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Momentum space and position space

  1. Aug 4, 2004 #1
    I am just starting reading some quantum stuff. There of course be many questions here and there.

    One thing comes to bother me now that it seems QM are treating this "state" in either momentum or position representation, if leave alone spin space. It seems to treat "momentum" and "position" as two basises in that both can span the complete Hilbert space and a unitary tranformation is used to translate the two basis.

    I vision now the Hilbert space can be constructed with a intertwining "momentum" and "position" not exactly as a direct sum or product; they might not be orthogonal to each others but either one will not be able to span the whole Hilbert space.

    Did I misunderstand QM at all?
     
  2. jcsd
  3. Aug 4, 2004 #2

    turin

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    One momentum component requires the specification of every single position component, and vice versa. For instance, a momentum of:

    p = hk

    corresponds to every position with the associated complex-valued weight:

    eikx.

    Therefore, while not just some momenta and positions can span the space, it is sufficient to specify only the momentum distribution or only the position distribution in terms of what is called the wave function. You cannot use only half momentum and half position to span your space because those halves can never truly be complementary in that sense.
     
  4. Aug 8, 2004 #3

    vanesch

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    This is correct. The momentum basis alone spans the whole state space, and so does the (other) position basis. They are indeed related by a unitary transformation (which is a Fourier transformation).

    I guess what troubles you is the fact that in classical mechanics, the phase space (p,q) is the state space, while in quantum theory, only one of both (q or p or a mixture) seems to be necessary.
    Well, that's the whole issue !
    If you would construct (you are allowed so) a quantum theory where p and q are commuting observables, so that you have a basis labeled in (p,q), you would find out that ALL states are stationary states !
    The simplest way to see this is in the Heisenberg picture:
    i hbar d O / dt = [O,H]. Because any sensible observable, as well as the hamiltonian, are made up of p and q, and all p and q commute, the commutator [O,H] = 0. So all d/dt = 0.
    So you are allowed to build such a quantum theory, but first of all it is a boring one, and second, most important: it doesn't go into the classical theory when h ->0.
    Dirac worked out that when we do things the way they have to be done (using q as a basis, or p as a basis) and having the commutation relations [q,p] = i hbar, we DO find back the classical mechanics theory when h->0 (the correspondence principle).

    cheers,
    Patrick.
     
  5. Aug 8, 2004 #4

    Haelfix

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    Another way of saying it, is that the quantum phase space is the restriction of the classical poisson bracket algebra to a commutator algebra on some hilbert space.
     
  6. Aug 9, 2004 #5
    The current definition of a "state" is a vector ( In another words, an element ) in the Hilbert space spanned by either position or momentum or energy.

    Now, if a "state" is not a vector but a "subspace" in a space spanned by momentum and position. Will that lead us to any places?

    This thought came when I found that postion or momentum eigenfunctions are actually a kind of dirac-delta function which actually carrys always an uncertainty to the so-called "definite" eigenvalue.
     
  7. Aug 9, 2004 #6
    More precisely, only the direction of the vector gives the physical state. One is forced to normalize the vector to unity, in order to respect the total probability=1 :
    if
    [tex]
    |\psi>=\sum_nc_n|n>
    [/tex]

    then
    [tex]
    1=<\psi|\psi>=\sum_n|c_n|^2
    [/tex]

    Either you consider that the vector must belong to the unit sphere, or you consider the projective space (the space of "directions")
     
    Last edited: Aug 9, 2004
  8. Aug 9, 2004 #7

    vanesch

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    Take a momentum p0 and a position x0. Then the subspace spanned by the two states |x0> (one specific vector of the position basis) and |p0> (one specific vector of the momentum basis) has the form:

    |psi> = a |x0> + b|p0>

    Well, that's a subspace of states like any other. Each element of it is not an eigenstate of the position operator X nor of the momentum operator P, except the cases a=0 or b=0. As a wave function, it is:

    a delta(x-x0) + b exp(i p0 x)

    You have the dual expression as the momentum wave function...
    You could see it as a rudimentary effort to construct a kind of wave packet :redface:


    So ? I don't see where this can lead us :confused:

    Eh? A dirac-delta function has no uncertainty ! (its width is 0).

    Patrick.
     
  9. Aug 10, 2004 #8
    Patrick,

    Actually this might not lead to any place. It could eventually just be a self-examination of QM.

    But, when we say an eigen solution as [tex] e^i(k_0 x+w_0 t) [/tex] , we really maen:

    [tex] \int \int \delta ( k - k_0 , w - w_0 ) e^{i(kx+wt)} dk dw [/tex].

    They are not completely the same.

    The simplified one in plain expression is actually a standing wave without any space lump.

    the integration function shows a space lump because the [tex] \delta ( k , w ) [/tex] has always a width no matter how small you squeeze it to.

    If I imagine a "state" as a compressible object with fixed volume in the momentum-position space, the I will see that momentum side will expand when you squeeze the position side. In other words, if I squeeze the "postion" to pinpoint the position, I will expand the "momentum" side so the "momentum" side has a wider dispersion. On the other hand, I squeeze the "momentum" then I will get a wider "position" dispersion.

    Also, the integration formula actually suggest a volume integration on the space of dk dw.
     
    Last edited: Aug 10, 2004
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