Temporary HW: Difference between revisions

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== Problem 1 ==


The Hamiltonian for an electron spin degree of freedom in the external magnetic field <math> \mathbf{B} </math> is given by:
<math> \hat{H} = - \mu_B  \vec{\sigma} \cdot \mathbf{B} </math>
where <math> \mu_B </math> is the Bohr magneton </math> and <math> \vec{\sigma}=(\hat{\sigma}_x,\hat{\sigma}_y,\hat{\sigma}_z) </math> is the vector of the Pauli matrices:
<math> \hat{\sigma}_x =
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
</math>
<math> \hat{\sigma}_x =
\begin{pmatrix}
0 & -i \\
i & 0
\end{pmatrix},
</math>
<math> \hat{\sigma}_x =
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}.
</math>
(a) In the quantum canonical ensemble, evaluate the density matrix if <math> \mathbf{B} </math> is along the ''z'' axis.
(b) Repeat the calculation from (a) assuming that <math> \mathbf{B} </math> points along the ''x'' axis.
(c) Calculate the average energy in each of the above cases.
== Problem 2 ==
Consider a single one-dimensional quantum harmonic oscillator described by the Hamiltonian:
<math> \hat{H}=\frac{\hat{p}^2}{2m} + \frac{m \omega^2 q^2}{2} </math>,
where <math> \hat{p} = \frac{\hbar}{i} \frac{d}{dq} </math>.
(a) Find the partition function <math> Z </math> in quantum canonical ensemble at temperature T.
(b) Using result from (a), calculate the averge energy <math> E =  \langle \hat{H} \rangle </math>.
(c) Write down the formal expression for the canonical density operator <math> \hat{\rho} </math> in terms of  eigenstates <math> |n\rangle </math> and
energy levels <math> \varepsilon_n = \hbar \omega (n + 1/2) </math>.
(d) Using result in (c), write down the density matrix in a coordinate representation <math> \langle q' |\hat{\rho}|q\rangle </math>.
(e) In the coordinate representation, calculate explicitly <math> \langle q' |\hat{\rho}|q\rangle </math> in the high temperature limit <math> T \rightarrow \infty </math>. HINT: One approach is to apply the following result
<math> e^{\beta \hat{A}) e^{\beta \hat{B}} = exp[\beta(\hat{A} + \hat{B}) + \beta^2[A,B]/2 + O(\beta^3)] </math>
which you can apply to the Boltzmann operator:
<math> exp(-beta\hat{H}) = exp(-\beta \frac{\hat{p}^2}{2m} - \beta  \frac{m\omega^2\q^2}{2}) <math>
while neglecting terms of order <math> \beta^2 </math>  and higher since <math> \beta </math> is very small in this limit.

Latest revision as of 15:48, 3 May 2011