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Question about wavefunction prob.

  1. Nov 20, 2004 #1

    Galileo

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    I don't like to delve to deep into this matter in such a way that this thread will be thrown into the philosophy forum, where I don't think it belongs.

    Take a particle and consider the state space. And let's call, say, the first tree stationary states [itex]|\psi_1 \rangle, |\psi_2 \rangle, |\psi_3 \rangle[/itex].

    The question is:
    Is there ANY (physical) difference between saying:
    "There is a 1/3 probabilty the particle is in the state [itex]|\psi_1 \rangle[/itex], a 1/6 probability it's in the state [itex]|\psi_2 \rangle[/itex] and a prob. of 1/2 that it's in the state [itex]|\psi_3 \rangle[/itex]."

    and saying

    "The particle is in the state
    [tex]\frac{1}{\sqrt{3}}|\psi_1 \rangle+\frac{1}{\sqrt{6}}|\psi_2 \rangle+\frac{1}{\sqrt{2}}|\psi_3 \rangle[/tex]"

    I'm pretty sure the answer is no, since any physical measurement will give the same predictions in both cases. Unless I missed something.
    Can someone answer this question?
    (Preferrably someone ho knows what he's talking about).
     
  2. jcsd
  3. Nov 20, 2004 #2

    jcsd

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    Surely though this is where Bell's theorum and the Aspect exepriment comes into play? Most of the time it seems to me hat there isn't much diffrence, but that's because we're not looking for the difference, howvere when we do start to see if the two situations are different we find that they are.
     
  4. Nov 20, 2004 #3
    Yes there is. The first state is called a mixed state. Essentially it says that the particle actually is in one of the states, just we don't know which one. The second state is a superposition of the states, which is intrinsically different from the three original states.

    The two possibilities can be distinguished experimentally. The superposition can give rise to interference effects (e.g. double slit experiment) while the mixed state doesn't.
     
  5. Nov 20, 2004 #4

    Galileo

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    Can you elaborate on that? I don't see how a single particle in itself can give rise to interference effects.

    Anyway, I realized that although they give the same predictions for energy measurements (as they are eigenstates of H), that's not necessarily the case for an arbitrary observable A. It was pretty dumb of me to miss that, I should've thought about it more before posting.

    If we want to measure A, we write:
    [tex]|\psi \rangle = \sum_n c_n |e_n \rangle [/tex]
    where [itex]\{ \itex |e_n \rangle \}[/itex] is the orthon. basis of e.v's of A.
    Then the probability of getting [itex]\lambda_n[/itex] (the eigenvalue of A corresponding to [itex]|e_n \rangle)[/itex] is [itex]|c_n|^2[/itex].
    Let
    [tex]|\psi_1 \rangle = \sum_n a_n |e_n \rangle [/tex]
    [tex]|\psi_2 \rangle = \sum_n b_n |e_n \rangle [/tex]
    [tex]|\psi_3 \rangle = \sum_n c_n |e_n \rangle [/tex]

    If [itex]|\psi \rangle = |\psi_1 \rangle[/itex], then the prob. is [itex]|a_n|^2[/itex]
    So for the first case the prob measuring [itex]\lambda_n[/itex] is
    [tex]1/3|a_n|^2+1/6|b_n|^2+1/2|c_n|^2[/tex]

    for the second case, where [itex]|\psi \rangle = \frac{1}{\sqrt{3}}|\psi_1 \rangle+\frac{1}{\sqrt{6}}|\psi_2 \rangle+\frac{1}{\sqrt{2}}|\psi_3 \rangle[/itex]:
    [tex]|\psi \rangle = \frac{1}{\sqrt{3}}\sum_n a_n |e_n \rangle+\frac{1}{\sqrt{6}}\sum_n b_n |e_n \rangle+\frac{1}{\sqrt{2}}\sum_n c_n |e_n \rangle = \sum_n \left( \frac{a_n}{\sqrt{3}}+\frac{b_n}{\sqrt{6}}+\frac{c_n}{\sqrt{2}}\right)|e_n \rangle[/tex]
    So the prob. of measuring [itex]\lambda_n[/itex] is [itex]|\frac{a_n}{\sqrt{3}}+\frac{b_n}{\sqrt{6}}+\frac{c_n}{\sqrt{2}}|^2[/itex]

    Right? :wink:
     
  6. Nov 23, 2004 #5
    You can do the double slit experiment with a single particle, in which case you find that it won't be detected at the minima of the interference pattern, so you can say that a single particle experiences interference effects.

    That looks OK to me. In particular, if you do a measurement to find whether the particle is in state [itex]|\psi_1 \rangle[/itex], then you would get the same answer for the mixed state and the superposition. However if you could do a measurement to find out whether the particle was in the state [tex]\frac{1}{\sqrt{3}}|\psi_1 \rangle+\frac{1}{\sqrt{6}}|\psi_2 \rangle+\frac{1}{\sqrt{2}}|\psi_3 \rangle[/tex]
    then if it was in that state then you would get the answer yes with probability 1. But if it was in the mixed state, then there would be a possibility of getting a no answer
     
  7. Nov 23, 2004 #6

    Tom Mattson

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    Let's call the following Condition 1.

    Let's call the following Condition 2.

    I think there is a physical difference. Say you have prepared an ensemble of particles that satisfy Condition 1. Now Condition 1 does not specify the phase factor exp(iδi) multiplying each eigenstate, but Condition 2 does. Specifically, Condition 2 sets δi=0 for i=1,2,3.

    Now if you take a beam of particles that satisfy Condition 2 and another beam of particles with, say, δi=π/2 for i=1,2,3, they will interfere. There is your physical difference.
     
    Last edited: Nov 23, 2004
  8. Nov 23, 2004 #7

    Galileo

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    Thanks Tom. I already found there was a difference though...
     
  9. Nov 23, 2004 #8
    If the Hamiltonian is time independent, then the probabilities for getting certain energies remain the same. You can see this from conservation of energy, where the average value of the energy should be the same:

    <E>=Sum[Prob(i)Ei]

    and since the allowed Ei doesn't change (by assumption), then the Prob(i) don't change.

    Also you know that the energy basis diagonalizes the Hamiltonian, and so putting a different phase factor in each coefficient of your superposition doesn't matter as far as energy measurements are concerned.

    An obvious example is if you consider the position basis, and multiply each

    a_i|x_i> by exp(ipx/h)

    this doesn't change the probability for position measurements, but shifts the average momentum over to the right by p.
     
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