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# Posts Tagged ‘probability’

## A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

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

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

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

## PHY452H1S Basic Statistical Mechanics. Problem Set 6: Max entropy, fugacity, and Fermi gas

Posted by peeterjoot on April 3, 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: Maximum entropy principle

Consider the “Gibbs entropy”

\begin{aligned}S = - k_{\mathrm{B}} \sum_i p_i \ln p_i\end{aligned} \hspace{\stretch{1}}(1.1)

where $p_i$ is the equilibrium probability of occurrence of a microstate $i$ in the ensemble.

For a microcanonical ensemble with $\Omega$ configurations (each having the same energy), assigning an equal probability $p_i= 1/\Omega$ to each microstate leads to $S = k_{\mathrm{B}} \ln \Omega$. Show that this result follows from maximizing the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1\end{aligned} \hspace{\stretch{1}}(1.2)

(for $p_i$ to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i,\end{aligned} \hspace{\stretch{1}}(1.3)

then fix $\alpha$ by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles $N$), maximize the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.4)

(for $p_i$ to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.5)

where $E_i$ is the energy of microstate $i$. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.6)

then fix $\alpha, \beta$ by demanding that the constraint be satisfied. What is the resulting $p_i$?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.7)

(for $p_i$ to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.8)

and a given fixed average particle number

\begin{aligned}\left\langle{{N}}\right\rangle = \sum_i N_i p_i.\end{aligned} \hspace{\stretch{1}}(1.9)

Here $E_i, N_i$ represent the energy and number of particles in microstate $i$. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i - \gamma \sum_i N_i p_i,\end{aligned} \hspace{\stretch{1}}(1.10)

then fix $\alpha, \beta, \gamma$ by demanding that the constrains be satisfied. What is the resulting $p_i$?

Writing

\begin{aligned}f = S - \alpha \sum_{j = 1}^\Omega p_j,= -\sum_{j = 1}^\Omega p_j \left( k_{\mathrm{B}} \ln p_j + \alpha \right),\end{aligned} \hspace{\stretch{1}}(1.11)

our unconstrained minimization requires

\begin{aligned}0 = \frac{\partial {f}}{\partial {p_i}}= -\left( k_{\mathrm{B}} \left( \ln p_i + 1 \right) + \alpha \right).\end{aligned} \hspace{\stretch{1}}(1.12)

Solving for $p_i$ we have

\begin{aligned}p_i = e^{-\alpha/k_{\mathrm{B}} - 1}.\end{aligned} \hspace{\stretch{1}}(1.13)

The probabilities for each state are constant. To fix that constant we employ our constraint

\begin{aligned}1 = \sum_{j = 1}^\Omega p_j= \sum_{j = 1}^\Omega e^{-\alpha/k_{\mathrm{B}} - 1}= \Omega e^{-\alpha/k_{\mathrm{B}} - 1},\end{aligned} \hspace{\stretch{1}}(1.14)

or

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \Omega.\end{aligned} \hspace{\stretch{1}}(1.15)

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

\begin{aligned}\boxed{p_i = e^{-\ln \Omega} = \frac{1}{{\Omega}}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

Using this we our Gibbs entropy can be summed easily

\begin{aligned}S &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega p_j \ln p_j \\ &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega \frac{1}{{\Omega}} \ln \frac{1}{{\Omega}} \\ &= -k_{\mathrm{B}} \frac{\Omega}{\Omega} \left( -\ln \Omega \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{S = k_{\mathrm{B}} \ln \Omega.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

For the “action” like quantity that we want to minimize, let’s write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j,\end{aligned} \hspace{\stretch{1}}(1.0.16)

for which we seek $\alpha$, $\beta$ such that

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\sum_j p_j\left( k_{\mathrm{B}} \ln p_j + \alpha + \beta E_j \right) \\ &= -k_{\mathrm{B}} (\ln p_i + 1) - \alpha - \beta E_i,\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( - \left( \alpha - \beta E_i \right) /k_{\mathrm{B}} - 1 \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability constraint is

\begin{aligned}1 &= \sum_j \exp\left( - \left( \alpha - \beta E_j \right) /k_{\mathrm{B}} - 1 \right) \\ &= \exp\left( - \alpha/k_{\mathrm{B}} - 1 \right)\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( \alpha/k_{\mathrm{B}} + 1 \right)=\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Taking logs we have

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

We could continue to solve for $\alpha$ explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

\begin{aligned}p_i = \exp\left( -\ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right) \right)\exp\left( - \beta E_i/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{p_i = \frac{ \exp\left( - \beta E_i/k_{\mathrm{B}} \right)}{ \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

To determine $\beta$ we must look implicitly to the energy constraint, which is

\begin{aligned}\left\langle{{E}}\right\rangle &= \sum_i E_i p_i \\ &= \sum_iE_i\left( \frac{ \exp\left( - \beta E_i/k_{\mathrm{B}} \right) } { \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right) } \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{\left\langle{{E}}\right\rangle = \frac{ \sum_i E_i \exp\left( - \beta E_i/k_{\mathrm{B}} \right)}{ \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The constraint $\beta$ ($=1/T$) is given implicitly by this energy constraint.

Again write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j - \gamma \sum_j N_j p_j.\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unconstrained minimization requires

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\left( k_{\mathrm{B}} (\ln p_i + 1) + \alpha + \beta E_i + \gamma N_i \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) \exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unit probability constraint requires

\begin{aligned}1 &= \sum_j p_j \\ &= \exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) \sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) =\frac{1}{{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability is then

\begin{aligned}\boxed{p_i = \frac{\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The average energy $\left\langle{{E}}\right\rangle = \sum_j p_j E_j$ and average number of particles $\left\langle{{N}}\right\rangle = \sum_j p_j N_j$ are given by

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{E_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

\begin{aligned}\left\langle{{N}}\right\rangle = \frac{N_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

The values $\beta$ and $\gamma$ are fixed implicitly by requiring simultaneous solutions of these equations.

## Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

\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}}(1.36)

where

\begin{aligned}\Gamma(\nu) = \int_0^\infty dy y^{\nu-1} e^{-y}\end{aligned} \hspace{\stretch{1}}(1.37)

denotes the gamma function.

Obtain the behavior of $f_\nu^\pm(z)$ for $z \rightarrow 0$ keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of $f_\nu^\pm(z)$ for $z \rightarrow \infty$ again keeping the two leading terms.

For Bosons, we must have $z \le 1$ (why?), obtain the leading term of $f_\nu^-(z)$ for $z \rightarrow 1$.

For $z \rightarrow 0$ we can rewrite the integrand in a form that allows for series expansion

\begin{aligned}\frac{x^{\nu - 1}}{z^{-1} e^x \pm 1} &= \frac{z e^{-x} x^{\nu - 1}}{1 \pm z e^{-x}} \\ &= z e^{-x} x^{\nu - 1}\left( 1 \mp z e^{-x} + (z e^{-x})^2 \mp (z e^{-x})^3 + \cdots \right)\end{aligned} \hspace{\stretch{1}}(1.38)

For the $k$th power of $z e^{-x}$ in this series our integral is

\begin{aligned}\int_0^\infty dx z e^{-x} x^{\nu - 1} (z e^{-x})^k &= z^{k+1}\int_0^\infty dx x^{\nu - 1} e^{-(k + 1) x} \\ &= \frac{z^{k+1}}{(k+1)^\nu}\int_0^\infty du u^{\nu - 1} e^{- u} \\ &= \frac{z^{k+1}}{(k+1)^\nu} \Gamma(\nu)\end{aligned} \hspace{\stretch{1}}(1.39)

Putting everything back together we have for small $z$

\begin{aligned}\boxed{f_\nu^\pm(z) =z\mp\frac{z^{2}}{2^\nu}+\frac{z^{3}}{3^\nu}\mp\frac{z^{4}}{4^\nu}+ \cdots}\end{aligned} \hspace{\stretch{1}}(1.40)

We’ll expand $\Gamma(\nu) f_\nu^+(e^y)$ about $z = e^y$, writing

\begin{aligned}F_\nu(e^y) &= \Gamma(\nu) f_\nu^+(e^y) \\ &= \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \int_0^y dx \frac{x^{\nu - 1}}{e^{x - y} + 1}+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.41)

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the $x y$ ranges. Observe that in the first integral we have

\begin{aligned}\frac{1}{{2}} \le \frac{1}{e^{x - y} + 1} \le \frac{1}{{1 + e^{-y}}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Since this term is of order 1, let’s consider the difference of this from $1$, writing

\begin{aligned}\frac{1}{e^{x - y} + 1} = 1 + u,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}u = \frac{1}{e^{x - y} + 1} - 1 &= \frac{1 -(e^{x - y} + 1)}{e^{x - y} + 1} \\ &= \frac{-e^{x - y} }{e^{x - y} + 1} \\ &= -\frac{1}{1 + e^{y - x}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives us

\begin{aligned}F_\nu(e^y) &= \int_0^y dx x^{\nu - 1} \left( 1 - \frac{ 1 } { 1 + e^{y - x} } \right)+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \frac{y^\nu}{\nu}-\int_0^y dx \frac{ x^{\nu - 1} } { 1 + e^{y - x} }+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Now let’s make a change of variables $a = y - x$ in the first integral and $b = x - y$ in the second. This gives

\begin{aligned}F_\nu(e^y) = \frac{y^\nu}{\nu}-\int_0^\infty da \frac{ (y - a)^{\nu - 1} } { 1 + e^{a} }+\int_0^\infty db \frac{(y + b)^{\nu - 1}}{e^{b} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

As $a$ gets large in the first integral the integrand is approximately $e^{-a} (y-a)^{\nu -1}$. The exponential dominates this integrand. Since we are considering large $y$, we can approximate the upper bound of the integral by extending it to $\infty$. Also expanding in series we have

\begin{aligned}F_\nu(e^y) &\approx \frac{y^\nu}{\nu}+\int_0^\infty da \frac{ (y + a)^{\nu - 1} -(y - a)^{\nu - 1} } { 1 + e^{a} } \\ &= \frac{y^\nu}{\nu}+\int_0^\infty da \frac{1}{{e^a + 1}}\left( \left( \frac{1}{{0!}} y^{\nu-1} a^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} a^2 + \cdots \right) - \left( \frac{1}{{0!}} y^{\nu-1} (-a)^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} (-a)^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} (-a)^2 + \cdots \right) \right) \\ &= \frac{y^\nu}{\nu}+ 2\int_0^\infty da \frac{1}{{e^a + 1}} \left( \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{3!}} (\nu-1) (\nu-2) (\nu - 3)y^{\nu-4} a^3 + \cdots \right) \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \left( \prod_{k = 1}^j (\nu-k) \right)\int_0^\infty da \frac{a^j}{e^a + 1} \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\int_0^\infty da \frac{a^j}{e^a + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For the remaining integral, Mathematica gives

\begin{aligned}\int_0^\infty da \frac{a^j}{e^a + 1}=\left( 1 - 2^{-j} \right) j! \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

where for $s > 1$

\begin{aligned}\zeta(s) = \sum_{k=1}^{\infty} k^{-s}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives

\begin{aligned}F_\nu(e^y) \approx \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} y^{\nu - 1 - j}\frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\left( 1 - 2^{-j} \right) \zeta(j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}f_\nu^+(e^y) &\approx y^\nu\left( \frac{1}{\nu \Gamma(\nu)} + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ 1 } {\Gamma(\nu - j) y^{j + 1} } \left( 1 - 2^{-j} \right) \zeta(j+1) \right) \\ &= \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ \Gamma(\nu + 1) } {\Gamma(\nu - j) } \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} } \right),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\boxed{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}}(1.0.42)

