Temporary HW

Problem 1

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

${\hat {H}}=-\mu _{B}{\vec {\sigma }}\cdot \mathbf {B}$

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

${\hat {\sigma }}_{x}={\begin{pmatrix}0&1\\1&0\end{pmatrix}},$

${\hat {\sigma }}_{x}={\begin{pmatrix}0&-i\\i&0\end{pmatrix}},$

${\hat {\sigma }}_{x}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}.$

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

(b) Repeat the calculation from (a) assuming that $\mathbf {B}$ 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:

${\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {m\omega ^{2}q^{2}}{2}}$,

where ${\hat {p}}={\frac {\hbar }{i}}{\frac {d}{dq}}$.

(a) Find the partition function $Z$ in quantum canonical ensemble at temperature T.

(b) Using result from (a), calculate the averge energy $E=\langle {\hat {H}}\rangle$.

(c) Write down the formal expression for the canonical density operator ${\hat {\rho }}$ in terms of eigenstates $|n\rangle$ and energy levels $\varepsilon _{n}=\hbar \omega (n+1/2)$.

(d) Using result in (c), write down the density matrix in a coordinate representation $\langle q'|{\hat {\rho }}|q\rangle$.

(e) In the coordinate representation, calculate explicitly $\langle q'|{\hat {\rho }}|q\rangle$ in the high temperature limit $T\rightarrow \infty$. HINT: One approach is to apply the following result

Failed to parse (syntax error): {\displaystyle e^{\beta \hat{A}) e^{\beta \hat{B}} = exp[\beta(\hat{A} + \hat{B}) + \beta^2 [\hat{A},\hat{B}]/2 + O(\beta^3)] }

which you can apply to the Boltzmann operator:

Failed to parse (unknown function "\q"): {\displaystyle 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 } and higher since $\beta$ is very small in this limit.