta \epsilon}}}{\partial {(-\beta \epsilon)}}\frac{\partial {}}{\partial {a}} \\ &= a\frac{\partial {}}{\partial {a}}\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that

\begin{aligned}-\frac{\partial {}}{\partial {\ln a}} \sum_{s = 1}^N a^s &= a \frac{d}{da} \left( \frac{a^{N+1} - a}{a - 1} \right) \\ &= a\left( \frac{(N+1) a^N - 1}{a - 1} - \frac{a^{N+1} - a} { (a - 1)^2 } \right) \\ &= \frac{a}{(a-1)^2}\left( \left( (N+1) a^N - 1 \right) (a - 1) - a^{N+1} + a \right) \\ &= \frac{a}{(a-1)^2}\left( N a^{N+1} -(N+1) a^N + 1 \right) \\ &= \frac{a}{(a-1)^2}\left( a^N ( N (a - 1) - 1 ) + 1 \right).\end{aligned} \hspace{\stretch{1}}(1.0.7)

The average number of links is thus

\begin{aligned}\left\langle{{k}}\right\rangle = \frac{a - 1}{a^{N+1} - 1}\frac{a}{(a-1)^2}\left( a^N ( N (a - 1) - 1 ) + 1 \right),\end{aligned} \hspace{\stretch{1}}(1.0.8)

or

\begin{aligned}\boxed{\left\langle{{k}}\right\rangle = \frac{1}{1 - e^{-\beta \epsilon(N+1)} }\frac{1}{e^{\beta \epsilon} - 1}\left( e^{-\beta \epsilon N} ( N (e^{-\beta \epsilon} - 1) - 1 ) + 1 \right).}\end{aligned} \hspace{\stretch{1}}(1.0.9)

In the very low temperature limit where \beta \epsilon \gg 1 (small T, big \beta), we have

\begin{aligned}\left\langle{{k}}\right\rangle \approx\frac{1}{e^{\beta \epsilon}}= e^{-\beta \epsilon},\end{aligned} \hspace{\stretch{1}}(1.0.10)

showing that on average no links are open at such low temperatures. An exact plot of \left\langle{{s}}\right\rangle for a few small N values is in fig. 1.2.

Fig 1.2: Average number of open links

 

References

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

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Pathria chapter 4 diatomic molecule problem

Posted by peeterjoot on March 18, 2013

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Question: Diatomic molecule ([1] pr 4.7)

Consider a classical system of non-interacting, diatomic molecules enclosed in a box of volume V at temperature T. The Hamiltonian of a single molecule is given by

\begin{aligned}H(\mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2) = \frac{1}{{2m}} \left( \mathbf{p}_1^2 + \mathbf{p}_2^2  \right)+\frac{1}{{2}} K \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Study the thermodynamics of this system, including the dependence of the quantity \left\langle{{r_{12}^2}}\right\rangle on T.

Answer

Partition function
First consider the partition function for a single diatomic pair

\begin{aligned}Z_1 &= \frac{1}{{h^6}} \int d^6 \mathbf{p} d^6 \mathbf{r} e^{-\beta \frac{ \mathbf{p}_1^2 + \mathbf{p}_2^2 }{2m}} e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}} \\ &= \frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{6/2}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Now we can make a change of variables to simplify the exponential. Let’s write

\begin{aligned}\mathbf{u} = \mathbf{r}_1 - \mathbf{r}_2\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{v} = \mathbf{r}_2,\end{aligned} \hspace{\stretch{1}}(1.0.3b)

or

\begin{aligned}\mathbf{r}_2 = \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\mathbf{r}_1=\mathbf{u} + \mathbf{v}.\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Our volume element is

\begin{aligned}d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 = d^3 \mathbf{u} d^3 \mathbf{v} \frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

It wasn’t obvious to me that this change of variables preserves the volume element, but a quick Jacobian calculation shows this to be the case

\begin{aligned}\frac{\partial(\mathbf{r}_1, \mathbf{r}_2)}{\partial(\mathbf{u}, \mathbf{v})} &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ \partial r_{21}/\partial u_1 & \partial r_{21}/\partial u_2 &\partial r_{21}/\partial u_3 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ \partial r_{22}/\partial u_1 & \partial r_{22}/\partial u_2 &\partial r_{22}/\partial u_3 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ \partial r_{23}/\partial u_1 & \partial r_{23}/\partial u_2 &\partial r_{23}/\partial u_3 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= \begin{vmatrix}\partial r_{11}/\partial u_1 & \partial r_{11}/\partial u_2 &\partial r_{11}/\partial u_3 &\partial r_{11}/\partial v_1 &\partial r_{11}/\partial v_2 &\partial r_{11}/\partial v_3 \\ \partial r_{12}/\partial u_1 & \partial r_{12}/\partial u_2 &\partial r_{12}/\partial u_3 &\partial r_{12}/\partial v_1 &\partial r_{12}/\partial v_2 &\partial r_{12}/\partial v_3 \\ \partial r_{13}/\partial u_1 & \partial r_{13}/\partial u_2 &\partial r_{13}/\partial u_3 &\partial r_{13}/\partial v_1 &\partial r_{13}/\partial v_2 &\partial r_{13}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{21}/\partial v_1 &\partial r_{21}/\partial v_2 &\partial r_{21}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{22}/\partial v_1 &\partial r_{22}/\partial v_2 &\partial r_{22}/\partial v_3 \\ 0 & 0 & 0 &\partial r_{23}/\partial v_1 &\partial r_{23}/\partial v_2 &\partial r_{23}/\partial v_3 \end{vmatrix} \\ &= 1.\end{aligned} \hspace{\stretch{1}}(1.0.6)

