Is there a potential energy for every wave function?

Question

In my early ventures of quantum mechanics, the pattern seems to be: choose a simple potential energy and try to solve for the corresponding wave function. (We can stick to a particle in a box for my purposes, I think.) At this stage, I'm trying to get a feel for which "parameters" can vary freely, and what the corresponding effects are. So if I fix the mass of a particle, and choose a "reasonable" wave function, then does there necessarily exist a potential energy that will yield the chosen wave function as a solution?

Are there some other variables that I need to fix here? Is there some sort of one to one mapping between potential energies and wave functions? Or am glossing over the fact that there's usually a whole set of wave functions for a given potential energy?

Answer 1

Hoping I understand correctly, you are given one wavefunction over a time interval and a space interval, and the mass. Then, by Schrödinger's equation we have the answer:

$\frac{ i\hbar\frac{\partial}{\partial t} \Psi(x,t) + \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)}{\Psi(x,t) } = V(x,t) .$

Such a potential would exist for a "well-behaved" wave function.

Since you mentioned the particle in a box, here is a report that calculates the "actual" potential function from its well-known solutions

Answer 2

What is a wave?

A physical observation of changes in space , as water waves, sound waves ....

What is a wave function?

A solution of differential equations called wave equations.

The wave equation is a second-order linear partial differential equation for the description of waves—as they occur in classical physics—such as mechanical waves (e.g. water waves, sound waves and seismic waves) or light waves.

Depending on the observations , the solutions can be used to model it mathematically, and those solutions can predict the behavior of the wave in time.

In order for the mathematical solutions to be descriptive and predictive extra axiom type postulates have to be imposed, to pick up those solutions that will be useful in modelling and predicting physics data.

We then come to the wave differential equations used for modeling behavior of matter at the quantum mechanical level. The extra axioms and assumptions in order to model data are seen here.

The simple particle in a box is a particular potential used in the Schrodinger wave equation.

Assume the potential U(x) in the time-independent Schrodinger equation to be zero inside a one-dimensional box of length L and infinite outside the box. For a particle inside the box a free particle wavefunction is appropriate, but since the probability of finding the particle outside the box is zero, the wavefunction must go to zero at the walls. This constrains the form of the solution to

Due to the boundary conditions the solutions are constrained mathematically for this simple case, because the boundary conditions are clear. (The infinite potential outside the wall, the zero inside).

In the general differential wave equations used for modeling quantum mechanical states, (Schrodinger,, Dirac,Klein Gordon, quantized Maxwell) if no potential exists, the solutions are plane waves .

Is there some sort of one to one mapping between potential energies and wave functions?

Take the Schrodinger solutions for the hydrogen atom . There are constraints on the wave equation that pick them up from the possible mathematical solutions:

These constraints are applied to the boundary conditions on the solutions, and in the process help determine the energy eigenvalues.

This should answer your question, because if these constraints are not imposed in picking up useful for modeling wavefunctions a large number of wavefunctions exist. It is not possible to pick the ones contributing to the hydrogen solutions ones by just looking at all the solutions.

I will quote from the link Zachos gave:

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum"

This should apply to your one to one mapping too.

Answer 3

Is there a potential energy for every wave function?

No, but for every external potential energy function $v_{ext}(\mathbf{r})$ one can in principle calculate the stationary state wave functions.

In other words, if we have stationary state wave functions $|\Psi_n\rangle$, which by definition solve the TISE: $$ \left(\hat T + \hat V_{ext} + \hat V_{interaction}\right)|\Psi_n\rangle = E_n |\Psi_n\rangle\;, $$ where $\hat T$ is the kinetic energy, $\hat V_{interaction}$ is the interaction energy of the dynamical quantum particles, and where $$ \hat V_{ext} = \sum_{i=1}^N v_{ext}(\hat{\mathbf{r}}_i) = \int d^3r v_{ext}(\mathbf{r}) \hat n(\mathbf{r})\;, $$ where $\hat n(\mathbf{r}) = \sum_{i}\delta(\mathbf{r} - \hat{\mathbf{r}}_i)$ is the density operatory and $v_{ext}$ is the single-particle external potential (e.g., $-Ze^2/r$ for an atom).

Thus we can write each of the stationary states as a functional of the external potential: $$ |\Psi_n\rangle = |\Psi_n[v_{ext}]\rangle\;, $$ since the other terms like $\hat T$ and $\hat V_{interaction}$ are universal for the system of interest (say, a system of interacting electrons).

In general the functional is quite complicated and hard to write down explicitly.

And, by the way, as it turns out, we can also write the ground state as a (generally very complicated) functional of the ground state density!