Evaluating the numerical portions explicitly, with

\begin{aligned}c(j) = 2 \left(1-2^{-j}\right) \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

\begin{aligned}\begin{aligned}c(1) &= \frac{\pi^2}{6} \\ c(3) &= \frac{7 \pi^4}{360} \\ c(5) &= \frac{31 \pi^6}{15120} \\ c(7) &= \frac{127 \pi^8}{604800},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.42)

so to two terms ($j = 1, 3$), we have

\begin{aligned}\boxed{f_\nu^+(e^y) \approx \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + \nu(\nu-1) \frac{\pi^2}{6 y^{2}} + \nu(\nu-1)(\nu-2)(\nu -3) \frac{7 \pi^4}{360 y^4} \right).}\end{aligned} \hspace{\stretch{1}}(1.0.42)

In order for the Boson occupation numbers to be non-singular we require $\mu$ less than all $\epsilon$. If that lowest energy level is set to zero, this is equivalent to $z < 1$. Given this restriction, a $z = e^{-\alpha}$ substitution is convenient for investigation of the $z \rightarrow 1$ case. Following the text, we'll write

\begin{aligned}G_\nu(e^{-\alpha})=\Gamma(\nu)f_\nu^-(e^{-\alpha}) = \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x + \alpha} - 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For $\nu = 1$, this is integrable

\begin{aligned}\frac{d}{dx} \ln\left( 1 - e^{-x - \alpha} \right) &= \frac{e^{-x - \alpha}}{ 1 - e^{-x - \alpha} } \\ &= \frac{1}{{ e^{x + \alpha} - 1}},\end{aligned} \hspace{\stretch{1}}(1.0.42)

so that

\begin{aligned}G_1(e^{-\alpha}) &= \int_0^\infty dx \frac{1}{e^{x + \alpha} - 1} \\ &= {\ln \left( 1 - e^{-x - \alpha} \right)}_{0}^{\infty} \\ &= \ln 1 - \ln \left( 1 - e^{- \alpha} \right) \\ &= -\ln \left( 1 - e^{- \alpha} \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Taylor expanding $1 - e^{-\alpha}$ we have

\begin{aligned}1 - e^{-\alpha} = 1 - \left( 1 - \alpha + \alpha^2/2 - \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Noting that $\Gamma(1) = 1$, we have for the limit

\begin{aligned}\lim_{\alpha \rightarrow 0} G_1(e^{-\alpha}) \rightarrow - \ln \alpha,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\lim_{z\rightarrow 1} f_\nu^-(z)= -\ln (-\ln z).\end{aligned} \hspace{\stretch{1}}(1.0.42)

For values of $\nu \ne 1$, the denominator is

\begin{aligned}e^{\alpha + x} - 1 = (\alpha + x) + (\alpha + x)^2/2 + \cdots\end{aligned} \hspace{\stretch{1}}(1.0.42)

To first order this gives us

\begin{aligned}f_\nu^-( e^{-\alpha} ) \approx \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Of this integral Mathematica says it can be evaluated for $0 < \nu < 1$, and has the value

\begin{aligned}\frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}=\frac{\pi}{\sin(\pi\nu)} \frac{1}{\alpha^{1 - \nu} \Gamma (\nu )}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

From [1] 6.1.17 we find

\begin{aligned}\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)},\end{aligned} \hspace{\stretch{1}}(1.0.42)

with which we can write

\begin{aligned}\boxed{f_\nu^-( e^{-\alpha} ) \approx \frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} }.}\end{aligned} \hspace{\stretch{1}}(1.0.42)

## Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number $A$. i.e., having $A$ total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin-$1/2$ (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius $R = r_0 A^{1/3}$, where $r_0 = 1.4 \times 10^{-13} \text{cm}$, calculate the Fermi energy and the average energy per nucleon in MeV.

Our nucleon particle density is

\begin{aligned}\rho &= \frac{N}{V} \\ &= \frac{A}{\frac{4 \pi}{3} R^3} \\ &= \frac{3 A}{4 \pi r_0^3 A} \\ &= \frac{3}{4 \pi r_0^3} \\ &= \frac{3}{4 \pi (1.4 \times 10^{-13} \text{cm})^3} \\ &= 8.7 \times 10^{37} (\text{cm})^{-3} \\ &= 8.7 \times 10^{43} (\text{m})^{-3}\end{aligned} \hspace{\stretch{1}}(1.67)

With $m$ for the mass of either the proton or the neutron, and $\rho_m = \rho_p = \rho/2$, the Fermi energy for these particles is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{6 \pi (\rho/2)}{2 S + 1} \right)^{2/3},\end{aligned} \hspace{\stretch{1}}(1.68)

With $S = 1/2$, and $2 S + 1 = 2(1/2) + 1 = 2$ for either the proton or the neutron, this is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2 m} \left( \frac{3 \pi^2 \rho}{2} \right)^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.69)

\begin{aligned}\begin{aligned}\hbar &= 1.05 \times 10^{-34} \,\text{m^2 kg s^{-1}} \\ m &= 1.67 \times 10^{-27} \,\text{kg}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.70)

This gives us

\begin{aligned}\epsilon_{\mathrm{F}} &= \frac{(1.05 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27}} \left( \frac{3 \pi^2 }{2} \frac{8.7 \times 10^{43} }{2} \right)^{2/3}\text{m}^4 \frac{\text{kg}^2}{s^2} \frac{1}{{\text{kg}}} \frac{1}{{\text{m}^2}} \\ &= 3.9 \times 10^{-12} \,\text{J} \times \left( 6.241509 \times 10^{12} \frac{\text{MeV}}{J} \right) \approx 24 \text{MeV}\end{aligned} \hspace{\stretch{1}}(1.71)

In lecture 16

we found that the total average energy for a Fermion gas of $N$ particles was

\begin{aligned}E = \frac{3}{5} N \epsilon_{\mathrm{F}},\end{aligned} \hspace{\stretch{1}}(1.72)

so the average energy per nucleon is approximately

\begin{aligned}\frac{3}{5} \epsilon_{\mathrm{F}} \approx 15 \,\text{MeV}.\end{aligned} \hspace{\stretch{1}}(1.73)

## Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at $T = 0$ moving in the gravitational field of a heavy point mass $M$ at the center. Show that the pressure $P$ obeys the equation

\begin{aligned}\frac{dP}{dr} = - \gamma M \frac{\rho(r)}{r^2},\end{aligned} \hspace{\stretch{1}}(1.74)

where $\gamma$ is the gravitational constant, $r$ is the distance from the center, and $\rho(r)$ is the density which only depends on distance from the center.

