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Help with simple question

  1. Apr 6, 2012 #1
    From the postulates of QM we know that any state |ψ> is the same as exp(iθ)|ψ>.

    Ok so take the spin state of an electron in the x direction and expand in the z basis to get

    |+x> = 1/sqrt2(|+> + |->)

    = 1/sqrt2(|+> + exp(iθ) |->) using the above,

    choosing theta =∏
    |+x> =1/sqrt2(|+> +(-1)|->)
    =1/sqrt2(|+> - |->)
    =|-x>
    which can't be right?
     
  2. jcsd
  3. Apr 6, 2012 #2
    It's not that [itex]|x\rangle[/itex] is actually equal to [itex]e^{i\theta}|x\rangle[/itex], its just that their magnitudes are the same. That is, if [itex]|y\rangle = e^{i\theta}|x\rangle[/itex], then [itex]\langle x|x\rangle = \langle y|y\rangle[/itex], because [itex]\langle y|y\rangle = \langle x e^{-i\theta}|e^{i\theta}x\rangle = \langle x| e^{i(\theta -\theta)}x\rangle = \langle x|x\rangle[/itex]. The reason people often say that the phase of a state doesn't matter is that all physically observable quantities are derived from the magnitude of the state only. So a complex rotation of a state does not have any physically observable consequences, but that doesn't mean that it's actually the same state.

    So in your example, you are correct that [itex]|\uparrow\rangle + |\downarrow\rangle \not= |\uparrow\rangle - |\downarrow\rangle[/itex], because you only rotated one of the component states. The magnitude will only be the same if you rotate all of the component states equally, that is [itex]e^{i\theta}(|\uparrow\rangle + |\downarrow\rangle) = -(|\uparrow\rangle + |\downarrow\rangle)[/itex].
     
  4. Apr 6, 2012 #3
    But interference effects do depend on relative phases. These phases can be introduced in experiments like in the standard measurement ansatz below

    Exp(iHt) |ψ_1>|a0> = exp(iθ_1)|ψ_1>|a_1>

    Exp(iHt) |ψ_2>|a0> = exp(iθ_2)|ψ_2>|a_2>

    Where the phases exp(iθ_i) are introduced (randomly?)by the experiment and the |a_i> are apparatus states that become correlated with the system states |ψ_i>. Thus for a superposition we would get:

    Exp(iHt) |ψ_0>|a0> =Exp(iHt) [ |ψ_1>+|ψ_2>] |a0>
    = exp(iθ_1)|ψ_1>|a_1> + exp(iθ_2)|ψ_2>|a_2>

    This would have a density matrix with off diagonal terms which would indicate the ability of the state to show interference effects which do have measureable consequences. These interference effects show up when beams of electrons are combined in mach zender type experiments using Stern Gerlach devices. see attached doc on p2-18.
    Sorry but I can't get the file to attach - keep getting upload errors. I have enclosed a word document with the argument though.

    Many thanks to anyone who can answer this query.
     

    Attached Files:

    Last edited: Apr 6, 2012
  5. Apr 7, 2012 #4

    mfb

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    And relative phases are physical, which means that you cannot choose them freely.
    The only freedom you have is a rotation of your full state (QFT has some more freedom, but that is another story).

    They are given by the setup of the experiment.
     
  6. Apr 7, 2012 #5
    MFB,

    Can you explain how the setup of the experiment determines the phases?
     
  7. Apr 7, 2012 #6

    mfb

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    Well, for example look at a double-slit and a detector somewhere at a screen behind it.

    You emit monochromatic photons from some source far away which can go through the double-slit. As the distances from the source to both slits are identical, the wave function arrives with the same phase at both. If a photon passes the slit, you can write it as superposition of two states, one for each slit. The initial phase for some t=0 is arbitrary, but you have to choose the same for both states. Now you can let these states evolve and look at the path to your detector: The phase there depends on the distance between detector and slit (it is the initial phase, shifted by -2pi*d/lambda with the distance d). The individual phases depend on your choice for the phase at the slits. The phase difference is fixed and given by the distance difference. Your amplitude just depends on this phase difference (and things like the slit width, which I ignored here).

    That is not the usual way to look at double-slits, as just taking the length difference without fixing any light phases is much easier. But it is possible to do this.
     
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