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!

Wave mechanics : operator problem

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Two operators , A and B , satisfy the equations

    [tex] A=B^{\dagger}B+3 and A= BB^{\dagger}+1[/tex]

    a)Show that A is self adjoint
    b)Find the commutator of [tex][B^{\dagger},B][/tex]
    c) Find the commutator of [tex][B,B^{\dagger}][/tex]
    d) Suppose [tex]\varphi[/tex] is an eigenfunction of A with eigenvalue a:

    A[tex]\varphi[/tex]=a[tex]\varphi[/tex]

    show that if B[tex]\varphi[/tex] =/ 0 then B[tex]\varphi[/tex] is an eigenfunction of A , and find the eigenvalue.

    2. Relevant equations



    3. The attempt at a solution

    I've only worked on the first part of the problem. I will address the remaining 3 parts later.

    [tex](A)^{\dagger}=(B^{\dagger}B+3)^{\dagger}=B^{\dagger}(B^{\dagger})^{\dagger}+(3)^{\dagger}=B^{\dagger}B+(3)^{\dagger}[/tex]. 3 is not an operator so I don't think you can take the adjoint of it.
     
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi noblegas! :smile:

    (LaTeX doesn't take any notice of spaces unless you put them inside \text{} :wink:)

    I assume they mean 3 times the identity, I.
     
  4. Sep 20, 2009 #3
    okak then [tex]B^{\dagger}B+(3I)^{\dagger}=B^{\dagger}B+3*I^{\dagger}=B^{\dagger}B+3*I=[/tex]

    Therefore [tex]A^{\dagger}=A[/tex]? Now proceeding to the next two parts of the problem;

    b) [tex][B,B^{\dagger}]=BB^{\dagger}-B^{\dagger}B[/tex]

    [tex] BB^{\dagger}=A-1, B^{\dagger}B=A-3[/tex], therefore

    [tex][B,B^{\dagger}]=BB^{\dagger}-B^{\dagger}B=A-1-(A-3)=A-1*I-(A-3*I)=2I[/tex]

    c)[tex][A,B]=AB-BA=(BB^{\dagger}+1)B-B(BB^{\dagger}+1)=BB^{\dagger}B+B-BBB^{\dagger}-B=BB^{\dagger}B-BBB^{\dagger}=B(BB^{\dagger}-BB^{\dagger})=B(2I)=2BI[/tex]
     
  5. Sep 21, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    You seem to have changed the questions :rolleyes:

    but it looks ok :smile:
     
  6. Sep 21, 2009 #5
    what do you mean? I was writing out my solutions for part b and c of my question; how should I start part d of the problem? Should I start by assuming that [tex]A
    * \varphi=a*\varphi=>A*\varphi-a*\varphi=0
    [/tex]? See my OP
     
    Last edited: Sep 21, 2009
  7. Sep 21, 2009 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi noblegas! :smile:

    (have a phi: φ and a dagger: † :wink:)
    I meant …
    For d), remember that when the examiner sets you a series of questions, the last one usually comes very easily from the preceding ones …

    in this case, just use c) :smile:
     
  8. Sep 21, 2009 #7
    oh I see then . [tex][B^{\dagger},B]=-2*I, and [A,B]=-2BI[/tex] correct?
     
  9. Sep 22, 2009 #8
    did you not understand my latest solution
     
  10. Sep 22, 2009 #9

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that looks ok (except that you needn't write -2BI, you can just write it as -2B). :smile:
     
  11. Sep 23, 2009 #10
    stuck on part d again: AB-BA=-2BI, ARe they saying [tex]B\varphi[/tex]=> 2BI [tex]\neq[/tex] 0
     
    Last edited: Sep 23, 2009
  12. Sep 24, 2009 #11

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    (what happened to that φ i gave you? :confused:)

    Sorry, I don't understand what you're asking. :redface:

    An eigenvector cannot be zero.

    d) says that if Aφ = aφ, and if Bφ ≠ 0, then A(Bφ) is a scalar multiple of Bφ …

    prove it using c).​
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Wave mechanics : operator problem
Loading...