Question
Using the result show that the expression of density matrix
of a
free particle in box of volume V in the canonical ensemble in the coordinate representation is given by
Answer :
Word Count : 430
We are given: $$ \hat{\rho} = \frac{e^{-\beta \hat{H}}}{\text{Tr}(e^{-\beta \hat{H}})} $$ We are to derive the expression: $$ \text{Tr}(e^{-\beta \hat{H}}) = V \left(\frac{m}{2\pi\beta \hbar^2}\right)^{3/2} $$ for a free particle in a box of volume $V$ in the canonical ensemble in coordinate representation. --- ### Step 1: Understand the Hamiltonian For a free particle, the Hamiltonian is purely kinetic: $$ \hat{H} = \frac{\hat{p}^2}{2m} $$ So the operator $e^{-\beta \hat{H}}$ becomes: $$ e^{-\beta \hat{H}} = e^{-\beta \frac{\hat{p}^2}{2m}} $$ --- ### Step 2: Evaluate the Trace Recall that the trace ___ _______ ______ _________ __________ _________ __________ _________ ________ _____ _____.
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We are given: $$ \hat{\rho} = \frac{e^{-\beta \hat{H}}}{\text{Tr}(e^{-\beta \hat{H}})} $$ We are to derive the expression: $$ \text{Tr}(e^{-\beta \hat{H}}) = V \left(\frac{m}{2\pi\beta \hbar^2}\right)^{3/2} $$ for a free particle in a box of volume $V$ in the canonical ensemble in coordinate representation. --- ### Step 1: Understand the Hamiltonian For a free particle, the Hamiltonian is purely kinetic: $$ \hat{H} = \frac{\hat{p}^2}{2m} $$ So the operator $e^{-\beta \hat{H}}$ becomes: $$ e^{-\beta \hat{H}} = e^{-\beta \frac{\hat{p}^2}{2m}} $$ --- ### Step 2: Evaluate the Trace Recall that the trace ___ _______ ______ _________ __________ _________ __________ _________ ________ _____ _____.
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