# Trace of density matrix for mixed state

$\DeclareMathOperator{\Tr}{Tr}$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?

• 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. – Valter Moretti Oct 31 '16 at 17:40
• Are you 100% sure that the link is correct? =D – Apogee Oct 31 '16 at 17:50
• Your 'online document' doesn't seem to say much at all about density matrices... – gj255 Oct 31 '16 at 18:22
• Oh my gosh I can't believe I accidentally put that link. XD My bad. It's changed now. – Arturo don Juan Oct 31 '16 at 18:59
• @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. – Arturo don Juan Oct 31 '16 at 20:02

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