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On page 174 of Townsend's "A Modern Approach to Quantum Mechanics", 2nd edition, it says the following:

"For a mixed state, one for which $p_k$ is the probability that a particle is in the state $|\psi^{(k)}\rangle$, then $\hat{\rho}=\sum_kp_k|\psi^{(k)}\rangle\langle\psi^{(k)}|$ where $\sum_kp_k=1$.

[...]

Since the density matrix is Hermitian ($\rho_{ij} = \rho_{ji}^*)$ [which I agree], the density matrix can always be diagonalized with diagonal matrix elements given by the probability $p_k$ [?!]. Thus $tr \hat{\rho}^2=\sum_k p_k^2 \leq 1$."

I do not believe the clause marked [?!] to be true. That would require choosing the states $|\psi^{(k)}\rangle$ as the "basis", but we know that the states $|\psi^{(k)}\rangle$ are neither orthonormal nor is it even a basis (for example, we could have more states than the dimension of the Hilbert space).

Am I correct in my critique above? Moreover, given that the density matrix is indeed Hermitian, what then is the "correct orthonormal basis of eigenvectors", and what do these eigenvectors and their corresponding eigenvalues physically mean?

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  • $\begingroup$ I agree, that $p_k$ in the density operator need not be the eigenvalues of the density matrix (also noted as $p_k$), if $\Psi^{(k)}$'s are not orthonormal. They are not the same $p_k$'s. Hard to say more, without seeing the whole text. $\endgroup$ – Mikael Kuisma Feb 20 '16 at 21:38
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Proposition: Let $\{ \left| \psi ^{(k)}\right> :k\}$ be any collection of normalized vectors in a hilbert space, let $0\leq p_k\leq 1$ be such that $\sum _kp_k=1$, and define $\rho :=\sum _kp_k\left| \psi ^{(k)}\right> \left< \psi ^{(k)}\right|$. Then, (i) $\operatorname{tr}(\rho )=1$ and (ii) $\operatorname{tr}(\rho ^2)\leq 1$.

Remark: "Any" really means any: this collection needs to be neither orthonormal nor a basis (though each $\left| \psi ^{(k)}\right>$ definitely needs to be normalized!).

Proof: Let $\{ \left| e_i\right> :i\}$ be an orthonormal basis of the hilbert space (not necessarily eigenvectors for $\rho$), and write $\left| \psi ^{(k)}\right> =\sum _ic^{(k)}_i\left| e_i\right>$. (Note that the index $i$ runs over (in general) very different index set that $k$ does.) Then, $$ \operatorname{tr}(\rho ):=\operatorname{tr}\left( \sum _kp_k\left| \psi ^{(k)}\right> \left< \psi ^{(k)}\right|\right) =\sum _kp_k\operatorname{tr}\left( \left| \psi ^{(k)}\right> \left< \psi ^{(k)}\right| \right)=\sum _kp_k\cdot 1=1, $$ where one can see that $\operatorname{tr}\left( \left| \psi ^{(k)}\right> \left< \psi ^{(k)}\right| \right) =1$ by picking any orthonormal basis of the hilbert space which contains $\left| \psi ^{(k)}\right>$ and using the definition of the trace.

Hence, $\rho$ is trace-class (Disclaimer: You actually need to check that $\operatorname{tr}(|\rho |)<\infty$, but it shouldn't be too hard to show that $|\rho |=\rho$.), hence compact, and so there is an orthonormal basis $\{ \left| f_i\right> :i\}$ consisting of eigenvectors of $\rho$: $\rho \left| f_i\right> =\lambda _i$. Expanding the equation $\lambda _i=\left< f_i\right| \rho \left| f_i\right>$ shows that $0\leq \lambda _i\leq 1$, and so $$ \operatorname{tr}(\rho ^2)=\sum _i\left< f_i\right| \rho ^2\left| f_i\right> =\sum _i\lambda _i^2\leq \sum _i\lambda _i=1. $$ $\square$

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  • $\begingroup$ The given form need not be an eigenvalue decomposition, it could just be a mixture of non-orthogonal states (and the $p_k$ were still probabilities!). $\endgroup$ – Norbert Schuch Feb 20 '16 at 11:32
  • $\begingroup$ @NorbertSchuch how is that going to work? if i then form the expectation of the density matrix with one of you non-orthogonal states I will not end up with the coefficient of that state... i.e. it's not the probability of that state. Furthermore eigenvectors of hermitian matrices are orthogonal anyway. $\endgroup$ – Wolpertinger Feb 20 '16 at 21:24
  • $\begingroup$ @Numrok I never claimed that. I just pointed out that if you have a set of states $\vert\psi_k\rangle$, each with probability $p_k$, the corresponding density matrix $\rho=\sum_k p_k \vert\psi_k\rangle\langle \psi_k\vert$ need not be an eigendecomposition. This relates to the fact that there are many ways to interpret a density matrix as a classical mixture of quantum states. $\endgroup$ – Norbert Schuch Feb 21 '16 at 0:30
  • $\begingroup$ @NorbertSchuch I don't really know what you're saying then, but I want to make clear that if $\rho :=\sum _kp_k\left| \psi ^{(k)}\right> \left< \psi ^{(k)}\right|$ then the states $\left| \psi ^{(k)}\right>$ have to be at least orthogonal, otherwise $p_k$ does NOT have the interpretation of a probability. Well, and then that really is the eigenvalue (or for more exotic cases like multi-particle spaces the Schmidt decomposition) of the density matrix. $\endgroup$ – Wolpertinger Feb 21 '16 at 12:34
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    $\begingroup$ @Numrok Let me give you an example: $\rho=\tfrac12|0\rangle\langle0|+\tfrac12|+\rangle\langle+|$ is a perfectly valid trace-normalized density matrix, and it can be interpreted as a 50-50 mixture of $|0\rangle$ and $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, which are not orthogonal. Nevertheless, the $\tfrac12$ can be naturally interpreted as the probabilities in the mixture. We can even mix more than 2 states (even infinitely many!). This relates to the fact that non-pure density matrices don't have a unique interpretation as a probabilistic mixture of other states. $\endgroup$ – Norbert Schuch Feb 21 '16 at 15:32
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The book is not so clear. I believe what he means is the following: Given a density operator

$$\sum_k p_k \left| \psi^{(k)} \right>\left< \psi^{(k)} \right|$$

we can always find a basis in which it is diagonal. Let's say

$$\left\{ \left| \phi_i \right> \right\}_{i=1}^{N},$$

where $N$ is the dimensionality of the problem. In this basis the density operator will be in the form

$$\sum_i q_i \left| \phi_i \right> \left< \phi_i \right|.$$

Then

$$\text{tr} \rho^2 = \sum_i q_i^2.$$

In fact the $q_i$'s are exactly the eigenvalues of the density matrix and from the fact that $\text{tr} \rho = 1$, we see that they must sum to one. We also know that they must be real. In order to prove that the trace above is less or equal to one, you must also show that the eigenvalues are non-negative (which will imply that $0\leq q_i \leq 1$).

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