PVMs of commuting observables

[adjklx]

Homework Statement

Consider two observables $$A$$ and $$B$$ such that $$[A,B]=0$$. Given the spectral resolutions of each operator:

$$A = \sum_k a_k P_k$$

$$B= \sum_j b_j Q_j$$

where $$P_k$$ and $$Q_j$$ are projectors onto the eigenstates of their respective operator.

Show that $$[P_k,Q_j]=0$$ for every $$k$$ and $$j$$

The Attempt at a Solution

I had two different ways of going about this problem:

1) Since $$A$$ and $$B$$ commute there exists a common complete set of eigenstates between them. Writing both of their spectral resolutions in terms of this basis gives:

$$[P_k,Q_j] = [P_k,P_j] = 0$$ for all k and j. This answer seems wrong to me because I'm ignoring all the other possible PVMs that I could use for these two operators.

2) $$[A,B] = \sum_{k,j} a_k b_j [P_k,Q_j] = 0$$ Once I have this expression I'm not quite sure what to do with it. One thing that has crossed my mind is this:

$$[A,B] = \sum_{k,j} a_k b_j [P_k,Q_j] = 0$$

$$\implies \sum_{k,j} a_k b_j P_k Q_j = (\sum_{k,j} a_k b_j P_k Q_j)*$$

but I'm not quite sure if this leads to anything. I also thought that there might be a reason that each term in the sum should go to zero individually and somehow it's only possible that the commutator in each term can go to zero. This didn't make much sense to me though because what if some of the eigenvalues $$a_k$$ and $$b_j$$ were zero.

Thanks for your help!

 

[Jhero]

Your second approach is on the right track. Here's how you can continue:

From [A,B]=0, we have:

\sum_{k,j} a_k b_j [P_k,Q_j] = 0

Since the commutator [P_k,Q_j] is a scalar, we can move it outside the summation:

\sum_{k,j} [P_k,Q_j] a_k b_j = 0

Now, since this holds for all values of a_k and b_j, we can say that each term in the sum must be equal to zero:

[P_k,Q_j] = 0 for all k and j

This shows that the commutator between the projectors P_k and Q_j is zero, and therefore [A,B]=0 implies [P_k,Q_j]=0.