Our remaining integral can now be evaluated

\begin{aligned}\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K\frac{ \left\lvert {\mathbf{r}_1 - \mathbf{r}_2} \right\rvert^2 }{2}}  &= \int d^3 \mathbf{u} d^3 \mathbf{v} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \int d^3 \mathbf{u} e^{-\beta K \left\lvert {\mathbf{u}} \right\rvert^2 /2 } \\ &= V \left( \frac{ 2 \pi }{ K \beta }  \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Our partition function is now completely evaluated

\begin{aligned}Z_1 = V\frac{1}{{h^6}} \left( \frac{2 \pi m}{\beta} \right)^{3}\left( \frac{ 2 \pi }{ K \beta }  \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

As a function of V and T as in the text, we write

\begin{aligned}Z_1 = V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}f(T) = \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

Gibbs sum

Our Gibbs sum, summing over the number of molecules (not atoms), is

\begin{aligned}Z_{\mathrm{G}} &= \sum_{N_r = 0}^\infty \frac{z^{N_r}}{N_r!} Z_1^{N_r} \\ &= e^{ z V f(T) },\end{aligned} \hspace{\stretch{1}}(1.0.10)

or

\begin{aligned}q &= \ln Z_{\mathrm{G}} \\ &= z V f(T) \\ &= P V \beta.\end{aligned} \hspace{\stretch{1}}(1.0.11)

The fact that we can sum this as an exponential series so nicely looks like it’s one of the main advantages to this grand partition function (Gibbs sum). We can avoid any of the large N! approximations that we have to use when the number of particles is explicitly fixed.

Pressure

The pressure follows

\begin{aligned}P &= z f(T) k_{\mathrm{B}} T \\ &= e^{\mu/k_{\mathrm{B}} T}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( k_{\mathrm{B}} T \right)^{11/2}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Average energy

\begin{aligned}\left\langle{{H}}\right\rangle &= -\frac{\partial {q}}{\partial {\beta}} \\ &= - z V \frac{9}{2} \frac{f(T)}{T} \frac{\partial {T}}{\partial {\beta}} \\ &= z V \frac{9}{2} \frac{f(T)}{T^3} \frac{1}{{k_{\mathrm{B}}}},\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}\left\langle{{H}}\right\rangle = e^{\mu/k_{\mathrm{B}} T} V \frac{9}{2} k_{\mathrm{B}}^2 \left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( k_{\mathrm{B}} T \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Average occupancy

\begin{aligned}\left\langle{{N}}\right\rangle &= z \frac{\partial {}}{\partial {z}} \ln Z_{\mathrm{G}} \\ &= z \frac{\partial {}}{\partial {z}} \left( z V f(T)  \right) \\ &= z V f(T)\end{aligned} \hspace{\stretch{1}}(1.0.15)

but this is just q, or

\begin{aligned}\left\langle{{N}}\right\rangle &= e^{\mu/k_{\mathrm{B}} T} V\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Free energy

\begin{aligned}F &= - k_{\mathrm{B}} T \ln \frac{ Z_{\mathrm{G}} }{z^N} \\ &= - k_{\mathrm{B}} T \left( q - N \ln z  \right) \\ &= N k_{\mathrm{B}} T \beta \mu - k_{\mathrm{B}} T q \\ &= z V f(T) \mu - k_{\mathrm{B}} T z V f(T) \\ &= z V f(T) \left( \mu - k_{\mathrm{B}} T  \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

\begin{aligned}F = e^{\mu/k_{\mathrm{B}} T} V \left( \mu - k_{\mathrm{B}} T  \right)\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( k_{\mathrm{B}} T \right)^{9/2}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

Entropy

\begin{aligned}S &= \frac{U - F}{T} \\ &= \frac{V}{T} e^{\mu/k_{\mathrm{B}} T} \left( k_{\mathrm{B}} T \right)^{3/2}\left( \frac{m }{h^2 } \sqrt{\frac{(2\pi)^3}{K}}  \right)^3\left( \frac{9}{2} k_{\mathrm{B}}^2 - \left( \mu - k_{\mathrm{B}} T  \right) \left( k_{\mathrm{B}} T \right)^3  \right).\end{aligned} \hspace{\stretch{1}}(1.0.19)

Expectation of atomic separation

The momentum portions of the average will just cancel out, leaving just

\begin{aligned}\left\langle{r_{12}^2}\right\rangle &= \frac{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }}{\int d^3 \mathbf{r}_1 d^3 \mathbf{r}_2 e^{-\beta K \left( \mathbf{r}_1 - \mathbf{r}_2 \right)^2 /2 }} \\ &= \frac{ \int d^3 \mathbf{u} \mathbf{u}^2 e^{-\beta K \mathbf{u}^2 /2 }}{\int d^3 \mathbf{u} e^{-\beta K \mathbf{u}^2 /2 }} \\ &= \frac{\int da db dc \left( a^2 + b^2 + c^2 \right) e^{-\beta K \left( a^2 + b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2}} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}\int db dc e^{-\beta K \left( b^2 + c^2 \right) /2}}{\int e^{-\beta K \left( a^2 + b^2 + c^2 \right)/2 }} \\ &= 3 \frac{\int da a^2 e^{-\beta K a^2/2}}{\int e^{-\beta K a^2/2}}\end{aligned} \hspace{\stretch{1}}(1.0.20)

