As I understand, a necessary and sufficient condition for a density matrix $P$ to be represented by a wavefunction $|\Psi\rangle$ is that it is idempotent, i.e. $P^2=P$. It is easy to see that if $P=|\Psi\rangle\langle \Psi|$ then $P$ is idempotent: $$ P^2=|\Psi\rangle\langle \Psi|\Psi\rangle\langle \Psi|=|\Psi\rangle\langle \Psi| $$ since by normalization $\langle \Psi|\Psi\rangle=1$. However, I have had trouble with the other direction - how can I show that if $P^2=P$ then $P=|\Psi\rangle\langle\Psi|$ for some $\Psi$?

I have been able to demonstrate that if $P^2=P$ then $$ P=\sum_{j} |\Psi_j\rangle\langle\Psi_j| $$ where the sum over $j$ has no more terms than the dimension of the density matrix $P$. My argument was analogous to the one found here. I suspect that I'm almost there - I think it's just a change of basis that I haven't seen yet. How can I complete this proof?


1 Answer 1


A density matrix $D$ is positive semidefinite, hermitian, and has trace one. Because of hermiticity we may assume that it is diagonal. Let's denote the eigenvalues with $\lambda_i$ . Because it is positive semidefinite, we have $0 \leq \lambda_i \leq 1$.

The matrix $D^2$ has eigenvalues $\lambda_i^2$. Since it is idempotent ($D = D^2$) and has trace one, we may write: $$\sum_i \lambda_i = \sum_i \lambda_i^2 = 1$$ The only possibility for this to be true is, if there is one and only one $j$ for which $\lambda_j = 1$.

So $D$ has one $1$ in the diagonal at the $j$-th column and all other values in the matrix are zero. You can easily represent this matrix with: $ e_j e_j^T$. ($e_j$ being the $j$-th unit vector.)

If you transform back to you original nondiagonal form, you obtain your $\Psi$.

  • $\begingroup$ Thank you! I see what I was missing - the key had been in the defining properties of the density matrix. $\endgroup$
    – BGreen
    Jun 29, 2019 at 21:56
  • $\begingroup$ I think that what you have described is a pure state, a stationary state (eigenstate of the time-independent Schr. Eq.). However if you have a superposition or coherence then I don't think D would be reducible to the form you describe, or? $\endgroup$
    – Buck Thorn
    Jun 30, 2019 at 9:27
  • 2
    $\begingroup$ If you have no pure state, then D would not be idempotent. Because lambda would not equal lambda^2. $\endgroup$
    – mcocdawc
    Jun 30, 2019 at 9:59

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