Observables and common eigenvectors

1. Nov 9, 2013

md.xavier

1. The problem statement, all variables and given/known data

In a given basis, the eigenvectors A and B are represented by the following matrices:

A = [ 1 0 0 ] B = [ 2 0 0 ]
[ 0 -1 0] [ 0 0 -2i ]
[ 0 0 -1] [ 0 2i 0 ]

What are A and B's eigenvalues?
Determine [A, B].
Obtain a set of eigenvectors common to A and B. Do they form a complete basis?

2. Relevant equations

(A - λI)x = 0
[A, B] = AB - BA

3. The attempt at a solution

Okay, so, I calculated the eigenvalues and the commutator quite easily.

For A, I got eigenvalues 1 and -1, with -1 having degeneracy 2.
For B, I got eigenvalues 2 and -2, with 2 having degeneracy 2.

The commutator was 0, so they commutate.

Now, as far as common eigenvectors go - I could only find one. [1 0 0] transposed.

Is this due to the eigenvalues having degeneracy? Does the fact that two observables commuting implies that they have a common complete basis of eigenvectors only hold up if they don't come from degenerate eigenvalues?

Thank you for your help -- the material given to me was not very clear regarding this particular case.

2. Nov 9, 2013

TSny

Can you tell us what you got for your three eigenvectors of B?

Last edited: Nov 9, 2013
3. Nov 9, 2013

md.xavier

I did not get B's eigenvectors -- I got A's, the three unitary vectors, and applied B on them. Only one gave me an eigenvalue of B's (I got [2 0 0] transposed from applying B to [1 0 0] transposed). The others gave me stuff like [0 0 2i] transposed and [0 -2i 0] transposed, which are NOT eigenvectors of B since the remaining eigenvalues are -2 and 2.

I did do the exercise a few weeks ago by calculating B's eigenvectors and got the same conclusions -- I have sadly lost the sheet in which I did them.

4. Nov 9, 2013

TSny

You have found two eigenvectors of A that have the same eigenvalue of -1. Any linear combination of these two vectors will still be an eigenvector of A with eigenvalue -1, as you can easily show. See if you can find a particular linear combination that will also be an eigenvector of B with eigenvalue 2. Then find another linear combination that will be an eigenvector of B with eigenvalue -2.

5. Nov 9, 2013

md.xavier

I just did that. Considering a vector a = β*[0 1 0] + μ*[0 0 1], (both transposed) and applying B to it, I get the vector [0 -2μi 2βi] transposed.

Plugging in the eigenvalues and equalizing them, the only solution is μ and β equal to zero for both of them... so I'm at a loss.

6. Nov 9, 2013

TSny

There are nonzero solutions for μ and β.

For example, suppose you want an eigenvector of B with eigenvalue 2. Then you need to find values of μ and β such that

B$\cdot$[0, β, μ]T = 2*[0, β, μ]T

or

[0, -2μi, 2βi] = [0, 2β, 2μ]

7. Nov 9, 2013

md.xavier

Oh. Thank you so much, I was messing up somewhere - I got exactly the same eigenvectors as I got for B. I probably should have gotten B's and compared them with A too.

Turns out they form a complete basis.

Thank you so much again!