# Determinant of density matrix

1. Nov 2, 2009

### Fredrik

Staff Emeritus
Why is the determinant of a mixed state density matrix always positive?

In the specific case of a 2-dimensional Hilbert space, the density matrix (as well as any other hermitian matrix) can be expressed as

$$\rho=\frac 1 2 (I+\vec r\cdot\vec \sigma)$$

so its determinant is

$$\det\rho=\frac 1 4(1-\vec r^2)[/itex] We have $|\vec r|=1$ if and only if we're dealing with a pure state, so we seem to need the condition $\det\rho\geq 0$ to see that the set of mixed states is the interior of the sphere rather than the exterior. 2. Nov 3, 2009 ### Petar Mali I think that if you have mixed states statistical operator you have [tex]{\hat{\rho}}^2 < \hat{\rho}$$

and of course for statistical operator $$Tr\hat{\rho}=1$$ and $$\hat{\rho}^{\dagger}=\hat{\rho}$$

I think that from that three conditions you have $$det\hat{\rho}>0$$

Last edited: Nov 3, 2009
3. Nov 3, 2009

### Fredrik

Staff Emeritus
Thanks, but now I have two problems instead of one. I don't see why $\rho^2<\rho$ for mixed states. (Do you mean that all expectation values of $\rho-\rho^2$ are positive?) I also don't see why these results imply that $\det\rho>0$.

4. Nov 3, 2009

### strangerep

The "state operators" in the original post don't necessarily qualify as such...

Summarizing from Ballentine section 2.3 (pp 50-51)...

An acceptable state operator (a) has unit trace, (b) is self-adjoint, and
(c) satisfies $\langle u | \rho | u \rangle \ge 0$, for all $| u \rangle$.

Property (b) implies that a state operator's eigenvalues $\rho_n$ are real.

Property (c) implies that a state operator's eigenvalues must satisfy $\rho_n \ge 0$.

Hence the determinant is real and non-negative, being the product of the eigenvalues.

[EDIT: Removed unnecessary silly mistake herein noted by Fredrik in post #5 below.]

Last edited: Nov 3, 2009
5. Nov 3, 2009

### Fredrik

Staff Emeritus
Thank you strangerep. That was surprisingly simple (at least when I only consider finite-dimensional Hilbert spaces, but that's the case I'm interested in for the moment). [Edit: Removed comment about silly mistake in #4 ]

Last edited: Nov 4, 2009
6. Nov 4, 2009

### Petar Mali

Just to show c)

$$\hat{\rho}=\sum^K_{k=1}w_k|\psi_k> <\psi_k|=\sum^K_{k=1}w_k\hat{P}_k$$

and

$$\left\langle \psi_n\left|\hat{P}_k\right|\psi_n\right\rangle=||\hat{P}_k\psi_n||^2$$