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Dirac delta wave function impossible?

  1. Jan 10, 2012 #1
    Hello,

    I was under the impression that a dirac delta was a "legitimate" state for a particle: maybe not mathematically, but least physically. But I was recently told by a post-doc in QM that if your particle is in a dirac delta state at one moment, the very next moment the particle is everywhere. I'm not sure how I can see this?

    PS: okay there is another problem, since you need to square it, and what is the square of a dirac delta? But this last problem doesn't seem very terrible, although I don't know what the best way is to circumvent it.

    Or is the correct answer: using the dirac delta for a state is completely hopeless? (The students certainly aren't given this impression.)
     
  2. jcsd
  3. Jan 10, 2012 #2
    Position eigenstates aren't eigenstates of the Hamiltonian, as you've always got a kinetic energy operator in there (which looks like the laplacian in a position basis). The Hamiltonian generates time evolution; if you're unfamiliar with this concept, it means that to shift a state you've defined at time t to [itex] t+\delta t [/itex], you act on it with [itex] 1-i\hat{H}\delta t [/itex]. The result of acting on a state that's not an eigenstate of H with H will in general be a superposition of all possible position states, and so will have a non-trivial wavefunction.

    As for the legitimacy or otherwise of delta-function descriptions: I think I'm right in saying that the dodge of using delta-function norms has a rigorous justification by appealing to the theory of rigged Hilbert spaces, in such a way that allows you to write physical states in terms of "position eigenstates" without actually admitting such states as physically real themselves. I'm not sure about this though- lots of people on here like the book on QM by Leslie Ballentine, which I believe discusses these things properly.
     
  4. Jan 10, 2012 #3
    I'm familiar with what you say in your first paragraph, but I don't seem to see how it tries to answer my question?

    Oh maybe you understood my "being everywhere" as "having a non-zero value everywhere". I understand that that will happen, but that's not what I meant. With being everywhere, I mean a constant. Ah perhaps the postdoc simply meant "a non-zero value everywhere"? That would make sense, but not in the context: I had just told him that I believed that if you start in a dirac delta-ish state, the state then evolves into a gaussian that keeps getting wider, upon which he gave me a frown and replied with what I quoted.
     
  5. Jan 10, 2012 #4
    Consider a Gaussian wave function of some width. The narrower the wave function, the larger the momentum uncertainty, and the faster the wave packet spreads out. In the limit that the Gaussian goes to a delta function (i.e., in the limit that the width goes to zero) the momentum uncertainty goes to infinity and the rate of spreading goes to infinity. So if you have a particle localized to an extremely small volume, extremely quickly it will spread out to a very large volume. If the wave function actually was a delta function, then the width would be zero and the particle would spread out into an arbitrarily large volume instantly.

    You can verify this using the explicit time dependent solution for a Gaussian wave packet here: http://en.wikipedia.org/wiki/Schrödinger_equation#Gaussian_wavepacket

    Delta function states are unphysical in exactly the same wave that infinite plane waves are unphysical (infinite plane waves are after all just delta functions in momentum space). Position delta functions might be a bit worse; they have infinite energy for instance.
     
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