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Normalisation of free particle wavefunction

  1. Nov 27, 2015 #1
    The wavefunction ##\Psi(x,t)## for a free particle is given by

    ##\Psi(x,t) = A e^{i(kx-\frac{\hbar k}{2m}t)}##

    This wavefunction is non-normalisable. Does this mean that free particles do not exist in nature?
     
  2. jcsd
  3. Nov 27, 2015 #2

    DrClaude

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    No, it means that a particle can't be found in that state. A free particle will be found in a superposition of such states, known as a wave packet, which is more constrained in space.

    That said, plane waves are very useful solutions to the Schrödinger equation, because they are stationary states. They are much used in scattering theory.
     
  4. Nov 27, 2015 #3
    Why use something as unphysical as plane waves, then, to derive predictions in scattering theory?

    If the predictions do agree with experiment (and they certainly do), then it's definitely got to be a hoax/sleight of fate, right?
     
  5. Nov 27, 2015 #4

    DrClaude

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    While a single plane wave is not a proper wave function, the plane waves still form a useful basis for calculations.

    For instance, you can use them to calculate the differential cross section for scattering from a given potential, and get it as a function of the wave vector k. But that doesn't mean that a given particle can be seen as having a definite value of k, the same way that you can't say that a particle has a definite position x.
     
  6. Nov 27, 2015 #5

    jtbell

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    No, it means that free particles with perfectly precise momentum (and energy) do not exist. However, free particles with imperfectly precise momentum (i.e. with some nonzero uncertainty in momentum) do exist, as wave packets: superpositions of precise-momentum states. Remember the Heisenberg uncertainty principle? :oldwink:
     
  7. Nov 27, 2015 #6

    bhobba

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    The answer to your question involves an advanced area of functional analysis called Rigged Hilbert Spaces:
    http://arxiv.org/abs/quant-ph/0502053

    To start with you need to come to grips with Distribution Theory:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    Distribution theory IMHO should be in the armoury of any applied mathematician, not just physicist. It makes, for example Fourier Transforms a snap - you don't become bogged down in tricky issues of convergence.

    Basically such states are physically unrealisable but introduced for mathematical convenience. Its like in differential equations you model a hammer strike by a Dirac Delta function. Such do not exist but its mathematically convenient to model it that way. Its one reason Distribution theory is so useful.

    Thanks
    Bill
     
  8. Nov 28, 2015 #7

    vanhees71

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    This is a very good question, and I think the answer is simply that many textbookwriters are just too lazy to explain it right ;-) (I'm kidding a bit, of course).

    As already emphasized, the plane-wave solutions of the Schrödinger equation of a free particle do not represent states. The reason is simply that they are generalized momentum eigenstates. With "generalized" I mean that they do not live in the Hilbert space of states (here represented by square-integrable functions) but in the dual space of a smaller dense subspace of the Hilbert space, where the momentum operator ##-\mathrm{i} \hbar \vec{\nabla}## are defined. Thus these eigenstates are in fact distributions with the domain of testfunctions defined in the smaller subspace of the Hilbert space. The modern way to express this "handwaving physicists' talk" is the formalism known as "rigged Hilbert-space formalism". A good introduction can be found in Ballentines textbook.

    Now to the scattering theory. Let's discuss the most simple case of scattering of a single particle on a potential. This is an approximation for the case that a light particle is scattering on a very heavy particle, which is effectively described by a potential of the force between the heavy and the light particle. The motion of the heavy particle is neglected. Further we assume that this interaction potential is "short ranged", i.e., it falls off faster than with ##1/r##, where ##r## is the distance of the light particle from the heavy particle (note that this excludes Coulomb scattering, which is a special case, because of the long-range nature of the Coulomb interaction but also because it can be solved exactly in this approximation).

    Then the physical picture is as follows: You start with the light particle far away from the heavy particle, so that it can be regarded as a free particle ("asymptotic free initial state") and ask about the probability to find it after a long time after the interaction with the heavy particle, when it can again regarded as free, because it went again far away from the heavy particle ("asymptotic free out state").

    Formally you calculate the transition probability from an asymptotic free in state into an asymptotic free out state. This is described in the interaction picture by the unitary time-evolution operator. Mapping directly the asymptotic free in-state via the unitary transformation to the asymptotic free out-state, which needs some careful "adiabatic switching" procedure to begin with (Gell-Mann-Low theorem), you define the socalled S-matrix elements. They are the matrix elements of the unitary operator mapping the in state directly to the out state.

