# Homework Help: 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.

Am I going about this the wrong way?

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!

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