# Subsequent measurements (of two different observables) of quantum system

A question I have gotten repeatedly wrong this semester looks like this:

There is an operator A corresponding to observable $$\alpha$$ with eigenvalues $$a_1$$ and $$a_2$$ and eigenfunctions X1 and X2. There is an operator B corresponding to observable $$\beta$$ with eigenvalues $$b_1$$ and $$b_2$$ and eigenfunctions W1 and W2. The eigenfunctions are related by:

$$X_1 = \frac{(2W_1 +3W_2)}{\sqrt{13}}$$ and $$X_2 = \frac{(3W_1 -2W_2)}{\sqrt{13}}$$

You measure $$\alpha$$ and obtain $$a_1$$, Then you measure $$\beta$$ and then measure $$\alpha$$ again after that. What is the probability that you get $$a_1$$ again?

Any help in understanding how to solve problems like this would be greatly appreciated.

If you measure $$\alpha=a_1$$ it means you are in state $$X_1$$ as it is the eigenstate with the corresponding eigenvalue. So, right after your measurement, you are in state $$X_1$$. Now, $$X_1$$ is not an eigenstate of $$\beta$$, it is composed of two parts $$X_1=(2W_1+3W_2)/\sqrt{13}$$ which means that when we measure $$\beta$$ the does not stay the same (because $$X_1$$ is not an eigenstate of $$\beta$$) but collapses either on status $$W_1$$ (if you get $$b_1$$) or $$W_2$$ (if you get $$W_2$$).
Using the expression of $$X_1$$ in terms of $$W_1$$ and $$W_2$$ we can compute that the state will collapse on $$W_1$$ with probabilit 4/13 and to $$W_2$$ with probability 9/13 (just taking the square product of the amplitudes of the two eigenstates of $$\beta$$ when we are in $$X_1$$).
Now, what you have to do (and I will leave it as exercise and solve it only on further request) is to re-write $$W_1$$ and $$W_2$$ in terms of $$X_1$$ and $$X_2$$, i.e. you have to invert the relationships you wrote in your post and get $$W_1=...$$ and $$W_2=..$$. From that, you get the probability of measuring $$\alpha=a_1$$ if before you collapsed on state $$W_1$$ and if before you collapsed on state $$W_2$$. Because you already know the probabilities of having collapsed on either $$W_1$$ or $$W_2$$ you will find the final probability of re-measuring $$a_1$$ by combining them appropriately.
The big point is that $$\alpha$$ and $$\beta$$, not having a common basis, change the state upon measurement, collapsing in on autostates which are not autostates of the other one so that the possibility of measuring $$a_1$$ again is changed by the fact that you measured $$\beta$$ between (otherwise, if you only measure $$\alpha$$ you always get $$a_1$$ after the first measurement).
• @AstroZ4ch This is a great and correct answer. Just to help be a bit more concise, let's say we express the final two measurements as a pair $(b_i,a_j)$. You need to find the probability of getting $(b_1,a_1)$ OR $(b_2,a_1)$. Since these outcomes are mutually exclusive, you can just add these probabilities together. – Aaron Stevens Dec 7 '18 at 2:05