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# Posts Tagged ‘binomial distribution’

## Final version of my phy452.pdf notes posted

Posted by peeterjoot on September 5, 2013

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

September 05, 2013 Large volume fermi gas density

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 24, 2013 Low temperature Fermi gas chemical potential

## Summary of statistical mechanics relations and helpful formulas (cheat sheet fodder)

Posted by peeterjoot on April 29, 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.)]

Central limit theorem

If $\left\langle{{x}}\right\rangle = \mu$ and $\sigma^2 = \left\langle{{x^2}}\right\rangle - \left\langle{{x}}\right\rangle^2$, and $X = \sum x$, then in the limit

\begin{aligned}\lim_{N \rightarrow \infty} P(X)= \frac{1}{{\sigma \sqrt{2 \pi N}}} \exp\left( - \frac{ (x - N \mu)^2}{2 N \sigma^2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}\left\langle{{X}}\right\rangle = N \mu\end{aligned} \hspace{\stretch{1}}(1.0.1b)

\begin{aligned}\left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = N \sigma^2\end{aligned} \hspace{\stretch{1}}(1.0.1c)

Binomial distribution

\begin{aligned}P_N(X) = \left\{\begin{array}{l l}\left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-X}{2}\right)!\left(\frac{N+X}{2}\right)!}& \quad \mbox{if X and N have same parity} \\ 0 & \quad \mbox{otherwise} \end{array},\right.\end{aligned} \hspace{\stretch{1}}(1.0.2)

where $X$ was something like number of Heads minus number of Tails.

Generating function

Given the Fourier transform of a probability distribution $\tilde{P}(k)$ we have

\begin{aligned}{\left.{{ \frac{\partial^n}{\partial k^n} \tilde{P}(k) }}\right\vert}_{{k = 0}}= (-i)^n \left\langle{{x^n}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.0.2)

Handy mathematics

\begin{aligned}\ln( 1 + x ) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\end{aligned} \hspace{\stretch{1}}(1.0.2)

\begin{aligned}N! \approx \sqrt{ 2 \pi N} N^N e^{-N}\end{aligned} \hspace{\stretch{1}}(1.0.5)

\begin{aligned}\ln N! \approx \frac{1}{{2}} \ln 2 \pi -N + \left( N + \frac{1}{{2}} \right)\ln N \approx N \ln N - N\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt\end{aligned} \hspace{\stretch{1}}(1.0.7)

\begin{aligned}\Gamma(\alpha) = \int_0^\infty dy e^{-y} y^{\alpha - 1}\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}\Gamma(\alpha + 1) = \alpha \Gamma(\alpha)\end{aligned} \hspace{\stretch{1}}(1.0.9)

\begin{aligned}\Gamma\left( 1/2 \right) = \sqrt{\pi}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\zeta(s) = \sum_{k=1}^{\infty} k^{-s}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\begin{aligned}\zeta(3/2) &\approx 2.61238 \\ \zeta(2) &\approx 1.64493 \\ \zeta(5/2) &\approx 1.34149 \\ \zeta(3) &\approx 1.20206\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}P(x, t) = \int_{-\infty}^\infty \frac{dk}{2 \pi} \tilde{P}(k, t) \exp\left( i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{P}(k, t) = \int_{-\infty}^\infty dx P(x, t) \exp\left( -i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.14b)

Heavyside theta

\begin{aligned}\Theta(x) = \left\{\begin{array}{l l}1 & \quad x \ge 0 \\ 0 & \quad x < 0\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}\frac{d\Theta}{dx} = \delta(x)\end{aligned} \hspace{\stretch{1}}(1.0.15b)

\begin{aligned}\sum_{m = -l}^l a^m=\frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}}\end{aligned} \hspace{\stretch{1}}(1.0.16.16)

\begin{aligned}\sum_{m = -l}^l e^{b m}=\frac{\sinh(b(l + 1/2))}{\sinh(b/2)}\end{aligned} \hspace{\stretch{1}}(1.0.16b)

\begin{aligned}\int_{-\infty}^\infty q^{2 N} e^{-a q^2} dq=\frac{(2 N - 1)!!}{(2a)^N} \sqrt{\frac{\pi}{a}}\end{aligned} \hspace{\stretch{1}}(1.0.17.17)

\begin{aligned}\int_{-\infty}^\infty e^{-a q^2} dq=\sqrt{\frac{\pi}{a}}\end{aligned} \hspace{\stretch{1}}(1.0.17.17)

\begin{aligned}\binom{-\left\lvert {m} \right\rvert}{k} = (-1)^k \frac{\left\lvert {m} \right\rvert}{\left\lvert {m} \right\rvert + k} \binom{\left\lvert {m} \right\rvert+k}{\left\lvert {m} \right\rvert}\end{aligned} \hspace{\stretch{1}}(1.0.18)

\begin{aligned}\int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1} =\frac{\pi ^4}{15 \beta ^4},\end{aligned} \hspace{\stretch{1}}(1.0.18)

volume in mD

\begin{aligned}V_m= \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}\end{aligned} \hspace{\stretch{1}}(1.0.20)

area of ellipse

\begin{aligned}A = \pi a b\end{aligned} \hspace{\stretch{1}}(1.0.21)

Radius of gyration of a 3D polymer

With radius $a$, we have

\begin{aligned}r_N \approx a \sqrt{N}\end{aligned} \hspace{\stretch{1}}(1.0.21)

Velocity random walk

Find

\begin{aligned}\mathcal{P}_{N_{\mathrm{c}}}(\mathbf{v}) \propto e^{-\frac{(\mathbf{v} - \mathbf{v}_0)^2}{2 N_{\mathrm{c}}}}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Random walk

1D Random walk

\begin{aligned}\mathcal{P}( x, t ) = \frac{1}{{2}} \mathcal{P}(x + \delta x, t - \delta t)+\frac{1}{{2}} \mathcal{P}(x - \delta x, t - \delta t)\end{aligned} \hspace{\stretch{1}}(1.0.23)

\begin{aligned}\frac{\partial {\mathcal{P}}}{\partial {t}}(x, t) =\frac{1}{{2}} \frac{(\delta x)^2}{\delta t}\frac{\partial^2 {{\mathcal{P}}}}{\partial {{x}}^2}(x, t) = D \frac{\partial^2 {{\mathcal{P}}}}{\partial {{x}}^2}(x, t) = -\frac{\partial {J}}{\partial {x}},\end{aligned} \hspace{\stretch{1}}(1.0.25)

The diffusion constant relation to the probability current is referred to as Fick’s law

\begin{aligned}D = -\frac{\partial {J}}{\partial {x}}\end{aligned} \hspace{\stretch{1}}(1.0.25)

with which we can cast the probability diffusion identity into a continuity equation form

\begin{aligned}\frac{\partial {\mathcal{P}}}{\partial {t}} + \frac{\partial {J}}{\partial {x}} = 0 \end{aligned} \hspace{\stretch{1}}(1.0.25)

In 3D (with the Maxwell distribution frictional term), this takes the form

\begin{aligned}\mathbf{j} = -D \boldsymbol{\nabla}_\mathbf{v} c(\mathbf{v}, t) - \eta \mathbf{v} c(\mathbf{v}, t)\end{aligned} \hspace{\stretch{1}}(1.0.28a)

\begin{aligned}\frac{\partial {}}{\partial {t}} c(\mathbf{v}, t) + \boldsymbol{\nabla}_\mathbf{v} \cdot \mathbf{j}(\mathbf{v}, t) = 0\end{aligned} \hspace{\stretch{1}}(1.0.28b)

Maxwell distribution

Add a frictional term to the velocity space diffusion current

\begin{aligned}j_v = -D \frac{\partial {c}}{\partial {v}}(v, t) - \eta v c(v).\end{aligned} \hspace{\stretch{1}}(1.0.29)

For steady state the continity equation $0 = \frac{dc}{dt} = -\frac{\partial {j_v}}{\partial {v}}$ leads to

\begin{aligned}c(v) \propto \exp\left(- \frac{\eta v^2}{2 D}\right).\end{aligned} \hspace{\stretch{1}}(1.0.30)

We also find

\begin{aligned}\left\langle{{v^2}}\right\rangle = \frac{D}{\eta},\end{aligned} \hspace{\stretch{1}}(1.0.30)

and identify

\begin{aligned}\frac{1}{{2}} m \left\langle{{\mathbf{v}^2}}\right\rangle = \frac{1}{{2}} m \left( \frac{D}{\eta} \right) = \frac{1}{{2}} k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.32)

Hamilton’s equations

\begin{aligned}\frac{\partial {H}}{\partial {p}} = \dot{x}\end{aligned} \hspace{\stretch{1}}(1.0.33a)

\begin{aligned}\frac{\partial {H}}{\partial {x}} = -\dot{p}\end{aligned} \hspace{\stretch{1}}(1.0.33b)

SHO

\begin{aligned}H = \frac{p^2}{2m} + \frac{1}{{2}} k x^2\end{aligned} \hspace{\stretch{1}}(1.0.34a)

\begin{aligned}\omega^2 = \frac{k}{m}\end{aligned} \hspace{\stretch{1}}(1.0.34b)

Quantum energy eigenvalues

\begin{aligned}E_n = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.35)

Liouville’s theorem

\begin{aligned}\frac{d{{\rho}}}{dt} = \frac{\partial {\rho}}{\partial {t}} + \dot{x} \frac{\partial {\rho}}{\partial {x}} + \dot{p} \frac{\partial {\rho}}{\partial {p}}= \cdots = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla}_{x,p} \cdot (\rho \dot{x}, \rho \dot{p})= \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{J}= 0,\end{aligned} \hspace{\stretch{1}}(1.0.35)

Regardless of whether we have a steady state system, if we sit on a region of phase space volume, the probability density in that neighbourhood will be constant.

