- #1
lokofer
- 106
- 0
I think i put this question before... but i can't be very sure.. let's suppose we have a Hamiltonian operator:
[tex] H= - \frac{d^2 }{dx^ 2}+V(x) [/tex] so its "energies" (eigenvalues)
satisfy that [tex] E(n)=-E(-n) [/tex] then here comes the question..is licit legal (at least as an approximation) to take:
[tex] Z(u)=\sum_{n=-\infty}^{n=\infty}e^{iuE(n)}\sim \iint dxdpe^{iup^2 +iuV(x)} [/tex] ??
The explanation is clear..you substitute a "discrete" sum over energies by a continuous sum over all the energies..classically the energy of the system is [tex] E=p^2 +V(x) [/tex] (time independent potential) , so it would be similar to take the "sum-integral" approximation (valid at least at first order ??) i know that perhaps using functional analysis you could justify my approach, as a physicist when dealing with Statistical mechanics we use it all the time since "sums" are very hard to evaluate, except when E(n)=log(n) or E(n)=n :grumpy: :grumpy: usually the exponential is "real" but i think that for this case this wouldn't be a problem..also if we kenw Z(u) we could obtain the inverse of the potential taking:
[tex] A\int_{-\infty}^{\infty}du \frac{Z(u)e^{-iut}}{\sqrt (u) } = V^{-1}(t) [/tex] :tongue2: :tongue2:
[tex] H= - \frac{d^2 }{dx^ 2}+V(x) [/tex] so its "energies" (eigenvalues)
satisfy that [tex] E(n)=-E(-n) [/tex] then here comes the question..is licit legal (at least as an approximation) to take:
[tex] Z(u)=\sum_{n=-\infty}^{n=\infty}e^{iuE(n)}\sim \iint dxdpe^{iup^2 +iuV(x)} [/tex] ??
The explanation is clear..you substitute a "discrete" sum over energies by a continuous sum over all the energies..classically the energy of the system is [tex] E=p^2 +V(x) [/tex] (time independent potential) , so it would be similar to take the "sum-integral" approximation (valid at least at first order ??) i know that perhaps using functional analysis you could justify my approach, as a physicist when dealing with Statistical mechanics we use it all the time since "sums" are very hard to evaluate, except when E(n)=log(n) or E(n)=n :grumpy: :grumpy: usually the exponential is "real" but i think that for this case this wouldn't be a problem..also if we kenw Z(u) we could obtain the inverse of the potential taking:
[tex] A\int_{-\infty}^{\infty}du \frac{Z(u)e^{-iut}}{\sqrt (u) } = V^{-1}(t) [/tex] :tongue2: :tongue2:
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