    Usually textbooks use right away the momentum eigenstates as the asymptotic in and out states, but that leads to a desaster, because they are distributions and not square integrable functions. Also in the here discussed situation energy-momentum conservation holds for the free in and out states, i.e., with plane waves as basis the S-matrix elements contain necessarily the corresponding Dirac ##\delta## functions, i.e.,
    $$S_{fi}=(2 \pi)^4 \delta^{(4)}(p_f-p_i) [1+\mathrm{i} T_{fi}].$$
    This is the probability amplitude! To get a probability you have to square the amplitude, and this you are not allowed to do, because you cannot square a Dirac ##\delta## distribution.

    There are two ways out of this dilemma, because for all practical puposes, you want to describe the particle as starting at a well-defined momentum and you'd like to know, the probability to find the particle with a given other momentum after the scattering again. The first way is physical and a bit more difficult than the second way to work out mathematically, but it's the intuitive way. So let's start to describe it in the physical way.

    The very fact that the momentum operator of a particle in free space has no square-integrable eigenvalues shows that you cannot prepare a particle with a sharply defined momentum value. This is, why the plane waves are not representing states, because the corresponding "wave functions" are not square integrable. So necessarily there's always some uncertainty in the momentum. You can determine it as good as you can with all kinds of modern accelerator technology, but you never can make it exact in the mathematical sense. So the true asymptotic free initial state has a finite spread of the momentum, which can be described, e.g., by a Gaussian wave function with a quite small width in momentum space. As you know from Heisenberg's uncertainty principle this implies that its position is rather indetermined, i.e., the wave function in position space (which is nothing else than the Fourier transform of the wave function in momentum space) is pretty broad. So you must find some compromise: On the one hand you want a rather well-defined momentum to define your scattering cross sections in a useful manner but on the other hand you want the wave packet in position space to be not so broad that it has already considerable overlap with the position of the target particle. In practice this is no problem because what's broad on a microscopic scale is not so broad on the macroscopic scale, i.e., you must only make sure that your projectile is initially far away from the target in the sense that it is far compared to the range of the potential. Now, if you calculate the S-matrix element with this wave packet as an initial asymptotic free state, you don't get a ##\delta## distribution anymore but something finite, and you can square it and calculate transition probability rates. After that you can take the limit of narrower and narrower initial wave functions in momentum space, i.e., the limit towards a plane wave. It doesn't matter too much which particular shape of the wave packet you choose, and the Gaussian wave packets are often convenient to take this limit (in fact a Gaussian in the limit of very small standard deviation is often a convenient way to regularize ##\delta## distributions). In almost all scattering experiment the particular shape of the wave packets is indeed irrelevant, and one can use the cross sections you get in this "plane-wave limit", but the limit has to be taken in the right order: First you calculate the S-matrix element with a true wave packet, then square it to calculate the transition probabilities, and then take the limit to very narrow and finally plane-wave "states".

    This indicates that there is a shortcut to make life easier to begin with. This is the second way to "tame" the ##\delta## distributions when taking the square of the S-matrix element. It's known as Fermi's trick. The idea is to look at the particle in a finite space-time region. You start with a "quantization volume", say a cube of length ##L##. The particular shape doesn't play an important role, and a cube is the most convenient. Since you want a well-defined momentum operator, you have to impose appriate boundary conditions to get well-defined wave functions. The momentum operator is the generator of spatial translations, and thus you impose periodic boundary conditions, i.e., you extend the wave functions beyond the cube by assuming that it's periodic in all three directions of the corresponding Cartesian coordinates with period ##L## (the length of the cube). Then you get a discrete set of allowed momentum eigenstates, ##\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^3##, and the corresponding plane waves are square integrable, because you integrate over the finite volume of the cube only. Now you calculate your S-matrix elements in this space, leading to finite results, because you work with true square integrable wave functions as asymptotic free in and out states. So you can square your S-matrix to get the transition probabilities and then take the limit ##L \rightarrow \infty##. The final result of this much simpler shortcut is the same as with the above described plane-wave limit of the wave-packet approach. Mathematically, it's just another equivalent way to regularize the energy-momentum conserving ##\delta## distributions in the S-matrix elements with plane waves.

    The upshot of both approaches finally is that you can calculate the S-matrix element with plane in and out states and then taking the square in the following way
    $$|S_{fi}|^2 = |T_{fi}|^2 (2 \pi)^4 \delta^{(4)}(p_f-p_i).$$
    That's why I said that many textbook authors are simply to lazy to carefully discuss, how this final recipe is mathematically derived, but I think, it's important to carefully think it over at least once! Of course, here we have assumed that the in and the out state are not chosen as the same state (indeed there is some probability that no scattering happens, and that's part of the famous "optical theorem").
     
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