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 ?

  • $\begingroup$ 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. $\endgroup$
    – user17581
    Jan 12, 2013 at 14:48
  • $\begingroup$ 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 ? $\endgroup$ Jan 12, 2013 at 14:56
  • $\begingroup$ 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. $\endgroup$ Jan 12, 2013 at 15:17
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    $\begingroup$ 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. $\endgroup$ Jan 12, 2013 at 15:46
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    $\begingroup$ 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. $\endgroup$
    – twistor59
    Jan 13, 2013 at 8:31

2 Answers 2


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?

The answer is Yes.

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.


My answer: not every density matrix is allowed to be written in this way. However, if you admit non normalized states, in finite dimensional Hilbert spaces, each density matrix can be expressed in this way and, moreover, there is an infinite number of decompositions with that form. (See for example the book Nielsen & Chuang "Quantum Computation and Quantum Information" )


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