let be a Hamiltonian in one dimension, i would like to evaluate the functional determinant $ det(E-H) $ in onde dimension

i believe that $ det(E-H)= Cexp(iN(E)) $ here $ N(E)$ is the number of energy levels less than a given number 'E'

my steps

  1. i use the identity $ logDet(E-H)=TrLog(E-H)$

  2. i replace the sum $\sum_{n} log(E-E_{n})$ by an integral over the phase space $ \iint_{D}dpdp log(E-p^{2}-V(x))$

  3. I take the derivative respect to 'E'

  4. I use the identity $ \int_{-\infty}^{\infty} \frac{dx}{x^{2}-a^{2}} = \frac{\pi i}{a}$

  5. I use the Bohr-Sommerfeld quantization condition $\int_{C}dx (E-V(x))^{1/2} = (n(E)+1/2)h$

  6. i use integration respect to 'E' again

7 i take the exponential

is this semiclassically correct :) thanks.


The formula doesn't work. Most of the manipulations are formally ok, although it is probably best to start right at step 3--- the derivative of the logarithm of the determinant is the (trace of the) Green's function, which is better behaved than the determinant itself.

Step 5 is incorrect--- there is no reduction using the WKB condition, because the quantity $\sqrt{E-V}$ is in the denominator, and the integration is unbounded. The correct semiclassical expansion for the Green's function is given by the Gutzwiller trace formula.

The best way to check all this is to try it out on the Harmonic oscillator. The formula you give doesn't work, although the semi-classical bit is nice. The semiclassical HO green's function is

$$\int dp dq {1\over E - p^2 - q^2} $$

Which is elementary (up to being divergent--- you can move E a little), and evaluates to $log(E) \pm i\pi (E>0)$ , where $\pm$ means either add, or subtract, or ignore depending on how you deal with the divergent point. plus a divergent constant, which is irrelevant.

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  • $\begingroup$ OK ron thanks... another question for the case of the Infinite potential well $ V=0 $ using the Gutzwiller formula should be true that $ det (E-H)= sin (E^{1/2}/E^{1/2} $ ?? $\endgroup$ – Jose Javier Garcia Nov 3 '11 at 9:16
  • $\begingroup$ @Jose: I didn't work out the exact Green's function--- I just used your method. It's interesting to do for the harmonic oscillator, and it would give you insight to how good your approximation is. $\endgroup$ – Ron Maimon Nov 3 '11 at 18:23

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