Ergodic

A system for which all accessible phase space is swept out by the trajectories. This and Liouville’s threorm allows us to assume that we can treat any given small phase space volume as if it is equally probable to the same time evolved phase space region, and switch to ensemble averaging instead of time averaging.

Thermodynamics

\begin{aligned}dE = T dS - P dV + \mu dN\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}\frac{1}{{T}} = \left({\partial {S}}/{\partial {E}}\right)_{{N,V}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}\frac{P}{T} = \left({\partial {S}}/{\partial {V}}\right)_{{N,E}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}-\frac{\mu}{T} = \left({\partial {S}}/{\partial {N}}\right)_{{V,E}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}P = - \left({\partial {E}}/{\partial {V}}\right)_{{N,S}}= - \left({\partial {F}}/{\partial {V}}\right)_{{N,T}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}\mu = \left({\partial {E}}/{\partial {N}}\right)_{{V,S}} = \left({\partial {F}}/{\partial {N}}\right)_{{V,T}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}T = \left({\partial {E}}/{\partial {S}}\right)_{{N,V}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}F = E - TS\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}G = F + P V = E - T S + P V = \mu N\end{aligned} \hspace{\stretch{1}}(1.0.37i)

\begin{aligned}H = E + P V = G + T S\end{aligned} \hspace{\stretch{1}}(1.0.37j)

\begin{aligned}C_{\mathrm{V}} = T \left({\partial {S}}/{\partial {T}}\right)_{{N,V}} = \left({\partial {E}}/{\partial {T}}\right)_{{N,V}} = - T \left( \frac{\partial^2 {{F}}}{\partial {{T}}^2} \right)_{N,V}\end{aligned} \hspace{\stretch{1}}(1.0.37k)

\begin{aligned}C_{\mathrm{P}} = T \left({\partial {S}}/{\partial {T}}\right)_{{N,P}} = \left({\partial {H}}/{\partial {T}}\right)_{{N,P}}\end{aligned} \hspace{\stretch{1}}(1.0.37l)

\begin{aligned}\underbrace{dE}_{\text{Change in energy}}=\underbrace{d W}_{\text{work done on the system}}+\underbrace{d Q}_{\text{Heat supplied to the system}}\end{aligned} \hspace{\stretch{1}}(1.0.38)

Example (work on gas): $d W = -P dV$. Adiabatic: $d Q = 0$. Cyclic: $dE = 0$.

Microstates

\begin{aligned}\beta = \frac{1}{k_{\mathrm{B}} T}\end{aligned} \hspace{\stretch{1}}(1.0.38)

\begin{aligned}S = k_{\mathrm{B}} \ln \Omega \end{aligned} \hspace{\stretch{1}}(1.0.40)

\begin{aligned}\Omega(N, V, E) = \frac{1}{h^{3N} N!} \int_V d\mathbf{x}_1 \cdots d\mathbf{x}_N \int d\mathbf{p}_1 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2 m} \cdots - \frac{\mathbf{p}_N^2}{2 m}\right)=\frac{V^N}{h^{3N} N!}\int d\mathbf{p}_1 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m} \cdots - \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.40)

\begin{aligned}\Omega = \frac{d\gamma}{dE}\end{aligned} \hspace{\stretch{1}}(1.0.42)

\begin{aligned}\gamma=\frac{V^N}{h^{3N} N!}\int d\mathbf{p}_1 \cdots d\mathbf{p}_N \Theta \left(E - \frac{\mathbf{p}_1^2}{2m} \cdots - \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.43)

quantum

\begin{aligned}\gamma = \sum_i \Theta(E - \epsilon_i)\end{aligned} \hspace{\stretch{1}}(1.0.44)

Ideal gas

\begin{aligned}\Omega = \frac{V^N}{N!} \frac{1}{{h^{3N}}} \frac{( 2 \pi m E)^{3 N/2 }}{E} \frac{1}{\Gamma( 3N/2 ) }\end{aligned} \hspace{\stretch{1}}(1.0.45)

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left(N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.46)

Quantum free particle in a box

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.0.47a)

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.47b)

\begin{aligned}\epsilon_k = \frac{\hbar^2 k^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.0.47b)

Spin

magnetization

\begin{aligned}\mu = \frac{\partial {F}}{\partial {B}}\end{aligned} \hspace{\stretch{1}}(1.0.48)

moment per particle

\begin{aligned}m = \mu/N\end{aligned} \hspace{\stretch{1}}(1.0.49)

spin matrices

\begin{aligned}\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50a)

\begin{aligned}\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50b)

\begin{aligned}\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50c)

$l \ge 0, -l \le m \le l$

\begin{aligned}\mathbf{L}^2 {\left\lvert {lm} \right\rangle} = l(l+1)\hbar^2 {\left\lvert {lm} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.51a)

\begin{aligned}L_z {\left\lvert {l m} \right\rangle} = \hbar m {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.51b)

\begin{aligned}S(S + 1) \hbar^2\end{aligned} \hspace{\stretch{1}}(1.0.51b)

Canonical ensemble

classical

\begin{aligned}\Omega(N, E) = \frac{ V }{ h^3 N} \int d\mathbf{p}_1 e^{\frac{S}{k_{\mathrm{B}}}(N, E)}e^{-\frac{1}{{k_{\mathrm{B}}}} \left( \frac{\partial {S}}{\partial {N}} \right)_{E, V} }e^{-\frac{\mathbf{p}_1^2}{2m k_{\mathrm{B}}}\left( \frac{\partial {S}}{\partial {E}} \right)_{N, V}}\end{aligned} \hspace{\stretch{1}}(1.0.53)

quantum

\begin{aligned}\Omega(E) \approx\sum_{m \in \text{subsystem}} e^{\frac{1}{{k_{\mathrm{B}}}} S(E)}e^{-\beta \mathcal{E}_m}\end{aligned} \hspace{\stretch{1}}(1.0.54.54)

\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}}(1.0.54b)

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{\int He^{- \beta H }}{\int e^{- \beta H }}\end{aligned} \hspace{\stretch{1}}(1.0.55a)

\begin{aligned}\left\langle{{E^2}}\right\rangle = \frac{\int H^2e^{- \beta H }}{\int e^{- \beta H }}\end{aligned} \hspace{\stretch{1}}(1.0.55b)

\begin{aligned}Z \equiv \frac{1}{{h^{3N} N!}}\int e^{- \beta H }\end{aligned} \hspace{\stretch{1}}(1.0.55c)

\begin{aligned}\left\langle{{E}}\right\rangle = -\frac{1}{{Z}} \frac{\partial {Z}}{\partial {\beta}} = - \frac{\partial {\ln Z}}{\partial {\beta}} =\frac{\partial {(\beta F)}}{\partial {\beta}}\end{aligned} \hspace{\stretch{1}}(1.0.55d)

\begin{aligned}\sigma_{\mathrm{E}}^2= \left\langle{{E^2}}\right\rangle - \left\langle{{E}}\right\rangle^2 =\frac{\partial^2 {{\ln Z}}}{\partial {{\beta}}^2} = k_{\mathrm{B}} T^2 \frac{\partial {\left\langle{{E}}\right\rangle}}{\partial {T}}= k_{\mathrm{B}} T^2 C_{\mathrm{V}} \propto N\end{aligned} \hspace{\stretch{1}}(1.0.55e)

\begin{aligned}Z = e^{-\beta (\left\langle{{E}}\right\rangle - T S) } = e^{-\beta F}\end{aligned} \hspace{\stretch{1}}(1.0.55f)

\begin{aligned}F = \left\langle{{E}}\right\rangle - T S = -k_{\mathrm{B}} T \ln Z\end{aligned} \hspace{\stretch{1}}(1.0.55g)

Grand Canonical ensemble

\begin{aligned}S = - k_{\mathrm{B}} \sum_{r,s} P_{r,s} \ln P_{r,s}\end{aligned} \hspace{\stretch{1}}(1.0.56)

\begin{aligned}P_{r, s} = \frac{e^{-\alpha N_r - \beta E_s}}{Z_{\mathrm{G}}}\end{aligned} \hspace{\stretch{1}}(1.0.57a)

\begin{aligned}Z_{\mathrm{G}} = \sum_{r,s} e^{-\alpha N_r - \beta E_s} = \sum_{r,s} z^{N_r} e^{-\beta E_s} = \sum_{N_r} z^{N_r} Z_{N_r}\end{aligned} \hspace{\stretch{1}}(1.0.57b)

\begin{aligned}z = e^{-\alpha} = e^{\mu \beta}\end{aligned} \hspace{\stretch{1}}(1.0.57c)

\begin{aligned}q = \ln Z_{\mathrm{G}} = P V \beta\end{aligned} \hspace{\stretch{1}}(1.0.57d)

\begin{aligned}\left\langle{{H}}\right\rangle = -\left({\partial {q}}/{\partial {\beta}}\right)_{{z,V}} = k_{\mathrm{B}} T^2 \left({\partial {q}}/{\partial {\mu}}\right)_{{z,V}} = \sum_\epsilon \frac{\epsilon}{z^{-1} e^{\beta \epsilon} \pm 1}\end{aligned} \hspace{\stretch{1}}(1.0.57e)

\begin{aligned}\left\langle{{N}}\right\rangle = z \left({\partial {q}}/{\partial {z}}\right)_{{V,T}} = \sum_\epsilon \frac{1}{{z^{-1} e^{\beta\epsilon} \pm 1}}\end{aligned} \hspace{\stretch{1}}(1.0.57f)

\begin{aligned}F = - k_{\mathrm{B}} T \ln \frac{ Z_{\mathrm{G}} }{z^N}\end{aligned} \hspace{\stretch{1}}(1.0.57g)

\begin{aligned}\left\langle{{n_\epsilon}}\right\rangle = -\frac{1}{{\beta}} \left({\partial {q}}/{\partial {\epsilon}}\right)_{{z, T, \text{other} \epsilon}} = \frac{1}{{z^{-1} e^{\beta \epsilon} \pm 1}}\end{aligned} \hspace{\stretch{1}}(1.0.57h)

\begin{aligned}\text{var}(N) = \frac{1}{{\beta}} \left({\partial {\left\langle{{N}}\right\rangle}}/{\partial {\mu}}\right)_{{V, T}} = - \frac{1}{{\beta}} \left({\partial {\left\langle{{n_\epsilon}}\right\rangle}}/{\partial {\epsilon}}\right)_{{z,T}} = z^{-1} e^{\beta \epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.57h)

