A problem in three sequential selective measurements in case of incompatible observables In case (a) the B filter takes care of B's eigenvalues and we sum all of their probabilities to calculate the probability of obtaining $$|c'\rangle$$.

In case (b) B filter is not used and this creates all the difference.

J. J. Sakurai says the probability of obtaining $$|c'\rangle$$ after obtaining $$|a'\rangle$$ won't depend upon the measurement through B filter if $$[A,B]=0$$ or $$[B,C]=0.$$

I think the fact that A and B would have the same set of eigenvectors if they commute may be used but I can't see how. I can prove it neither mathematically nor intuitively. How can it be done?

• It is important to realize that in case (a), we get a different probability of obtaining $|c'\rangle$ depending on the choice of $|b'\rangle$, and we take the sum over all possible $|b'\rangle$. You should maybe add that to the question, because it is not obvious from the figure. – Noiralef Sep 29 '18 at 8:24
• Yeah, I should add that. – Asit Srivastava Oct 1 '18 at 16:46

The probability of obtaining the final state $$| c' \rangle$$ from an initial state $$| a' \rangle$$ is given by:

$${|\langle c' | b' \rangle|}^2 {|\langle b' | a' \rangle|}^2$$

Consider the first experiment where the $$B$$ filter is present. Let's say the orthonormal eigenstates $$| b' \rangle$$ are $$| 0 \rangle, | 1 \rangle, | 2 \rangle, \cdots$$. The $$B$$ filter selects one eigenstate $$| b' \rangle$$ per experiment. We wish to know what the probability will be when any $$| b' \rangle$$ is selected. Then, the probability of measuring any $$| b' \rangle$$ with the $$B$$ filter amounts to:

\begin{equation} \begin{aligned} \text{Measuring any | b' \rangle} =& \text{ Measuring | 0 \rangle or Measuring | 1 \rangle or Measuring | 2 \rangle or} \cdots\\ =& \ {|\langle c' | 0 \rangle|}^2 {|\langle 0 | a' \rangle|}^2 + {|\langle c' | 1 \rangle|}^2 {|\langle 1 | a' \rangle|}^2 + {|\langle c' | 2 \rangle|}^2 {|\langle 2 | a' \rangle|}^2 + \cdots\\ =& \ \sum_{b'} {|\langle c' | b' \rangle|}^2 {|\langle b' | a' \rangle|}^2\\ =& \ \sum_{b'} \langle c' | b' \rangle \langle b' | a' \rangle \langle a' | b' \rangle \langle b' | c' \rangle \qquad ---(1) \end{aligned} \end{equation}

Note that there is a single summation over the eigenstates $$| b' \rangle$$.

When there is no $$B$$ filter present, the probability is simply:

\begin{equation} \begin{aligned} \text{Measuring no | b' \rangle} =& \ |\langle c' | a' \rangle|^2\\ =& \ \sum_{b'} \sum_{b''} \langle c' | b' \rangle \langle b' | a' \rangle \langle a' | b'' \rangle \langle b'' | c' \rangle \qquad ---(2) \end{aligned} \end{equation}

where the completeness relation $$\displaystyle \sum_{b'} |b' \rangle \langle b' |$$ has been inserted twice.

Say:

$$A | a' \rangle = a' |a' \rangle$$

$$B | b' \rangle = b' |b' \rangle$$

$$C | c' \rangle = c' |c' \rangle$$

Now, assume that $$A$$ and $$B$$ are compatible observables: $$[A,B] = 0$$. (We also assume absence of any degeneracy.) Then $$A$$ and $$B$$ have simultaneous eigenkets. We rewrite $$(2)$$ in the following way:

\begin{equation} \begin{aligned} \sum_{b'} \langle c' | b' \rangle \langle b' | a' \rangle \sum_{b''} \langle a' | b'' \rangle \langle b'' | c' \rangle \end{aligned} \end{equation}

Since the eigenkets of $$A$$ and $$B$$ are simultaneous, $$| a' \rangle$$ and $$| b' \rangle$$ span the same eigenspace, and only one of the inner products $$\langle b' | a' \rangle$$ in the first sum over $$b'$$ will be nonzero. Let's assume that the nonzero inner product occurs for $$b' = e$$. In the second sum over $$b''$$, the inner product $$\langle a' | b'' \rangle$$ will be nonzero only for $$b'' = e$$. Removing the contributions that are zero, we see that we are able to impose $$b' = b''$$, while being able to remove the second sum. $$(2)$$ then becomes:

$$\sum_{b'} \langle c' | b' \rangle \langle b' | a' \rangle \langle a' | b' \rangle \langle b' | c' \rangle$$

which is equal to $$(1)$$. Actually, the sum over $$b'$$ could also be removed because $$\langle b' | a' \rangle = 0$$ for all $$b'$$ except $$b' = e$$.

The same result holds when $$[B,C] = 0$$.

Intuitively, if $$[A,B] = 0$$, then the eigenstate $$| a' \rangle$$ does not get 'destroyed' when it comes out through filter $$B$$. So for the final measurement, it does not matter whether the $$B$$ filter was present or not.