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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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Noiralef Sep 29 '18 at 8:24
  • $\begingroup$ Yeah, I should add that. $\endgroup$ – 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.


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

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

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

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.


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