In the grand canonical scheme the pressure for a Fermion system is given by

\begin{aligned}\beta P V &= \ln Z_{\mathrm{G}} \\ &= \ln \prod_\epsilon \sum_{n = 0}^1 \left( z e^{-\beta \epsilon} \right)^n \\ &= \sum_\epsilon \ln \left( 1 + z e^{-\beta \epsilon} \right).\end{aligned} \hspace{\stretch{1}}(1.75)

The kinetic energy of the particle is adjusted by the gravitational potential

\begin{aligned}\epsilon &= \epsilon_\mathbf{k}- \frac{\gamma m M}{r} \\ &= \frac{\hbar^2 \mathbf{k}^2}{2m}- \frac{\gamma m M}{r}.\end{aligned} \hspace{\stretch{1}}(1.76)

Differentiating eq. 1.75 with respect to the radius, we have

\begin{aligned}\beta V \frac{\partial {P}}{\partial {r}} &= -\beta \frac{\partial {\epsilon}}{\partial {r}}\sum_\epsilon \frac{z e^{-\beta \epsilon}}{ 1 + z e^{-\beta \epsilon} } \\ &= -\beta\left( \frac{\gamma m M}{r^2} \right)\sum_\epsilon \frac{1}{ z^{-1} e^{\beta \epsilon} + 1} \\ &= -\beta\left( \frac{\gamma m M}{r^2} \right)\left\langle{{N}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.77)

Noting that $\left\langle{{N}}\right\rangle m/V$ is the average density of the particles, presumed radial, we have

\begin{aligned}\boxed{\frac{\partial P}{\partial r} &= -\frac{\gamma M}{r^2} \frac{m \left\langle N \right\rangle}{V} \\ &= -\frac{\gamma M}{r^2} \rho(r).}\end{aligned} \hspace{\stretch{1}}(1.0.78)

# 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] Kerson Huang. Introduction to statistical physics. CRC Press, 2001.

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

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

Posted by peeterjoot on March 27, 2013

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

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

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

## probability forms of entropy

Posted by peeterjoot on March 16, 2013

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

## Question: Entropy as probability

[1] points out that entropy can be written as

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

where

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

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

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

In terms of the free and average energies, we have

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

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

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

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

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

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

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

and show

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

With

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

the free energy takes the form

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

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

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

# References

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

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

## 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 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].

## PHY452H1S Basic Statistical Mechanics. Lecture 2: Probability. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 11, 2013

# Disclaimer

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

# Probability

The discrete case is plotted roughly in fig 1.

Fig1: Discrete probability distribution

\begin{aligned}P(x) \ge 1\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}\sum_x P(x) = 1\end{aligned} \hspace{\stretch{1}}(1.0.1b)

A continuous probability distribution may look like fig 2.

Fig2: Continuous probability distribution

\begin{aligned}\mathcal{P}(x) > 0\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\int \mathcal{P}(x) dx = 1\end{aligned} \hspace{\stretch{1}}(1.0.2b)

Probability that event is in the interval $x_1 - x_0 = \Delta x$ is

\begin{aligned}\int_{x_0}^{x_1} \mathcal{P}(x) dx\end{aligned} \hspace{\stretch{1}}(1.0.3)

# Central limit theorem

\begin{aligned}x \leftrightarrow P(x)\end{aligned} \hspace{\stretch{1}}(1.0.4)

Suppose we construct a sum of random variables

\begin{aligned}X = \sum_{i = 1}^N x_i\end{aligned} \hspace{\stretch{1}}(1.0.5)

## Gambling, coin toss

\begin{aligned}x \rightarrow \left\{\begin{array}{l l}+1 & \quad \mbox{Heads} \\ -1 & \quad \mbox{Tails} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.6)

If we ask the question about what the total number of heads minus the total number of tails (do we have excess heads, and by how much).
}

Given an average of

\begin{aligned}\left\langle{{x}}\right\rangle = \mu\end{aligned} \hspace{\stretch{1}}(1.0.7)

and a variance (or squared standard deviation) of

\begin{aligned}\left\langle{{x^2}}\right\rangle - \left\langle{{x}}\right\rangle^2 = \sigma^2\end{aligned} \hspace{\stretch{1}}(1.0.8)

we have for the sum of random variables

\begin{aligned}\lim_{N \rightarrow \infty} P(X)= \frac{1}{{\sigma \sqrt{2 \pi}}} \exp\left( - \frac{ x - N \mu}{2 \sigma^2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}\left\langle{{X}}\right\rangle = N \mu\end{aligned} \hspace{\stretch{1}}(1.0.9b)

\begin{aligned}\left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = N \sigma^2\end{aligned} \hspace{\stretch{1}}(1.0.9c)

To be proven in the notes not here.

## Coin toss

Given

\begin{aligned}P(\text{Heads}) = \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(1.0.10a)

\begin{aligned}P(\text{Tails}) = \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(1.0.10b)

Our probability distribution may look like fig 3.

