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).

  • $\begingroup$ @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. $\endgroup$ – Aaron Stevens Dec 7 '18 at 2:05
  • $\begingroup$ I confirm Aaron Stevens' comment, try to work out the final result by yourself now $\endgroup$ – JalfredP Dec 7 '18 at 9:13

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