# Sakurai - Modern Quantum Mechanics - page 34, equation (1.4.49)

1. Apr 2, 2013

### omoplata

1. The problem statement, all variables and given/known data

From page 34 of 'Modern Quantum Mechanics' by J.J. Sakurai,

The book considers 2 cases of sequential Sern-Gerlach like selective measurements.

First case:
There are 3 filters. The first (A) filter selects $\mid a' \rangle$ and rejects all others, second (B) filter selects $\mid b' \rangle$ and rejects all others, third (C) filter selects $\mid c' \rangle$ and rejects all others. The probability of obtaining $\mid c' \rangle$ is,
$${| \langle c' \mid b' \rangle |}^2 {| \langle b' \mid a' \rangle |}^2$$
Then we sum over $b'$ to consider the total probability of going through all possible $b'$ routes.
$$\Sigma_{b'} {| \langle c' \mid b' \rangle |}^2 {| \langle b' \mid a' \rangle |}^2 = \Sigma_{b'} \langle c' \mid b' \rangle \langle b' \mid a' \rangle \langle a' \mid b' \rangle \langle b' \mid c' \rangle$$

Second case:

The B filter is removed. There are only A and C filters now. The probability of obtaining $\mid c' \rangle$ is,
$${| \langle c' \mid a' \rangle |}^2 = {| \Sigma_{b'} \langle c' \mid b' \rangle \langle b' \mid a' \rangle |}^2 = \Sigma_{b'} \Sigma_{b''} \langle c' \mid b' \rangle \langle b' \mid a' \rangle \langle a' \mid b'' \rangle \langle b'' \mid c' \rangle$$

The book states that the probabilities for finding |c'> in both cases become equal when $$[A,B]=0$$ or $$[B,C]=0$$, and asks the reader to prove it.

2. Relevant equations

See above.

3. The attempt at a solution

I have no idea how to connect the operators A, B and C to the state kets |a'>, |b'> and |c'>.

2. Apr 2, 2013

### naele

You're not connecting the operators to the kets. When [A,B]=0 is true, this places conditions on |a> and |b>. You should consider equations 1.4.34a and 1.4.34b as I think it should be helpful to solve the problem.

3. Apr 2, 2013

### omoplata

OK, equations 1.4.34a and 1.3.34b are for simultaneous eigenkets, right? But as far as I know, in general, no pair from |a'>, |b'> and |c'> are simultaneous eigenkets of each other.

What I meant by 'connecting the operators to the state kets' is, how do I use [A,B]=0 to prove $\Sigma_{b'} \langle c' \mid b' \rangle \langle b' \mid a' \rangle \langle a' \mid b' \rangle \langle b' \mid c' \rangle = \Sigma_{b'} \Sigma_{b''} \langle c' \mid b' \rangle \langle b' \mid a' \rangle \langle a' \mid b'' \rangle \langle b'' \mid c' \rangle$?

Let's assume the initial state ket is |d>. Once it goes through the filter A, it changes in to |a'>. Does this mean, A|d> = |a'> ? Or does it mean, A|d> = p|a'>, where p is some constant? If it is, I guess I can use that?

4. Apr 2, 2013

### naele

If [A,B]=0 then |a> and |b> are simultaneous eigenstates of A and B. Given this, can you make the argument you have the freedom to choose |a'> and |b'> such that |a'> = |a,b> and |b'>= |a',b'> ?

edit: sorry i don't have Sakurai in front of me anymore so I'm making some mistakes trying to copy his notation.

5. Apr 2, 2013

### omoplata

OK, thanks! I completely forgot that!

But I still have trouble building up the proof.

If |a'> = |a',b'> and |b'> = |a',b'>, does that mean |a'> = |b'> ? If not, why do we use the same symbol to represent two different things?

In Ʃb'<c'|b'><b'|a'><a'|b'><b'|c'> , can we replace |a'> and |b'> with |a',b'> and <a'| and <b'| with <a',b'| ?

6. Apr 3, 2013

### naele

It's more like |a'> = |a',b'> and |b'> = |a'',b''>. I frankly find Sakurai's notation a little confusing/cumbersome so I prefer not to use it. Instead, if |a'> and |b'> are simultaneous eigenstates of each other's operators, what does this tell you about <a'|b'> ?

7. Apr 4, 2013

### omoplata

Ok. I guess this is wrong, but I'll write it down anyway.

