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PHY452H1S Basic Statistical Mechanics. Lecture 12: Helmholtz free energy. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 28, 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

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

Canonical partition

We found

\begin{subequations}

\begin{aligned}\frac{\sigma_{\mathrm{E}}}{E} \propto \frac{T \sqrt{C_V}}{E} k_{\mathrm{B}}^2\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}Z = \sum_{\{c\}} e^{-\beta E(c)}\end{aligned} \hspace{\stretch{1}}(1.0.1b)

\begin{aligned}C_V \sim N\end{aligned} \hspace{\stretch{1}}(1.0.1c)

\begin{aligned}E \sim N\end{aligned} \hspace{\stretch{1}}(1.0.1d)

\end{subequations}

where the partition function \index{partition function} acts as a probability distribution so that we can define an average as

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{\sum_{\{c\}} A(c) e^{-\beta E(c)}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.2)

If we suppose that the energy is typically close to the average energy as in fig. 1.1.

Fig 1.1: Peaked energy distribution

, then we can approximate the partition function as

\begin{aligned}Z \approx e^{-\beta \left\langle{{E}}\right\rangle} \sum_{\{c\}} \delta_{E, \bar{E}}= e^{-\beta \left\langle{{E}}\right\rangle} e^S/k_{\mathrm{B}},\end{aligned} \hspace{\stretch{1}}(1.0.4)

where we’ve used S = k_{\mathrm{B}} \ln \Omega to express the number of states where the energy matches the average energy \Omega = \sum \delta_{E, \bar{E}}.

This gives us

\begin{aligned}Z = e^{-\beta (\left\langle{{E}}\right\rangle - k_{\mathrm{B}} T S/k_{\mathrm{B}}) } = e^{-\beta (\left\langle{{E}}\right\rangle - T S) } \end{aligned} \hspace{\stretch{1}}(1.0.4)

or

\begin{aligned}\boxed{Z = e^{-\beta F},}\end{aligned} \hspace{\stretch{1}}(1.0.5)

where we define the Helmholtz free energy F as

\begin{aligned}\boxed{F = \left\langle{{E}}\right\rangle - T S.}\end{aligned} \hspace{\stretch{1}}(1.0.6)

Equivalently, the log of the partition function provides us with the partition function

\begin{aligned}F = - k_{\mathrm{B}} T \ln Z.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Recalling our expression for the average energy, we can now write that in terms of the free energy

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{\sum_{\{c\}} E(c) e^{-\beta E(c)}}{\sum_{\{c\}} e^{-\beta E(c)}}= -\frac{\partial {}}{\partial {\beta}}\ln Z=\frac{\partial {(\beta F)}}{\partial {\beta}}\end{aligned} \hspace{\stretch{1}}(1.0.8)

Quantum mechanical picture

Consider a subsystem as in fig. 1.2 where we have states of the form

Fig 1.2: subsystem in heat bath

\begin{aligned}{\left\lvert {\Psi_{\text{full}}} \right\rangle} = {\left\lvert {\chi_{\text{subsystem}}} \right\rangle} {\left\lvert {\phi_{\text{bath}}} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.9)

and a total Hamiltonian operator of the form

\begin{aligned}H_{\text{full}} = H_{\text{subsystem}} + H_{\text{bath}} (+ H_{\text{coupling}})\end{aligned} \hspace{\stretch{1}}(1.0.10)

where the total energy of the state, given energy eigenvalues \mathcal{E}_n and \lambda_n for the states {\left\lvert {\chi_{\text{subsystem}}} \right\rangle} and {\left\lvert {\phi_{\text{bath}}} \right\rangle} respectively, is given by the sum

\begin{aligned}E = \mathcal{E}_m + \lambda_n.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Here \mathcal{E}_m, \lambda_n are many body energies, so that \delta E \sim \#e^{-\#N}.

