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I've been reading Weinberg's Lectures on Quantum Mechanics and the topic of the density operator (or density matrix, whatever you may call it) baffles me.

Basically we have,

$\hat{\rho}=\sum_ip_i\left|\psi_i\right>\left<\psi_i\right|$.

So it's just a simple matrix, right? Meaning any matrix can be a density matrix?

On Weinberg's take on density matrices he says to take

$\hat{\rho}=\begin{bmatrix}a&b&0\\c&d&0\\0&0&e\end{bmatrix}$

as an example and determine under what conditions this type of matrix is the density operator of a pure state.

But here is my problem: As I see it any matrix can be seen as a density matrix. At least that's what I think when I look at the general formula for the density matrix, since any type of matrix can be expressed with that formula.

So I assume the conditions come from the pure state. A pure state is just one which corresponds to vectors in Hilbert Space, right? Thus we can get any type of matrix and $\hat{\rho}=\begin{bmatrix}a&b&0\\c&d&0\\0&0&e\end{bmatrix}$ therefore should be a density operator of a pure state.

Any ideas?

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  • $\begingroup$ Recall for a pure state the density matrix is idempotent, $\rho^2=\rho$, basically a projection operator. So, upon diagonalization, given the normalization condition, it is basically diag(1,0,0,...). Now do the Weinberg exercise. $\endgroup$ – Cosmas Zachos Jul 1 '16 at 19:24
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Being a density matrix is equivalent to describing a physical state. A density matrix need to be: 1)(Normalized)$$\text{Tr}(\rho)=1$$ 2)(Hermitian) $$\rho=\rho^{\dagger}$$ 3) (Positive Semidefinite) $$\rho > 0$$

This is not the same as being a pure state, the pure state are a subset of physical states, but there are others the so called mixed states.

Note that Linden has a nice condition for pure states being physical $$1=\text{Tr}(\rho)=\text{Tr}(\rho^{2})=\text{Tr}(\rho^{3})$$.

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