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$\DeclareMathOperator{\Tr}{Tr}$On page 5 of this online document, it states a seemingly trivial fact: that if we have a density-matrix for a mixed state defined by

$$\hat{\rho}=\sum_kp_k|\psi_k\rangle\langle\psi_k|$$

where $\{|\psi_k\rangle\}$ are (not-necessarily orthogonal) pure states, then we have the following double-sided implication:

$$\Tr (\hat{\rho})=1~~~\iff~~~\sum_kp_k=1$$

This seems intuitively clear to me, but when I try to go from the left-side to the right-side I get stuck. Here's what I mean:

$$\begin{align} \Tr(\hat{\rho})&=\sum_m \langle\psi_m| \hat{\rho }| \psi_m \rangle \\ &=\sum_{m} \langle\psi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \psi_m \rangle\\ &=\sum_k p_k \sum_m |\langle \psi_m |\psi_k\rangle |^2 \end{align}$$

Now, if $\{| \psi_k\rangle \}$ is orthogonal, then $|\langle \psi_m |\psi_k\rangle |^2=\delta_{mk}$ and everything works out easily - but, they aren't orthogonal. So what do I do?

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    $\begingroup$ Your procedure to compute the trace is wrong! If the $\psi_n$s are not orthonormal vectors $tr(\rho) \neq \sum_n \langle \psi_n| \rho \psi_n\rangle$...Starting from $\hat{\rho}=\sum_kp_k|\psi_k\rangle\langle\psi_k|$, you should use another orthonormal basis to compute the trace, and your procedure, taking this into account, gives rise to the wanted result. $\endgroup$ – Valter Moretti Oct 31 '16 at 17:40
  • $\begingroup$ Are you 100% sure that the link is correct? =D $\endgroup$ – Apogee Oct 31 '16 at 17:50
  • $\begingroup$ Your 'online document' doesn't seem to say much at all about density matrices... $\endgroup$ – gj255 Oct 31 '16 at 18:22
  • $\begingroup$ Oh my gosh I can't believe I accidentally put that link. XD My bad. It's changed now. $\endgroup$ – Arturo don Juan Oct 31 '16 at 18:59
  • $\begingroup$ @ValterMoretti I thought the trace was basis independent, and so it didn't matter which basis I chose to sum over so long as the density matrix is put in terms of that same basis. $\endgroup$ – Arturo don Juan Oct 31 '16 at 20:02
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Let us focus on your chain of identities. $$\begin{align} \text{Tr}(\hat{\rho})&=\sum_m \langle\psi_m| \hat{\rho }| \psi_m \rangle \\ &=\sum_{m} \langle\psi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \psi_m \rangle\\ &=\sum_k p_k \sum_m |\langle \psi_m |\psi_k\rangle |^2 \end{align}$$ The point in the implications above is that the first line is the correct definition of trace if and only if the vectors $| \psi_m \rangle$ form a orthonormal basis. Otherwise the right-hand side is not the trace of $\hat{\rho}$ and the reasoning stops there.

If the vectors $| \psi_m \rangle$ are normalized but are not mutually orthogonal and $$\hat{\rho} :=\sum_k p_k|\psi_k\rangle\langle\psi_k|\:,$$ then the correct procedure is to pick out an orthonormal basis of vectors $| \phi_m \rangle$ and then $$\begin{align} \text{Tr}(\hat{\rho})&=\sum_m \langle\phi_m| \hat{\rho }| \phi_m \rangle \\ &=\sum_{m} \langle\phi_m|\left(\sum_k p_k|\psi_k\rangle\langle\psi_k|\right)| \phi_m \rangle\\ &=\sum_k p_k \sum_m |\langle \phi_m |\psi_k\rangle |^2 = \sum_k p_k |||\psi_k\rangle||^2 = \sum_k p_k 1 = \sum_k p_k \end{align}$$ we therefore have the wanted double-sided implication you mention:

$$ \text{Tr} (\hat{\rho})=1~~~\iff~~~\sum_kp_k=1\:.$$

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  • $\begingroup$ Ah okay, this is crystal clear. Then my confusion stemmed from the fact that, as you pointed out in your final comment, the author of that paper didn't specify that in the definition of trace the sum was taken over an orthonormal basis. $\endgroup$ – Arturo don Juan Oct 31 '16 at 23:16
  • $\begingroup$ I am happy that you got the point. $\endgroup$ – Valter Moretti Nov 1 '16 at 7:52

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