Temporary HW

Problem 1

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

$\hat{H}=-{\mu }_B\overrightarrow{\sigma }\cdot \mathbf{𝐁}$

where ${\mu }_B$ is the Bohr magneton </math> and $\overrightarrow{\sigma }=({\hat{\sigma }}_x,{\hat{\sigma }}_y,{\hat{\sigma }}_z)$ is the vector of the Pauli matrices:

${\hat{\sigma }}_x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),$

${\hat{\sigma }}_x=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),$

${\hat{\sigma }}_x=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).$

(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:

Failed to parse (unknown function "\q"): {\displaystyle \hat{H}=\frac{\hat{p}^2}{2m} + \frac{m\omega^2\q^2}{2} } ,

where $p=\frac{\hslash }{i}\frac{d}{dq}.(a)Findthepartitionfunction<math>Z$ in quantum canonical ensemble at temperature T.

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

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

(d) Using result in (c), write down the density matrix in a coordinate representation $⟨q^{\prime }|\hat{\rho }|q⟩$.

(e) In the coordinate representation, calculate explicitly $⟨q^{\prime }|\hat{\rho }|q⟩$ in the high temperature limit $T\to \mathrm{\infty }$. 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[A,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.