Expanding the numerator by parts we have

\begin{aligned}\int da a^2 e^{-\beta K a^2/2} \\ &= \int a d\frac{ e^{-\beta K a^2/2}}{- 2 \beta K/2} \\ &= \frac{1}{\beta K}\int e^{-\beta K a^2/2}.\end{aligned} \hspace{\stretch{1}}(1.0.21)

This gives us

\begin{aligned}\boxed{\left\langle r_{12}^2 \right\rangle = \frac{3}{\beta K} = \frac{3 k_{\mathrm{B}} T}{K}.}\end{aligned} \hspace{\stretch{1}}(1.0.22)

References

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

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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.

Answer

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)

Answer

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.

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Quantum anharmonic oscillator

Posted by peeterjoot on March 13, 2013

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Question: Quantum anharmonic oscillator ([1] pr 3.30)

The energy levels of a quantum-mechanical, one-dimensional, anharmonic oscillator may be approximated as

\begin{aligned}\epsilon_n = \left( n + \frac{1}{{2}} \right) \hbar \omega - x \left( n + \frac{1}{{2}} \right)^2 \hbar \omega\qquad n = 0, 1, 2, \cdots\end{aligned} \hspace{\stretch{1}}(1.0.1)

The parameter x, usually \ll 1, represents the degree of anharmonicity. Show that, to the first order in x and the fourth order in u \equiv \hbar \omega/k_{\mathrm{B}} T, the specific heat of a system of N such oscillators is given by

\begin{aligned}C = N k_{\mathrm{B}} \left( \left( 1 - \frac{1}{{12}} u^2 + \frac{1}{{240}} u^4 \right) + 4 x \left( \frac{1}{{u}} + \frac{1}{{80}} u^3 \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.2)

Answer

We can expand the partition function in a first order Taylor series about x = 0, then evaluate the sums

\begin{aligned}Z_1 &= \sum_{n = 0}^\infty \exp\left( -\beta \left( n + \frac{1}{{2}} \right) \hbar \omega + \beta x \left( n + \frac{1}{{2}} \right)^2 \hbar \omega \right) \\ &= \sum_{n = 0}^\infty \exp\left( - \left( n + \frac{1}{{2}} \right) u + x \left( n + \frac{1}{{2}} \right)^2 u \right)\approx\sum_{n = 0}^\infty e^{ - \left( n + \frac{1}{{2}} \right) u }\left( 1 + x u \left( n + \frac{1}{{2}} \right)^2 \right).\end{aligned} \hspace{\stretch{1}}(1.0.3)

The quadratic sum can be evaluated indirectly as it can be expressed as a derivative

\begin{aligned}Z_1 &= \left( 1 + x u \frac{d^2}{du^2} \right)\sum_{n = 0}^\infty e^{ - \left( n + \frac{1}{{2}} \right) u } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)e^{-u/2}\sum_{n = 0}^\infty e^{ - n u } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)e^{-u/2}\frac{ 1 }{1 - e^{-u} } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)\frac{ 1 }{e^{u/2} - e^{-u/2} } \\ &= \left( 1 + x u \frac{d^2}{du^2} \right)\frac{ 1 }{2 \sinh(u/2)}\end{aligned} \hspace{\stretch{1}}(1.0.4)

Finally, evaluation of the derivatives gives us

\begin{aligned}Z_1=\frac{ 1 }{\sinh(u/2)}\left( 1 + x u \frac{2 \coth^2(u/2) - 1}{8} \right).\end{aligned} \hspace{\stretch{1}}(1.0.5)

Now we’d like to compute the specific heat in terms of derivatives of u. First, for the average energy

\begin{aligned}\left\langle{{H}}\right\rangle &= -N \frac{\partial {}}{\partial {\beta}} \ln Z_1 \\ &= -N \hbar \omega \frac{\partial {}}{\partial {u}} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.0.6)

The specific heat is

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}} \\ &= \frac{\partial {u}}{\partial {T}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= \frac{\hbar \omega}{k_{\mathrm{B}}} \frac{\partial {(1/T)}}{\partial {T}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= -\frac{\hbar \omega}{k_{\mathrm{B}} T^2} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}} \\ &= -\frac{k_{\mathrm{B}} u^2}{\hbar \omega} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {u}},\end{aligned} \hspace{\stretch{1}}(1.0.7)

or

\begin{aligned}C_{\mathrm{V}} = N k_{\mathrm{B}} u^2 \frac{\partial^2 {{}}}{\partial {{u}}^2} \ln Z_1.\end{aligned} \hspace{\stretch{1}}(1.0.8)

Actually computing that is messy algebra (See \nbref{pathria_3_30.nb}), and the result isn’t particularily interesting looking. The plot fig. 1.1 is interesting though and shows negative heat capacities near zero and a funny little jog near C_{\mathrm{V}} = 0.

Fig 1.1: Quantum anharmonic heat capacity

 

Also confirmed in the Mathematica notebook is equation eq. 1.0.2, which follows by first doing a first order series expansion in x, then a subsequent series expansion in u.

