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Quantum-mechanical three-state system

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose your quantum mechanics professor can be modeled as a three-state quantum system:
    [tex] |{\rm{professor}}\rangle = \tfrac{7}{10} |{\rm{Bohr}}\rangle + \tfrac{7}{10} |{\rm{Dirac}}\rangle + \tfrac{\sqrt{2}}{10} |{\rm{student}}\rangle [/tex]
    The three states have the usual properties: [itex] \langle\textrm{Bohr}|\textrm{Dirac}\rangle = \langle\textrm{Bohr}|\textrm{student}\rangle = \langle\textrm{Dirac}|\textrm{student}\rangle = 0 [/itex] and
    [itex] \langle\textrm{Bohr}|\textrm{Bohr}\rangle = \langle\textrm{Dirac}|\textrm{Dirac}\rangle = \langle\textrm{student}|\textrm{student}\rangle = 1 [/itex].

    (a) If I add up the coefficients, I get: [itex]\tfrac{7}{10}+\tfrac{7}{10}+\tfrac{\sqrt{2}}{10}=1.54.[/itex] A classmate says that quantum mechanical probabilities must add up to 1 so [itex]|\textrm{professor}\rangle[/itex] is not a quantum state. What's the flaw in that argument?

    (b) If we take a "Bohr-meter" and measure the professor's "Bohr-ness" on any particular day, what do we find? What's the amplitude for the measurement result?

    (c) What's the probability of finding [itex]|\textrm{professor}\rangle[/itex] "Bohr"?

    (d) Finally, if we measure the professor to be "Bohr", in what state do we leave him? After we measure him to be "Bohr", what happens if we immediately ask again whether he's "Bohr"?

    2. Relevant equations

    3. The attempt at a solution

    Here's what I think the answers are; but I'm not currently in a setting where I can ask anybody if I have them right/complete. I'd just like it if somebody could look over my work and/or provide additional insight I may be missing here.

    (a) There's no rule about quantum mechanical amplitudes having to add up to anything — in fact, in general amplitudes can be imaginary, in which case there's no way they'd add up to 1. Quantum mechanical probabilities, equal to the square of the amplitudes, must sum to 1, and indeed they do here:
    [tex] \left( \tfrac{7}{10} \right)^2 + \left( \tfrac{7}{10} \right)^2 + \left( \tfrac{\sqrt{2}}{10} \right)^2 = \tfrac{49}{100} + \tfrac{49}{100} + \tfrac{2}{100} = 1. [/tex]

    (b) and (c) The amplitude to measure the professor to be "Bohr" is [itex]\frac{7}{10}[/itex]. However, after a measurement with the "Bohr-meter", we will only find him either to be 100% Bohr or 0% Bohr: [itex]\frac{49}{100}[/itex] of the time we will measure him to be "Bohr", and [itex]\frac{51}{100}[/itex] of the time, we will not.

    (d) Immediately after measuring the professor to be "Bohr", he is no longer in his original state. Instead he is in the state
    [tex] |{\rm{professor}}\rangle = 1|{\rm{Bohr}}\rangle + 0|{\rm{Dirac}}\rangle + 0|{\rm{student}}\rangle .[/tex]
    So if we measure him again immediately, we will find him to be "Bohr". However, over time his state will evolve such that if we measure him again after some time has passed, it is no longer certain whether he will be "Bohr".
     
  2. jcsd
  3. Feb 1, 2012 #2
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