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Statistical Mechanics - Physics at Oregon State University

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A.6. SOLUTIONS FOR CHAPTER 6. 259<br />

for any variable X. Using R, R∗ , and x for X and setting the result to zero<br />

gives:<br />

<br />

0<br />

0 = T r<br />

0<br />

1<br />

0<br />

<br />

<br />

0<br />

log(ρ) + T r<br />

0<br />

1<br />

0<br />

<br />

<br />

0<br />

0 = T r<br />

1<br />

0<br />

0<br />

<br />

<br />

0<br />

log(ρ) + T r<br />

1<br />

0<br />

0<br />

<br />

<br />

1<br />

0 = T r<br />

0<br />

<br />

<br />

0<br />

1<br />

log(ρ) + T r<br />

−1<br />

0<br />

<br />

0<br />

−1<br />

or<br />

Therefore<br />

and hence<br />

(log(ρ)) 21 = 0<br />

(log(ρ)) 12 = 0<br />

(log(ρ)) 11 − (log(ρ)) 22 = 0<br />

log(ρ) =<br />

ρ =<br />

y 0<br />

0 y<br />

z 0<br />

0 z<br />

Using the fact th<strong>at</strong> the trace is one we finally get:<br />

<br />

1<br />

ρ = 2 0<br />

<br />

0 1<br />

2<br />

or R = 0,x = 1<br />

2 .<br />

The same can be obtained by finding the eigenvalues of ρ.<br />

The second order deriv<strong>at</strong>ives are all neg<strong>at</strong>ive or zero. Use result (9,33) for any<br />

of the elements and change to coordin<strong>at</strong>es R, R ∗ , and x.<br />

Problem 6.<br />

A quantum mechanical system is described by a simple Hamiltonian H, which<br />

obeys H 2 = 1. Evalu<strong>at</strong>e the partition function for this system. Calcul<strong>at</strong>e the<br />

internal energy for T → 0 and T → ∞.<br />

Z = T r<br />

∞<br />

n=0<br />

<br />

<br />

1<br />

n! (−1)n β n H n

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