References

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

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PHY452H1S Basic Statistical Mechanics. Lecture 14: Grand canonical ensemble. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 13, 2013

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Disclaimer

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

This lecture had a large amount of spoken content not captured in these notes. Reference to section 4 [1] was made for additional details.

Grand canonical ensemble

Fig 1.1: Ensemble pictures

We are now going to allow particles to move to and from the system and the reservoir. The total number of states in the system is

\begin{aligned}\Omega_tot (N, V, E) =\sum_{N_S, E_S} \Omega_S(N_S, V_S, E_S)\Omega_R(N - N_S, V_R, E - E_S),\end{aligned} \hspace{\stretch{1}}(1.2.1)

so for N_S \ll N, and E_S \ll E, we have

\begin{aligned}\Omega_R &= \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N- N_S, V_R, E - E_S) \right) \\ &\approx \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N, V_R, E) - \frac{N_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {N}}\right)_{{V, E}} - \frac{E_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {E}}\right)_{{N, V}}  \right) \\ &\propto \Omega_S(N_S, V_S, E_S)e^{-\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} },\end{aligned} \hspace{\stretch{1}}(1.2.2)

where the chemical potential \index{chemical potential} and temperature \index{temperature} are defined respectively as

\begin{aligned}\frac{\mu}{T} = -\left({\partial {S_R}}/{\partial {N}}\right)_{{V,E}}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\frac{1}{T} = \left({\partial {S_R}}/{\partial {E}}\right)_{{N,V}}.\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\begin{aligned}\mathcal{P} \propto e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.4)

With \{c\} as the set of all possible configuration pairs \{N_S, E_S\}, we define the grand partition function

\begin{aligned}Z_{\mathrm{G}} = \sum_{\{c\}}e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.5)

So that the probability of finding a given state with energy and particle numbers \{E_S, N_S\} is

\begin{aligned}\mathcal{P}(E_S, N_S) = \frac{e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }}{Z_{\mathrm{G}}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

For a classical system we have

\begin{aligned}\{ c \} \rightarrow \{ x \} \{ p \},\end{aligned} \hspace{\stretch{1}}(1.0.7)

whereas in a quantum content we have

\begin{aligned}\{ c \} \rightarrow \text{eigenstate}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}Z_{\mathrm{G}}^{\mathrm{\mathrm{\mathrm{QM}}}} = {\text{Tr}}_{\{\text{energy}, N\}} \left( e^{ -\beta (\hat{H} - \mu \hat{N} } \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

We want to do this because the calculation of the number of states

\begin{aligned}\int_{\{ x \} \{ p \}} \delta\left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right),\end{aligned} \hspace{\stretch{1}}(1.0.10)

can quickly become intractable. We want to go to the canonical ensemble was because the partition function

\begin{aligned}Z_c = \int_{\{ x \} \{ p \}}e^{-\beta \left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right)},\end{aligned} \hspace{\stretch{1}}(1.0.11)

yields the same results, but can be much easier to compute. We have a similar reason to go to the grand canonical ensemble, because this computation, once we allow the number of particles to vary also becomes very hard.

We are now going to define a notion of equilibrium so that it includes

  1. All forces are equal (mechanical equilibrium)
  2. Temperatures are equal (no net heat flow)
  3. Chemical potentials are equal (no net particle flow)

We’ll isolate a subsystem, containing a large number of particles fig. 1.2.

Fig 1.2: A subsystem to and from which particle motion is allowed

When we think about Fermions we have to respect the “Pauli exclusion” principle \index{Pauli exclusion principle}.

Suppose we have just a one dimensional Fermion system for some potential as in fig. 1.3.

Fig 1.3: Energy level filling in a quantum system

For every momentum k there are two possible occupation numbers n_k \in \{0, 1\}

our partition function is

\begin{aligned}Z_c = \sum_{n_k,\sum_k n_k = N} e^{-\beta \sum_k \epsilon_k n_k}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

We’d find that this calculation with this \sum_k n_k = N constraint becomes essentially impossible.

We’ll see that relaxing this constraint will allow this calculation to become tractable.

References

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

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Midterm II reflection, take II, with approximate anharmonic oscillator solution

Posted by peeterjoot on March 11, 2013

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Question: Perturbation of classical harmonic oscillator (2013 midterm II p2)

Consider a single particle perturbation of a classical simple harmonic oscillator Hamiltonian

\begin{aligned}H = \frac{1}{{2}} m \omega^2 \left( {x^2 + y^2} \right) + \frac{1}{{2 m}} \left( {p_x^2 + p_y^2} \right) + a x^4 + by^6\end{aligned} \hspace{\stretch{1}}(1.0.12)

Calculate the canonical partition function, mean energy and specific heat of this system.

This problem can be attempted in two ways, the first of which was how I did it on the midterm, differentiating under the integral sign, leaving the integrals in exact form, but not evaluated explicitly in any way.

That solution was posted previously.

Alternately, by Taylor expanding around c = 0 and d = 0 with those as the variables in the Taylor expansion (as now done in the Pathria 3.29 problem), we can form a solution in short order. Given my low midterm mark, it seems very likely that this was what was expected.

