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QM:Expectation values and calculating probabilities

  1. May 18, 2014 #1
    1. The problem statement, all variables and given/known data

    An operator [itex]\mathbf{A}[/itex], corresponding to a physical quantity [itex]\alpha [/itex], has two normalized eigenfunctions [itex]\psi_1(x)\quad \text{and}\quad \psi_2(x)[/itex], with eigenvalues [itex]a_1 \quad\text{and}\quad a_2[/itex]. An operator [itex]\mathbf{B}[/itex], corresponding to another physical quantity [itex]\beta[/itex], has normalized eigenfunctions [itex]\phi_1(x)\quad \text{and}\quad \phi_2(x)[/itex], with eigenvalues [itex]b_1 \quad\text{and}\quad b_2[/itex]. [itex]\alpha [/itex] is measured and the value [itex]a_1[/itex] is obtained. If [itex]\beta[/itex] is then measured and then [itex]\alpha [/itex] again, show that the probability of obtaining [itex]a_1[/itex] a second time is [itex]\frac{97}{169}[/itex].

    2. Relevant equations

    The eigenfunctions are related via:

    [itex] \psi_1 = \frac{(2 \phi_1+3 \phi_2)}{\sqrt{13}}[/itex]
    [itex]\psi_2 = \frac{(3 \phi_1-2 \phi_2)}{\sqrt{13}}[/itex]

    3. The attempt at a solution

    Okay now I know I can represent [itex]|\psi\rangle[/itex] by:
    [itex]|\psi\rangle = \frac{1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}|\phi\rangle[/itex]

    I also know that initially:

    [itex]\mathbf{A}|\psi\rangle = a_1|\psi\rangle = \frac{a_1}{\sqrt{13}}\begin{pmatrix}2&3\\3&-2\end{pmatrix}|\phi\rangle[/itex] which I can then bra through by [itex]\langle \psi|[/itex] in order to get [itex] a_1 \langle \psi |\psi \rangle [/itex]

    Here's where I'm stuck, but I think maybe I should repeat the above process with the respective operators to get something like [itex]\langle\mathbf{A}_{\alpha}\rangle \langle\mathbf{B}_{\beta}\rangle \langle\mathbf{A}_{\alpha}\rangle[/itex]

    However, I'm unsure because I'm not very familiar with QM and I'm trying to prepare for the class before it begins this fall. Thanks for your help everyone.
     
  2. jcsd
  3. May 18, 2014 #2

    strangerep

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    It's a lot simpler than that -- I suspect you're over-thinking the problem.

    I'm not sure how much I should give away in a first response, so I'll just offer an initial hint:

    What is the probability that a state prepared in state ##\psi_i## will be detected as state ##\phi_j## ?
    In other words, what is ##P(\phi_j|\psi_i)## (which is read as "probability of ##\phi_j##, given ##\psi_i##) ?

    BTW, which textbook(s) are you working from? If you have Ballentine, then you might be able to deduce the answer to my question from his eq(2.28).
     
    Last edited: May 18, 2014
  4. May 19, 2014 #3
    Well the text I have is Modern Quantum Mechanics 2nd Edition by J.J. Sakurai; however, I'm just picking random problems from different sources to get a feel for QM. This particular problem is from here: http://farside.ph.utexas.edu/teaching/qmech/Quantum/node44.html

    In regards to the problem:[itex] P(\phi_j|\psi_i) = |\langle \phi_j|\psi_i\rangle|^2 [/itex] correct? So then the probability of measuring [itex] b_1[/itex] when the system is in the state [itex] \psi_1 [/itex] is [itex] |\langle \phi_1|\psi_1\rangle|^2=\frac{4}{13} [/itex]
     
  5. May 19, 2014 #4

    CAF123

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    Correct, so having measured ##b_1##, the system is now in the state ##\phi_1##. Then a third measurement is made, this time for A. What is the probability that, for your system in the state ##\phi_1##, that you measure ##a_1##?
     
  6. May 19, 2014 #5
    Ah okay that would be [itex] |\langle \psi_1|\phi_1\rangle|^2 =\frac{4}{13}[/itex]
     
  7. May 19, 2014 #6

    CAF123

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    Correct, can you finish the problem?
     
  8. May 19, 2014 #7
    I think so because I think now I need to multiply the two probabilities together to get [itex] \frac{16}{169} [/itex] Then I need to repeat the process for the state [itex] \phi_2 [/itex]:

    [itex] |\langle \phi_2|\psi_1\rangle|^2 = \frac{9}{13} [/itex]

    Measuring [itex] \mathbf{A} [/itex] again:

    [itex] |\langle \psi_1|\phi_2\rangle|^2 = \frac{9}{13} [/itex]

    Multiplying together: [itex] \frac{81}{169}[/itex]

    Now adding the two together yields [itex] \frac{16+81}{169} = \frac{97}{169} [/itex]
     
  9. May 19, 2014 #8

    CAF123

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    Yes, looks fine.
     
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