Temporary HW: Difference between revisions

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(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  
(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 [\hat{A},\hat{B}]/2 + O(\beta^3)] </math>
<math> e^{\beta \hat{A}} e^{\beta \hat{B}} = exp[\beta(\hat{A} + \hat{B}) + \beta^2 [\hat{A},\hat{B}]/2 + O(\beta^3)] </math>


which you can apply to the Boltzmann operator:
which you can apply to the Boltzmann operator:

Revision as of 14:31, 23 February 2011

Problem 1

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

H^=μBσ𝐁

where μB is the Bohr magneton </math> and σ=(σ^x,σ^y,σ^z) is the vector of the Pauli matrices:

σ^x=(0110),

σ^x=(0ii0),

σ^x=(1001).

(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.


Problem 2

Consider a single one-dimensional quantum harmonic oscillator described by the Hamiltonian:

H^=p^22m+mω2q22,

where p^=iddq.

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

(b) Using result from (a), calculate the averge energy E=H^.

(c) Write down the formal expression for the canonical density operator ρ^ in terms of eigenstates |n and energy levels εn=ω(n+1/2).

(d) Using result in (c), write down the density matrix in a coordinate representation q|ρ^|q.

(e) In the coordinate representation, calculate explicitly q|ρ^|q in the high temperature limit T. HINT: One approach is to apply the following result

eβA^eβB^=exp[β(A^+B^)+β2[A^,B^]/2+O(β3)]

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 β is very small in this limit.