Performing a two variable Taylor expansion of Z, about (c, d) = (0, 0) we have

\begin{aligned}Z \approx\frac{2 \pi m}{\beta}\int dx dye^{- \beta m \omega^2 x^2/2}e^{- \beta m \omega^2 y^2/2}\left( 1 - \beta a x^4 - \beta b y^6  \right)=\frac{2 \pi m}{\beta}\frac{ 2 \pi}{\beta m \omega^2}\left( 1 - \beta a \frac{3!!}{(\beta m \omega^2)^2} - \beta b \frac{5!!}{(\beta m \omega^2)^3}  \right),\end{aligned} \hspace{\stretch{1}}(1.0.22)

or

\begin{aligned}\boxed{Z \approx\frac{(2 \pi/\omega)^2}{\beta^2}\left( 1 - \frac{3 a }{\beta (m \omega^2)^2} - \frac{15 b }{\beta^2 (m \omega^2)^3}  \right).}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Now we can calculate the average energy

\begin{aligned}\left\langle{{H}}\right\rangle = - \frac{\partial {}}{\partial {\beta}}\ln Z= - \frac{\partial {}}{\partial {\beta}}\left( -2 \ln \beta + \ln \left( 1 - \frac{3 a }{\beta (m \omega^2)^2} - \frac{15 b }{\beta^2 (m \omega^2)^3}  \right)  \right)=\frac{2 \beta}-\frac{    \frac{3 a }{\beta^2 (m \omega^2)^2}+    \frac{30 b }{\beta^3 (m \omega^2)^3}}{   1    - \frac{3 a }{\beta (m \omega^2)^2}   - \frac{15 b }{\beta^2 (m \omega^2)^3}}.\end{aligned} \hspace{\stretch{1}}(1.0.24)

Dropping the c, d terms of the denominator above, we have

\begin{aligned}\boxed{\left\langle{{H}}\right\rangle=\frac{2 \beta}-    \frac{3 a }{\beta^2 (m \omega^2)^2}-    \frac{30 b }{\beta^3 (m \omega^2)^3}.}\end{aligned} \hspace{\stretch{1}}(1.0.25)

The heat capacity follows immediately

\begin{aligned}\boxed{C_{\mathrm{V}} = \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}}= 2 - \frac{6 a k_{\mathrm{B}} T}{(m \omega^2)^2} - \frac{90 k_{\mathrm{B}}^2 T^2 b }{(m \omega^2)^3}.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

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Heat capacity of perturbed harmonic oscillator

Posted by peeterjoot on March 11, 2013

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This problem was suggested as prep for the second midterm, but I spent too much time on my problem set. That’s pretty unfortunate since this showed exactly the approach that was expected for the second midterm problem. Not hard, just not obvious in the heat of the moment how to do that Taylor expansion.

Question: Anharmonic oscillator ([1] pr 3.29)

The potential energy of a one-dimensional, anharmonic oscillator may be written as

\begin{aligned}V(q) = c q^2 - g q^3 - f q^4,\end{aligned} \hspace{\stretch{1}}(1.1)

where c, g, and f are positive constant; quite generally, g and f may be assumed to be very small in value.

Show that the leading contribution of anharmonic terms to the heat cpacity of the oscillator, assumed classical, is given by

\begin{aligned}\frac{3}{2} k_{\mathrm{B}}^2 \left( {\frac{f}{c^2} + \frac{5}{4} \frac{g^2}{c^3} } \right) T,\end{aligned} \hspace{\stretch{1}}(1.2)

To the same order, show that the mean value of the position coordinate q is given by

\begin{aligned}\frac{3}{4} \frac{g k_{\mathrm{B}} T}{c^2}.\end{aligned} \hspace{\stretch{1}}(1.3)

Answer

Our partition function is

\begin{aligned}Z = \int dp dq e^{-\beta p^2/2m} e^{ -\beta \left( { c q^2 - g q^3 - f q^4} \right) }= \sqrt{\frac{2 \pi m}{\beta} }\int dq e^{ -\beta \left( { c q^2 - g q^3 - f q^4} \right) }\end{aligned} \hspace{\stretch{1}}(1.4)

How to expand this wasn’t immediately clear to me (as it wasn’t on the midterm either). We can’t Taylor expand in q, because there’s no single position q that is of interest to expand around (we are integrating over all q). What we can do though is Taylor expand about the values f and g, which are assumed to be small. Here’s the two variable Taylor expansion of this perturbed harmonic oscillator exponential. With

\begin{aligned}A(f, g) = e^{ -\beta \left( { c q^2 - g q^3 - f q^4} \right) }\end{aligned} \hspace{\stretch{1}}(1.5)

The expansion to second order is

\begin{aligned}A(f, g) &= A(0, 0) + f {\left. \frac{\partial A}{\partial f} \right\vert}_{f = 0} + g {\left. \frac{\partial A}{\partial g} \right\vert}_{g = 0} + \frac{1}{2} f^2 {\left. \frac{\partial^2 A}{\partial f^2} \right\vert}_{f = 0} + \frac{1}{2} g^2 {\left. \frac{\partial^2 A}{\partial g^2} \right\vert}_{g = 0} + f g {\left. \frac{\partial^2 A}{\partial g \partial f} \right\vert}_{f, g = 0} + \cdots \\ &= e^{ -\beta c q^2 } \left( 1 + g \beta q^3 + f \beta q^4+ \frac{1}{2} g^2 \left( \beta q^3 \right)^2 + \frac{1}{2} f^2 \left( \beta q^4 \right)^2 + f g \left( \beta q^3 \right) \left( \beta q^4 \right) + \cdots \right) \\ &= e^{ -\beta c q^2 } \left( 1 + g \beta q^3 + f \beta q^4+ \frac{1}{2} g^2 \beta^2 q^6+ f g \beta^2 q^7+ \frac{1}{2} f^2 \beta^2 q^8 + \cdots \right) \end{aligned} \hspace{\stretch{1}}(1.6)

