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My professor tells me that the following wave function can not be normalized, therefore it does not represent a particle.

$\psi(x) = Ae^{ikx}$

However, he goes on to say that the wave function can be thought of as being a beam of particles by using fourier series, however I don't understand how this is even possible and wondered if anyone could provide perhaps some proof of sorts?

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  • $\begingroup$ Perhaps what your professor means is that the wave function, which represents the probability amplitude to observe a particle, needs to be normalized. Otherwise, it cannot give the correct probability interpretation. One can represent such a wave function as a spectrum of plane waves. $\endgroup$ – flippiefanus Dec 11 '20 at 9:40
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There's a fairly good discussion of the free particle case here. I'm assuming you have shown that $\psi$ is not normalizable.

Why does $\psi$ not being normalizable mean that it cannot represent a particle? Well, this represents a plane wave with constant amplitude everywhere (only the phase changes). As the amplitude of the wave function tells you about the probability to find a particle at a given location, you can think of this as implying that the particle has the same probability to be anywhere in the universe, which does not really correspond to our idea of a particle. In purely mathematical terms, we usually restrict our attention to solutions of the Schrödinger equation which are square-integrable, which this solution is not. This means that this solution is not part of our Hilbert space. Why does this happen? Note that $\psi$ has a single $k$-component, which corresponds to the momentum. Thus, it has an infinitely sharp momentum (it's momentum is exactly $k$), and therefore an infinitely smeared out position by the Heisenberg uncertainty principle.

However, that does not mean that this solution is not important! The Schrödinger equation is a linear differential equation, so the principle of superposition holds. This means that if $\psi_1$ satisfies the Schrödinger equation and $\psi_2$ satisfies the Schrödinger equation, then $\psi_1 + \psi_2$ will also satisfy the Schrödinger equation!

Why is this important? Well, we can build normalizable solutions by taking linear combinations of free particles, the so called wave-packets. Let us take some function $f(k)$ which describes how our amplitudes fluctuates in $k$. This corresponds to taking multiple components of momentum, e.g. introducing some "spread" in momentum value. Then form a linear superposition as: $$ \psi_3(x) = A\int_{-\infty}^{\infty} dk\ f(k) e^{ikx} $$ This is called a wavepacket solution. Note that this is a solution of the Schrödinger equation, by linearity (think of the integral as an infinite sum). However, for appropriate choices of $f(k)$, $\psi_3$ can be made normalizable! Essentially, you are summing multiple particles of different momentum $k$, which is what your professor means by a "beam" of particles. If you want to think about it in physical terms, then we're making the wavefunction less localized in momentum, and thereby achieve more localization in position.

So how is this related to Fourier series? Well, note that $\psi_3(x)$ above is (up to constant factors) nothing but the Fourier tranform of the function $f(k)$. This is useful once you solve the time-dependent Schrödinger equation, as the free particle solution has the simplest possible time-evolution, and any reasonable initial state of the system can be Fourier transformed. Thus, one way to get time-evolution in quantum mechanics is to Fourier transform the initial wavefunction and add the time-evolution. This is equivalent to the seperation of variables method for solving PDEs.

There are other options for fixing the free-particle solution of course. One is to require that the system lives in some (arbitrarily large) finite box. Then the solution is normalizable, and lives in a Hilbert space. This is an interesting solution, because it clearly leads to problems with relativity if the box is large enough. This, and other related problems, were what lead to the development of relativistic quantum mechanics, and ultimately quantum field theory.

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  • $\begingroup$ Apologies for the late reply.I have marked the question as answered however I would like to clear a misunderstanding.Correct me if I am wrong, if I let $f(k) = Ae^{ikx}$ would $\psi_{3}(x)$ become a sum of dirac delta functions. If so, how does this represent a beam of particles? $\endgroup$ – Hello123 Dec 12 '20 at 8:39
  • $\begingroup$ Actually, choosing $f(k)$ this way, would still lead you to the same problem, as you get the integral of $e^{i2kx}$, which is no more normalizable than $e^{ikx}$. In general though, you cannot choose any $f(k)$, the requirement is that $f(k)$ makes $\psi_3$ normalizable, so that it lives in the Hilbert space. The "beam of particles" may be a confusing way to look at it, because we're working with a single particle Hamiltonian. But you can think of $f(k)$ as representing a sum of waves which interfere. If you think of the particle-wave duality, this can be though of as a sum of particles. $\endgroup$ – G.Lang Dec 12 '20 at 11:01
  • $\begingroup$ If you can't decompose $\psi (x) = Ae^{ikx}$ into a sum by taking the Fourier transform. How can it possibly be considered a sum of waves, therefore particles? $\endgroup$ – Hello123 Dec 12 '20 at 14:19

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