\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}}(1.0.59.59)

\begin{aligned}Z_{\mathrm{G}}= \sum_{N=0}^\infty e^{\beta \mu N}\sum_{n_k, \sum n_m = N} e^{-\beta \sum_m n_m \epsilon_m}=\prod_{k} \left( \sum_{n_k} e^{-\beta(\epsilon_k - \mu) n_k} \right)\end{aligned} \hspace{\stretch{1}}(1.0.59b)

\begin{aligned}Z_{\mathrm{G}}^{\mathrm{QM}} = {\text{Tr}}_{\{\text{energy}, N\}} \left( e^{ -\beta (\hat{H} - \mu \hat{N} ) } \right)\end{aligned} \hspace{\stretch{1}}(1.0.59b)

\begin{aligned}P V = \frac{2}{3} U\end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}f_\nu^\pm(z) = \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{x^{\nu - 1}}{z^{-1} e^x \pm 1}\end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}f_\nu^\pm(z \approx 0) =z\mp\frac{z^{2}}{2^\nu}+\frac{z^{3}}{3^\nu}\mp\frac{z^{4}}{4^\nu}+ \cdots \end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}z \frac{d f_\nu^{\pm}(z) }{dz} = f_{\nu-1}^{\pm}(z)\end{aligned} \hspace{\stretch{1}}(1.0.61)

\begin{aligned}\frac{d f_{3/2}^{\pm}(z) }{dT} = -\frac{3}{2T} f_{3/2}^{\pm}(z)f_{\nu-1}^{\pm}(z)\end{aligned} \hspace{\stretch{1}}(1.0.62)

Fermions

\begin{aligned}\sum_{n_k = 0}^1 e^{-\beta(\epsilon_k - \mu) n_k}=1 + e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.0.62)

\begin{aligned}N = (2 S + 1) V \int_0^{k_{\mathrm{F}}} \frac{4 \pi k^2 dk}{(2 \pi)^3}\end{aligned} \hspace{\stretch{1}}(1.0.64)

\begin{aligned}k_{\mathrm{F}} = \left( \frac{ 6 \pi^2 \rho }{2 S + 1} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{6 \pi \rho}{2 S + 1} \right)^{2/3}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}} + \cdots \end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\lambda \equiv \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\frac{N}{V}=\frac{g}{\lambda^3} f_{3/2}(z)=\frac{g}{\lambda^3} \left( e^{\beta \mu} - \frac{e^{2 \beta \mu}}{2^{3/2}} + \cdots \right) \end{aligned} \hspace{\stretch{1}}(1.0.68)

(so $n = \frac{g}{\lambda^3} e^{\beta \mu}$ for large temperatures)

\begin{aligned}P \beta = \frac{g}{\lambda^3} f_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}U= \frac{3}{2} N k_{\mathrm{B}} T \frac{f_{5/2}(z)}{f_{3/2}(z) }.\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}f_\nu^+(e^y) \approx\frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \nu \sum_{j = 1, 3, 5, \cdots } (\nu-1) \cdots (\nu - j) \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} } \right)\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}\frac{C}{N} = \frac{\pi^2}{2} k_{\mathrm{B}} \frac{ k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}}}\end{aligned} \hspace{\stretch{1}}(1.0.71.71)

\begin{aligned}A = N k_{\mathrm{B}} T \left( \ln z - \frac{f_{5/2}(z)}{f_{3/2}(z)} \right)\end{aligned} \hspace{\stretch{1}}(1.0.71.71)

Bosons

\begin{aligned}Z_{\mathrm{G}} = \prod_\epsilon \frac{1}{{ 1 - z e^{-\beta \epsilon} }}\end{aligned} \hspace{\stretch{1}}(1.0.72)

\begin{aligned}P \beta = \frac{1}{{\lambda^3}} g_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.73)

\begin{aligned}U = \frac{3}{2} k_{\mathrm{B}} T \frac{V}{\lambda^3} g_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.74)

\begin{aligned}N_e = N - N_0 = N \left( \frac{T}{T_c} \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.0.75)

For $T < T_c$, $z = 1$.

\begin{aligned}g_\nu(1) = \zeta(\nu).\end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}\sum_{n_k = 0}^\infty e^{-\beta(\epsilon_k - \mu) n_k} =\frac{1}{{1 - e^{-\beta(\epsilon_k - \mu)}}}\end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}f_\nu^-( e^{-\alpha} ) = \frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} } + \cdots \end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}\rho \lambda^3 = g_{3/2}(z) \le \zeta(3/2) \approx 2.612\end{aligned} \hspace{\stretch{1}}(1.0.79.79)

\begin{aligned}k_{\mathrm{B}} T_{\mathrm{c}} = \left( \frac{\rho}{\zeta(3/2)} \right)^{2/3} \frac{ 2 \pi \hbar^2}{m}\end{aligned} \hspace{\stretch{1}}(1.0.79.79)

BEC

\begin{aligned}\rho= \rho_{\mathbf{k} = 0}+ \frac{1}{{\lambda^3}} g_{3/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.80.80)

\begin{aligned}\rho_0 = \rho \left(1 - \left( \frac{T}{T_{\mathrm{c}}} \right)^{3/2}\right)\end{aligned} \hspace{\stretch{1}}(1.0.80b)

\begin{aligned}\frac{E}{V} \propto \left( k_{\mathrm{B}} T \right)^{5/2}\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

\begin{aligned}\frac{C}{V} \propto \left( k_{\mathrm{B}} T \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

\begin{aligned}\frac{S}{N k_{\mathrm{B}}} = \frac{5}{2} \frac{g_{5/2}}{g_{3/2}} - \ln z \Theta(T - T_c)\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

Density of states

Low velocities

\begin{aligned}N_1(\epsilon)=V \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon}}\end{aligned} \hspace{\stretch{1}}(1.0.82a)

\begin{aligned}N_2(\epsilon)=V \frac{m}{\hbar^2}\end{aligned} \hspace{\stretch{1}}(1.0.82b)

\begin{aligned}N_3(\epsilon)=V \left( \frac{2 m}{\hbar^2} \right)^{3/2} \frac{1}{{4 \pi^2}} \sqrt{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.82c)

relativistic

\begin{aligned}\mathcal{D}_1(\epsilon)=\frac{2 L}{ c h } \frac{ \sqrt{ \epsilon^2 - \left( m c^2 \right)^2} }{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

\begin{aligned}\mathcal{D}_2(\epsilon)=\frac{2 \pi A}{ (c h)^2 } \frac{ \epsilon^2 - \left( m c^2 \right)^2 }{ \epsilon }\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

\begin{aligned}\mathcal{D}_3(\epsilon)=\frac{4 \pi V}{ (c h)^3 } \frac{\left( \epsilon^2 - \left( m c^2 \right)^2 \right)^{3/2}}{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

## 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.

## 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

## 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)

## 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

## PHY452H1S Basic Statistical Mechanics. Problem Set 3: State counting

Posted by peeterjoot on February 10, 2013

# Disclaimer

## Question: Surface area of a sphere in d-dimensions

Consider a $d$-dimensional sphere with radius $R$. Derive the volume of the sphere $V_d(R)$ and its surface area $S_d(R)$ using $S_d(R) = dV_d(R)/dR$.

1. $1$-sphere : 2D circle, with “volume” $V_2$
2. $2$-sphere : 3D sphere, with volume $V_3$
3. $3$-sphere : 4D Euclidean hypersphere, with “volume” $V_4$
4. $4$-sphere : 5D Euclidean hypersphere, with “volume” $V_5$
5. $\cdots$

To calculate the volume, we require a parameterization allowing for expression of the volume element in an easy to integrate fashion. For the $1$-sphere, we can use the usual circular and spherical coordinate volume elements

\begin{aligned}V_2(R) = 4 \int_0^R r dr \int_0^{\pi/2} d\theta = 4 \frac{R^2}{2} \frac{\pi}{2} = \pi R^2\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}V_3(R) = 8 \int_0^R r^2 dr \int_0^{\pi/2} \sin\theta d\theta \int_0^{\pi/2} d\phi= 8 \frac{R^3}{3} (1) \frac{\pi}{2}= \frac{4}{3} \pi R^3\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Here, to simplify the integration ranges, we’ve calculated the “volume” integral for just one quadrant or octet of the circle and sphere respectively, as in (Fig 1)

Fig1: Integrating over just one quadrant of circle or octet of sphere

How will we generalize this to higher dimensions? To calculate the volume elements in a systematic fashion, we can introduce a parameterization and use a Jacobian to change from Cartesian coordinates. For the $1$-volume and $2$-volume cases those parameterizations were the familiar

\begin{aligned}\begin{aligned}x_1 &= r \cos\theta \\ x_0 &= r \sin\theta\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\begin{aligned}x_2 &= r \cos\theta \\ x_1 &= r \sin\theta \cos\phi \\ x_0 &= r \sin\theta \sin\phi\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2b)

