# Products of observables

1. Jan 25, 2010

### Fredrik

Staff Emeritus
I have to ask a question that looks very simple, and perhaps is very simple, but for some reason I can't answer it in a way that I'm satisfied with.

If A and B are bounded self-adjoint operators that correspond to two different equivalence classes of measuring devices in the real world, and [A,B]≠0, then what device or procedure in the real world does AB correspond to? I thought it would be something like "measure B first and write down the result, then measure A and multiply the two results". But I see several reasons why that can't be correct: 1) The average result of that procedure isn't equal to the expectation value of AB. 2) If the procedure ends with a measurement of A, the system should end up in an eigenstate of A, but a measurement of AB should leave the system in an eigenstate of AB.

It's actually pretty embarassing that I don't know this.

2. Jan 25, 2010

### meopemuk

There are quantum observables, which are expressed as rather complicated functions of other (non-commuting) observables. For example, the observables of (Newton-Wigner) position and spin are functions of Poincare generators (which are observables of energy, momentum, angular momentum, and boost). However, actual measurements of position do not require us to measure all these generators and combine the results. So, my opinion is that functional relationships between various observabls are rather formal and they are not related to the measurement process in any way.

Eugene.

3. Jan 25, 2010

### peteratcam

AB is Hermitian if and only if [A,B] = 0. I think this affects your question.

4. Jan 25, 2010

### Fredrik

Staff Emeritus
I actually didn't even think of that. Now I'm even more embarrassed then before. Thanks both of you.

Last edited: Jan 25, 2010
5. Jan 25, 2010

### SpectraCat

One other salient point is a measurement is *not* equivalent to applying an operator to a quantum state, as you seem to suggest in your OP. Applying an operator (matrix) to a quantum state (vector) produces another quantum state. In order to generate a measurement (scalar), you need to take the inner product of the result with another vector in the same basis.

6. Jan 25, 2010

### Fredrik

Staff Emeritus
I know, and I certainly didn't mean to suggest that. My question was motivated by the fact that the algebraic approach to QM starts with the axiom that equivalence classes of measuring devices can be represented mathematically by members of a C*-algebra (which makes me wonder which members actually correspond to a device or a procedure in the real world). It's sometimes argued that the reason why we need a theory that's constructed from a non-abelian C*-algebra is that measurements disturb the system. See e.g. the last sentence of page 41 in this book.

7. Jan 25, 2010

### peteratcam

Maybe its worth thinking about what measurement does to a system:

The process of measuring corresponds to adding an extra term to the hamiltonian which couples the system and the measuring device.
(1)The extra term should be hermitian since we want a hermitian hamiltonian.
(2)Operators acting on the system and operators acting on the measuring device necessarily commute.

Typical measurement terms look like
$$H_{\rm measure} = g(t) A_m \otimes A_s$$
where the experimentalist has control of g(t). A_m is a hermitian operator acting on the measuring device, whereas A_s acts on the system.
If g(t) is very short but strong compared to other timescales, then the joint system approximately evolves (unitarily) exactly according to $$H_{\rm measure}$$.
The state space for the measuring device and the initial state should be chosen so that H_measure gives a 'good' measurement.

You can then introduce two measuring devices, and ask questions about the final state of the system and the results of the two measurements in the interesting case when [A,B]=/= 0.
$$H_{\rm measure} = g_1(t) A_{m1} \otimes A_s + g_2(t) B_{m2} \otimes B_s$$
In this case, you could imagine two experimentalists who don't like talking to each other. They accidentally try to measure two non-commuting observables at once - what happens? (Discuss how the answer depends on the measuring protocols g_1(t), g_2(t))

I have designed a workable example at undergrad level using the description above which shows how the intensity of the interference pattern in a double slit experiment is related to how 'good' a measurement you can make of which slit the particle goes through. The explanation is of course entanglement with measuring apparatus.