Fig3: Discrete probability distribution for Heads and Tails coin tosses

Aside: continuous analogue

Note that the continuous analogue of a distribution like this is

\begin{aligned}\mathcal{P}(x) = \frac{1}{{2}} \delta(x - 1) + \frac{1}{{2}} \delta(x + 1)\end{aligned} \hspace{\stretch{1}}(1.0.11)

2 tosses:

\begin{aligned}(x_1, x_2) \in \{(1, 1), (1, -1), (-1, 1), (-1, -1)\}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}X \rightarrow 2, 0, 0, 2\end{aligned} \hspace{\stretch{1}}(1.0.13)

Fig4: 2 tosses distribution

3 tosses

\begin{aligned}(x_1, x_2, x_3) \in \{(1, 1, 1), \cdots (-1, -1, -1)\}\end{aligned} \hspace{\stretch{1}}(1.0.14)

1. $X = 3$ : 1 way
2. $X = 1$ : 3 ways
3. $X = -1$ : 3 ways
4. $X = -3$ : 1 way

Fig5: 3 tosses

N tosses

We want to find $P_N(X)$. We have

\begin{aligned}\text{Tails} - \text{Heads} = X \end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}\text{Total tosses} = \text{Tails} + \text{Heads} = N \end{aligned} \hspace{\stretch{1}}(1.0.16)

So that

1. Heads: $\frac{N - X}{2}$
2. Tails: $\frac{N + X}{2}$

How many ways can we find a specific event such as the number of ways we find $2$ heads and $1$ tail? We can enumerate these $\{ (H, H, T), (H, T, H), (T, H, H)\}$.

The number of ways of choosing just that combination is

\begin{aligned}P_N(X) = \left(\frac{1}{{2}}\right)^N \binom{N}{\frac{N-X}{2}} \quad\mbox{or}\quad\left(\frac{1}{{2}}\right)^N \binom{N}{\frac{N+X}{2}}\end{aligned} \hspace{\stretch{1}}(1.0.17)

we find the 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}}(1.0.18)

We’ll use the Stirling formula 1.0.36, to find

\begin{aligned}\begin{aligned}P_N(X) &= \left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-X}{2}\right)!\left(\frac{N+X}{2}\right)!} \\ &\approx\left( \frac{1}{{2}} \right)^N \frac{ e^{-N} N^N \sqrt{ 2 \pi N} }{ e^{-\frac{N+X}{2}} \left( \frac{N+X}{2}\right)^{\frac{N+X}{2}} \sqrt{ 2 \pi \frac{N+X}{2}} e^{-\frac{N-X}{2}} \left( \frac{N-X}{2}\right)^{\frac{N-X}{2}} \sqrt{ 2 \pi \frac{N-X}{2}} } \\ &=\left( \frac{1}{{2}} \right)^N \frac{ 2 N^N \sqrt{ N} }{ \sqrt{2 \pi}\left( \frac{N+X}{2}\right)^{\frac{N+X}{2}} \sqrt{ N^2 - X^2} \left( \frac{N-X}{2}\right)^{\frac{N-X}{2}} } \\ &=\frac{ 2 N^N \sqrt{ N} }{ \sqrt{2 \pi}\left( N^2 - X^2 \right)^{N/2 + 1/2}\left( \frac{N+X}{N-X}\right)^{X/2} }\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.19)

This can be apparently be simplified to

\begin{aligned}P_N(X) = \frac{2}{\sqrt{2 \pi N}} \exp\left( -\frac{X^2}{2N} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

}

# Stirling formula

To prove the Stirling formula we’ll use the Gamma (related) function

\begin{aligned}I(\alpha) = \int_0^\infty dy e^{-y} y^\alpha\end{aligned} \hspace{\stretch{1}}(1.0.21)

Observe that we have

\begin{aligned}I(0) = \int_0^\infty dy e^{-y} = 1,\end{aligned} \hspace{\stretch{1}}(1.0.22)

and

\begin{aligned}I(\alpha + 1) = \int_0^\infty dy e^{-y} y^{\alpha + 1}= \int_0^\infty d \left( \frac{e^{-y}}{-1} \right) y^{\alpha + 1} = -\int_0^\infty dy \left( \frac{e^{-y}}{-1} \right) (\alpha + 1) y^{\alpha}= (\alpha + 1)I(y).\end{aligned} \hspace{\stretch{1}}(1.0.23)

This induction result means that

\begin{aligned}I(\alpha = N) = N!,\end{aligned} \hspace{\stretch{1}}(1.0.24)

so we can use the large $\alpha$ behaviour of this function to find approximations of the factorial. What does the innards of this integral (integrand) look like. We can plot these fig 7, and find a hump for any non-zero value of $\alpha$ ($\alpha = 0$ is just a line)

Fig8: Some values of Stirling integrand

There’s a peak for large alpha that can be approximated by a Gaussian function. When $\alpha$ is large enough then we can ignore the polynomial boundary effects. We want to look at where this integrand is peaked. We can write

\begin{aligned}I(\alpha) = \int_0^\infty dy e^{-y + \alpha \ln y} = \int_0^\infty dy f(y),\end{aligned} \hspace{\stretch{1}}(1.0.25)

and look for where $f(y)$ is the largest. We’ve set

\begin{aligned}f(y) = -y + \alpha \ln y,\end{aligned} \hspace{\stretch{1}}(1.0.26)

and want to look at where

\begin{aligned}0 = {\left.{{f'(y)}}\right\vert}_{{y^{*}}}= -1 + \frac{\alpha}{y^{*}}\end{aligned} \hspace{\stretch{1}}(1.0.27)

so that the peak value

\begin{aligned}y^{*} = \alpha.\end{aligned} \hspace{\stretch{1}}(1.0.28)

We now want to expand the integrand around this peak value

\begin{aligned}I(\alpha) = \int_0^\infty\exp\left(f(y^{*}) + {\left.{{\frac{\partial {f}}{\partial {y}}}}\right\vert}_{{y^{*}}} (y - y^{*}) + \frac{1}{{2}}{\left.{{\frac{\partial^2 {{f}}}{\partial {{y}}^2}}}\right\vert}_{{y^{*}}}(y - y^{*})^2+ \cdots\right)\end{aligned} \hspace{\stretch{1}}(1.0.29)

