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\sect{Photon gas as collection of harmonic oscillators}


\[H=\frac{\hat{p}^{2}}{2m}+\frac{1}{2}mw^{2}\hat{x}^{2}\]
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we define ${\hat{n}=a^{+}a, \,a=\sqrt{\frac{w}{2}}\hat{x}-i
\frac{1}{\sqrt{2w}}\hat{p}}$, and then it's possible to 
write ${H=w\hat{n}+ const}$.
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If ${const=\frac{1}{2}w}$, ${\hat{n}}$ is an operator with self
states ${|n>}$ and the matching eigenvalues are ${n=0,1,2...}$.
therefore, the stationary states of the oscillator are
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\[|n>,\,E_{n}=nw\]
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The next reinterpretation is acceptable for the states ${|n>}$:
${|n>}$ is a state where ${n}$ particles ("bozons") occupies
uniparticle level, with energy ${w}$.
When the system, described as ${H}$ is in equilibrium with the
environment
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\[P_{n}=\frac{1}{Z\left(\beta\right)}e^{-\beta
E_{n}}=\frac{1}{Z\left(\beta\right)}e^{-\beta wn}\]
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Prove ${Z\left(\beta\right)=\frac{1}{1-e^{-\beta w}}}$
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\[\langle n\rangle=\frac{1}{e^{\beta w}-1}\equiv f\left(w\right)\]
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${f\left(w\right)}$ is called Bose's occupation function.



The hamiltonian that describes the electromagnetic field, if we use
'normal coordinates', is a collection of independent oscillators.
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\[H=\sum_{k\alpha}w_{k\alpha}\hat{n_{k\alpha}}\]
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We give the hamiltonian the next reinterpretation: ${|k,\alpha>}$ is
a uniparticle state of photon with ${k}$ momentum and ${\alpha}$
polarization. Periodic boundary conditions 
are ${\vec{k}\rightarrow\frac{2\pi}{L}\left(m_{1},m_{2},m_{3}\right)}$. 
Two transversal polarization directions ${\alpha=1,2}$.  
The photon energy in the state ${|k,\alpha>}$ is ${w_{k\alpha}=|\vec{k}|}$. 
The state ${|...n_{k\alpha}...>}$ is a multiparticle occupation state 
with ${n_{k\alpha}}$ photons, settelling the uniparticle state ${|k\alpha>}$. 
The occupation states ${|...n_{k\alpha}...>}$ are the stationary
states of the system, and we get
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\[E_{\left(...n_{k\alpha}...\right)}=\sum_{k\alpha}n_{k\alpha}w_{k\alpha}\]
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In thermal equilibrium  of the electromagnetic field, described with
${H}$, with the environment
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\[P\left(...n_{k\alpha}...\right)=\frac{1}{Z}e^{-\beta
E\left(...n_{k\alpha}...\right)}\]



\begin {itemize}

\item[(a)]
Calculate the average number of photons ${N\left(T,V\right)}$.
The temperature and the volume of the medium are given.

\item[(b)] 
Calculate the distribution function ${Z\left(\beta,V\right)}$.
Use factorization and integration in parts.

\item[(c)] 
Calculate the thermic energy of the field in two ways. show
that you get
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\[E\left(T,V\right)=\frac{\pi^{2}}{15}V T^{4}\]
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\item[(d)]
Find the radiation pressure. Is it possible to use the results from
exercise 17 with ${\nu=1}$. Show that any way,

Pressure${=\frac{1}{3}\left(\frac{E}{V}\right)}$.

\end {itemize}


To solve paragraphs (a) - (d) use the next hints. always remember
that photon gas\,${=}$\,electromagnetic field\,${=}$\,oscillator's
collection, so there's no need to panic...

Hints to solve (a) - (d)


\begin {itemize}

\item[(a)]
The photon's number ${N\left(T,V\right)=\sum_{k\alpha}\langle n_{k\alpha}\rangle}$
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\[\sum_{k\alpha}\rightarrow 2+S\left(\frac{L}{2\pi}\right)^{2}d^{2}k, \langle
n_{k\alpha}\rangle=f\left(w_{k\alpha}\right)\]
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\[\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=2\zeta
\left(3\right)=2\left(\frac{1}{1^{3}}+\frac{1}{2^{3}}+\frac{1}{3^{3}}...\right)\]
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\item[(b)]
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\[\ln Z\left(\beta\right)=\sum_{k\alpha}\ln Z^{\left
(k,\alpha\right)}\left(\beta\right)=...\]
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\[\int_{0}^{\infty}\frac{x^{2}dx}{e^{x}-1}=6
\zeta\left(4\right)=\frac{\pi}{15}^{4}\]
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\item[(c)]
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\[E\left(T,V\right)=\sum_{k\alpha}w_{k\alpha}\langle
n_{k\alpha}\rangle=...\]
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\[E\left(T,V\right)=-\frac{\partial\ln Z}{\partial\beta}=...\]
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\item[(e)]
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\[{P}\left(T,V\right)=\frac{1}{\beta}\,\frac{\partial\ln
Z}{\partial V}=...\]
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\end {itemize}

Additional materials about photon gas will be learned in the frame
of the grand canonical ensemble. Formally, photon
gas is like Bose gas with chemical potential ${\mu=0}$.




