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Eigenvalues and operators, step the involves switching and substitutin

  1. Sep 15, 2014 #1
    Operator C = I+><-I + I-><+I

    Wavefunction PSI = Q I+> +V I->

    C PSI = Q I-> + V I+>

    note the I is just a straight line (BRAKET vectors), the next step is where I get confused, p is subbed in and the ket vectors switch places...

    C PSI = pQ I+> + pV I-> <---- why??

    therefore

    V = pQ and Q = pV

    which gives p^2 Q = Q therefore p = +/- 1

    why can p be subbed in and the ket vectors switch around, i.e. I-> no longer with Q and I+> no longer with V

    thanks
     
  2. jcsd
  3. Sep 15, 2014 #2

    kith

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    What is p? What are you even trying to do? What source of information are you using?
     
  4. Sep 15, 2014 #3
    Hi, im reviewing last semesters quantum mechanics, the complete question is:

    Determine the eigenvalues of the operator C = I+><-I + I-><+I. You may use PSI = Q I+> + V I-> as an eigenvector.

    so p would be the eigenvalues of the operator C which are +1 and -1

    the solution introduces p to C PSI = Q I-> + V I+> by switching the ket vectors around and putting p with each term, I don't understand why this is.

    thanks for the reply
     
  5. Sep 15, 2014 #4

    Fredrik

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    Please don't write I. There's a vertical line on the keyboard that you can use. Or you can copy and paste this one: |

    I assume that Q and V are just numbers. What is p and how did it enter the calculation?

    Oh wait, p is the eigenvalue, right? Then by definition of eigenvalue and eigenvector, you have ##C|\psi\rangle =p|\psi\rangle##. Now you can just use the definition of ##|\psi\rangle## to get
    $$C|\psi\rangle =p|\psi\rangle =p(Q|+\rangle+V|-\rangle)=pQ|+\rangle+pV|-\rangle.$$ Since you already know that
    $$C|\psi\rangle=Q|-\rangle+V|+\rangle,$$ this implies that pQ=V and pV=Q.
     
    Last edited: Sep 15, 2014
  6. Sep 15, 2014 #5
    oh dear I missed something in the solution:

    Recall the eigenvalue equation: C |PSI> = p |PSI>

    that would explain where the p comes from, but it doesn't explain why the ket vectors switch around... or does it?

    Yes Q and V are just numbers.

    Apologies on the I.
     
  7. Sep 15, 2014 #6

    Fredrik

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    If p is the eigenvalue, then by definition of eigenvalue and eigenvector, you have ##C|\psi\rangle =p|\psi\rangle##. Now you can just use the definition of ##|\psi\rangle## to get
    $$C|\psi\rangle =p|\psi\rangle =p\big(Q|+\rangle+V|-\rangle\big)=pQ|+\rangle+pV|-\rangle.$$ Since you already know that
    $$C|\psi\rangle=Q|-\rangle+V|+\rangle,$$ this implies that
    $$(pQ-V)|+\rangle +(pV-Q)|-\rangle=0.$$ Since the basis vectors are linearly independent, this implies that pQ-V=0 and pV-Q=0.
     
  8. Sep 15, 2014 #7
    Ahhh I see now, that's great!! Thanks very much for your time, really appreciated!
     
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