This can now be integrated by parts, where any odd powers are killed. For even powers we have

\begin{aligned}\int q^{2 N} e^{-a q^2} dq &= \int q^{2 N - 1} d \frac{ e^{-a q^2} }{-2 a} \\ &= \frac{2 N - 1}{2a} \int q^{2 (N - 1)} e^{-a q^2} dq \\ &= \frac{(2 N - 1)!!}{(2a)^N} \sqrt{\frac{\pi}{a}}.\end{aligned} \hspace{\stretch{1}}(1.7)

This gives us

\begin{aligned}Z &= \sqrt{ \frac{\pi}{\beta c} }\sqrt{\frac{2 \pi m}{\beta} }\left( {1 + f \beta \frac{3!!}{(2 \beta c)^2}+ \frac{1}{{2}} g^2 \beta^2 \frac{5!!}{(2 \beta c)^3}+ \frac{1}{{2}} f^2 \beta^2 \frac{7!!}{(2 \beta c)^4}+ \cdots} \right) \\ &= \frac{\pi}{\beta}\sqrt{ \frac{2 m}{c} }\left( {1 + \frac{3 f}{4 c^2 \beta}+ \frac{15 g^2}{16 c^3 \beta}+ \frac{105 f^2}{32 c^4 \beta}+ \cdots} \right).\end{aligned} \hspace{\stretch{1}}(1.8)

Retaining only the first two terms of the expansion, we have

\begin{aligned}\boxed{Z \approx\frac{\pi}{\beta}\sqrt{ \frac{2 m}{c} }\left( {1 + \frac{3 f}{4 c^2 \beta}+ \frac{15 g^2}{16 c^3 \beta}} \right).}\end{aligned} \hspace{\stretch{1}}(1.9)

Our average energy, in this approximation

\begin{aligned}\left\langle{{H}}\right\rangle &= -\frac{\partial {}}{\partial {\beta}} \ln Z \\ &\approx -\frac{\partial {}}{\partial {\beta}}\left( {- \ln \beta + \ln \left( 1 + \frac{3 f}{4 c^2 \beta}+ \frac{15 g^2}{16 c^3 \beta} \right) } \right) \\ &= \frac{1}{{\beta}} + \frac{1}{{\beta^2}} \frac{\frac{3 f}{4 c^2} + \frac{15 g^2}{16 c^3} }{1 + \frac{1}{{\beta}}\left( {\frac{3 f}{4 c^2} + \frac{15 g^2}{16 c^3} } \right)} \\ &= k_{\mathrm{B}} T + k_{\mathrm{B}}^2 T^2\left( { \frac{3 f}{4 c^2} + \frac{15 g^2}{16 c^3} } \right)\left( {1 - k_{\mathrm{B}} T \left( \frac{3 f}{4 c^2} + \frac{15 g^2}{16 c^3}  \right) + \cdots} \right).\end{aligned} \hspace{\stretch{1}}(1.10)

So to first order in T our specific heat is

\begin{aligned}C_{\mathrm{V}} = \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}}\approx1+ 2 k_{\mathrm{B}} T \left( {\frac{3 f}{4 c^2} + \frac{15 g^2}{16 c^3} } \right)\end{aligned} \hspace{\stretch{1}}(1.11)

or

\begin{aligned}\boxed{C_{\mathrm{V}} = 1+ k_{\mathrm{B}} T \left( {\frac{3 f}{2 c^2} + \frac{15 g^2}{8 c^3} } \right)+ \cdots}\end{aligned} \hspace{\stretch{1}}(1.12)

For the coordinate

\begin{aligned}\left\langle{{q}}\right\rangle &= \sqrt{\frac{2 \pi m}{\beta} }\frac{1}{{Z}}\int q e^{ -\beta \left( { c q^2 - g q^3 - f q^4} \right) } dq \\ &= \sqrt{\frac{2 \pi m}{\beta} }\frac{   \int q    e^{ -\beta c q^2 }   \left( { 1 + g \beta q^3 + f \beta q^4 } \right) dq}{   \frac{\pi}{\beta}   \sqrt{ \frac{2 m}{c} }   \left( { 1 + \frac{3 f}{4 c^2 \beta} + \frac{15 g^2}{16 c^3 \beta} } \right)} \\ &\approx \sqrt{\frac{\not{{2 \pi m}}}{\not{{\beta}}} }g \not{{\beta}} \frac{3!!}{(2 \beta c)^2} \sqrt{\frac{\not{{\pi}}}{\not{{\beta}} \not{{c}}}}\frac{1}{{   \frac{\not{{\pi}}}{\beta}   \sqrt{ \frac{\not{{2 m}}}{\not{{c}}} }}},\end{aligned} \hspace{\stretch{1}}(1.13)

or

\begin{aligned}\boxed{\left\langle{{q}}\right\rangle \approx\frac{3 g k_{\mathrm{B}} T}{4 c^2}.}\end{aligned} \hspace{\stretch{1}}(1.14)

Compare this to the expectation of the coordinate for an unperturbed harmonic oscillator

\begin{aligned}\left\langle{{q}}\right\rangle = \frac{\int q e^{ -\beta c q^2}}{\int e^{-\beta c q^2}} = 0.\end{aligned} \hspace{\stretch{1}}(1.15)

We now have a temperature dependence to the expecation of the coordinate that we didn’t have for the harmonic oscillator.