Some reflection shows that this generalizes nicely. Let’s use shorthand

\begin{aligned}C_k = \cos\theta_k\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}S_k = \sin\theta_k,\end{aligned} \hspace{\stretch{1}}(1.0.3b)

and pick $V_5$, say, a dimension bigger than the 2D or 3D cases that we can do by inspection, we can parameterize with

\begin{aligned}\begin{aligned}x_4 &= r C_4 \\ x_3 &= r S_4 C_3 \\ x_2 &= r S_4 S_3 C_2 \\ x_1 &= r S_4 S_3 S_2 C_1 \\ x_0 &= r S_4 S_3 S_2 S_1.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.4)

Our volume element

\begin{aligned}dV_5 = \underbrace{\frac{\partial(x_4, x_3, x_2, x_1, x_0)}{\partial(r, \theta_4, \theta_3, \theta_2, \theta_1)} }_{\equiv J_5}dr d\theta_4 d\theta_3 d\theta_2 d\theta_1\end{aligned} \hspace{\stretch{1}}(1.0.5)

That Jacobian is

\begin{aligned}J_5 = \left\lvert\begin{array}{lllll}C_4 & - r S_4 & \phantom{-} 0 & \phantom{-} 0 & \phantom{-} 0 \\ S_4 C_3 & \phantom{-} r C_4 C_3 & - r S_4 S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_4 S_3 C_2 & \phantom{-} r C_4 S_3 C_2 & \phantom{-} r S_4 C_3 C_2 & - r S_4 S_3 S_2 & \phantom{-} 0 \\ S_4 S_3 S_2 C_1 & \phantom{-} r C_4 S_3 S_2 C_1 & \phantom{-} r S_4 C_3 S_2 C_1 & \phantom{-} r S_4 S_3 C_2 C_1 & - r S_4 S_3 S_2 S_1 \\ S_4 S_3 S_2 S_1 & \phantom{-} r C_4 S_3 S_2 S_1 & \phantom{-} r S_4 C_3 S_2 S_1 & \phantom{-} r S_4 S_3 C_2 S_1 & \phantom{-} r S_4 S_3 S_2 C_1 \end{array}\right\rvert=r^4 S_4^3 S_3^2 S_2^1 S_1^0\left\lvert\begin{array}{lllll} \begin{array}{l} C_4 \end{array} \begin{array}{l} \phantom{S_3 S_2 S_1} \end{array} & \begin{array}{l} -S_4 \end{array} \begin{array}{l} \phantom{S_3 S_2 S_1} \end{array} & \begin{array}{l} \phantom{-}0 \end{array} & \begin{array}{l} \phantom{-}0 \end{array} & \begin{array}{l} \phantom{-}0 \end{array}\\ \begin{array}{l}S_4 \\ S_4 \\ S_4 \\ S_4 \end{array}\boxed{\begin{array}{l} C_3 \\ S_3 C_2 \\ S_3 S_2 C_1 \\ S_3 S_2 S_1 \end{array} }&\begin{array}{l}\phantom{-}C_4 \\ \phantom{-}C_4 \\ \phantom{-}C_4 \\ \phantom{-}C_4 \end{array}\boxed{\begin{array}{l}C_3 \\ S_3 C_2 \\ S_3 S_2 C_1 \\ S_3 S_2 S_1 \end{array}}&\begin{array}{l} - S_3 \\ \phantom{-} C_3 C_2 \\ \phantom{-} C_3 S_2 C_1 \\ \phantom{-} C_3 S_2 S_1 \end{array}&\begin{array}{l}\phantom{-} 0 \\ - S_2 \\ \phantom{-} C_2 C_1 \\ \phantom{-} C_2 S_1 \end{array}&\begin{array}{l}\phantom{-}0 \\ \phantom{-}0 \\ - S_1 \\ \phantom{-}C_1 \end{array}\end{array}\right\rvert\end{aligned} \hspace{\stretch{1}}(1.0.6)

Observe above that the cofactors of both the $1,1$ and the $1,2$ elements, when expanded along the first row, have a common factor. This allows us to work recursively

\begin{aligned}J_5 = r^4 S_4^3 S_3^2 S_2^1 S_1^0(C_4^2 - - S_4^2)\left\lvert\begin{array}{llll}C_3 & - S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_3 C_2 & \phantom{-} C_3 C_2 & - S_2 & \phantom{-} 0 \\ S_3 S_2 C_1 & \phantom{-} C_3 S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ S_3 S_2 S_1 & \phantom{-} C_3 S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert=r S_4^3 J_4\end{aligned} \hspace{\stretch{1}}(1.0.7)

Similarly for the 4D volume

\begin{aligned}J_4 = r^3 S_3^2 S_2^1 S_1^0\left\lvert\begin{array}{llll}C_3 & - S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_3 C_2 & \phantom{-} C_3 C_2 & - S_2 & \phantom{-} 0 \\ S_3 S_2 C_1 & \phantom{-} C_3 S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ S_3 S_2 S_1 & \phantom{-} C_3 S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r^3 S_3^2 S_2^1 S_1^0 (C_2^2 + S_2^2)\left\lvert\begin{array}{lll}\phantom{-} C_2 & - S_2 & \phantom{-} 0 \\ \phantom{-} S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ \phantom{-} S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_3^2 J_3\end{aligned} \hspace{\stretch{1}}(1.0.8)

and for the 3D volume

\begin{aligned}J_3 = r^2 S_2^1 S_1^0\left\lvert\begin{array}{lll}\phantom{-} C_2 & - S_2 & \phantom{-} 0 \\ \phantom{-} S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ \phantom{-} S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r^2 S_2^1 S_1^0 (C_2^2 + S_2^2)\left\lvert\begin{array}{ll}C_1 & - S_1 \\ S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_2^1 J_2,\end{aligned} \hspace{\stretch{1}}(1.0.9)

and finally for the 2D volume

\begin{aligned}J_2= r S_1^0\left\lvert\begin{array}{ll}C_1 & - S_1 \\ S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_1^0.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Putting all the bits together, the “volume” element in the n-D space is

\begin{aligned}dV_n = dr d\theta_{n-1} \cdots d\theta_{1} r^{n-1} \prod_{k=0}^{n-2} (\sin\theta_{k+1})^{k}, \end{aligned} \hspace{\stretch{1}}(1.0.11)

and the total volume is

\begin{aligned}V_n(R) = 2^n \int_0^R r^{n-1} dr \prod_{k=0}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta= 2^n \frac{R^{n}}{n}\prod_{k=0}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta= \frac{4 R^2}{n} (n-2) 2^{n-2} \frac{R^{n-2}}{n-2}\prod_{k=n-3}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta\prod_{k=0}^{n-4} \int_0^{\pi/2} d\theta \sin^k \theta=4 R^2 \frac{n-2}{n} V_{n-2}(R)\int_0^{\pi/2} d\theta \sin^{n-2} \theta\int_0^{\pi/2} d\theta \sin^{n-3} \theta=4 R^2 \frac{n-2}{n} V_{n-2}(R)\frac{\pi}{2 (n-2)}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Note that a single of these sine power integrals has a messy result (see: statMechProblemSet3.nb)

\begin{aligned}\int_0^{\pi/2} d\theta \sin^{k} \theta=\frac{\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right)}{2 \Gamma \left(\frac{k}{2}+1\right)},\end{aligned} \hspace{\stretch{1}}(1.0.13)

but the product is simple, and that result, computed in statMechProblemSet3.nb, is inserted above, providing an expression for the n-D volume element as a recurrence relation

\begin{aligned}\boxed{V_{n} = \frac{2 \pi}{n} R^2 V_{n-2}(R)}\end{aligned} \hspace{\stretch{1}}(1.0.14)

With this recurrence relation we can find the volume $V_{2n}$ or $V_{2n+1}$ in terms of $V_2$ and $V_3$ respectively. For the even powers we have

\begin{aligned}\begin{array}{lll}V_{2(2)} &= \frac{2 \pi R^2}{4} V_{2} &= \frac{2^1 \pi R^2}{2(2)} V_{2} \\ V_{2(3)} &= \frac{(2 \pi R^2)^2}{(6)(4)} V_{2} &= \frac{2^2 (\pi R^2)^2}{2^2 (3)(2)} V_{2} \\ V_{2(4)} &= \frac{(2 \pi R^2)^3}{(8)(6)(4)} V_{2} &= \frac{2^3 (\pi R^2)^3}{2^3 (4)(3)(2)} V_{2} \end{array}\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}V_{2(n)} = \frac{2^n (\pi R^2)^{n-1}}{(2n)!!} V_{2} = \frac{(\pi R^2)^{n-1}}{(n)!} V_{2} = \frac{(\pi R^2)^{n-1}}{(n)!} \pi R^2\end{aligned} \hspace{\stretch{1}}(1.0.16)

This gives us a closed form expression for the even powers for $n \ge 2$

\begin{aligned}\boxed{V_{2(n)} = \frac{(\pi R^2)^{n}}{n!}.}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Observe that this also conveniently gives $V_2 = \pi R^2$, so is actually valid for $n \ge 1$. Now for the odd powers

\begin{aligned}\begin{array}{lll}V_{2(2)+1} &= \frac{2 \pi R^2}{5} V_{3} &= 3 \frac{2^1 \pi R^2}{5!!} V_{3} \\ V_{2(3)+1} &= 3 \frac{(2 \pi R^2)^2}{7 5 3} V_{3} &= 3 \frac{2^2 (\pi R^2)^2}{7!!} V_{2} \\ V_{2(4)+1} &= 3 \frac{(2 \pi R^2)^3}{(9)(7)(5)(3)} V_{2} &= 3 \frac{2^3 (\pi R^2)^3}{9!!} V_{2} \end{array}\end{aligned} \hspace{\stretch{1}}(1.0.18)

\begin{aligned}V_{2(n)+1} = 3 \frac{2^{n-1} (\pi R^2)^{n-1}}{(2n +1)!!} V_{2} = 3 \frac{2^{n-1} (\pi R^2)^{n-1}}{(2n +1)!!} \frac{4}{3} \pi R^3\end{aligned} \hspace{\stretch{1}}(1.0.19)

So, for $n \ge 2$ we have

\begin{aligned}\boxed{V_{2(n)+1} = \frac{2^{n+1} \pi^n R^{2 n + 1}}{(2n +1)!!}.}\end{aligned} \hspace{\stretch{1}}(1.0.20)

As with the even powered expression 1.0.17 we see that this is also good for $n = 1$, yielding $V_3 = 4 \pi R^2/3$ as required for the 3D spherical volume.

