1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Observables and common eigenvectors

  1. Nov 9, 2013 #1
    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. jcsd
  3. Nov 9, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Can you tell us what you got for your three eigenvectors of B?
     
    Last edited: Nov 9, 2013
  4. Nov 9, 2013 #3
    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.
     
  5. Nov 9, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  6. Nov 9, 2013 #5
    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?
     
  7. Nov 9, 2013 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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]
     
  8. Nov 9, 2013 #7
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Observables and common eigenvectors
Loading...