Commuting Operators in Sequential Stern Gerlach Experiment

[maverick280857]

Hi everyone

How do I show that the expression

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

equals the expression

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

when either

$$[A,B] = 0$$
or
[tex][B,C] = 0[/itex]

I tried this but the algebra didn't work out. Do I just equate the summands or is there something else to be done?

If I do that, I get just

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

Thanks in advance.
Cheers

 

[maverick280857]

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

This gives

$$\langle a'|c'\rangle = \langle a'|b'\rangle \langle b'|c'\rangle$$

Anyone?

 

[strangerep]

Maverick,

You seem not to have specified the problem in enough detail.
What is the relationship between your A,B,C operators
and the a,b,c,etc in your bras and kets? And do the latter
form a complete orthonormal set, or what?

 

[maverick280857]

Hi, yes, sorry for that. $\{|a'\rangle\}$, $\{|b'\rangle\}$ and $\{|c' \rangle\}$ are eigenkets of operators A, B and C respectively.

The set $\{|a'\rangle\}$ is normalized to unity.

(PS--This is a situation described in JJ Sakurai's book.)

 

[malawi_glenn]

I think you should use the fact that two compatible observables are diagonal in the same eigenbase. Was a long time since I did this stuff and I have my notebooks 1500km away from me.

You must use the fact that A&B or B&C commutes.

 

[strangerep]

Hi, yes, sorry for that. $\{|a'\rangle\}$, $\{|b'\rangle\}$ and $\{|c' \rangle\}$ are eigenkets of operators A, B and C respectively.

The set $\{|a'\rangle\}$ is normalized to unity.

(PS--This is a situation described in JJ Sakurai's book.)

Still not enough detail for me, since I don't have a copy of Sakurai.
I have quite a few others, but I suppose I should order Sakurai since lots of
people seem to use it.

In the meantime, I guess you'll either have to spell out the problem
far more explicitly, or else wait for someone else to answer.

 

[maverick280857]

You must use the fact that A&B or B&C commutes.

But that's precisely what I have to show, i.e. the the results of the two probability measurements are identical (which means the presence or absence of B leaves the measurement unaffected) when either A and B commute or B and C commute.

strangerep, basically we have 3 Stern Gerlach filters (if that's what they're called). The first one, A (with eigenkets normalized to unity) allows a state |a'> to pass, the second one, B allows a certain state |b'> to pass and C allows |c'> to pass (each filter supresses all the other states). We're trying to show that if the filter B is physically present, the probability of obtaining |c'> from a state |a'> is different from the same probability when the filter B is absent or non-operative. In the former case, the probability is simply the sum over b' terms of the form

|<c'|b'>|^2 |<b'|a'>|^2

Operationally this means that the experiment is repeated for every b' keeping all the others blocked.

In the second experiment, when filter B is either absent or non-operative, the probability is simply |<c'|a'>|^2 and now we expand |a'> in terms of the complete set of eigenkets |b'>.

Now, we want to show that the two measurements will yield identical probabilities for obtaining |c'> from |a'> provided either [A, B] = 0 or [B, C] = 0. I am stuck with the algebra. Thats the gist of the problem.

 

[malawi_glenn]

But you must use the commutator relations, otherwise you get *stuck* in the algebra. Have you used that fact yet?

 

[mrandersdk]

Lets take the case where $[A,B] = 0$ then you can choose |a> and |b> such that they both have the form

$$A |a,b> = a |a,b>$$
$$B |a,b> = b |a,b>$$

I now make the notation

$$|a,b> = |K>$$

then you have

$$\sum_{b'} <c'|b'><b'|a'><a'|b'><b'|c'> = \sum_{K'} <c'|K'><K'|K><K|K'><K'|c'> = \sum_{K'} <c'|K'>\delta_{K,K'}<K'|c'> =$$
$$<c'|K><K|c'> = |<K|c'> |^2 = |<a'|c'> |^2$$

is this what you need?

 

[maverick280857]

Ok mrandersdk, I think I get it. If you take [A,B] = 0 then you find that the two expressions are identical. What I want to show is the converse, i.e. [A,B] = 0 or [B,C] = 0 follows from the equality of those two summations.

malawi_glenn, could you please elaborate on the commutator relations you are referring to?

 

[mrandersdk]

Okay, but if you read your first post you actually ask for the opposite of what you want.

