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

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?

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
$$|0\rangle = \alpha^*|\psi\rangle + \beta|\psi^\perp\rangle$$
• After numerous edits to this comment I think I actually get it now. If I measure $|\psi\rangle$ in the $\{|0\rangle, |1\rangle\}$ basis its obvious that this collapses to one of those basis states with the $|\alpha|^2$ or $|\beta|^2$ probability. However it works in the reverse manner as well, I just did not write $|0\rangle$ in the measurement basis of $|\psi\rangle$. Thanks