QM - Can We Conclude [A,B]=0 with {|a',b'>}?

[indigojoker]

Homework Statement

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&#039;,b&#039;&gt;=a&#039;b&#039;|a&#039;,b&#039;&gt; <-- skipped several steps
BA|a&#039;,b&#039;&gt;=a&#039;b&#039;|a&#039;,b&#039;&gt;

AB|a&#039;,b&#039;&gt;-BA|a&#039;,b&#039;&gt;=0
[AB-BA]|a&#039;,b&#039;&gt;=0
[A,B]|a&#039;,b&#039;&gt;=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&#039;,b&#039;&gt;} form a complete orthonormal set of base ket.

was this thought process correct?

 

[dextercioby]

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

 

[meopemuk]

[A,B] |a&#039;,b&#039;\rangle =0...(1)
[A,B]=0.....(2)

It is important to mention that vectors |a&#039;,b&#039; \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.

 

[indigojoker]

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

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

 

[dextercioby]

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

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.

 

[indigojoker]

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

Yes, but the observables are incompatible, not the vectors.

 

[dextercioby]

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

 

[indigojoker]

just wonder, how would this change if |a&#039; , b&#039; \rangle did not form a complete orthonormal set of base ket