References

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

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Midterm II reflection

Posted by peeterjoot on March 10, 2013

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Here’s some reflection about this Thursday’s midterm, redoing the problems without the mad scramble. I don’t think my results are too different from what I did in the midterm, even doing them casually now, but I’ll have to see after grading if these solutions are good.

Question: Magnetic field spin level splitting (2013 midterm II p1)

A particle with spin S has 2 S + 1 states -S, -S + 1, \cdots S-1, S. When exposed to a magnetic field, state splitting results in energy E_m = \hbar m B. Calculate the partition function, and use this to find the temperature specific magnetization. A “sum the geometric series” hint was given.

Answer

Our partition function is

\begin{aligned}Z &= \sum_{m = -S}^S e^{-\hbar \beta m B} \\ &= e^{-\hbar \beta S B}\sum_{m = -S}^S e^{-\hbar \beta (m + S) B} \\ &= e^{\hbar \beta S B}\sum_{n = 0}^{2 S} e^{-\hbar \beta n B}.\end{aligned} \hspace{\stretch{1}}(1.0.1)

Writing

\begin{aligned}a = e^{-\hbar \beta B},\end{aligned} \hspace{\stretch{1}}(1.0.2)

that is

\begin{aligned}Z &= a^{-S}\sum_{n = 0}^{2 S} a^n \\ &= a^{-S} \frac{ a^{2 S + 1} - 1 }{a - 1} \\ &= \frac{ a^{S + 1} - a^{-S} }{a - 1} \\ &= \frac{ a^{S + 1/2} - a^{-S - 1/2} }{a^{1/2} - a^{-1/2}}.\end{aligned} \hspace{\stretch{1}}(1.0.3)

Substitution of a gives us

\begin{aligned}\boxed{Z = \frac{ \sinh( \hbar \beta B (S + 1/2) ) }{ \sinh( \hbar \beta B /2 )  }.}\end{aligned} \hspace{\stretch{1}}(1.0.4)

To calculate the magnetization M, I used

\begin{aligned}M = -\left\langle{{H}}\right\rangle/B.\end{aligned} \hspace{\stretch{1}}(1.0.5)

As [1] defines magnetization for a spin system. It was pointed out to me after the test that magnetization was defined differently in class as

\begin{aligned}\mu = \frac{\partial {B}}{\partial {F}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

These are, up to a sign, identical, at least in this case, since we have \beta and B travelling together in the partition function. In terms of the average energy

\begin{aligned}M &= -\frac{\left\langle{{H}}\right\rangle}{B} \\ &= \frac{1}{{B}} \frac{\partial {}}{\partial {\beta}} \ln Z(\beta B) \\ &= \frac{1}{{Z B}} \frac{\partial {}}{\partial {\beta}}Z(\beta B) \\ &= \frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta B)}} Z(\beta B)\end{aligned} \hspace{\stretch{1}}(1.0.7)

Compare this to the in-class definition of magnetization

\begin{aligned}\mu &= \frac{\partial {F}}{\partial {B}} \\ &= \frac{\partial {}}{\partial {B}} \left( - k_{\mathrm{B}} T \ln Z(\beta B)  \right) \\ &= -\frac{\partial {}}{\partial {B}} \frac{\ln Z (\beta B)}{\beta} \\ &= -\frac{1}{{\beta Z}} \frac{\partial {}}{\partial {B}} Z(\beta B) \\ &= -\frac{1}{{Z}} \frac{\partial {}}{\partial {(\beta B)}} Z(\beta B).\end{aligned} \hspace{\stretch{1}}(1.0.8)

For this derivative we have

\begin{aligned}\frac{\partial {}}{\partial {(\beta B)}} \ln Z \\ &= \frac{\partial {}}{\partial {(\beta B)}} \ln \frac{ \sinh( \hbar \beta B (S + 1/2) ) }{ \sinh( \hbar \beta B /2 )  } \\ &= \frac{\partial {}}{\partial {(\beta B)}} \left( \ln \sinh( \hbar \beta B (S + 1/2) ) - \ln \sinh( \hbar \beta B /2 )  \right) \\ &= \frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 )  \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

This gives us

\begin{aligned}\mu &= -\frac{1}{{Z}} \frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 )  \right) \\ &= -\frac{ \sinh( \hbar \beta B /2 )  }{ \sinh( \hbar \beta B (S + 1/2) ) }\frac{\hbar }{2}\left( (2 S + 1) \coth( \hbar \beta B (S + 1/2) ) - \coth( \hbar \beta B /2 )  \right)\end{aligned} \hspace{\stretch{1}}(1.0.10)

After some simplification (done offline in \nbref{midtermTwoQ1FinalSimplificationMu.nb}) we get

\begin{aligned}\boxed{\mu = \hbar \frac{(s+1) \sinh(\hbar \beta B s)-s \sinh(\hbar \beta B (s+1))}{\cosh(\hbar \beta B(2 s+1)) - 1}.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

I got something like this on the midterm, but recall doing it somehow much differently.

