Commutators, mutual eigenkets, and observables

In summary, the conversation discusses the relationship between the commutation of operators and the existence of a mutual complete set of eigenkets. It also addresses the concept of hermitian operators and their role in determining observable eigenvalues. A theorem is referenced which states that the eigenvalues of self-adjoint operators are all real, making them suitable for observation in quantum mechanics. The conversation also touches on the issue of degeneracy in eigenvalues and its implications in observing quantum phenomena.
  • #1
unscientific
1,734
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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?)
 
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  • #2
unscientific said:
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.

unscientific said:
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.
 
  • #3
a fast way (you may find also the answer in simultaneously diagonalization):
[itex] [x,y] |a> = xy |a> - yx|a> =0 [/itex]
I used the fact that they commute. So
[itex] xy|a>=yx|a> [/itex]
Suppose now that |a> is an eigenstate of y... then you have:
[itex] A_{y} x|a> = yx|a> [/itex]
so what we get?
[itex] y (x|a>)= A_{y} (x|a>)[/itex]
so [itex]x|a>[/itex] is eigenstate of y, with eigenvalue the same as [itex]|a>[/itex]
So we know that
[itex] x|a> \propto |a>[/itex]
So [itex] |a>[/itex] 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
 
  • #4
ChrisVer said:
So we know that
[itex] x|a> \propto |a>[/itex]
So [itex] |a>[/itex] 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>## ??
 
  • #5
Because [itex] y|a>=A_{y}|a>[/itex]
 
  • #6
unscientific said:
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.
 
  • #7
then say that [itex]x|a> \propto |a'>[/itex]
then [itex] y(x|a>)= y|a'>= A'_{y} |a'>[/itex]
and only for [itex]A'_{y}=A_{y}[/itex] the above would hold? while it holds for every |a>?
 
  • #8
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
 

1. What is a commutator?

A commutator is a mathematical operation between two operators that represents the difference between applying them in one order versus the other. It is denoted by [A,B] and is equal to AB - BA.

2. What are mutual eigenkets?

Mutual eigenkets are vectors that are simultaneously eigenkets of two or more operators. This means that when these operators act on the vector, the result is a scalar multiple of the same vector. They are used to find the common eigenvalues of the operators.

3. How are commutators related to observables?

Commutators are related to observables in that they represent the uncertainty or incompatibility between two observables. If the commutator of two operators is non-zero, it means that the two observables cannot be measured simultaneously with precision.

4. What is the significance of eigenvalues and eigenkets in quantum mechanics?

Eigenvalues and eigenkets are important in quantum mechanics because they represent the physical quantities that can be measured in a quantum system. The eigenvalues correspond to the possible outcomes of a measurement, while the eigenkets represent the states of the system in which the measurement will yield that particular outcome.

5. How are mutual eigenkets used to find the eigenvalues of observables?

Mutual eigenkets are used to find the eigenvalues of observables by applying the operators to the eigenkets and setting the resulting equation equal to the eigenvalue multiplied by the eigenket. This yields an eigenvalue equation that can be solved to find the possible eigenvalues of the observable.

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