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How can I ignore information in my wave function?

  1. Jan 24, 2012 #1
    Hello,

    Say I have a system with a spatial part and a spin degree of freedom, hence the wavefunction generally looks like [itex]\psi_+(\textbf r) |+\rangle + \psi_-(\textbf r) |- \rangle[/itex] w.r.t. for example the z-axis.

    Now what if I'm simply interested in the spatial part? Can I perform an operation on this wavefunction such that I get a new wavefunction that is only spatial, but gives the same predictions for any observable that only depends on space variables? (e.g. [itex]L_z[/itex])

    I feel this must be possible, and that it shouldn't even be hard, but I can't seem to think of the appriopriate operation right away and I also can't google it because I don't know what such an operation would be called.
     
  2. jcsd
  3. Jan 24, 2012 #2
    I'm probably way off here, but oh well...

    Doesn't this depend on if you are considering the Schrodinger picture or the Heisenberg picture?

    If it is the former then H is time-independent, in the latter it is time-dependent. The time- dependence or independence of H will transfer to the time- dependence or independence of the observables.

    It sounds like you are talking about the Schrodinger picture.

    Also, you might want to look up the "propagator" which is probably the operator you are after.
     
    Last edited: Jan 24, 2012
  4. Jan 24, 2012 #3
    Hm, I'm indeed talking about the Schrödinger picture, although I'm not sure how it would change matters, but then again I'm not too familiar with the Heisenberg picture, so yes for my answer one can restrict himself to the Schrödinger picture.
     
  5. Jan 24, 2012 #4
    Well those things aside, would just dividing by the time part of your wavefunction give your new desired wavefunction? Or do you want a single operator which does it for all wavefunctions?
     
  6. Jan 24, 2012 #5
    The time part of my wavefunction? I'm afraid you have misunderstood my question.

    I'm looking for a (mathematical) way to erase the spin information in [itex]\psi_+(\textbf r) |+\rangle + \psi_-(\textbf r) |- \rangle[/itex] to get a purely spatial wavefunction. This is not time-related. It would be the wave function I would see if I had never heard of spin.

    More practical, it would be useful if I were asked the mean position of the particle, or if I were asked the possible outcomes of a L_z measurement, I would just have to decompose this new spatial wavefunction into eigenfunctions of L_z and etc via the usual method.
     
  7. Jan 24, 2012 #6
    Ok I probably can't help you. I thought spin is basically the same as angular momentum but just the ones with l = half-integer values??
     
  8. Jan 24, 2012 #7

    Bill_K

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    Your wavefunction has two "spatial parts": ψ+(r) if the spin is up and ψ-(r) if the spin is down. If you want to ignore the spin you can't get a wavefunction, the best you can do is a probability, |ψ|2 = |ψ+|2 + |ψ-|2.
     
  9. Jan 24, 2012 #8

    Rap

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    I would think the answer would be Aψ+(r) +B ψ-(r) where A and B are constants which keep normalization. I think this wavefunction will predict the same results as the original wave function, as long as no spin-detecting measurements were made.
     
  10. Jan 25, 2012 #9

    kith

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    If you want to describe measurements which don't affect the spin, I think you can use a reduced density matrix for the spatial part.

    In basis-independent form, your state is |ψ+>⊗|+> + |ψ->⊗|->. The general approach would be to construct the full density matrix from that and trace over the spin. Then you get the reduced density matrix for the spatial part ρspatial, which can be used to calculate every relevant quantity for measurements which affect only the spatial part. These are of the type A⊗1. Expectation values can then be calculated by <A>=tr{Aρspatial}.

    In your initial post, your state is given in the position basis. It is not immediately clear to me, how all of this looks in this "half" basis-dependent form, but this is probably just a technical thing.

    If you want to do a simple position measurement, I agree with Bill. I think you have to add probabilities here, not probability amplitudes.
     
    Last edited: Jan 25, 2012
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