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Can a wave function be anything? such as a parabola. I know it is used to measure the probability of where a particle is (or so i think) in a 1 dimensional plane.

What is an example of a wave function that can be graphed?

Qmechanic
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    Does this answer your question? Allowed Wave Functions of System – BioPhysicist Mar 16 '21 at 22:33
  • @BioPhysicist the present question is definitely not a duplicate of the one you linked. In that case, there is an explicit reference to a Hamiltonian. A generic wavefunction has no reason to be an eigenstate of any Hamiltonian. This comment applies as well to the previous comment by Triatticus. – GiorgioP-DoomsdayClockIsAt-90 Mar 16 '21 at 22:52
  • @GiorgioP The question doesn't specify a specific Hamiltonian beyond it being single-particle, and the answer doesn't say wave functions have to be eigenstates of the Hamiltonian. – BioPhysicist Mar 16 '21 at 22:54
  • @BioPhysicist. That question made an explicit reference to boundary conditions given by $H$. Even if there is no specific Hamiltonian, the question was about eigenfunctions or a combination of eigenfunctions of a given Hamiltonian. However, it is perfectly possible to produce states which are not eigenstates of any Hamiltonian and the present question is clearly free from the constraints of the previous one. – GiorgioP-DoomsdayClockIsAt-90 Mar 16 '21 at 23:05
  • @GiorgioP The link doesn't say all states are eigenstates of a Hamiltonian. I think the link answers the question, so I voted accordingly. If no one else does that is fine. – BioPhysicist Mar 17 '21 at 03:47

2 Answers2

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Free wavefunctions (corresponding to a free particle) have to be continuous and smooth (the derivative has to exist everywhere, so, for example, no saw form function is allowed), and if you integrate the square of the function all over space this must result in the value 1: it's certain that you find the particle connected to the wavefunction somewhere in space, i.e., the probability is 1. We say that the wavefunction is normalized. An example of such a wavefunction (which can be plotted in a one-dimensional graph):

enter image description here

In general (when the particle is not free), the wavefunction must be continuous and square-integrable only. As is said in the comment below by @J.Murray.

Deschele Schilder
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    Wavefunctions need only be square-integrable, they don't need to be smooth. Even energy eigenstates don't need to be smooth; their second derivatives inherit whatever discontinuities may be present in the potential. – J. Murray Mar 16 '21 at 22:36
  • @J.Murray You are right. I was thinking about free wavefunctions only. – Deschele Schilder Mar 16 '21 at 22:43
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The wave function $\psi(x,t)$ for a specific time $t=t_0$ can be almost anything. $\psi(x,t_0)$ represents the initial condition of the quantum mechanical system. However, the squared magnitude $\psi^*\psi$ of the wave function also needs to be integrable (normalizable) over space, otherwise you cannot interpret it as a probability density (which must add up to one). That is why a parabola defined all over space cannot be a valid wave function, because a parabola is not normalizable. But, of course, the wave function could be a parabola on a finite domain if it drops off sufficiently fast outside that domain. In this case it would be normalizable again.

Due to the Schrödinger equation, the wave function at a different time $t=t_1$ is not arbitrary anymore, but given by the time evolution operator acting on the initial state: $$\psi(x,t)=\exp\left(-{\frac{i}{\hbar}H\cdot (t-t_0)}\right)\psi(x,t_0)$$ Don't be put off by the operator exponential, it is basically just a mathematical shorthand notation for "solving the Schrödinger equation".

You have to distinguish carefully between the above meaning of the wave function as the momentary state of the system and the eigen-states of the time-independent Schrödinger equation. The latter just serve as a very convenient basis for representing actual wave functions. The system need never be in any of the possible eigen-states, although it can be shown, that when considering radiation, the eigen-states are particularly stable, and hence, tend to be assumed asymptotically by the system.

Think of the Schrödinger equation as analogous to the wave equations that describe a drumhead. The drumhead can assume any shape (including initial velocities) you like at a given instant in time. But from there the further movement of the membrane is determined (at least until you hit the drum externally). The drumhead also has certain eigen-frequencies and corresponding eigen-shapes, which you may be able to excite selectively by carefully poking it with the drum stick at very specific locations. These are the analogs of the solutions to the time-independent Schrödinger equation. Any real excitation results in a superposition of the eigen-states combined with their respective time evolution (eigen-frequencies).

The crucial difference between the drum and a quantum mechanical system, however, is that for the drum you need to specify initial displacements and velocities, while the quantum mechanical wave function only needs to specify the initial state. This is a consequence of the Schrödinger equation being first order in the time-derivative, while the wave equation of the drumhead is second order in time.

oliver
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    A generic wavefunction is not twice-differentiable and does not solve the Schrodinger equation. – J. Murray Mar 16 '21 at 22:38
  • @J.Murray: I am not into the mathematical ramifications so much. But I think the OP might be happy with the specific restriction that the application of the Laplacian is working if the corresponding second derivatives exist and are finite. – oliver Mar 16 '21 at 22:46
  • @J.Murray: ... and nowhere did I say that it solves the Schrödinger equation. If you specify the wave function arbitrarily at initial time, you can develop a solution of the Schrödinger equation from it. – oliver Mar 16 '21 at 22:52
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    Okay, but the phase "Usually it is even differentiable twice for realistic potentials" makes it sound like an arbitrary wavefunction has anything whatsoever to do with the Hamiltonian or the potential, which it doesn't. I'm also not sure I understand your first comment about what the OP would be happy with, because your initial answer that a wave function must be twice differentiable was simply incorrect. It would be true if you were talking about solutions to the Schrodinger equation with a continuous potential, hence my original comment. – J. Murray Mar 16 '21 at 23:00
  • @J.Murray: fair enough. I just wanted to remind the OP that although the initial conditions are arbitrary for all practical purposes, they have to fulfill some mathematical prerequisites in order to even apply the Schrödinger equation to it (like differentiating twice in the operator of kinetic energy/Laplacian). Since I got this wrong, obviously, I will remove this sentence, because it does not add any benefit to the general idea that the initial conditions are arbitrary. – oliver Mar 17 '21 at 07:07
  • There is a more fundamental difference between a QM wavefunction and the membrane of a drum. I.e. the intrinsic constraint of continuity of the membrane displacement and its velocity field. This is directly related to the properties of the surface and not to the underlying dynamic equation. Similar continuity requirements are missing tn the case of a QM wavefunction. One may start with a discontinuous square-integrable wavefunction. Smooth evolution and continuity only apply to the individual basis wavefunctions once the eigenvectors of a specific Hamiltonian have been chosen as the basis set – GiorgioP-DoomsdayClockIsAt-90 Mar 17 '21 at 13:15
  • @GiorgioP: absolutely. The drum analogy wasn't meant as a perfect correspondence but rather just a general visualization of what eigenvectors are in the context of wave equations. – oliver Mar 17 '21 at 15:01