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I have a problem in understanding why we can write the expectation value of an operator $\hat{O}$ as the trace of $\hat{\rho}\hat{O}$ where $\hat{\rho}$ is the density matrix defined for pure state.

In this link: http://www.cithep.caltech.edu/~fcp/physics/quantumMechanics/densityMatrix/densityMatrix.pdf

at page 2, after equation $(8)$, I read the following: $$\langle\hat{O}\rangle= \sum_{m}^{} \sum_{n}^{} a_m^*a_n Q_{mn}= \sum_{m}^{} \sum_{n}^{} \rho_{nm} Q_{mn}= \sum_{n}^{} [\rho Q]_{nn} \quad .$$

I can't understand the last passage, why do we pass from a double sum to a single sum just over $n$?

Here $Q_{mn}$ are the matrix elements of the operator $Q$ with respect to an orthonormal basis ${u_m,u_n,...}$ and $\rho_{nm}$ are the matrix elements of the density matrix again with respect to the same basis, $a_m=\langle u_m|\psi\rangle$ and $u_n=\langle\psi|u_n\rangle$ and $\rho=|\psi\rangle\langle\psi|$.

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    $\begingroup$ Hint: Make use of the completeness relation for a ONB (in Dirac Notation): $\mathbb I =\sum_m |m\rangle \langle m|$. $\endgroup$ May 13 at 19:28

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For two matrices $M$ and $N$, the definition of matrix multiplication is $$[MN]_{ab}= \sum_c M_{ac} N_{cb}$$

In this case, $$\sum_n \underbrace{\left(\sum_m \rho_{nm}Q_{mn}\right)}_{= [\rho Q]_{nn}}= \sum_n [\rho Q]_{nn} = \mathrm{Tr}(\rho Q)$$

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    $\begingroup$ With $[MN]_{ab}$ do you mean the $ab$ element of the matrix that is the product of $M$ and $N$? $\endgroup$
    – Salmone
    May 13 at 20:53
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    $\begingroup$ @Salmone Yes, that’s right. $\endgroup$
    – J. Murray
    May 13 at 20:54

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