# Commutator question

1. Mar 6, 2013

### LagrangeEuler

For every operator $A$, $[A,A^n]=0$. And if operators commute they have complete eigen- spectrum the same. But if I look for $p$ and $p^2$ in one dimension $sin kx$ is eigen- function of $p^2$, but it isn't eigen-function of $p$.
$$p^2 \sin kx=number \sin kx$$
$$p\sin kx \neq number \sin kx$$
where
$$p=-i\hbar\frac{d}{dx}$$
$$p^2=-\hbar^2\frac{d^2}{dx^2}$$

2. Mar 6, 2013

### Einj

When two operators commute with each other this doesn't mean that they have the same set of eigenstates. This means that you can alway find a set of eigenstates common to both the operators.
In you particular case the set of common eigenstates must be composed of some other functions.

3. Mar 6, 2013

### DocZaius

If I understand your post correctly, you are saying that if two operators commute, then it is wrong to conclude that every eigenstate of one operator is necessarily an eigenstate of the other operator. However it is right to conclude that at least one set of eigenstates of one is a set of eigenstates of the other.

And so the set of eigenstates the original poster found for p^2 just happened to be one of those sets that isn't a set of eigenstates for p.

Is that right?

4. Mar 6, 2013

### The_Duck

As a simple example of what is going on here, consider two matrices: diag(1, 1) and diag(1, -1). These matrices commute (after all, one is the identity). Yet the column vector (1 1) is a eigenvalue of the first matrix, but not of the second. The fact that the matrices commute only tells us that there is some basis of common eigenvectors; in this case, the vectors (1, 0) and (0, 1).

5. Mar 6, 2013

### Einj

That's correct