Commutators, mutual eigenkets, and observables

[unscientific]

I have two quick questions:

1. Why if say [x,y] = 0, it implies that there is a mutual complete set of eigenkets?

where x and y can be anything, like momentum, position operators.


2. If an operator is not hermitian, why isn't it an observable? (More specifically, why isn't its eigenvalue an observable?)

 

[Ravi Mohan]

1. Why if say [x,y] = 0, it implies that there is a mutual complete set of eigenkets?

where x and y can be anything, like momentum, position operators.

You can find a proof in Modern Quantum Mechanics by J. J. Sakurai, Chapter 1, Compatible observables.

2. If an operator is not hermitian, why isn't it an observable? (More specifically, why isn't its eigenvalue an observable?)

The term "Self-adjoint" must be used instead of "Hermitian".
It is a postulate of Quantum Mechanics. Unless you find an experiment violating this postulate, you must consider it to be true.

There is a theorem (http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Spectral_theorem.html) which says that the eigen-values of self-adjoint operators are all real (which is required for an observable outcome). Also, their eigenvectors form a complete set.

 

[ChrisVer]

a fast way (you may find also the answer in simultaneously diagonalization):
[x,y] |a> = xy |a> - yx|a> =0
I used the fact that they commute. So
xy|a>=yx|a>
Suppose now that |a> is an eigenstate of y... then you have:
A_{y} x|a> = yx|a>
so what we get?
y (x|a>)= A_{y} (x|a>)
so x|a> is eigenstate of y, with eigenvalue the same as |a>
So we know that
x|a> \propto |a>
So |a> must also be eigenstate of x...


As for your other question, hermitian operators ensure that you'll find real eigenvalues... In physics we are not used in observing i-s

 

[unscientific]

So we know that
x|a> \propto |a>
So |a> must also be eigenstate of x...As for your other question, hermitian operators ensure that you'll find real eigenvalues... In physics we are not used in observing i-s

I don't see how you can say $x|a> \propto |a>$ ??

 

[ChrisVer]

Because y|a>=A_{y}|a>

 

[Fredrik]

I don't see how you can say $x|a> \propto |a>$ ??

This step is valid if the subspace that consists of all eigenvectors of y with eigenvalue $A_y$ is known to be 1-dimensional. This isn't true for an arbitrary y.

 

[ChrisVer]

then say that x|a> \propto |a'>
then y(x|a>)= y|a'>= A'_{y} |a'>
and only for A'_{y}=A_{y} the above would hold? while it holds for every |a>?

 

[bhobba]

The following may help:
http://www.glue.umd.edu/afs/glue.umd.edu/department/phys/courses/Phys622/public_html/ji/lecture5.pdf

If there is no degeneracy (ie the eigenvalues are distinct) its easy AB |a> = BA |a> = Ba|a> = a B|a> so that B|a> is also an eigenvector of A with eigenvalue a. But since there is no degeneracy B|a> must be a multiple of |a> ie B|a> = b|a>.

When there is degeneracy things are more difficult - but the link gives the detail.

Physically though it's not hard - changing the eigenvalue simply changes the value of the observation outcome, so nothing physically really changes by altering the observables to be non degenerate to prove it, then changing back.

Thanks
Bill