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\sect{Electron gas in magnetic field}

The interaction of the electron with the electromagnetic field is
described by the hamiltonian ${\left(e=-|e|\right)}$:
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\[\hat{H}=\frac{1}{2m}\left(\hat{p}-e\hat{A}\right)^{2}-rB\cdot
\hat{\sigma}\]
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For an homogenic magnetic field along axis ${Z}$ it's possible to
choose ("Landau scaling"):
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\[\overrightarrow{A}=\left(0,Bx,0\right)\]
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\[\overrightarrow{B}=\nabla \times A=\left(0,0,B\right)\]
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Notice that the degree of freedom of the movement along axis ${Z}$
doesn't depend on both degrees of freedom of the orbital movement in
the plane vertical to axis ${Z}$.
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The self states of the hamiltonian are
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\[\hat{H}|p_{2}nr\sigma_{2}>=E_{p_{2}nr\sigma_{2}}|p_{2}nr\sigma_{2}>\]
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\[E_{p_{2}nr\sigma_{2}}=\frac{p_{2}^{2}}{2m}+\left(\frac{1}{2}+n\right)w_{c}-\gamma
B \sigma_{2}\]
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\[w_{c}\equiv \frac{eB}{m},-\infty \leq p_{2}\leq \infty\]
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\[n=0,1,2...,\sigma_{2}=\pm 1\]
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\[r=1,2...\frac{L^{2}}{2\pi}eB \,\,\mbox{(Landau's direction)}\]
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Calculate the distribution function ${Z\left(\beta\right)}$ for the
electron gas, assuming that it's possible to  relate to the
electrons like Boltzman's particles.
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\[Z_{N}=\frac{1}{N!}Z_{1}^{N}\]
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From here, derive expressions for the magnetization ${\tilde{M}}$
and for the susceptibility ${\chi}$. Notice the expressions you got
includes two terms



\begin {itemize}
\item[(a)]
Diamagnetic term because of the "orbital" interaction of the
electron with the field.

\item[b]
Paramagnetic term because of the interaction of the electron spin
with the field.

\end {itemize}

Diamagnetism is a quantal effect. To prove this argument, take the
hamiltonian ${\hat{H}}$ omitting the spin's degree of freedom and
calculate the distribution function by the definition of the
classical statistical mechanics.
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\[Z_{1}\left(\beta,B\right)\equiv \int\frac{d^{3}x
d^{2}p}{\left(2\pi\right)^{3}}e^{-\beta}H\left(\vec{x},\vec{p}\right)
\]




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{\bf Exercise appendix}

This appendix includes a simple way to get to the expression for the
self energies of the Hamiltonian ${\hat{H}}$ given in the exercise.
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\[\hat{H}=H\left(\hat{Z}\hat{P}_{2};\hat{y}\hat{p}_{y},
\hat{X}\hat{P}_{X};\hat{\sigma}_{2}\right)\]
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We'll define a new set of coordinates
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\[\hat{\Pi}_{x}=m\frac{d\hat{x}}{dt}=mi[\hat{H},\hat{x}]=\hat{p}_{x}-e\hat{A}_{x}\]
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\[\hat{\Pi}_{y}=m\frac{d\hat{y}}{dt}=mi[\hat{H},\hat{y}]=\hat{p}_{y}-e\hat{A}_{y}\]
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\[\hat{X}\equiv \hat{X}+\frac{1}{mw_{2}}\hat{\Pi}_{y}\]
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\[\hat{Y}\equiv \hat{y}+\frac{1}{mw_{2}}\hat{\Pi}_{x}\]
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Classical ${\Pi_{x}\Pi_{y}}$ is the mechanical momentum of the
electron, and ${XY}$ are the expressions for the center of the
helix.

note: ${[\hat{\Pi}_{x},\hat{\Pi}_{y}]=imw_{2}}$
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\[\left[\hat{X},\hat{Y}\right]=-i\frac{1}{mw_{2}}\]
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${X,Y}$ commute with ${\hat{\pi}_{x},\hat{\pi}_{y}}$, therefore,
${\frac{1}{mw_{c}}\hat{\pi}_{y}}$ is the conjugate of
${\hat{\pi}_{x}}$ and ${mw_{c}\hat{X}}$ is the conjugate of
${\hat{Y}}$.

We'll write the hamiltonian through the canonical new coordinates.
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\[\hat{H}=H
\left(\hat{z},\hat{p}_{z};\hat{\pi}_{x},\frac{1}{mw_{c}}\hat{\pi}_{y};
\hat{Y},mw_{c}\hat{X};\hat{\sigma}_{z}\right)=\]
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\[=\frac{\hat{p}_{2}^{2}}{2m}+\left(\frac{\hat{\pi}_{2}^{2}}{2m}+\frac{1}{2}mw_{c}^{2}\left(\frac{\hat{\pi}}{mw_{c}}\right)^{2}\right)-\gamma B\hat{\sigma}_{z}\]
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notice that the freedom degree ${\hat{X}\hat{Y}}$ doesn't appear in
${\hat{H}}$ (as expected) because ${\hat{X}\hat{Y}}$ are movement
constants.

immediately we see that ${\hat{H}}$ includes transeltore freedom
degree (direction ${z}$), harmonically freedom degree, spin's
freedom degree and degenerated ${\hat{X}\hat{Y}}$ freedom degree.
There for the self energies of the hamiltonian are:
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\[E_{p_{2}nr\sigma_{1}}=\frac{p_{2}^{2}}{2m}+\left(\frac{1}{2}+n\right)w_{c}-rB\sigma_{z}\]
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It's possible to set Landau's degeneration
${r=1,2...\frac{L^{2}}{2\pi}eB}$ from the semy classical equation
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\[\mathcal{N}=\int\int \frac{d\left(mw_{c}X\right)dY}{2\pi}\]




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