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?


1 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.)

  • $\begingroup$ 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 $\endgroup$
    – TTa
    Aug 23, 2023 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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