Is there a quantum Bayes rule?

Question

This question has been bothering me for a while. Wading through the internet hasn't turned up any answers that I have been able to understand.

First some motivation: Let $S = \{s_1,s_2,s_3\}$ be a set and consider a random variable $x \in S$ with distribution $p(x=s_i) = p_i$. Suppose that we have an observable $f$ with $f(s_1) = f(s_2) = \lambda$ and $f(s_3) = \mu$. If we measure $f = \lambda$, then the distribution of $x$ collapses to a vector proportional to $(p_1,p_2,0)$. This is just a convoluted way of stating Bayes rule.

Now suppose that $x \in S$ is a "quantum particle". It is specified by a wave function $\psi(s_i) = a_i \in \mathbb{C}$. Consider the observable $$f = \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \mu \end{pmatrix}.$$ If we measure $f = \lambda$, then the wave function $\psi$ should collapse to something. It seems reasonable to take something proportional to $(a_1,a_2,0)$ in accordance with Bayes rule, but I have never seen anyone explicitly say this. Is this correct?

Answer 1

According to von Neumann's description of measurement, if you measure an observable $f$ when the wave-function is $\psi$, obtaining value $\lambda$ which is an eigenvalue of $f$, the resulting wave-function will be $$ \frac{E_\lambda \psi}{\|E_\lambda \psi\|}$$ where $E_\lambda$ is the orthogonal projection on the eigenspace for $f$ with eigenvalue $\lambda$.

So in this case with $$ \psi = \pmatrix{a_1\cr a_2\cr a_3}, \ E_\lambda = \pmatrix{1 & 0 & 0\cr 0 & 1 & 0\cr 0 & 0 & 0\cr}$$ you should indeed get $$ \pmatrix{a_1\cr a_2\cr 0\cr}/\sqrt{|a_1|^2 + |a_2|^2}$$