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In Quantum Mechanics, we know that given a potential we can solve the eigen value problem to find out the energy eigen values and eigen functions. Now suppose in an experiment we have information only about the energy levels of a system and not the potential. Can we do the reverse and find out the nature of the Potential of the System uniquely?

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marked as duplicate by Qmechanic quantum-mechanics Feb 7 at 8:08

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  • $\begingroup$ Well, one may construct a reflectionless potential given the bound state energies, and, of course, using supersymmetric quantum mechanics "inverse scattering", which was done extensively in the first years of charmonium. But it is quite a long story. And virtually never uniquely. $\endgroup$ – Cosmas Zachos Feb 6 at 16:16
  • $\begingroup$ The classic accessible review of the industry is Kwong and Rosner 1986. $\endgroup$ – Cosmas Zachos Feb 6 at 21:07
  • $\begingroup$ Yes, indeed it is possible. It's a typical inverse problem. $\endgroup$ – lcv Feb 7 at 7:14
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/13480/2451 and links therein. $\endgroup$ – Qmechanic Feb 7 at 8:08
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No. There are isospectral potentials that have the same eigenvalues, at least in the one- and two-dimensional cases. You can generate them with a "Darboux transformation". Here's how it works in one-dimension:

Start with the Schrodinger equation

$$-\frac{\hbar^2}{2m}\Psi''+V\Psi=E\Psi\tag{1}.$$

We'd like to find another potential $\tilde{V}$ with wavefunctions $\tilde{\Psi}$ that satisfy

$$-\frac{\hbar^2}{2m}\tilde{\Psi}''+\tilde{V}\tilde{\Psi}=E\tilde{\Psi}\tag{2}$$

where $E$ is the same in both equations.

Make the ansatz

$$\tilde{\Psi}=\Psi'+A\Psi\tag{3}$$

where $A$ is a function of $x$ to be determined.

Substitute (3) into (2), and reduce the $\Psi''$ and $\Psi'''$ that appear using (1). The result is an equation in which only $\Psi$ and $\Psi'$ appear:

$$\left[\frac{\hbar^2}{m}A'+(V-\tilde{V})\right]\Psi'+\left[\frac{\hbar^2}{2m}A''+(V-\tilde{V})\right]\Psi=0\tag{4}.$$

We can satisfy this equation by choosing $A$ and $\tilde{V}$ to make the coefficient of both terms zero:

$$\frac{\hbar^2}{m}A'+(V-\tilde{V})=0\tag{5a}$$ $$\frac{\hbar^2}{2m}A''+V'+A(V-\tilde{V})=0\tag{5b}.$$

To solve these, first eliminate $V-\tilde{V}$ to get

$$\frac{\hbar^2}{2m}(A''-2AA')+V'=0\tag{6}.$$

The left side is a total derivative, so we can integrate it to get

$$\frac{\hbar^2}{2m}(A'-A^2)+V=C\tag{7}$$

where $C$ is a constant of integration.

This is a Riccati equation that can be linearized by setting

$$A=-(\log{\zeta})'\tag{8}$$

where $\zeta$ is an unknown function of $x$ to get

$$-\frac{\hbar^2}{2m}\zeta''+V\zeta=C\zeta\tag{9}$$.

This is simply Schrodinger equation (1) with potential $V$ and energy $C$!

So, take any eigenstate $\zeta$ of potential $V$, satisfying (1). Then, if $\Psi$ is an eigenstate of $V$ with energy $E$,

$$\tilde{\Psi}=\Psi'-(\log{\zeta})'\Psi\tag{10}$$

is an eigenstate of a different potential,

$$\tilde{V}=V-\frac{\hbar^2}{m}(\log{\zeta})''\tag{11},$$

with exactly the same energy $E$. You can get a whole set of isospectral potentials by taking $\zeta$ in the Darboux transformation to be different eigenstates.

This explanation was adapted from a paper which also addresses the two-dimensional case.

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  • $\begingroup$ In general link only answers are frowned on, as if the link goes dead, so does the answer. If you could flesh out the answer with an outline of the idea in the paper it would be better. $\endgroup$ – StephenG Feb 7 at 0:30
  • $\begingroup$ @StephenG I've given the details for the one-dimensional case. $\endgroup$ – G. Smith Feb 7 at 6:24
  • $\begingroup$ @G.Smith this is true but the fact that there is more than one solution, doesn't mean that there is no solution. In fact, given some real numbers that do not grow too fast, it is generally possible to find a potential that has those as eigenvalues. $\endgroup$ – lcv Feb 7 at 7:13

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