The even and odd power expressions don’t look quite different on the surface, but can be put into a consistent form, when expressed in terms of the gamma function.

For the even powers, using a substitution $2 n = m$ we have for even values of $m \ge 2$

\begin{aligned}V_{m} = \frac{\pi^{m/2} R^m}{\Gamma(n + 1)}= \frac{\pi^{m/2} R^m}{\Gamma(m/2 + 1)}.\end{aligned} \hspace{\stretch{1}}(1.0.21)

For the even powers, with the help of [1] we find

\begin{aligned}(2 n + 1)!! = \frac{2^{n+1} \Gamma\left(n + \frac{3}{2}\right)}{\sqrt{\pi}}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This gives us

\begin{aligned}V_{2(n)+1} = \frac{\pi^{n + 1/2} R^{2 n + 1}}{\Gamma\left(n + \frac{3}{2}\right)}= \frac{\pi^{n + 1/2} R^{2 n + 1}}{\Gamma\left(\frac{(2 n + 1) + 2}{2}\right)}.\end{aligned} \hspace{\stretch{1}}(1.0.23)

Writing $m = 2 n + 1$, or $n = (m - 1)/2$ we have for odd values of $m \ge 3$ a match with the even powered expression of 1.0.21

\begin{aligned}\boxed{V_{m} = \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

We’ve shown that this is valid for any dimension $m \ge 2$.

Tabulating some values of these for $n \in [2, 7]$ we have respectively

\begin{aligned}\pi r^2,\frac{4 \pi r^3}{3},\frac{\pi ^2 r^4}{2},\frac{8 \pi ^2 r^5}{15},\frac{\pi ^3 r^6}{6},\frac{16 \pi ^3 r^7}{105},\end{aligned} \hspace{\stretch{1}}(1.0.25)

The only task left is computation of the surface area. That comes by inspection and is

\begin{aligned}S_{m}(R) =\frac{d V_m(R)}{d R}= \frac{\pi^{m/2} m R^{m-1}}{\Gamma\left( m/2 + 1 \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

Again for $m \in [2, 7]$ we have

\begin{aligned}2 \pi r,4 \pi r^2,2 \pi ^2 r^3,\frac{8 \pi ^2 r^4}{3},\pi ^3 r^5,\frac{16 \pi ^3 r^6}{15}.\end{aligned} \hspace{\stretch{1}}(1.0.27)

## Question: State counting – polymer

A typical protein is a long chain molecule made of very many elementary units called amino acids – it is an example of a class of such macromolecules called polymers. Consider a protein made $N$ amino acids, and assume each amino acid is like a sphere of radius a. As a toy model assume that the protein configuration is like a random walk with each amino acid being one “step”, i.e., the center-to-center vector from one amino acid to the next is a random vector of length $2a$ and ignore any issues with overlapping spheres (so-called “excluded volume” constraints). Estimate the spatial extent of this protein in space. Typical proteins assume a compact form in order to be functional. In this case, taking the constraint of nonoverlapping spheres, estimate the radius of such a compact protein assuming it has an overall spherical shape with fully packed amino acids (ignore holes in the packing, and use only volume ratios to make this estimate). With $N = 300$ and $a \approx 5 \text{\AA}$, estimate these sizes for the random walk case as well as the compact globular case.

We are considering a geometry like that of (Fig 2), depicted in two dimensions for ease of illustration.

Fig2: Touching “spheres”

From the geometry, if $\mathbf{c}_k$ is the vector to the center of the $k$th sphere, we have for some random unit vector $\hat{\mathbf{r}}_k$

\begin{aligned}\mathbf{c}_{k+1} = \mathbf{c}_k + 2 a \hat{\mathbf{r}}_k\end{aligned} \hspace{\stretch{1}}(1.0.28)

Proceeding recursively, writing $d_N$, and $n = N -1 > 0$, we have for the difference of the positions of the first and last centers of the chain

\begin{aligned}d_N = \left\lvert {\mathbf{c}_N - \mathbf{c}_1} \right\rvert = 2 a \left\lvert { \sum_{k = 1}^{n} \hat{\mathbf{r}}_k} \right\rvert= 2 a \left( \sum_{k,m = 1}^{n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}= 2 a \left( n + 2 \sum_{1 \le k < m \le n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}= 2 a \sqrt{n} \left( 1 + \frac{2}{n} \sum_{1 \le k < m \le n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(1.0.29)

The $\hat{\mathbf{r}}_k$‘s clearly cannot be completely random since we have a constraint that $\hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_{k+1} > -\cos\pi/3$, or else two adjacent spheres will overlap. There will also be overlapping constraints for longer portions of the chain that are harder to express. We are ignoring both such constraints, and seek the ensemble average of all systems of the form 1.0.29.

Employing random azimuthal and polar angular variables $\theta_k$, and $\phi_k$, we have

\begin{aligned}\hat{\mathbf{r}}_k = \begin{bmatrix}\sin\theta_k \cos\phi_k \\ \sin\theta_k \sin\phi_k \\ \cos\theta_k,\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.30)

so that the average polymer length is

\begin{aligned}\begin{aligned}\left\langle{{d_N }}\right\rangle &=2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \int_{\theta_j \in [0, \pi]}d\theta_1 d\theta_2 \cdots d\theta_n \int_{\phi_j \in [0, 2 \pi]}d\phi_1 d\phi_2 \cdots d\phi_n \times \\ &\quad \left( 1 + \frac{2}{n} \sum_{1 \le k < m \le n} \sin\theta_k \cos\phi_k \sin\theta_m \cos\phi_m + \sin\theta_k \sin\phi_k \sin\theta_m \sin\phi_m + \cos\theta_k \cos\theta_m \right)^{1/2}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.31)

Observing that even $\int \sqrt{1 + a \cos\theta} d\theta$ is an elliptic integral, we don’t have any hope of evaluating this in closed form. However, to first order, we have

\begin{aligned}\begin{aligned}d_N&\approx2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \int_{\theta_j \in [0, \pi]}d\theta_1 d\theta_2 \cdots d\theta_n \int_{\phi_j \in [0, 2 \pi]}d\phi_1 d\phi_2 \cdots d\phi_n \times \\ &\quad \left( 1 + \frac{1}{n} \sum_{1 \le k < m \le n} \sin\theta_k \cos\phi_k \sin\theta_m \cos\phi_m + \sin\theta_k \sin\phi_k \sin\theta_m \sin\phi_m + \cos\theta_k \cos\theta_m \right) \\ &=2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \left( 2 \pi^2 \right)^n \\ &\quad +2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \left( 2 \pi^2 \right)^{n - 2} \left( \frac{1}{{2}} n(n+1) \right) \frac{1}{{n}} \times \\ &\quad \int_0^{\pi}d\theta \int_0^{\pi}d\theta'\int_0^{2 \pi}d\phi\int_0^{2 \pi}d\phi'\left( \sin\theta \cos\phi \sin\theta' \cos\phi' + \sin\theta \sin\phi \sin\theta' \sin\phi' + \cos\theta \cos\theta' \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.32)

The $\int_0^{2 \pi} \cos\phi$ integrals kill off the first term, the $\int_0^{2 \pi} \sin\phi$ integrals kill of the second, and the $\int_0^\pi \cos\theta$ integral kill of the last term, and we are left with just

\begin{aligned}d_N \approx 2 a \sqrt{N-1}.\end{aligned} \hspace{\stretch{1}}(1.0.33)

Ignoring the extra $2 \times a$ of the end points, and assuming that for large $N$ we have $\sqrt{N-1} \approx \sqrt{N}$, the spatial extent of the polymer chain is

\begin{aligned}\boxed{d_N \approx 2 a \sqrt{N}.}\end{aligned} \hspace{\stretch{1}}(1.0.34)

Spherical packing

Assuming the densest possible spherical packing, a face centered cubic [3], as in (Fig 3), we see that the density of such a spherical packing is

Fig3: Element of a face centered cubic

\begin{aligned}\eta_{\mathrm{FCC}} =\frac{\left( 8 \times \frac{1}{{8}} + 6 \times \frac{1}{{2}} \right) \frac{4}{3} \pi r^3}{ \left( \sqrt{8 r^2} \right)^3 }= \frac{\pi}{\sqrt{18}}.\end{aligned} \hspace{\stretch{1}}(1.0.35)

With a globular radius of $R$ and an atomic radius of $a$, and density $\eta$ we have

\begin{aligned}\eta \frac{4}{3} \pi R^3 = N \frac{4}{3} \pi a^3,\end{aligned} \hspace{\stretch{1}}(1.0.36)

so that the globular radius $R$ is

\begin{aligned}\boxed{R_N(\eta) = a \sqrt[3]{\frac{N}{\eta}}.}\end{aligned} \hspace{\stretch{1}}(1.0.37)

Some numbers

With $N = 300$ and $a \approx 5 \text{\AA}$, and ignoring spaces (i.e. $\eta = 1$, for a non-physical infinite packing), our globular diameter is approximately

\begin{aligned}2 \times 5 \text{\AA} \sqrt[3]{300} \approx 67 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.38)

This is actually not much different than the maximum spherical packing of an FCC lattice, which results a slightly larger globular cluster diameter

\begin{aligned}2 \times 5 \text{\AA} \sqrt[3]{300 \sqrt{18}/\pi} \approx 74 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.39)

Both however, are much less than the end to end length of the random walk polymer chain

\begin{aligned}2 (5 \text{\AA}) \sqrt{300} \approx 173 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.40)

## Question: State counting – spins

Consider a toy model of a magnet where the net magnetization arises from electronic spins on each atom which can point in one of only two possible directions – Up/North or Down/South. If we have a system with $N$ spins, and if the magnetization can only take on values $\pm 1$ (Up = $+1$, Down = $-1$), how many configurations are there which have a total magnetization $m$, where $m = (N_\uparrow - N_\downarrow)/N$ (note that $N_\uparrow + N_\downarrow = N$)? Simplify this result assuming $N \gg 1$ and a generic $m$ (assume we are not interested in the extreme case of a fully magnetized system where $m = \pm 1$). Find the value of the magnetization $m$ for which the number of such microscopic states is a maximum. For $N = 20$, make a numerical plot of the number of states as a function of the magnetization (note: $-1 \le m \le 1$) without making the $N \gg 1$ assumption.

