functional determinant and WKB approximation

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.

-
 Interesting idea. Thanks. – Ron Maimon Nov 3 '11 at 8:57

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.
$$\int dp dq {1\over E - p^2 - q^2}$$
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.
 OK ron thanks... another question for the case of the Infinite potential well $V=0$ using the Gutzwiller formula should be true that $det (E-H)= sin (E^{1/2}/E^{1/2}$ ?? – Jose Javier Garcia Nov 3 '11 at 9:16 @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. – Ron Maimon Nov 3 '11 at 18:23