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## Midterm II reflection

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.)]

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

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

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

Our partition function is

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

Writing

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

that is

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

Substitution of $a$ gives us

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

To calculate the magnetization $M$, I used

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

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

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

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

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

Compare this to the in-class definition of magnetization

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

For this derivative we have

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

This gives us

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

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

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

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

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

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

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

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

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

The canonical partition function is

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

With

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

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

the momentum integrals are

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

Writing

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

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

we have

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

The mean energy is

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

The specific heat follows by differentiating once more

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

Differentiating the integral terms we have, for example,

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

so that the specific heat is

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

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

# References

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