Let's say |a'> = |a',b''>, and |b'> = |a'',b'> .

Assume A and B are Hermitian. Then their eigenvalues are real (Theorem in page 17).

Then,
<a'|b'> = <a',b''|a'',b'> = a'<a',b''|a'',b'>/a' = <a',b''|a'|a'',b'>/a' = <a',b''|A|a'',b'> /a' = <a',b''|a''|a'',b'>/a' = (a''/a')<a',b''|a'',b'>

Therefore, a' = a''

Similarly we can prove that b' = b''.

Therefore, |a',b'> = |a'> = |b'>.

So, <a'|b'> = <a'|a'>

We have, Ʃb'<c'|b'><b'|a'><a'|b'><b'|c'> = <c'|a'><a'|a'><a'|a'><a'|c'>

Similarly we can say,
Ʃb'Ʃb''<c'|b'><b'|a'><a'|b''><b''|c'> = <c'|a'><a'|a'><a'|a'><a'|c'>

So the two expressions are equal.

This must be wrong somewhere, but this is all I could come up with.

8. Apr 4, 2013

### vela

Staff Emeritus
One problem is that if a' and a'' are different, the inner product is 0. From (a''/a')0 = 0, you can't conclude anything about how a'' and a' are related.

9. Apr 4, 2013

### omoplata

Oh, of course! That is given in the theorem in page 17! How could I be so blind!

OK, here's my new solution.

We assume A and B are Hermitian (I guess we can do this because it says in page 18 that all observables have real eigenvalues. If the operator is Hermitian that is SUFFICIENT for this to happen. But I don't know if it is NECESSARY for the operator to be Hermitian ).

Let's say |a'> = |a',b''>, and |b'> = |a'',b'> .

If a' ≠ a'', then <a'|b'> = 0. Therefore, Ʃb'<c'|b'><b'|a'><a'|b'><b'|c'> = 0 and Ʃb'Ʃb''<c'|b'><b'|a'><a'|b''><b''|c'> = 0. Therefore the two expressions are equal. But the probability for obtaining |c'> is 0.

The a' = a'' case happens only when |a'> = |b'>. That means that no change has been made to the state. This means that the filter B is either the identity operator (no filter) or B = A ? So it is trivial that the probability of obtaining |c'> is the same in case 1 and case 2.

Is this solution correct?

10. Apr 4, 2013

### vela

Staff Emeritus
No, you're overcomplicating the problem. You're given [A,B]=0 and that the eigenkets of A are non-degenerate, so the theorem on page 30 applies. Read that page again.

A second suggestion is to use slightly different notation to avoid confusion with the primes. Sakurai was using the convention that $a'$ and $b'$ are the eigenvalues, respectively, of operators $\hat{A}$ and $\hat{B}$ corresponding to the simultaneous eigenket $\lvert a', b' \rangle$. Later, when he talks about the Stern-Gerlach setup, $\hat{A}$ and $\hat{B}$ and no longer assumed to be compatible, and $a'$ and $b'$ are simply labels for the eigenkets of $\hat{A}$ and $\hat{B}$.

For example, I'd drop the prime on b' in equation 1.4.46:
$$\sum_b \lvert \langle c' \vert b \rangle \rvert^2 \lvert \langle b \vert a' \rangle \rvert^2 = \cdots$$

11. Apr 6, 2013

### omoplata

OK. I read page 30. So what I should be using is equation (1.4.32) ? Since |b'> = B|a'>, we can say <a'|b'> = <a'|B|a'> = b' ?

12. Apr 6, 2013

### vela

Staff Emeritus
No, that's not right. You can't just insert B in the middle like that.

What would you say is the gist of page 30? That is, if A and B commute, what do you know about the eigenstates of A and B?

13. Apr 7, 2013

### omoplata

The inner product of the eigenstates is non-zero only if they are the same?

14. Dec 2, 2016

### jrlon

Can anyone offer some clarification? I understand that qualitatively if A and B commute, then we can find a simultaneous eigenket of A and B, i.e. |a', b'> in Sakurai's notation. So in the first scenario measurement (A) puts us into this ket, and so the state is unchanged by filter (B) (we are already in an eigenket of B).

This makes sense, but I'm struggling to show mathematically that equations (1.4.46) and (1.4.47) are the same in this case.