Temporary HW: Difference between revisions

Revision as of 14:13, 23 February 2011

The Hamiltonian for an electron spin degree of freedom in the external magnetic field $𝐁$ is given by:

$\hat{H} = - μ_{B} \vec{σ} \cdot 𝐁$

where $μ_{B}$ is the Bohr magneton </math> and $\vec{σ} = ({\hat{σ}}_{x}, {\hat{σ}}_{y}, {\hat{σ}}_{z})$ is the vector of the Pauli matrices:

${\hat{σ}}_{x} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}),$

${\hat{σ}}_{x} = (\begin{matrix} 0 & - i \\ i & 0 \end{matrix}),$

${\hat{σ}}_{x} = (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) .$

(a) In the quantum canonical ensemble, evaluate the density matrix if $𝐁$ is along the z axis.

(b) Repeat the calculation from (a) assuming that $𝐁$ points along the x axis.

(c) Calculate the average energy in each of the above cases.

@@ Line 5: / Line 5: @@
 <math> \hat{H} = - \mu_B  \vec{\sigma} \cdot \mathbf{B} </math>
-where <math> \mu_B 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:
+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 =
@@ Line 33: / Line 33: @@
 (c) Calculate the average energy in each of the above cases.
 == Problem 2 ==