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?)

Related Quantum Physics News on Phys.org

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 (Spectral Theorem) 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

Gold Member
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

Gold Member
Because $y|a>=A_{y}|a>$

Fredrik

Staff Emeritus
Gold Member
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

Gold Member
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

Mentor
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

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving