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Measuring the physical quantity corresponding to an operator.

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=58272&stc=1&d=1367083476.png

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I am having trouble getting my head around these questions, the first part a) wasn't too tricky, I used the fact that eigenfunctions of a Hermitian operator [itex]\hat{O}[/itex] are orthogonal and got my normalisation constant = 1/14.

    However, I'm having trouble understanding what part b) c) and d) are talking about, here's my understanding:

    b) So, I have three eigenfunction equations involving [itex]\hat{O}[/itex]: [itex]\hat{O} \phi_{1}(x) = \phi_{1}(x)[/itex], [itex]\hat{O} \phi_{2}(x) = 5\phi_{2}(x)[/itex], and [itex]\hat{O} \phi_{3}(x) = 9\phi_{3}(x)[/itex] but what does it mean by physical quantity corresponding to [itex]\hat{O}[/itex] in the state [itex]\phi (x)[/itex]? Does it mean the eigenvalues of each equation? If so, would the 'possible' results just be 1, 5, & 9? If so, why? Then how do I go about calculating the probability of each outcome? I am very confused!

    c) Now, I understand that [itex]<\hat{O}> = \int_{-∞}^{∞} \psi^{*}(x) \hat{O} \psi (x)[/itex] but I don't understand how this will lead to what the question is asking for..

    d) I just don't understand this.

    Basically, I'm having trouble understanding the questions, what they mean, and what they're asking me to do! Any help would be much appreciated!

    Thank you
     

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    Last edited by a moderator: Apr 27, 2013
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  3. Apr 27, 2013 #2

    micromass

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    These questions depend on what theory you've seen.

    But for (b). Given an hermitian operator O that you want to observe. What are the possible outcomes for O? With what probabilities do these outcomes occur?? This is something that should be covered in the theory.

    For (c), you essentially need to calculate ##<\psi|O|\psi>##. Use your knowledge of the bra-ket notation (for example, it is linear and so on).

    For (d), what state do you collapse in after measuring ##O##. This should be covered in the theory.
     
  4. Apr 27, 2013 #3

    vela

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    This question covers basic concepts in quantum mechanics. I'm sure your textbook has examples very similar to this, covering "taking a measurement" corresponds to mathematically.

    By the way, the normalization constant should be ##1/\sqrt{14}##.
     
  5. Apr 27, 2013 #4
    Thank you both for your replies, I can see I am out of my depth at the moment, the course notes don't provide very much information on this but I believe I have now found a relevant chapter in one of my textbooks which I will peruse. I don't think we've covered bra-ket notation yet but we've done inner products, operators and commutators.

    vela, thank you, I haven't had much practice normalising wavefunctions and I can see now why you would have a constant^2 in front of the normalisation integral.

    I will try these again after a read and see if I can do them.
     
  6. Apr 28, 2013 #5
    After some reading

    So I feel silly for not properly researching this before resorting to PF, and for that I apologise. However, would anyone now be able to confirm what I have below?

    a) I got (1/√14) as the normalisation coefficient for ψ(x).

    b) So the eigenvalues of a Hermitian operator O^ represent the possible results of carrying out a measurement of the value of the dynamical variable that is being represented by O^, so we have 1,5, & 9. Then, I understand that ψ(x) = Ʃann(x) and that the probability of a result occurring is |an|2 so the probabilities = (1/14), (4/14), and (9/14).

    c) Expectation value <O^> = ∫ψ* O^ ψ dx , sub in eigenfunction expansion and replace O^ ∅n with λnn and use orthonormality of eigenfunctions again to give <O^> = 7.29

    d) If O^ is measured & λ2 = 5 obtained then the wavefunction collapses to the eigenfunction corresponding to the eigenvalue λ2 = 5, so ψ(x) = ∅2(x)

    Any insight/nod of approval/pointing out flaws would be much appreciated thank you!
     
  7. Apr 28, 2013 #6

    micromass

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    I agree with everything except with (c). Could you provide more details for (c) perhaps?
     
  8. Apr 28, 2013 #7
    So [tex]<\hat{O}> =\int \psi^{*}(x) \hat{O} \psi(x) dx = \frac{1}{14}\int[\phi^{*}_{1}(x) + 2\phi^{*}_{2}(x)+3\phi^{*}_{3}(x)][\hat{O}\phi_{1}(x) + 2\hat{O}\phi_{2}(x)+3\hat{O}\phi_{3}(x)]dx [/tex]

    Now since [itex]\hat{O}\phi_{n}(x) = λ_{n}\phi_{n}[/itex] I replace [itex]\hat{O}\phi_{1}(x)[/itex] with [itex]λ_{1}\phi_{1}[/itex], [itex]2\hat{O}\phi_{2}(x)[/itex] with [itex]2λ_{2}\phi_{2}[/itex] and [itex]3\hat{O}\phi_{1}(x)[/itex] with [itex]3λ_{3}\phi_{1}[/itex]

    Which gives: [tex]\frac{1}{14}\int[\phi^{*}_{1}(x) + 2\phi^{*}_{2}(x)+3\phi^{*}_{3}(x)][\phi_{1}(x) + 10\phi_{2}(x)+27\phi_{3}(x)]dx [/tex]

    Then using the orthogonality of the eigenfunctions we know that the integrals where [itex]n\not=m[/itex] will be 0 and those where [itex]n=m[/itex] will be 1 so we have:

    [tex]\frac{1}{14}[1 + 20 + 81] = \frac{102}{14} \approx 7.29[/tex]
     
  9. Apr 28, 2013 #8

    micromass

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    Weird. I must have made some computation error somewhere. It looks like your solution is correct after all.
     
  10. Apr 29, 2013 #9

    dextercioby

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    For point d) you miss a numerical factor in front of phi_2. I think this is 2/sqrt(14).
     
  11. Apr 29, 2013 #10
    Ah I see, makes sense thank you!
     
  12. Apr 29, 2013 #11

    vela

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    Given that ##\phi_2## is normalized, your original answer to (d) was fine. Including the constant would simply give you an unnormalized state. There's no reason to include it.
     
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