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 $\rho=|\psi\rangle\langle\psi|$.

  • 2
    $\begingroup$ Hint: Make use of the completeness relation for a ONB (in Dirac Notation): $\mathbb I =\sum_m |m\rangle \langle m|$. $\endgroup$ Commented May 13, 2022 at 19:28

1 Answer 1


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

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

  • 1
    $\begingroup$ With $[MN]_{ab}$ do you mean the $ab$ element of the matrix that is the product of $M$ and $N$? $\endgroup$
    – Salmon
    Commented May 13, 2022 at 20:53
  • 2
    $\begingroup$ @Salmone Yes, that’s right. $\endgroup$
    – J. Murray
    Commented May 13, 2022 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.