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, 2022 at 6:10

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