What is the probability of a state $|0\rangle$ being in another state $\alpha|0\rangle+\beta|1\rangle$?

Question

I am trying to calculate the probability of a state (density matrix) being in a specific other state.

Lets say I have a 2-dimensional state with the states given by the orthonormal basis states $|0\rangle, |1\rangle$.

Say I have a density matrix $\hat{\rho} = |0\rangle\langle0|$

I want to calculate, what is the chance of $\hat{\rho}$ being in the superposition state $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, and $\hat{\sigma} = |\psi\rangle\langle\psi|$.

From what I've read, this is simply calculating the following:

$\operatorname{Tr}({\hat{\sigma}\hat{\rho}}) = \langle\psi|\hat{\rho}|\psi\rangle$

In my simple example, $\langle\psi|\hat{\rho}|\psi\rangle = |\alpha|^2$. This is not the probability of $\hat{\rho}$ being in the state $|\psi\rangle$ as that should be equal to $0$ as there is no $|1\rangle$ component in $\hat{\rho}$.

To me, this only makes sense when $|\psi\rangle$ is a basis state of $\hat{\rho}$. So what is it that I've just calculated and how would I obtain what I am looking for?

Answer 1

This is the probability of measuring $|\psi\rangle$ from $\rho = |0\rangle \langle 0|$. The measurement apparatus you described consists of projecting onto the orthogonal states $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ and (say) $|\psi^\perp\rangle = \beta^*|0\rangle - \alpha^*|1\rangle$. Then with probability $|\alpha|^2$, the measurement projects from $|0\rangle$ to $|\psi\rangle$, and with probability $\mathrm{tr}(|\psi^\perp\rangle \langle\psi^\perp| \rho) = |\beta|^2$ it projects onto an orthogonal state $|\psi^\perp\rangle$.

You shouldn't think about this measurement as "there is no $|1\rangle$ component in $\hat{\rho}$". You are measuring in the basis of $\{|\psi\rangle, |\psi^\perp\rangle\}$, so you should be asking "is there a component of $|\psi\rangle$ in $\rho$?" And indeed there is, because we can write

$$|0\rangle = \alpha^*|\psi\rangle + \beta|\psi^\perp\rangle$$

(as a simple exercise, I recommend verifying this yourself.)