# 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