For the first couple values of $N$, lets enumerate the spin sample spaces, their magnetization.

$N = 1$

1. $\uparrow$ : $m = 1$
2. $\downarrow$, $m = -1$

$N = 2$

1. $\uparrow \uparrow$ : $m = 1$
2. $\uparrow \downarrow$ : $m = 0$
3. $\downarrow \uparrow$, $m = 0$
4. $\downarrow \downarrow$, $m = -1$

$N = 3$

1. $\uparrow \uparrow \uparrow$ : $m = 1$
2. $\uparrow \uparrow \downarrow$ : $m = 1/3$
3. $\uparrow \downarrow \uparrow$ : $m = 1/3$
4. $\uparrow \downarrow \downarrow$ : $m = -1/3$
5. $\downarrow \uparrow \uparrow$ : $m = 1/3$
6. $\downarrow \uparrow \downarrow$ : $m = -1/3$
7. $\downarrow \downarrow \uparrow$ : $m = -1/3$
8. $\downarrow \downarrow \downarrow$ : $m = 1$

The respective multiplicities for $N = 1,2,3$ are $\{1\}$, $\{1, 2, 1\}$, $\{1, 3, 3, 1\}$. It’s clear that these are just the binomial coefficients. Let’s write for the multiplicities

\begin{aligned}g(N, m) = \binom{N}{i(m)}\end{aligned} \hspace{\stretch{1}}(1.0.41)

where $i(m)$ is a function that maps from the magnetization values $m$ to the integers $[0, N]$. Assuming

\begin{aligned}i(m) = a m + b,\end{aligned} \hspace{\stretch{1}}(1.0.42)

where $i(-1) = 0$ and $i(1) = N$, we solve

\begin{aligned}\begin{aligned}a (-1) + b &= 0 \\ a (1) + b &= N,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.43)

so

\begin{aligned}i(m) = \frac{N}{2}(m + 1)\end{aligned} \hspace{\stretch{1}}(1.0.44)

and

\begin{aligned}g(N, m) = \binom{N}{\frac{N}{2}(1 + m)} = \frac{N!}{\left(\frac{N}{2}(1 + m)\right)!\left(\frac{N}{2}(1 - m)\right)!}\end{aligned} \hspace{\stretch{1}}(1.0.45)

From

\begin{aligned}\begin{aligned}2 m &= N_{\uparrow} - N_{\downarrow} \\ N &= N_{\uparrow} + N_{\downarrow},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.46)

we see that this can also be written

\begin{aligned}\boxed{g(N, m) = \frac{N!}{(N_{\uparrow})!(N_{\downarrow})!}}\end{aligned} \hspace{\stretch{1}}(1.0.47)

Simplification for large $N$

Using Stirlings approximation

\begin{aligned}\begin{aligned}\ln g(N, m) &= \ln N! - \ln N_{\downarrow}! - \ln N_{\uparrow}! \\ &\approx \left( N + \frac{1}{{2}} \right) \ln N - \not{{N}} + \not{{ \frac{1}{{2}} \ln 2 \pi }} \\ &\quad-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow} + \not{{N_{\uparrow}}} - \not{{\frac{1}{{2}} \ln 2 \pi}} \\ &\quad-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow} + \not{{N_{\downarrow}}} - \frac{1}{{2}} \ln 2 \pi \\ &=\left( N_{\uparrow} + N_{\downarrow} + \frac{1}{{2}} + \underbrace{\frac{1}{{2}}}_{\text{add}} \right) \ln N -\frac{1}{{2}} \ln 2 \pi - \underbrace{\frac{1}{{2}} \ln N}_{\text{then subtract}} \\ &\quad -\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow} \quad -\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow} \\ &=-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow}/N-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow}/N- \frac{1}{{2}} \ln 2 \pi N \\ &=-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln \frac{1}{{2}}( 1 + m )-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln \frac{1}{{2}} ( 1 - m )- \frac{1}{{2}} \ln 2 \pi N \\ &\approx\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln 2+\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln 2 - \frac{1}{{2}} \ln 2 \pi N \\ &\quad -\left( N_{\uparrow} + \frac{1}{{2}} \right) \left(m - \frac{1}{{2}} m^2 \right) \\ &\quad -\left( N_{\downarrow} + \frac{1}{{2}} \right) \left( -m - \frac{1}{{2}} m^2 \right) \\ &= (N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N+ m ( N_{\downarrow} - N_{\uparrow} )+ \frac{1}{{2}} m^2 ( N + 1 ) \\ &= (N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N- N m^2+ \frac{1}{{2}} m^2 ( N + 1 ) \\ &\approx(N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N- \frac{1}{{2}} m^2 N\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.48)

This gives us for large $N$

\begin{aligned}g(N, m) \approx 2^N \sqrt{\frac{2}{ \pi N}} e^{ -m^2 N/2 }\end{aligned} \hspace{\stretch{1}}(1.0.49)

For $N = 20$ this approximation and the exact expression are plotted in (Fig 4).

Fig4: Distribution of number of configurations for N = 20 magnets as a function of magnetization

With the large scales of this extremely peaked function, no visible difference can be seen. That difference does exist, and is plotted in (Fig 5)

Fig5: N = 20 differences between the exact binomial expression and the Gaussian approximation

# References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[2] Wikipedia. N-sphere — Wikipedia, The Free Encyclopedia, 2013\natexlab{a}. URL http://en.wikipedia.org/w/index.php?title=N-sphere&oldid=534164100. [Online; accessed 26-January-2013].

[3] Wikipedia. Sphere packing — wikipedia, the free encyclopedia, 2013\natexlab{b}. URL http://en.wikipedia.org/w/index.php?title=Sphere_packing&oldid=535578971. [Online; accessed 31-January-2013].

## Limit of unfair coin distribution, the hard way

Posted by peeterjoot on February 7, 2013

We calculated the distribution for the sum of random variables associated with $N$ unfair coin tosses, where the probabilities were $r$, and $s = 1 - r$ for heads and tails respectively. Assigning heads and tails values of $-1$ and $+1$ respectively, the probability distribution of the sum $X$ of the total numbers of heads and tails values for $N$ such tosses was found to be

\begin{aligned}P_N(r, k) = \binom{N}{k} r^{N-k} s^k,\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}k = \frac{N + X}{2}\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Part of the problem was to calculate the limit for $N \gg 1$ and $N \gg X$. I did this with the central limit theorem, but we were apparently not allowed to do it that way. Here’s a bash at it the hard way, using Stirling’s approximation and Taylor series expansion for the logs.

Application of Stirling’s approximation gives us

\begin{aligned}P_N(r, k) &\approx \frac{ \sqrt{2 \pi N} \not{{e^{-N}}} N^N }{\sqrt{2 \pi (N - k)} \not{{e^{-N + k}}} (N - k)^{N-k}\sqrt{2 \pi k} \not{{e^{-k}}} k^{k}}r^{N-k} s^k \\ &= \sqrt{\frac{N}{2 \pi k(N-k)}}N^{N\underbrace{-k + k}_{\text{Add and subtract}}}\left( \frac{r}{N-k} \right)^{N-k} \left( \frac{s}{k} \right)^k \\ &= \sqrt{\frac{N}{2 \pi k(N-k)}}\left( \frac{N r}{N-k} \right)^{N-k} \left( \frac{N s}{k} \right)^k\end{aligned} \hspace{\stretch{1}}(1.0.2)

The $N r/(N-k)$ term looks like it can probably be coersed into $1/(1 - y/N)$ form that will allow for Taylor expansion of the log. With that change of variables, we find

\begin{aligned}k = N(1 - r) - y r = N s + y r\end{aligned} \hspace{\stretch{1}}(1.0.3)

so

\begin{aligned}\frac{k}{Ns} = 1 + \frac{y r}{N s}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\frac{N - k}{N r} = 1 - \frac{y}{N}\end{aligned} \hspace{\stretch{1}}(1.0.4b)

This is a bit unsymmetrical, so let’s write $y r = x$ so that

\begin{aligned}\frac{N - k}{N r} = 1 - \frac{x}{N r}\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\frac{k}{Ns} = 1 + \frac{x}{N s}\end{aligned} \hspace{\stretch{1}}(1.0.5b)

We’ve also got terms in $k$ and $N - k$ above that we need to express. With $k = N s + x$, we have $N - k = N r - x$, and

\begin{aligned}\frac{(N - k) k}{N} &= \frac{1}{{N}}(N r - x)(N s + x) \\ &= \frac{1}{{N}}( -x^2 + N^2 r s + x N(r - s) ) \\ &= -\frac{x^2}{N} + N r s + x( 2 r - 1 ) \\ &\approx N r s\end{aligned} \hspace{\stretch{1}}(1.0.6)

