# QM - observables

1. Sep 15, 2007

### indigojoker

1. The problem statement, all variables and given/known data
If A and B were observables, and say the simultaneous eigenkets of A and B $${|a',b'>}$$ form a complete orthonormal set of base ket. Can we conclude that $$[A,B]=0$$?

2. The attempt at a solution

Assume $${|a',b'>}$$ is incompatible:

$$AB|a',b'>=a'b'|a',b'>$$ <-- skipped several steps
$$BA|a',b'>=a'b'|a',b'>$$

$$AB|a',b'>-BA|a',b'>=0$$
$$[AB-BA]|a',b'>=0$$
$$[A,B]|a',b'>=0$$
$$[A,B]=0$$

and so we reach a contradiction. Therefore, we conclude that $$[A,B]=0$$ assuming the simultaneous eigenkets of A and B $${|a',b'>}$$ form a complete orthonormal set of base ket.

2. Sep 15, 2007

### dextercioby

Yes, but the observables are incompatible, not the vectors. And your LaTex would look better by using "$$\rangle$$" instead of ">".

3. Sep 15, 2007

### meopemuk

It is important to mention that vectors $|a',b' \rangle$ form a full basis. Therefore any vector in the Hilbert space can be represented as a linear combination of these basis vectors. Therefore, by linearity, your eq. (1) is valid for any vector $|\psi \rangle$

$$[A,B] |\psi \rangle =0$$

Then you can conclude that eq. (2) holds.

Eugene.

4. Sep 15, 2007

### indigojoker

Sorry, you mean I should assume that A and B are incompatible and the do the proof by contradiction right?

5. Sep 15, 2007

### dextercioby

Nope, compatibility of the observables is no issue here. And you don't need to use any "proof by contradiction". A direct proof is enough. And that's what you did in post #1 in this thread.

6. Sep 15, 2007

### indigojoker

Then I'm not sure what you mean by this:

7. Sep 15, 2007

### dextercioby

I was correcting your expression: the observables are compatible/incompatible, and not the vectors, since you cannot measure vectors, but only observables.

8. Sep 16, 2007

### indigojoker

just wonder, how would this change if $$|a' , b' \rangle$$ did not form a complete orthonormal set of base ket