Simultaneous measuring of two operators

[Caulfield]

Hello everybody.

I am new here, and also new to quantum mechanics. This is the question to which I can't answer neither in mathematical nor physical way.

a,b → observables (like position and momentum)
A,B → corresponding operators.

"It is possible for particles to be in a state of definite a and b at the same time only if the corresponding operators A and B commute."

Why?

 

[atyy]

When you measure something, you collapse its state into an eigenstate of whatever you are measuring. The state remains undisturbed if it is already an eigenstate of whatever you are measuring. This is why a particle in an eigenstate of A is said to have a definite value of A. A particle can be in an eigenstate of both A and B only if A and B commute.

The other way to say it is that if you measure an observable A on a particle in a state, you will get a particular result with certainty only if the state is an eigenstate of A. This is why we say an eigenstate of A has a definite value of A. Again, a particle can only be in an eigenstate of both A and B only if A and B commute.

 

[Caulfield]

if they don't commute, for example: [A,B]=-iħ

A(ψ)=a(ψ)

BA(ψ)=Ba(ψ)

-iħA(B(ψ))=aB(ψ)

A(B(ψ))=(i/ħ)aB(ψ)

but A(ψ)=aψ, and (i/ħ) is not equal to a.

This means we have two different wave functions as eigenfunctions of the operator A. (So eigenstate is disturbed)

How this imply that B(ψ) is not equal to bB(ψ)? (that is, how does it imply that b is not definite?)

 

[atyy]

Let ψ be an eigenstate of A and of B.

Aψ=aψ

Bψ=bψ

[A,B]ψ
=ABψ-BAψ
=Abψ-Baψ
=bAψ-aBψ
=baψ-abψ
=0

So if ψ is an eigenstate of A and B, then A and B will commute. (It may depend on the subspace, and some other conditions, which I don't remember.)

 

[bhobba]

"It is possible for particles to be in a state of definite a and b at the same time only if the corresponding operators A and B commute."

Its a theorem on non commuting observables:
http://www.pa.msu.edu/~mmoore/Lect23_HeisUncPrinc.pdf

Also any good book on QM will detail it - eg see section 8.4 - Ballentine - Quantum Mechanics - A Modern Development.

But seeing why commuting observables have no issues isn't that hard. Basically it means they have the same set of eigenvectors, which roughly translates to they are really the same observation in disguise - note to the more knowledgeable reading this - I said ROUGHLY. Technically such are said to be compatible.

Thanks
Bill