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?



Is there a quantum Bayes rule?
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
$$ \pmatrix{a\_1\cr a\_2\cr 0\cr}/\sqrt{|a\_1|^2 + |a\_2|^2}$$


