# Is it possible to write a Density Matrix in the following form?

Is it possible to write an arbitrary density matrix $\hat{\rho}$ in the following form ?

$$\hat{\rho} ~=~ \frac{1}{N} \sum_{\ell=1}^N \left|x_{\ell}\right\rangle \left\langle x_{\ell}\right|,$$

where $\left\{\left|x_{\ell}\right\rangle\right\}_{\ell = 1}^{N}$ are normalized states (but not necessarily orthogonal).

If yes, how can one prove this ?

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I'm not answering right away because right now I can't think of a way to mathematically prove it (and I'm a lil' bit busy ATM), but I'm sure that to construct a density matrix you don't need the states to be orthogonal to each other. –  user17581 Jan 12 '13 at 14:48
yeah, that's right but the coefficients of $|x_{\ell}>$ here are all the same and we have factorized them out as $\frac{1}{N}$, besides $|x_{\ell}>$ are normal states! how can it be possible ? –  physics_xyz Jan 12 '13 at 14:56
It's just the diagonalization of the density matrix, a Hermitian matrix, isn't it? $N$ must be chosen to be nothing else than the dimension of the matrix for generic ones, otherwise the $x$-vectors wouldn't be orthogonal to each other. –  Luboš Motl Jan 12 '13 at 15:17
I don't think so, because here $\left\{|x_{\ell}>\right\}_{\ell = 1}^{N}$ don't form a basis for space they are just an ensemble. –  physics_xyz Jan 12 '13 at 15:46
You said that the $|x_l\rangle$ are not necessarily orthonormal, but do they span the (presumably finite dimensional) space? If not, then you can't construct an arbitrary density matrix in this way. –  twistor59 Jan 13 '13 at 8:31
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Let us reformulate OP's question (v3) as follows:

Let $H$ be an $N$-dimensional Hilbert space. Is it possible to write an arbitrary density operator $$\tag{1} \hat{\rho}~\in~ B(H)~\cong~ {\rm Mat}_{N\times N}(\mathbb{C})$$ on the form $$\tag{2} \hat{\rho} ~=~ \frac{1}{N} \sum_{m=1}^N |m) (m|,$$ where $\left\{|m) \right\}_{m = 1}^{N}$ are normalized states $$\tag{3}(m|m) ~=~1,$$ but not necessarily orthogonal?

Proof: Because $\hat{\rho}$ is a positive operator, it may be diagonalized wrt. an orthonormal basis. Hence there exists an orthonormal basis $\left\{|n\rangle \right\}_{n = 1}^{N}$, and eigenvalues $\lambda_1, \ldots, \lambda_N \geq 0$, such that

$$\tag{4} \hat{\rho} ~=~ \sum_{n=1}^N \lambda_n|n\rangle \langle n|,$$

and with unit trace

$$\tag{5} \sum_{n=1}^N \lambda_n~=~ {\rm tr} \hat{\rho}~=~1.$$

Now define

$$\tag{6} |m)~:=~ \sum_{n=1}^N \exp\left(\frac{2\pi i}{N} mn \right) \sqrt{\lambda_n} |n\rangle .$$

It is straightforward to check that eqs. (2) and (3) are satisfied.

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