College Math Teaching

July 25, 2011

Quantum Mechanics and Undergraduate Mathematics V: compatible observables

This builds on our previous example. We start with a state \psi and we will make three successive observations of observables which have operators A and B in the following order: A, B, A . The assumption is that these observations are made so quickly that no time evolution of the state vector can take place; all of the change to the state vector will be due to the effect of the observations.

A simplifying assumption will be that the observation operators have the following property: no two different eigenvectors have the same eigenvalues (e. g., the eigenvalue uniquely determines the eigenvector up to multiplication by a constant of unit modulus).

First of all, this is what “compatible observables” means: two observables A, B are compatible if, upon three successive measurements A, B, A the first measurement of A is guaranteed to be the second measurement of A . That is, the state vector after the first measurement of A is the same state vector after the second measurement of A .

So here is what the compatibility theorem says (I am freely abusing notation by calling the observable by the name of its associated operator):

Compatibility Theorem
The following are equivalent:

1. A, B are compatible observables.
2. A, B have a common eigenbasis.
3. A, B commute (as operators)

Note: for this discussion, we’ll assume an eigenbasis of \alpha_i for A and \beta_i for B .

1 implies 2: Suppose the state of the system is \alpha_k just prior to the first measurement. Then the first measurement is a_k . The second measurement yields b_j which means the system is in state \beta_j , in which case the third measurement is guaranteed to be a_k (it is never anything else by the compatible observable assumption). Hence the state vector must have been \alpha_k which is the same as \beta_j . So, by some reindexing we can assume that \alpha_1 = \beta_1 . An argument about completeness and orthogonality finishes the proof of this implication.

2 implies 1: after the first measurement, the state of the system is \alpha_k which, being a basis vector for observable B means that the system after the measurement of B stays in the same state, which implies that the state of the system will remain \alpha_k after the second measurement of A . Since this is true for all basis vectors, we can extend this to all state vectors, hence the observables are compatible.

2 implies 3: a common eigenbasis implies that the operators commute on basis elements so the result follows (by some routine linear-algebra type calculations)

3 implies 2: given any eigenvector \alpha_k we have AB \alpha_k = BA \alpha_k = a_k B \alpha_k which implies that B \alpha_k is an eigenvector for A with eigenvalue \alpha_k . This means that B \alpha_k = c \alpha_k where c has unit modulus; hence \alpha_k must be an eigenvector of B . In this way, we establish a correspondence between the eigenbasis of B with the eigenbasis of A .

Ok, what happens when the observables are NOT compatible?

Here is a lovely application of conditional probability. It works this way: suppose on the first measurement, a_k is observed. This puts us in state vector \alpha_k . Now we measure the observable B which means that there is a probability |\langle \alpha_k, \beta_i \rangle|^2 of observing eigenvalue b_i . Now \beta_i is the new state vector and when observable A is measured, we have a probability |\langle \alpha_j, \beta_i \rangle|^2 of observing eigenvalue a_j in the second measurement of observable A .

Therefore given the initial measurement we can construct a conditional probability density function p(a_j|a_k) = \sum_i p(b_i|a_k)p(a_j|b_i)= \sum_i |\langle \alpha_k, \beta_i \rangle| |^2 |\langle \beta_i, \alpha_j |^2

Again, this makes sense only if the observations were taken so close together so as to not allow the state vector to undergo time evolution; ONLY the measurements changes the state vector.

Next: we move to the famous Heisenberg Uncertainty Principle, which states that, if we view the interaction of the observables A and B with a set state vector and abuse notation a bit and regard the associated density functions (for the eigenvalues) by the same letters, then V(A)V(B) \geq (1/4)|\langle \psi, [AB-BA]\psi \rangle |^2.

Of course, if the observables are compatible, then the right side becomes zero and if AB-BA = c for some non-zero scalar c (that is, (AB-BA) \psi = c\psi for all possible state vectors \psi ), then we get V(A)V(B) \geq (1/4)|c|^2 which is how it is often stated.


1 Comment »

  1. thank you

    Comment by sani — May 13, 2012 @ 7:32 pm

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