We’ll drop all but the quadratic term, and first need the second derivative

\begin{aligned}f''(y) = \frac{d}{dy} \left(-1 + \alpha \frac{1}{{y}}\right)= -\alpha \frac{1}{{y^2}},\end{aligned} \hspace{\stretch{1}}(1.0.30)

at $y = y^{*} = \alpha$ we have $f''(y^{*}) = -\frac{1}{{\alpha}}$ and

\begin{aligned}I(\alpha \gg 1) \approx e^{f(y^{*})}\int_0^\infty\exp\left(\frac{1}{{2}}{\left.{{\frac{\partial^2 {{f}}}{\partial {{y}}^2}}}\right\vert}_{{y^{*}}}(y - y^{*})^2\right)=e^{f(\alpha)}\int_0^\infty dy e^{-\frac{(y - \alpha)^2}{2 \alpha}}\end{aligned} \hspace{\stretch{1}}(1.0.31)

For the integral Mathematica gives

\begin{aligned}\int_0^\infty dy e^{-\frac{(y - \alpha)^2}{2 \alpha}}=\sqrt{\frac{\pi \alpha }{2}} \left(\text{erf} \left(\sqrt{\frac{\alpha }{2}}\right)+1\right).\end{aligned} \hspace{\stretch{1}}(1.0.32)

From fig 8 observe that $\text{erf}(x) \rightarrow 1$ in the limit

So we have for large $\alpha$

\begin{aligned}\int_0^\infty dy e^{-\frac{(y - \alpha)^2}{2 \alpha}}\approx \sqrt{2 \pi \alpha}\end{aligned} \hspace{\stretch{1}}(1.0.33)

\begin{aligned}e^{f(\alpha)} = e^{-\alpha + \alpha \ln \alpha}\end{aligned} \hspace{\stretch{1}}(1.0.34)

We have for $\alpha \gg 1$

\begin{aligned}I(\alpha) \approx e^{-\alpha} e^{\alpha \ln \alpha} \sqrt{2 \pi \alpha}\end{aligned} \hspace{\stretch{1}}(1.0.35)

This gives us the Stirling approximation

\begin{aligned}\boxed{N! \approx \sqrt{ 2 \pi N} N^N e^{-N},}\end{aligned} \hspace{\stretch{1}}(1.0.36)

## PHY452H1S Basic Statistical Mechanics. Lecture 1: What is statistical mechanics and equilibrium. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 10, 2013

[Click here for a PDF of this post with nicer formatting and figures if the post had any]

# Disclaimer

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

# Equilibrium

This is the study of systems in equilibrium. What is equilibrium?

1. Mechanical equilibrium example: ball in bowl
2. Chemical equilibrium: when all the reactants have been consumed, or rates of forward and backwards reactions have become more constant.

More generally, equilibrium is a matter of time scales!

Feynman
“Fast things have happened and all slow things have not”

## Water in a cup

After 1 min – 1 hour, when any sloshing has stopped, we can say it’s in equilibrium. However, if we consider a time scale like 10 days, we see that there are changes occurring (evaporation).

## Hot water in a cup

Less than 10 minute time scale: Not in equilibrium (evaporating). At a longer time scale we may say it’s reached equilibrium, but again on a, say, 10 day time scale, we’d again reach the conclusion that this system is also not in equilibrium.

## Window glass

1. $\sim 10$ years $\rightarrow$ equilibrium.
2. $\sim 100-1000$ years $\rightarrow$ not in equilibrium.

While this was given in class as an example, [1] refutes this.

## Battery and resistor

Steady current over a small time scale, but we are persistently generating heat and draining the battery.

How do we reach equilibrium?

1. We’ll be looking at small systems connected to a very much larger “heat bath” or “environment”. Such a system will eventually, after perhaps exchange of particles, radiation, … will eventually take on a state that is dictated by the state of the environment.
2. Completely isolated system! Molecules in a box, say, may all initially be a corner of a box. We’ll have exchange of energy and momentum, and interaction with the walls of the box. The “final” state of the system will be determined by the initial state of the molecules. Experiments of this sort have recently been performed and studied in depth (actually a very tricky problem).

# Probabilities

Why do I need probabilities?

1. QM: Ultimately there’s an underlying quantum state, and we can only talk about probabilities. Any measurement has uncertainties, and this microscopic state forces us to use statistical methods.
2. Classical chaos and unpredictability: In a many particle system, or even a system of a few interacting particles with non-linear interaction, even given an initial state to 10 decimal places, we’ll end up with uncertainties. Given enough particles, even with simple interactions, we’ll be forced to use statistical methods.
3. Too much information: Even if we suppose that we could answer the question of where is and how fast every particle in a large collection was, what would we do with that info. We don’t care about such a fine granularity state. We want to know about things like gas pressure. We are forced to discard information.

We need a systematic way of dealing with and discarding specific details. We want to use statistical methods to throw away useless information.

Analogy: Election

We can poll a sample of the population if we want to attempt to predict results. We want to know things, but not what every person thinks.

# References

[1] YM Stokes. Flowing windowpanes: fact or fiction? Proceedings of the Royal Society of London. Series A: Mathematical, Physical and Engineering Sciences, 455 (1987): 2751–2756, 1999. URL http://albertorojo.com/BlogsTN/GlassStokes.pdf.

## My submission for PHY356 (Quantum Mechanics I) Problem Set 3.

Posted by peeterjoot on November 30, 2010

# Problem 1.

## Statement

A particle of mass $m$ is free to move along the x-direction such that $V(X)=0$. The state of the system is represented by the wavefunction Eq. (4.74)

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k x} e^{- i \omega t} f(k)\end{aligned} \hspace{\stretch{1}}(1.1)

with $f(k)$ given by Eq. (4.59).

\begin{aligned}f(k) &= N e^{-\alpha k^2}\end{aligned} \hspace{\stretch{1}}(1.2)

Note that I’ve inserted a $1/\sqrt{2\pi}$ factor above that isn’t in the text, because otherwise $\psi(x,t)$ will not be unit normalized (assuming $f(k)$ is normalized in wavenumber space).

\begin{itemize}
\item
(a) What is the group velocity associated with this state?
\item
(b) What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(c) What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?
\item
(d) Explain the physical meaning of the above results.
\end{itemize}

## Solution

### (a). group velocity.

To calculate the group velocity we need to know the dependence of $\omega$ on $k$.