Are you even sure it is true what you want?

 

[maverick280857]

Hi everyone

How do I show that the expression

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

equals the expression

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

when either

$$[A,B] = 0$$
or
[tex][B,C] = 0[/itex]

Okay, so mathematically, I was imprecise. I should have said the following:

Prove that

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

$$\implies [A,B] = 0$$ or $$[B,C] = 0$$.

Are you even sure it is true what you want?

This is a problem mentioned in the text "Modern Quantum Mechanics" by JJ Sakurai. I am trying to prove it.

 

[malawi_glenn]

And if they commute, what relation do you have between the eigenvectors?

 

[maverick280857]

And if they commute, what relation do you have between the eigenvectors?

If operators A and B commute, then every eigenket $|a'\rangle$ of A will be a simultaneous eigenket of B.

 

[maverick280857]

If operators A and B commute, then every eigenket $|a'\rangle$ of A will be a simultaneous eigenket of B.

EDIT: Okay well, I think I see what your point is. I could basically follow a reverse derivation of the one you gave me and then infer that either [A, B] = 0 or [B, C] = 0 for this to happen. I was kind of hoping for a more linear approach.

 

[CarlB]

It's not how you are to do the problem, but as long as I'm pushing density matrix formalism, I should write up the solution to the problem in density operator formalism.

The first expession in density matrix form:
$$\Sigma_b <c'|b'><b'|a'><a'|b'><b'|c'> = tr( \Sigma_b( B A B C) )$$

If A commutes with B, then write:
$$tr( \Sigma_b( B A B C) ) = tr(\Sigma_b ( A B B C) = tr(\Sigma_b ( A B C) )$$
where recourse has been made to the idempotency relation, BB = B. Now, since the set of B are complete, the sum over them is unity: $$\Sigma B = 1$$, hence the final result.

If C commutes with B, then
$$tr( \Sigma_b( B A B C) ) = tr(\Sigma_b ( B A C B)$$
The useful property of the trace allows us to change the order of the trace of a product by making what was first become last (or moving the last become the first). So the following traces are equal:
tr( J K L M) = tr( K L M J) = tr( L M J K) = tr( M J K L).
And the trace commutes with the summation. With this change, the above becomes:
$$= tr(\Sigma_b ( A C B B) ) = tr(\Sigma_b(A C B))$$
and then, as before, the sum over B reduces to unity and you have the required result.

The general case, where B does commutes with neither A nor C, is also instructive. The product B A B, for any pure state B, is always a complex multiple of B. Written in state vector form:
$$BAB = <b|A|b> B$$.
For the case of the operator A being a pure state, i.e. A = |a><a|, BAB is a real multiple of A, that is
$$BAB = <b|a><a|b>B = |<b|a>|^2 B$$,
which is a real multiple of B. With this simplification, the first expression reduces to:
$$tr (\Sigma_b <b|A|b> BC )$$.
By the property of the trace that allows you to move stuff around, this is equal to
$$tr (\Sigma_b <b|C|b> BA )$$

In general, any function of bras and kets that is an observable can be written as the trace of a product of density matrices. Under the assumption that density operators are fundamental, one defines "observable" for bras and kets as "any function that can be written in density operator form". Once one has an observable in density operator form (as a trace), then one can use the properties of the trace to manipulate the observable. For those who like to work in geometric algebra (Clifford algebra), it is common to replace "trace" with something like "scalar part", that is, one splits the object X into "blades" and keeps the scalar blade, $$<X>_0$$.

Contrary to what some people sometimes say, there do exist observables that are complex. The best example is quantum phase, also called Berry-Pancharatnam phase. It arises when a pure state is sent through a sequence of at least three pure states. That is, given three pure states A, B, and C, the observable
tr(A B C) = tr (B C A) = tr (C A B)
is, in general, complex. For the Pauli algebra, the complex phase is equal to half the (oriented) surface area of the spherical triangle defined by the spin vector directions A, B, and C. Since the total surface area of the [Bloch] sphere is 4 pi, the total complex phase is half this, which is conveniently 2 pi.

If one wishes to write these complex numbers as complex multiples of a state (as discussed in the general case described above), then one can write, for example,
A B C A = complex multiple of A,
A B C A = tr(A B C) A.

 

[maverick280857]

Hi, thanks for this detailed reply. I'll go through it and post back if/when I have any questions.