Taking logs of 1.0.2 we have

\begin{aligned}\ln P_N(r, k) &\approx \ln \sqrt{\frac{1}{2 \pi N r s}}- (N r - x)\ln \left( 1 - \frac{x}{N r} \right)- (N s + x)\ln \left( 1 + \frac{x}{N s} \right) \\ &\approx\ln \sqrt{\frac{1}{2 \pi N r s}}+ (x - N r)\left( - \frac{x}{N r} - \frac{1}{{2}} \left( \frac{x}{N r} \right)^2 \right)- (x + N s)\left( \frac{x}{N s} - \frac{1}{{2}} \left( \frac{x}{N s} \right)^2 \right) \\ &= \ln \sqrt{\frac{1}{2 \pi N r s}}+ \frac{1}{{2}} \frac{x^2}{N}\left( \frac{1}{{r}} + \frac{1}{{s}} \right)- \frac{x^2}{N}\left( \frac{1}{{r}} + \frac{1}{{s}} \right)+ \frac{x^3}{(N)^2}\left( \frac{1}{{r^2}} - \frac{1}{{s^2}} \right)\end{aligned} \hspace{\stretch{1}}(1.0.7)

Dropping the $O(1/N^2)$ term and noting that

\begin{aligned}-\frac{1}{{2}} \left( \frac{1}{{r}} + \frac{1}{{s}} \right)=-\frac{1}{{2 r s}} (r + (1 - r)=-\frac{1}{{2 r s}}\end{aligned} \hspace{\stretch{1}}(1.0.8)

We have

\begin{aligned}P_N(r, k) \approx\frac{1}{{\sqrt{2 \pi N r s}}} \exp\left( -\frac{x^2}{2 N r s} \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

With

\begin{aligned}x &= k - N s \\ &= \frac{1}{{2}} ( N + X ) - N s \\ &= \frac{1}{{2}} ( N + X - 2 N s ) \\ &= \frac{1}{{2}} ( X + N ( 1 - 2 s) ) \\ &= \frac{1}{{2}} ( X - N ( 1 - 2 r) ),\end{aligned} \hspace{\stretch{1}}(1.0.10)

we have

\begin{aligned}\boxed{P_N(r, k) \approx\frac{1}{{\sqrt{2 \pi N r s}}} \exp\left( -\frac{ \left( X - N (1 - 2r) \right)^2 }{8 N r s} \right). }\end{aligned} \hspace{\stretch{1}}(1.0.11)

This recovers the result obtained with the central limit theorem (after that result was adjusted to account for parity).

## One dimensional random walk

Posted by peeterjoot on February 3, 2013

## Question: One dimensional random walk

Random walk in 1D by unit steps. With the probability to go right of $p$ and a probability to go left of $1 -p$ what are the first two moments of the final position of the particle?

This was a problem from the first midterm. I ran out of time and didn’t take the answer as far as I figured I should have. Here’s a more casual bash at the problem.

First we need the probabilities.

One step: $N = 1$

Our distance (from the origin) can only be $X = \pm 1$.

\begin{aligned}P_{X = -1} = p^{0} (1 -p)^1\end{aligned} \hspace{\stretch{1}}(1.0.1)

\begin{aligned}P_{X = 1} = p^1 (1-p)^{1 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Two steps: $N = 2$

We now have three possibilities

\begin{aligned}P_{X = -2} = p^{0} (1 -p)^{2 - 0}\end{aligned} \hspace{\stretch{1}}(1.0.3)

\begin{aligned}P_{X = 0} = 2 p^1 (1-p)^{2 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.4)

\begin{aligned}P_{X = 2} = p^2 (1-p)^{2 - 2}\end{aligned} \hspace{\stretch{1}}(1.0.5)

Three steps: $N = 3$

We now have three possibilities

\begin{aligned}P_{X = -3} = p^{0} (1 - p)^{3 - 0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}P_{X = -1} = 3 p^1 (1 - p)^{3 - 1}\end{aligned} \hspace{\stretch{1}}(1.0.7)

\begin{aligned}P_{X = 1} = 3 p^2 (1 - p)^{3 - 2}\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}P_{X = 3} = p^3 (1-p)^{3 - 3}\end{aligned} \hspace{\stretch{1}}(1.0.9)

General case

The pattern is pretty clear, but we need a mapping from the binomial index to the the final distance. With an index $k$, and a guess

\begin{aligned}D(k) = a k + b,\end{aligned} \hspace{\stretch{1}}(1.0.10)

where

\begin{aligned}D(0) = -N = b\end{aligned} \hspace{\stretch{1}}(1.0.11)

\begin{aligned}D(N) = a N + b = (a - 1)N = N.\end{aligned} \hspace{\stretch{1}}(1.0.12)

So

\begin{aligned}D(k) = 2 k - N,\end{aligned} \hspace{\stretch{1}}(1.0.13)

and

\begin{aligned}k = \frac{D + N}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Our probabilities are therefore

\begin{aligned}\boxed{P_{X = D} = \binom{N}{(N + D)/2} p^{(N + D)/2}(1 - p)^{(N - D)/2}.}\end{aligned} \hspace{\stretch{1}}(1.0.15)

First moment

For the expectations let’s work with $k$ instead of $D$, so that the expectation is

\begin{aligned}\left\langle{{X}}\right\rangle &= \sum_{k = 0}^N (2 k - N) \binom{N}{k} p^k (1 - p)^{N - k} \\ &= 2 \sum_{k = 0}^N k \frac{N!}{(N-k)!k!} p^k (1 - p)^{N - k} - N \\ &= 2 N p \sum_{k = 1}^N \frac{(N-1)!}{(N - 1 - (k - 1))!(k-1)!} p^{k-1} (1 - p)^{N - 1 - (k - 1)} - N \\ &= 2 N p \sum_{s = 0}^{N-1} \binom{N-1}{s} p^{s} (1 - p)^{N -1 - s} - N.\end{aligned} \hspace{\stretch{1}}(1.0.16)

This gives us

\begin{aligned}\boxed{\left\langle{{X}}\right\rangle = N( 2 p - 1 ).}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Second moment

\begin{aligned}\left\langle{{X^2}}\right\rangle &= \sum_{k = 0}^N (2 k - N)^2 \binom{N}{k} p^k (1 - p)^{N - k} \\ &= 4 \sum_{k = 0}^N k^2 \binom{N}{k} p^k (1 - p)^{N - k}- 4 N^2 p+ N^2 \\ &= 4 N p \sum_{k = 1}^N k \frac{(N-1)!}{(N - 1 - (k - 1))! (k-1)!} p^{k-1} (1 - p)^{N - k} + N^2(1 - 4 p) \\ &= 4 N p \sum_{s = 0}^N (s + 1) \frac{(N-1)!}{(N - 1 - s)! s!} p^s (1 - p)^{N - 1 - s} + N^2(1 - 4 p) \\ &= 4 N p ((N-1) p + 1) + N^2(1 - 4 p) \\ &= N^2 ( 1 - 4 p + 4 p^2 ) + 4 N p ( 1 - p ).\end{aligned} \hspace{\stretch{1}}(1.0.18)

So the second moment is

\begin{aligned}\boxed{\left\langle{{X^2}}\right\rangle = N^2 ( 1 - 2 p )^2 + 4 N p ( 1 - p )}\end{aligned} \hspace{\stretch{1}}(1.0.19)

From this we see that the variance is just this second term

\begin{aligned}\sigma^2 = \left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = 4 N p ( 1 - p ).\end{aligned} \hspace{\stretch{1}}(1.0.20)

## PHY452H1S Basic Statistical Mechanics. Problem Set 1: Binomial distributions

Posted by peeterjoot on January 20, 2013

# Disclaimer

## Question: Limiting form of the binomial distribution

Starting from the simple case of the binomial distribution

\begin{aligned}P_N(X) = 2^{-N} \frac{N!}{\left(\frac{N + X}{2}\right)!\left(\frac{N - X}{2}\right)!}\end{aligned} \hspace{\stretch{1}}(1.0.1)

derive the Gaussian distribution which results when $N \gg 1$ and ${\left\lvert{X}\right\rvert} \ll N$.

We’ll work with the logarithms of $P_N(X)$.