Let’s step back and consider the time evolution action on $\psi(x,0)$. For the free particle case we have

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} = -\frac{\hbar^2}{2m} \partial_{xx}.\end{aligned} \hspace{\stretch{1}}(1.3)

Writing $N' = N/\sqrt{2\pi}$ we have

\begin{aligned}-\frac{i t}{\hbar} H \psi(x,0) &= \frac{i t \hbar }{2m} N' \int_{-\infty}^\infty dk (i k)^2 e^{i k x - \alpha k^2} \\ &= N' \int_{-\infty}^\infty dk \frac{-i t \hbar k^2}{2m} e^{i k x - \alpha k^2}\end{aligned}

Each successive application of $-iHt/\hbar$ will introduce another power of $-it\hbar k^2/2 m$, so once we sum all the terms of the exponential series $U(t) = e^{-iHt/\hbar}$ we have

\begin{aligned}\psi(x,t) =N' \int_{-\infty}^\infty dk \exp\left( \frac{-i t \hbar k^2}{2m} + i k x - \alpha k^2 \right).\end{aligned} \hspace{\stretch{1}}(1.4)

Comparing with 1.1 we find

\begin{aligned}\omega(k) = \frac{\hbar k^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.5)

This completes this section of the problem since we are now able to calculate the group velocity

\begin{aligned}v_g = \frac{\partial {\omega(k)}}{\partial {k}} = \frac{\hbar k}{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

## (b). What is the probability for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

In order to evaluate the probability, it looks desirable to evaluate the wave function integral 1.4.
Writing $2 \beta = i/(\alpha + i t \hbar/2m )$, the exponent of that integral is

\begin{aligned}-k^2 \left( \alpha + \frac{i t \hbar }{2m} \right) + i k x&=-\left( \alpha + \frac{i t \hbar }{2m} \right) \left( k^2 - \frac{i k x }{\alpha + \frac{i t \hbar }{2m} } \right) \\ &=-\frac{i}{2\beta} \left( (k - x \beta )^2 - x^2 \beta^2 \right)\end{aligned}

The $x^2$ portion of the exponential

\begin{aligned}\frac{i x^2 \beta^2}{2\beta} = \frac{i x^2 \beta}{2} = - \frac{x^2 }{4 (\alpha + i t \hbar /2m)}\end{aligned}

then comes out of the integral. We can also make a change of variables $q = k - x \beta$ to evaluate the remainder of the Gaussian and are left with

\begin{aligned}\psi(x,t) =N' \sqrt{ \frac{\pi}{\alpha + i t \hbar/2m} } \exp\left( - \frac{x^2 }{4 (\alpha + i t \hbar /2m)} \right).\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that from 1.2 we can compute $N = (2 \alpha/\pi)^{1/4}$, which could be substituted back into 1.7 if desired.

Our probability density is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 &=\frac{1}{{2 \pi}} N^2 {\left\lvert{ \frac{\pi}{\alpha + i t \hbar/2m} }\right\rvert} \exp\left( - \frac{x^2}{4} \left( \frac{1}{{(\alpha + i t \hbar /2m)}} + \frac{1}{{(\alpha - i t \hbar /2m)}} \right) \right) \\ &=\frac{1}{{2 \pi}} \sqrt{\frac{2 \alpha}{\pi} } \frac{\pi}{\sqrt{\alpha^2 + (t \hbar/2m)^2 }} \exp\left( - \frac{x^2}{4} \frac{1}{{\alpha^2 + (t \hbar/2m)^2 }} \left( \alpha - i t \hbar /2m + \alpha + i t \hbar /2m \right)\right) \\ &=\end{aligned}

With a final regrouping of terms, this is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 =\sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right).\end{aligned} \hspace{\stretch{1}}(1.8)

As a sanity check we observe that this integrates to unity for all $t$ as desired. The probability that we find the particle at position $x > x_0$ is then

\begin{aligned}P_{x>x_0}(t) = \sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\int_{x=x_0}^\infty dx \exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right)\end{aligned} \hspace{\stretch{1}}(1.9)

The only simplification we can make is to rewrite this in terms of the complementary error function

\begin{aligned}\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty e^{-t^2} dt.\end{aligned} \hspace{\stretch{1}}(1.10)

Writing

\begin{aligned}\beta(t) = \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 },\end{aligned} \hspace{\stretch{1}}(1.11)

we have

\begin{aligned}P_{x>x_0}(t_0) = \frac{1}{{2}} \text{erfc} \left( \sqrt{\beta(t_0)/2} x_0 \right)\end{aligned} \hspace{\stretch{1}}(1.12)

Sanity checking this result, we note that since $\text{erfc}(0) = 1$ the probability for finding the particle in the $x>0$ range is $1/2$ as expected.