Question: Pertubation of classical harmonic oscillator (2013 midterm II p2)

Consider a single particle perturbation of a classical simple harmonic oscillator Hamiltonian

\begin{aligned}H = \frac{1}{{2}} m \omega^2 \left( x^2 + y^2 \right) + \frac{1}{{2 m}} \left( p_x^2 + p_y^2 \right) + a x^4 + b y^6\end{aligned} \hspace{\stretch{1}}(1.0.12)

Calculate the canonical partition function, mean energy and specific heat of this system.

There were some instructions about the form to put the integrals in.

Answer

The canonical partition function is

\begin{aligned}Z &= \int dx dy dp_x dp_y e^{-\beta H} \\ &= \int dx e^{-\beta \left( \frac{1}{{2}} m \omega^2 x^2 + a x^4  \right)}\int dy e^{-\beta \left( \frac{1}{{2}} m \omega^2 y^2 + b y^6  \right)}\int dp_x dp_y e^{-\beta p_x^2/2 m} e^{-\beta p_y^2/2 m}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

With

\begin{aligned}u = \sqrt{\frac{\beta}{2m}} p_x\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}v = \sqrt{\frac{\beta}{2m}} p_y,\end{aligned} \hspace{\stretch{1}}(1.0.14b)

the momentum integrals are

\begin{aligned}\int dp_x dp_y e^{-\beta p_x^2/2 m} e^{-\beta p_y^2/2 m} \\ &= \frac{2m}{\beta}\int du du e^{- u^2 - v^2} \\ &= \frac{m}{\beta}2 \pi\int 2 r dr e^{- r^2} \\ &= \frac{2 \pi m}{\beta}.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Writing

\begin{aligned}f(x) = \frac{1}{{2}} m \omega^2 x^2 + a x^4\end{aligned} \hspace{\stretch{1}}(1.0.16a)

\begin{aligned}g(x) = \frac{1}{{2}} m \omega^2 y^2 + b y^4,\end{aligned} \hspace{\stretch{1}}(1.0.16b)

we have

\begin{aligned}\boxed{Z = \frac{2 \pi m}{\beta}\int dx e^{- \beta f(x)}\int dy e^{- \beta g(y)}.}\end{aligned} \hspace{\stretch{1}}(1.0.17)

The mean energy is

\begin{aligned}\left\langle{{H}}\right\rangle &= \frac{\int H e^{-\beta H}}{\int e^{-\beta H}} \\ &= -\frac{\partial {}}{\partial {\beta}} \ln \int e^{-\beta H} \\ &= \frac{\partial {}}{\partial {\beta}} \left( \ln \beta -\ln \int dx e^{- \beta f(x)} -\ln \int dy e^{- \beta g(y)}  \right) \\ &= \frac{1}{{\beta}} + \frac{\int dx f(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}+ \frac{\int dy g(y) e^{- \beta g(y)}}{\int dy e^{- \beta g(y)}}.\end{aligned} \hspace{\stretch{1}}(1.0.18)

The specific heat follows by differentiating once more

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {T}} \\ &= \frac{\partial {\beta}}{\partial {T}}\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= -\frac{1}{{k_{\mathrm{B}} T^2}}\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= -k_{\mathrm{B}} \beta^2\frac{\partial {\left\langle{{H}}\right\rangle}}{\partial {\beta}} \\ &= - k_{\mathrm{B}} \beta^2\left( -\frac{1}{{\beta^2}} + \frac{\partial {}}{\partial {\beta}} \left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} } + \frac{ \int dy g(y) e^{- \beta g(y)} } { \int dy e^{- \beta g(y)} }  \right)  \right).\end{aligned} \hspace{\stretch{1}}(1.0.19)

Differentiating the integral terms we have, for example,

\begin{aligned}\frac{\partial {}}{\partial {\beta}} \frac{\int dx f(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}=-\frac{\int dx f^2(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}-\left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} }  \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.20)

so that the specific heat is

\begin{aligned}\boxed{C_{\mathrm{V}} =k_{\mathrm{B}} \left(1 + \frac{\int dx f^2(x) e^{- \beta f(x)}}{\int dx e^{- \beta f(x)}}+\left( \frac{ \int dx f(x) e^{- \beta f(x)} } { \int dx e^{- \beta f(x)} }  \right)^2+ \frac{\int dy g^2(y) e^{- \beta g(y)}}{\int dy e^{- \beta g(y)}}+\left( \frac{ \int dy g(y) e^{- \beta g(y)} } { \int dy e^{- \beta g(y)} }  \right)^2\right).}\end{aligned} \hspace{\stretch{1}}(1.0.21)

That’s as far as I took this problem. There was a discussion after the midterm with Eric about Taylor expansion of these integrals. That’s not something that I tried.

References

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

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , | 1 Comment »

PHY452H1S Basic Statistical Mechanics. Problem Set 5: Temperature

Posted by peeterjoot on March 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.)]

Disclaimer

This is an ungraded set of answers to the problems posed.

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”.)

Answer

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.

Answer

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.

Answer

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)

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