Recently, I came across density matrix. Given a qubit $|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$, we can find its density matrix by computing $\rho \equiv |\psi \rangle \langle \psi |$.

My question is how can one revert this process? That is given $\rho$, how can we find its initial quantum states that resulted in such $\rho$ (regardless of whether $\rho$ is pure or mixed)?

As an example, consider

$$\rho = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} $$

how can we deduce $\rho = 0.5 |0\rangle \langle0| + 0.5 |1\rangle \langle 1|?$

This might be a rather straightforward example, but I am curious to know the process for a random density matrix. Does it produce unique results?


1 Answer 1


In general, for a given $\rho$ there are many ensembles, i.e. sets of pure states $|\psi_i\rangle$ and probabilities $p_i$ such that $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$. For example, if $\rho=\mathrm{diag}(c^2, s^2)$ where $c,s\in\mathbb{R}_+$ with $c^2+s^2=1$, then

$$ \rho=c^2|0\rangle\langle 0|+s^2|1\rangle\langle 1|=\frac12|\psi_+\rangle\langle \psi_+|+\frac12|\psi_-\rangle\langle\psi_-| $$

where $|\psi_{\pm}\rangle=c|0\rangle\pm s|1\rangle$.

Still, you can obtain an ensemble corresponding to $\rho$ by computing its eigendecomposition.

  • $\begingroup$ It should be noted that if the state is pure then we can recover the original ket vector (up to global phase). $\endgroup$
    – Rammus
    Jun 20 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.