# Density of states via a Laplace transform

Is this formula correct?

$$\frac{-1}{\pi}Im\int_{0}^{\infty}\!dt~\exp(Et)\Theta (t)\exp(i \epsilon t) ~=~ \sum _{n}\delta (E-E_{n}),$$

$$\Theta (t)~=~ \sum_{n}\exp(-tE_{n}) ,$$

and I have used Shokhotsky's formula for the Dirac delta function

$$\delta (x)~=~ \frac{-1}{\pi}Im \frac{1}{x+i\epsilon},$$

where $\epsilon \rightarrow 0^+,$

so I can obtain the density of states from the Heat kernel $Tr(e^{-tH})$.

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As Qmechanic points out, your expressions are potentially troublesome even if they are formally ok. Can you provide context for them? What are you using them for? What assumptions can you make about the spectrum $\{E_n\}_n$? – Emilio Pisanty Jul 4 '12 at 20:49
i am working on the context to find a Hamiltonian with a potential $f(x)$ so the Energies are $E_{n}=1/4+k_{n}^{2}$ with $\xi(1/2+ik_{m})=0$ , for the Theta kernel i can evaluate it by the Semiclassical double integral $\iint dxdp exp(-tp^{2}-tf(x))$ and $f^{-1} (x)$ is proportional to the eigenvalue staricase $N(E)$ – Jose Javier Garcia Jul 4 '12 at 21:06
@JoseJavierGarcia: You are trying to find the semiclassical spectrum, and you are doing it in a crazy roundabout way that is not going to work, because the integral is not going to be delta-functions classically, just a classical smooth density of states which you get by broadening each delta-function to be wider than the distance between successive delta functions. If you want the discrete levels semiclassically, you should use the Bohr-Sommerfeld quantization. – Ron Maimon Jul 5 '12 at 2:41

Qmechanic's answer is correct, but you can say more now that you provided more detail in the comment.

The function $\theta(t)$ is supposed to be the quantum partition function at imaginary time t, considered as a function of time, and zero for so that it decays according to the gap between the first excited state and the ground state (assuming you add a constant to H to set the ground state energy to zero). This thing decays exponentially as the energy gap between the ground state and the first excited state.

The only convergence issue is that you have a growing exponential multiplying a function $\theta(t)$ which is only decaying with a fixed exponential decay. So you will get a divergence when E is bigger than the decay rate. You introduced a small oscillatory term in $\exp(i\theta)$ to deal with this, and this works formally, but it is conceptually a little roundabout. The oscillations are growing exponentially even with $\epsilon$, and although you get the right answer, it is required to justify this mathematically, and for your own peace of mind.

To do this, you should do an analytic continuation from a purely oscillatory Fourier transform. Consider the function $\theta(t)$ defined as you did for t>0 and defined to be 0 for t<0. Then take it's Fourier transform. The analytic continuation of the FT to the imaginary axis is the Laplace transform, but now since you are explicitly analytically continuing from a convergent expression.

In this case, you can find that the FT is derivable from the FT of a decaying exponential for positive times:

$$\int_0^\infty e^{iEt} e^{-E_i t} dt = {1\over iE-E}$$

$$\tilde\theta(E) = \sum_i {1\over iE-E_i}$$

So that all the poles lie on the imaginary axis, at the position you expect. Taking the imaginary part along the imaginary axis gives you the density of states, and now it is seen to be a sequence of delta-functions without any formal trickery, just by analytically continuing from the FT, where the integrals are manifestly convergent.

### Relation of classical and quantum density of states

You want to semiclassically quantize and find discrete levels using classical heat kernels. This doesn't work. To show why, consider the explicit form for the classical $\theta$ for a one dimensional harmonic oscillator:

$$H= {p^2\over 2} + {q^2\over 2}$$

$$\theta(t) = \int e^{-tH(p,q)} dp dq = {2\pi\over t}$$

Or for a free particle confined by hard-walls to a length L

$$\theta(t) = L {\sqrt{2\pi\over t}}$$

In both cases, you see that the thing doesn't decay exponentially at long times, but as a power law. This is because the classical system has states of arbitrarily low energy.

The classical density of states can be derived from this by your method (at least formally, the integrals are divergent and tricky, because of the power-law blowup at small t, this is the high-temperature partition function). but the classical answer is just the total phase space area contained between energy E and E+dE, and this is the density of states you will find if you do it correctly by any method.

For the Harmonic oscillator, the area in phase space contained in the motion of area E is $2\pi E$, so that the area density in between any equal increments dE is the same regardless of E. This is the classical version of the statement that the Harmonic oscillator has a constant density of states. The Bohr-Sommerfeld (or, in this case, Planck) quantization tells you that you are supposed to gather up this uniform density of states into delta-function lumps so that each lump occurs with a separation of area hbar.

For the free particle of unit mass, the area in phase space between E and E+dE is given by the range in momentum between these, or $p=\sqrt{2E}$. The area increment is ${dE L \over \sqrt{2E}}$. So classically, the density of states is a smooth $L\over\sqrt{2E}$, with no delta functions. Quantum mechanically, you gather this smooth density into delta-function lumps which each contain h area of phase space, by the Bohr quantization rule. So the level spacing is one level per $L\over \sqrt{2E}$, so that the n-th level is at $n^2\over 2\pi L^2$, as is correct semiclassically (and quantum mechanically too).

The point is that you don't get quantum states without a quantum assumption, and the correct quantum assumption is the Bohr rule: the classical states corresponding to quantum states are separated by area h in phase space. Your method will not produce discrete delta functions from any classically computable phase space integral.

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OP's algebraic manipulations are formally correct.

But the formal calculation could be wrong for many reasons. For instance:

1. If the summation $\sum_n$ in $\Theta$ is not convergent.

2. If one is not allowed to change order of summation $\sum_n$ and integration $\int_{0}^{\infty}\!dt$.

3. If the Sokhotsky distribution formula is applied to a singular expression.

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