Note that the logarithm of the Stirling approximation takes the form

\begin{aligned}\ln a! &\approx \ln \sqrt{2\pi} + \frac{1}{{2}} \ln a + a \ln a - a \\ &=\ln \sqrt{2\pi} + \left( a + \frac{1}{{2}} \right) \ln a - a\end{aligned} \hspace{\stretch{1}}(1.0.2)

Using this we have

\begin{aligned}\ln \left((N + X)/2\right)!=\ln \sqrt{2 \pi}+\left(\frac{N + 1 + X}{2} \right)\left(\ln \left(1 + \frac{X}{N}\right)+ \ln \frac{N}{2}\right)- \frac{N + X}{2}\end{aligned} \hspace{\stretch{1}}(1.0.3)

Adding $\ln \left( (N + X)/2 \right)! + \ln \left( (N - X)/2 \right)!$, we have

\begin{aligned}2 \ln \sqrt{2 \pi}-N+\left(\frac{N + 1 + X}{2} \right)\left(\ln \left(1 + \frac{X}{N}\right)+ \ln \frac{N}{2}\right)+\left(\frac{N + 1 - X}{2} \right)\left(\ln \left(1 - \frac{X}{N}\right)+ \ln \frac{N}{2}\right)=2 \ln \sqrt{2 \pi}-N+\left(\frac{N + 1}{2} \right)\left(\ln \left(1 - \frac{X^2}{N^2}\right)+ 2 \ln \frac{N}{2}\right)+\frac{X}{2}\left( \ln \left( 1 + \frac{X}{N} \right)- \ln \left( 1 - \frac{X}{N} \right)\right)\end{aligned} \hspace{\stretch{1}}(1.0.4)

Recall that we can expand the log around $1$ with the slowly converging Taylor series

\begin{aligned}\ln( 1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\ln( 1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4},\end{aligned} \hspace{\stretch{1}}(1.0.5b)

but if $x \ll 1$ the first order term will dominate, so in this case where we assume $X \ll N$, we can approximate this sum of factorial logs to first order as

\begin{aligned}2 \ln \sqrt{2 \pi} -N+\left(\frac{N + 1}{2} \right)\left(- \frac{X^2}{N^2}+ 2 \ln \frac{N}{2}\right)+\frac{X}{2}\left( \frac{X}{N} + \frac{X}{N}\right) &= 2 \ln \sqrt{2 \pi} -N+ \frac{X^2}{N} \left( - \frac{N + 1}{2N} + 1\right)+ (N + 1) \ln \frac{N}{2} &\approx 2 \ln \sqrt{2 \pi} -N+ \frac{X^2}{2 N} + (N + 1) \ln \frac{N}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

Putting the bits together, we have

\begin{aligned}\ln P_N(X) &\approx - N \ln 2 + \left( N + \frac{1}{{2}}\right) \ln N - \not{{N}} - \ln \sqrt{2 \pi} + \not{{N}} -\frac{X^2}{2N} - (N + 1) \ln \frac{N}{2} \\ &= \left(-\not{{N}} + (\not{{N}} + 1) \ln 2\right)+\left(\not{{N}} + \frac{1}{{2}} - \not{{N}} - 1\right) \ln N- \ln \sqrt{2 \pi} - \frac{X^2}{2N} \\ &= \ln \left(\frac{2}{\sqrt{2 \pi N}}\right)-\frac{X^2}{2 N}\end{aligned} \hspace{\stretch{1}}(1.0.7)

Exponentiating gives us the desired result

\begin{aligned}\boxed{P_N(X) \rightarrow \frac{2}{\sqrt{2 \pi N}} e^{-\frac{X^2}{2 N}}.}\end{aligned} \hspace{\stretch{1}}(1.0.8)

## Question: Binomial distribution for biased coin

Consider the more general case of a binomial distribution where the probability of a head is $r$ and a tail is $(1 - r)$ (a biased coin). With $\text{head} = -1$ and $\text{tail} = +1$, obtain the binomial distribution $P_N(r,X)$ for obtaining a total of $X$ from $N$ coin tosses. What is the limiting form of this distribution when $N \gg 1$ and $\left\langle{X - \left\langle{X}\right\rangle}\right\rangle \ll N$? The latter condition simply means that I need to carry out any Taylor expansions in X about its mean value $\left\langle{{X}}\right\rangle$. The mean $\left\langle{{X}}\right\rangle$ can be easily computed first in terms of “r”.

Let’s consider 1, 2, 3, and N tosses in sequence to understand the pattern.

1 toss

The base case has just two possibilities

1. Heads, $P = r$, $X = -1$
2. Tails, $P = (1 - r)$, $X = 1$

If $k = 0,1$ for $X = -1, 1$ respectively, we have

\begin{aligned}P_1(r, X) = r^{1 - k} (1 - r)^{k}\end{aligned} \hspace{\stretch{1}}(1.0.9)

As a check, when $r = 1/2$ we have $P_1(X) = 1/2$

2 tosses

Our sample space is now a bit bigger

1. $(h,h)$, $P = r^2$, $X = -2$
2. $(h,t)$, $P = r (1 - r)$, $X = 0$
3. $(t,h)$, $P = r (1 - r)$, $X = 0$
4. $(t,t)$, $P = (1 - r)^2$, $X = 2$

Here $P$ is the probability of the ordered sequence, but we are interested only in the probability of each specific value of $X$. For $X = 0$ there are $\binom{2}{1} = 2$ ways of picking a heads, tails combination.

Enumerating the probabilities, as before, with $k = 0, 1, 2$ for $X = -1, 0, 1$ respectively, we have

\begin{aligned}P_2(r, X) = r^{2 - k} (1 - r)^{k} \binom{2}{k}\end{aligned} \hspace{\stretch{1}}(1.0.10)

3 tosses

Increasing our sample space by one more toss our possibilities for all ordered triplets of toss results is

1. $(h,h,h)$, $P = r^3$, $X = -3$
2. $(h,h,t)$, $P = r^2(1 - r)$, $X = -1$
3. $(h,t,h)$, $P = r^2(1 - r)$, $X = -1$
4. $(h,t,t)$, $P = r(1 - r)^2$, $X = 1$
5. $(t,h,h)$, $P = r^2(1 - r)$, $X = -1$
6. $(t,h,t)$, $P = r(1 - r)^2$, $X = 1$
7. $(t,t,h)$, $P = r(1 - r)^2$, $X = 1$
8. $(t,t,t)$, $P = r (1 - r)$, $X = 0$
9. $(t,t,t)$, $P = (1 - r)^3$, $X = 3$

Here $P$ is the probability of the ordered sequence, but we are still interested only in the probability of each specific value of $X$. We see that we have
$\binom{3}{1} = \binom{3}{2} = 3$ ways of picking some ordering of either $(h,h,t)$ or $(t,t,h)$

Now enumerating the possibilities with $k = 0, 1, 2, 3$ for $X = -3, -1, 1, 3$ respectively, we have

\begin{aligned}P_3(r, X) = r^{3 - k} (1 - r)^{k} \binom{3}{k}\end{aligned} \hspace{\stretch{1}}(1.0.11)

n tosses

To generalize we need a mapping between our random variable $X$, and the binomial index $k$, but we know what that is from the fair coin problem, one of $(N-X)/2$ or $(N + X)/2$. To get the signs right, let’s evaluate $(N \pm X)/2$ for $N = 3$ and $X \in \{3, -1, 1, 3\}$

Mapping between $k$ and $(N \pm X)/2$ for $N = 3$:

 X (N-X)/2 (N+X)/2 -3 3 0 -1 2 1 1 1 2 3 0 3

Using this, we see that the generalization to unfair coins of the binomial distribution is

\begin{aligned}\boxed{P_N(r, X) = r^{\frac{N-X}{2}} (1 - r)^{\frac{N+X}{2}} \frac{N!}{\left(\frac{N + X}{2}\right)!\left(\frac{N - X}{2}\right)!}}\end{aligned} \hspace{\stretch{1}}(1.0.12)

Checking against the fair result, we see that we have the $1/2^N$ factor when $r = 1/2$ as expected. Let’s check for $X = -1$ (two heads, one tail) to see if the exponents are right. That is

\begin{aligned}P_3(r, -1) = r^{\frac{3 + 1}{2}} (1 - r)^{\frac{3 - 1}{2}} \frac{3!}{\left(\frac{3 - 1}{2}\right)!\left(\frac{3 + 1}{2}\right)!}=r^2 (1-r) \frac{3!}{1! 2!}= r^2 (1 - r)\end{aligned} \hspace{\stretch{1}}(1.0.13)

Good, we’ve got a $r^2$ (two heads) term as desired.

Limiting form

To determine the limiting behavior, we can utilize the Central limit theorem. We first have to calculate the mean and the variance for the $N=1$ case. The first two moments are

\begin{aligned}\left\langle{{X}}\right\rangle &= -1 r + 1 (1-r) \\ &= 1 - 2 r\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\left\langle{{X^2}}\right\rangle &= (-1)^2 r + 1^2 (1-r) \\ &= 1\end{aligned} \hspace{\stretch{1}}(1.0.14b)

and the variance is

\begin{aligned}\left\langle{{X^2}}\right\rangle -\left\langle{{X}}\right\rangle^2 &= 1 - (1 - 2r)^2 \\ &= 1 - ( 1 - 4 r + 4 r^2 ) \\ &= 4 r - 4 r^2 \\ &= 4 r ( 1 - r )\end{aligned} \hspace{\stretch{1}}(1.0.15)

The Central Limit Theorem gives us

\begin{aligned}P_N(r, X) \rightarrow \frac{1}{{ \sqrt{8 \pi N r (1 - r) }}} \exp\left(- \frac{( X - N (1 - 2 r) )^2}{8 N r ( 1 - r )}\right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

however, we saw in [1] that this theorem was derived for continuous random variables. Here we have random variables that only take on either odd or even integer values, with parity depending on whether $N$ is odd or even. We’ll need to double the CLT result to account for this. This gives us

\begin{aligned}\boxed{P_N(r, X) \rightarrow \frac{1}{ \sqrt{2 \pi N r (1 - r) }} \exp\left(- \frac{( X - N (1 - 2 r) )^2}{8 N r ( 1 - r )}\right)}\end{aligned} \hspace{\stretch{1}}(1.0.17)

As a check we note that for $r = 1/2$ we have $r(1-r) = 1/4$ and $1 - 2r = 0$, so we get

\begin{aligned}P_N(1/2, X) \rightarrow \frac{2}{ \sqrt{2 \pi N }} \exp\left(- \frac{ X^2}{2 N }\right).\end{aligned} \hspace{\stretch{1}}(1.0.18)

Observe that both this and 1.0.8 do not integrate to unity, but to $2$. This is expected given the parity of the discrete random variable $X$. An integral normalization check is really only approximating the sum over integral values of our discrete random variable, and here we want to skip half of those values.

# References

[1] Peter Young. Proof of the central limit theorem in statistics, 2009. URL http://physics.ucsc.edu/ peter/116C/clt.pdf. [Online; accessed 13-Jan-2013].