Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

let be $ N(x)= \sum_{n} H(x-E_{n}) $ the eingenvalue 'staircase' function

and let be a system so $ V(x)=V(-x)$ and $ V^{-1}(x)=\sqrt \pi \frac{d^{1/2}}{dx^{1/2}} N(x) $

then would it be true that the two function

$\sum_{n}exp(-tE_{n})=Z(t)= \int_{0}^{\infty}dN(x)exp(-tx) $

and $ \int_{0}^{\infty}dx\int_{0}^{\infty}dpexp(-tp^{2}-tV(x))=U(t) $

would be equal ?? i have just compared the two results

$\int_{0}^{\infty}dnN(x)exp(-tx)=\int_{0}^{\infty}dx\int_{0}^{\infty}dpexp(-tp^{2}-tV(x)) $

i have taken the Laplace transform inside and get the desired result assuming that the potential V(x) is EVEN

share|improve this question

1 Answer 1

The answer is in general No, for various reasons.

  1. The WKB approximation is generally not exact for a finite energy-level $N$.

  2. Point 1 is true even if we include the metaplectic correction from the Maslov index.

  3. OP's sought-for formula is not exact for a finite energy-level $N$ even for the simple quantum harmonic oscillator (SHO). (This is despite the fact that the WKB approximation with the metaplectic correction famously yields the exact SHO spectrum.) To derive the correct parabolic potential via the fractional derivative formula$^1$, one should use an $N(E)$ function that is affine in the energy $E$, not the staircase function $N(E)=\sum_n \Theta(E-E_n)$. (Of course in the harmonic oscillator case, the staircase function can be approximated by an affine function in the continuum limit.)

--

$^1$See e.g. this answer for more details.

share|improve this answer
    
umm is still 'incorrect' even if we add the SMOOTH and FLUCTUATING correction to the eigenvalue staircase ??, but as an approximation this is still valid isn't it ? thanks :) –  Jose Javier Garcia Apr 8 '12 at 19:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.