We can now write the total number of states as

\begin{aligned}\Omega(E) &= \underbrace{\sum_m}_{\text{subsystem}}\underbrace{\sum_n}_{\text{bath}}\delta(E - \mathcal{E}_m -\lambda_n)\\ &= \sum_m e^{\frac{1}{{k_{\mathrm{B}}}} S(E - \mathcal{E}_m)} \\ &\approx \sum_m e^{\frac{1}{{k_{\mathrm{B}}}} S(E)}e^{\beta \mathcal{E}_m}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}Z = \sum_m e^{-\beta \mathcal{E}_m} = \text{Tr} \left( e^{-\beta \hat{H}_{\text{subsystem}}} \right)\end{aligned} \hspace{\stretch{1}}(1.0.13)

We’ve ignored the coupling term in eq. 1.0.10. This is actually a problem in quantum mechanics since we require this coupling to introduce state changes.

Example: Spins

Given N spin 1/2 objects \uparrow, \downarrow, satisfying

\begin{aligned}S_z = \pm \frac{1}{{2}} \hbar\end{aligned} \hspace{\stretch{1}}(1.0.14)

Dropping \hbar we have

\begin{aligned}S_z \rightarrow \pm \frac{1}{{2}} \sigma\end{aligned} \hspace{\stretch{1}}(1.0.15)

Our system has a state {\left\lvert {\sigma_1, \sigma_2, \cdots \sigma_N} \right\rangle} where \sigma_i = \pm 1. The total number of states is 2^N.

Our Hamiltonian is

\begin{aligned}\hat{H} = - B \sum_i \hat{S}_{z_i}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

This is the associated with the Zeeman effect, where states can be split by a magnetic field, as in fig. 1.3.

Fig 1.3: Zeeman splitting

Our minimum and maximum energies are

\begin{subequations}

\begin{aligned}E_{\mathrm{min}} = -\frac{B}{2} N\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}E_{\mathrm{max}} = -\frac{B}{2} N\end{aligned} \hspace{\stretch{1}}(1.0.17b)

\end{subequations}

The total energy difference is

\begin{aligned}\Delta E = B N,\end{aligned} \hspace{\stretch{1}}(1.0.23)

and the energy differences are

\begin{aligned}\delta E \sim \frac{B N}{2^N} \sim \# e^{-\# N}.\end{aligned} \hspace{\stretch{1}}(1.0.23)

This is a measure of the average energy difference between two adjacent energy levels. In a real system we cannot assume that we have non-interacting spins. Any weak interaction will split our degenerate energy levels as in fig. 1.4.

Fig 1.4: Interaction splitting

We can now express the partition function

\begin{aligned}Z &= \sum_{\{\sigma\}} e^{-\beta \left( -\frac{B}{2} \sum_i \sigma_i \right)} \\ &= \left( \sum_{\{\sigma_1\}} \exp \left( -\frac{\beta B}{2} \sigma_i \right) \right)\left( \sum_{\{\sigma_2\}} \exp \left( -\frac{\beta B}{2} \sigma_i \right) \right)\cdots \\ &= \left( \exp \left( -\frac{\beta B}{2} (1) \right) + \exp \left( -\frac{\beta B}{2} (-1) \right) \right)^N \\ &= \left( 2 \cosh\left( \frac{B}{2 k_B T} \right) \right)^N\end{aligned} \hspace{\stretch{1}}(1.0.23)

Our free energy is

\begin{aligned}F = - k_B T N \ln \left( 2 \cosh \left( \frac{B}{2 k_B T} \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

For the expected value of the spin we find

\begin{aligned}\left\langle{{S_z}}\right\rangle = \sum_i \left\langle{{S_{z_i}}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.0.23)

\begin{aligned}\left\langle{{S_{z_i}}}\right\rangle=\frac{1}{{2}} \frac{\sum_\sigma \sigma e^{\beta B \sigma/2}}{\sum_\sigma e^{\beta B \sigma/2}}= \frac{1}{{2}} \tanh \left( \frac{B}{2 k_B T} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

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