# How to find initial quantum states from the density matrix?

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?

• – glS
Jun 20, 2022 at 8:10

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.