Inverse spectral problem in QM

[zetafunction]

I have taken QM , and i find it very interesting but my question is , we have the Hamiltonian eigenvalue problem

$$i \partial _t \Psi (x,t) = \lambda _n \Psi (x,t)=(p^2+V(x))\Psi(x,t)$$

of course in general, we know the potential V(x) but my question is the inverse, if we knew how the spectrum is or for example the (approximate) value of $$\sum_n e^{it \lambda _n}$$

for example for Harmonic oscillator since all energies are lineal we know that

$$\sum_n e^{it \lambda _n}=(1-exp(it/2))^{-1}$$ from this could we deduce that V(x) is a quadratic potential x*x

 

[alxm]

I don't think you can (although I can't be bothered to try and prove it mathematically.. might be related to the v-representability problem though.).

If you knew the eigenfunctions, then you could expand the potential on some basis and then treat it as a minimization problem. (an 'inverse Ritz variational method' if you like)

 

[meopemuk]

Knowing energy eigenvalues is not sufficient to restore the Hamiltonian. As alxm said, you need to know all eigenfunctions too.

Two different Hamiltonians $$H$$ and $$H' = UHU^{-1}$$ have exactly the same energy spectra if $$U$$ is any unitary operator.

 

[Ilja]

Knowing energy eigenvalues is not sufficient to restore the Hamiltonian. As alxm said, you need to know all eigenfunctions too.

Two different Hamiltonians $$H$$ and $$H' = UHU^{-1}$$ have exactly the same energy spectra if $$U$$ is any unitary operator.

But H' would be most probable not of the form $$H' = \Delta + V(x)$$, so that this simple consideration is not sufficient to answer the question if it is possible to identify V(x).

It is not possible, and I have used this as an argument against many worlds, see http://arxiv.or/abs/arXiv:0901.3262" and the references therein.

 

[meopemuk]

But H' would be most probable not of the form $$H' = \Delta + V(x)$$, so that this simple consideration is not sufficient to answer the question if it is possible to identify V(x).
QUOTE]

If you choose a unitary operator $$U$$ commuting with momentum, then the transformed Hamiltonian is of the form $$H' = \Delta + V(x,p)$$. I don't see anything wrong with having potential V that depends on both position and momentum of the particle.

 

[zetafunction]

a potential $$V(x,p)$$ that depend on momentum is not common in physics since you should modifie Newton law or E-L equation

the idea is that sometimes you only known the spectrum but not the Eigenfunctions , perhaps using WKB approximation , you would know that eigenfunctions depend on the potential $$\Psi (x) = exp (i \oint (E-V(x))^{1/2})$$

but in general you only have information about the espectrum