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Homework Help: Quantum Operators (or just operators in general)

  1. Feb 15, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]\phi_1[/tex] and [tex]\phi_2[/tex] are normalized eigenfunctions of observable A which are degenerate, and hence not necessarily orthogonal, if <[tex]\phi_1[/tex] | [tex]\phi_2[/tex]> = c and c is real, find linear combos of [tex]\phi_1[/tex] and [tex]\phi_2[/tex] which are normalized and orthogonal to: a) [tex]\phi_1[/tex]; b) [tex]\phi_1[/tex]+[tex]\phi_2[/tex]

    2. Relevant equations

    Non?

    3. The attempt at a solution

    attempt at a)
    from the fact that <[tex]\phi_1[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 0

    i got a + bc = 0

    if I let b = 1, then a = -c,

    then if i use the fact that it must be normalised:
    <a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 1,

    both equations cant be satisfied simultaneously? Where am i going wrong? thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 15, 2010 #2

    kuruman

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    You are forgetting that the wavefunction a|φ1>+b|φ2> must be normalized. If you let b = 1, then necessarily a = 0.
     
  4. Feb 15, 2010 #3

    vela

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    You could try using Gram-Schmidt to solve the problem.
     
  5. Feb 15, 2010 #4

    Isn't the normalization taken into account when I did:
    <a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex] | a[tex]\phi_1[/tex] + b[tex]\phi_2[/tex]> = 1


    Sorry I havent really gotten grips with this operator stuff.
     
  6. Feb 15, 2010 #5

    kuruman

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    It has nothing to do with operators; it has to do with normalization. For a two-component linear combination, a normalized wavefunction (with real coefficients) is always

    [tex]a|\phi_1>+\sqrt{1-a^2}| \phi_2>[/tex]

    Now pick a = 1 and see what happens to this linear combination.
     
  7. Feb 15, 2010 #6


    How is this proved?
     
  8. Feb 15, 2010 #7

    vela

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    That's only if they're orthogonal, right?
     
  9. Feb 15, 2010 #8

    kuruman

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    Correct. That's only if they are orthogonal. My mistake.

    *** Correction ***

    Should read orthonormal.
     
    Last edited: Feb 15, 2010
  10. Feb 15, 2010 #9
    Orthonormal?
     
  11. Feb 15, 2010 #10

    kuruman

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    However, you have

    a + bc = 0 and the normalization condition

    a2 + b2 = 1

    Two equations and two unknowns. It is picking b =1 that you cannot do.
     
  12. Feb 15, 2010 #11

    a2 + b2 = 1

    Where did you get that from? the cross terms in normalization don't vanish?
     
  13. Feb 15, 2010 #12

    kuruman

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    I seem to have a one-track mind here. Yes, throw in the cross term 2ab (for real coefficients). You still have two equations and two unknowns.
     
  14. Feb 15, 2010 #13
    How do we know for sure the coefficients are real?
     
  15. Feb 15, 2010 #14

    vela

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    You have two choices: If you set b=1, then a=-c, and the linear combination you have is orthogonal, but not normalized. So you have to normalize -c|1>+|2>. If you don't set b=1, you can eliminate a using a+bc=0 and then solve for b using the normalization condition. What you can't do is require b=1 in the normalized solution. That's where you're running into problems.
     
  16. Feb 15, 2010 #15


    Yeah, I realize now that I cannot do that, is it possible to do it without resorting to the Gram-Schmit process? Kuruman was suggesting using simultaneous equations, but are the coefficients a and b necessarily real?
     
  17. Feb 15, 2010 #16

    vela

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    You don't need to use Gram-Schmidt. It just seemed like an obvious method to me when I first read the problem.

    Just pick an approach and normalize the resulting state the usual way.
     
  18. Feb 15, 2010 #17
    I end up with conjugates of the coefficients, so I don't have to worry about them? meaning they're real then? If so, how come?
     
  19. Feb 15, 2010 #18

    vela

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    The normalization condition always turns out to be real because the squared terms always involve the multiplication of a complex number by its conjugate and the cross terms are complex conjugates of each other so when they combine, the imaginary part cancels out.

    In this problem, because c is assumed to be real, a and b can be real simultaneously. You could always multiply the state by an arbitrary phase and get complex coefficients if you want, but the relative phase of a and b will remain unchanged.
     
  20. Feb 15, 2010 #19
    Thanks guys! have been very helpful!
     
  21. Feb 15, 2010 #20

    kuruman

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    They don't have to be. You can use coefficients

    [tex]ae^{i\phi_a}; \:be^{i\phi_b}[/tex]

    where a and b are real. Then the normalization condition will be in terms of a, b and the (arbitrary) phase difference φab. Still two equations and two unknowns.
     
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