Commutative operators and obserables

[blakegriffin1]

I'm just revising some Quantum Mech and I have two questions.

I know that if two operators commute say for instance [$\hat{A}$,$\hat{B}$] = [$\hat{B}$,$\hat{A}$] = 0 Then the observables that the operators extract from the wavefunction can be measured exactly (without losing information about the other).
so that B $\psi$ = b$\psi$ and A $\psi$ = a $\psi$

Question 1) Does this mean that the expectation value of $\hat{B}$ and $\hat{A}$ will be "b" and "a" respectively ? or does this mean that after each and every measurement the measured value of $\hat{B}$ and $\hat{A}$ will be b and a ?

Secondly
Now if two operators don't commute then they can't both be measured simultaneously.
So my question is basically let's assume we have the position and momentum operators and we wish to measure their observables. Assume we know the momentum of the particle exactly. Then we have:
$\hat{p}$ $\psi$ = j $\psi$
(where j is the exact value of the momentum)

Now Clearly since the "x" operator simply acts by "x" we have:
$\hat{x}$ $\psi$ = x $\psi$
So they are measured exactly ?
I know my understanding is off somewhere can someone point out to me where ?
Thanks in advance

 

[chogg]

I'm just revising some Quantum Mech and I have two questions.

I know that if two operators commute say for instance [$\hat{A}$,$\hat{B}$] = [$\hat{B}$,$\hat{A}$] = 0 Then the observables that the operators extract from the wavefunction can be measured exactly (without losing information about the other).
so that B $\psi$ = b$\psi$ and A $\psi$ = a $\psi$

Question 1) Does this mean that the expectation value of $\hat{B}$ and $\hat{A}$ will be "b" and "a" respectively ? or does this mean that after each and every measurement the measured value of $\hat{B}$ and $\hat{A}$ will be b and a ?

Both. (Of course, the second implies the first: if you always get b, then the expectation value is b too.)

Secondly
Now if two operators don't commute then they can't both be measured simultaneously.
So my question is basically let's assume we have the position and momentum operators and we wish to measure their observables. Assume we know the momentum of the particle exactly. Then we have:
$\hat{p}$ $\psi$ = j $\psi$
(where j is the exact value of the momentum)

Now Clearly since the "x" operator simply acts by "x" we have:
$\hat{x}$ $\psi$ = x $\psi$
So they are measured exactly ?
I know my understanding is off somewhere can someone point out to me where ?
Thanks in advance

The equation
$$\hat{x} \psi = x \psi$$
is only true if $\psi$ is an eigenstate of the $\hat{x}$ operator. Since it's an eigenstate of $\hat{p}$, it couldn't possibly be an eigenstate of $\hat{x}$.

 

[blakegriffin1]

Right but isn't any wavefunction $\psi$ an eigenfunction of $\hat{x}$
as $\hat{x}$ $\psi$ = x $\psi$
This should work for basically everything right since the operator $\hat{x}$ doesn't really do anything besides apply "x" ?

 

[dextercioby]

The x operator acts by multiplication with x only in the coordinate representation. Choosing another representation of the commutation relation means the operator will act in a different way.