## (c). What is the probability per unit length for measuring the particle at position $x=x_0>0$ at time $t=t_0>0$?

This unit length probability is thus

\begin{aligned}P_{x>x_0+1/2}(t_0) - P_{x>x_0-1/2}(t_0) &=\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0+\frac{1}{{2}} \right) \right) -\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0-\frac{1}{{2}} \right) \right) \end{aligned} \hspace{\stretch{1}}(1.13)

## (d). Explain the physical meaning of the above results.

To get an idea what the group velocity means, observe that we can write our wavefunction 1.1 as

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k (x - v_g t)} f(k)\end{aligned} \hspace{\stretch{1}}(1.14)

We see that the phase coefficient of the Gaussian $f(k)$ “moves” at the rate of the group velocity $v_g$. Also recall that in the text it is noted that the time dependent term 1.11 can be expressed in terms of position and momentum uncertainties $(\Delta x)^2$, and $(\Delta p)^2 = \hbar^2 (\Delta k)^2$. That is

\begin{aligned}\frac{1}{{\beta(t)}} = (\Delta x)^2 + \frac{(\Delta p)^2}{m^2} t^2 \equiv (\Delta x(t))^2\end{aligned} \hspace{\stretch{1}}(1.15)

This makes it evident that the probability density flattens and spreads over time with the rate equal to the uncertainty of the group velocity $\Delta p/m = \Delta v_g$ (since $v_g = \hbar k/m$). It is interesting that something as simple as this phase change results in a physically measurable phenomena. We see that a direct result of this linear with time phase change, we are less able to find the particle localized around it’s original time $x = 0$ position as more time elapses.

# Problem 2.

## Statement

A particle with intrinsic angular momentum or spin $s=1/2$ is prepared in the spin-up with respect to the z-direction state ${\lvert {f} \rangle}={\lvert {z+} \rangle}$. Determine

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.16)

and

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.17)

and explain what these relations say about the system.

## Solution: Uncertainty of $S_z$ with respect to ${\lvert {z+} \rangle}$

Noting that $S_z {\lvert {f} \rangle} = S_z {\lvert {z+} \rangle} = \hbar/2 {\lvert {z+} \rangle}$ we have

\begin{aligned}{\langle {f} \rvert} S_z {\lvert {f} \rangle} = \frac{\hbar}{2} \end{aligned} \hspace{\stretch{1}}(2.18)

The average outcome for many measurements of the physical quantity associated with the operator $S_z$ when the system has been prepared in the state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$ is $\hbar/2$.

\begin{aligned}\Bigl(S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \Bigr) {\lvert {f} \rangle}&= \frac{\hbar}{2} {\lvert {f} \rangle} -\frac{\hbar}{2} {\lvert {f} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.19)

We could also compute this from the matrix representations, but it is slightly more work.

Operating once more with $S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ on the zero ket vector still gives us zero, so we have zero in the root for 2.16

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = 0\end{aligned} \hspace{\stretch{1}}(2.20)

What does 2.20 say about the state of the system? Given many measurements of the physical quantity associated with the operator $V = (S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1})^2$, where the initial state of the system is always ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, then the average of the measurements of the physical quantity associated with $V$ is zero. We can think of the operator $V^{1/2} = S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1}$ as a representation of the observable, “how different is the measured result from the average ${\langle {f} \rvert} S_z {\lvert {f} \rangle}$”.

So, given a system prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$, and performance of repeated measurements capable of only examining spin-up, we find that the system is never any different than its initial spin-up state. We have no uncertainty that we will measure any difference from spin-up on average, when the system is prepared in the spin-up state.

## Solution: Uncertainty of $S_x$ with respect to ${\lvert {z+} \rangle}$

For this second part of the problem, we note that we can write

\begin{aligned}{\lvert {f} \rangle} = {\lvert {z+} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ).\end{aligned} \hspace{\stretch{1}}(2.21)

So the expectation value of $S_x$ with respect to this state is

\begin{aligned}{\langle {f} \rvert} S_x {\lvert {f} \rangle}&=\frac{1}{{2}}( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) S_x ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) \\ &=\hbar ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( {\lvert {x+} \rangle} - {\lvert {x-} \rangle} ) \\ &=\hbar ( 1 + 0 + 0 -1 ) \\ &= 0\end{aligned}

After repeated preparation of the system in state ${\lvert {f} \rangle}$, the average measurement of the physical quantity associated with operator $S_x$ is zero. In terms of the eigenstates for that operator ${\lvert {x+} \rangle}$ and ${\lvert {x-} \rangle}$ we have equal probability of measuring either given this particular initial system state.

For the variance calculation, this reduces our problem to the calculation of ${\langle {f} \rvert} S_x^2 {\lvert {f} \rangle}$, which is

\begin{aligned}{\langle {f} \rvert} S_x^2 {\lvert {f} \rangle} &=\frac{1}{{2}} \left( \frac{\hbar}{2} \right)^2 ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( (+1)^2 {\lvert {x+} \rangle} + (-1)^2 {\lvert {x-} \rangle} ) \\ &=\left( \frac{\hbar}{2} \right)^2,\end{aligned}

so for 2.22 we have

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = \frac{\hbar}{2}\end{aligned} \hspace{\stretch{1}}(2.22)

The average of the absolute magnitude of the physical quantity associated with operator $S_x$ is found to be $\hbar/2$ when repeated measurements are performed given a system initially prepared in state ${\lvert {f} \rangle} = {\lvert {z+} \rangle}$. We saw that the average value for the measurement of that physical quantity itself was zero, showing that we have equal probabilities of measuring either $\pm \hbar/2$ for this experiment. A measurement that would show the system was in the x-direction spin-up or spin-down states would find that these states are equi-probable.

I lost one mark on the group velocity response. Instead of 3.23 he wanted

\begin{aligned}v_g = {\left. \frac{\partial {\omega(k)}}{\partial {k}} \right\vert}_{k = k_0}= \frac{\hbar k_0}{m} = 0\end{aligned} \hspace{\stretch{1}}(3.23)

since $f(k)$ peaks at $k=0$.

I’ll have to go back and think about that a bit, because I’m unsure of the last bits of the reasoning there.

I also lost 0.5 and 0.25 (twice) because I didn’t explicitly state that the probability that the particle is at $x_0$, a specific single point, is zero. I thought that was obvious and didn’t have to be stated, but it appears expressing this explicitly is what he was looking for.

Curiously, one thing that I didn’t loose marks on was, the wrong answer for the probability per unit length. What he was actually asking for was the following

\begin{aligned}\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 - \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x_0, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.24)

That’s a whole lot more sensible seeming quantity to calculate than what I did, but I don’t think that I can be faulted too much since the phrase was never used in the text nor in the lectures.