# Density of states with delta function

• I
Hello,
I'm stuck with this exercise, so I hope anyone can help me.

It is to prove, that the density of states of an unknown, quantum mechanical Hamiltonian ##\mathcal{H}##, which is defined by
$$\Omega(E)=\mathrm{Tr}\left[\delta(E1\!\!1-\boldsymbol{H})\right]$$
is also representable as
$$\Omega(E)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\mathrm{d}k\ e^{ikE}\mathrm{Tr}\left[e^{-ik\boldsymbol{H}}\right]$$
when using the definition of the Dirac delta function
$$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\mathrm{d}k\ e^{ikx}$$
I really don't know what to do. Is it necessary to change to a dirac notation or is this just a simple representation of the Trace, which i don't know yet?

It would be great if anyone can give me a hand with that.
Thanks a lot,
Arnd

blue_leaf77
Homework Helper
If you replace ##x## in the definition of Dirac delta function with ##EI-H## you get ##e^{ik(EI-H)}##. ##EI## commutes with ##H##, thus ##e^{ik(EI-H)}= e^{ikEI}e^{-ikH}##. Now simplify ##e^{ikEI}## such that the identity operator "goes down" out of the exponent.

Arnd Obert
When multiplying $$e^{ikE1} \cdot e^{-ikH}$$ you get $$e^{ikE} e^{-ikH}$$ because ##e^{ikE1}## is diagonal, isn't it? And then you can pull it out of the trace?

Thank you a lot, but now it got even worse:
This Hamiltonian should be inserted
$$H = \sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)$$
and result in the density of states
$$\Omega(E)=\frac{1}{2\pi}\int\mathrm{d}k\ e^{ikE}\left(\sum_{n=0}^{\infty}e^{-i\hbar k\omega(n+1/2)}\right)^N$$

I don't know how to handle the ladder operators. Is there a good lecture?
Thanks a lot.

blue_leaf77
Homework Helper
because eikE1eikE1e^{ikE1} is diagonal, isn't it? And then you can pull it out of the trace?
Yes.

Note that
$$e^{ikH} = \exp \left( ik\sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)\right) \\ = \exp \left( ik\hbar\omega(a_1^{\dagger}a_1 + 1/2)\right) \otimes \exp \left( ik\hbar\omega(a_2^{\dagger}a_2 + 1/2)\right) \otimes \ldots \otimes \exp \left( ik\hbar\omega(a_N^{\dagger}a_N + 1/2)\right)$$
Then use the property ##\textrm{Tr }(A\otimes B) = (\textrm{Tr }A)(\textrm{Tr }B)##.

Arnd Obert
vanhees71
Gold Member
What are the tensor products doing there? Isn't this just the usual calculation of the microcanonical partition function in terms of a Fourier transform of the canonical one? As far as I can see, at the end of #3 you already have the correct result.

The End of #3 is the correct result but I have to find the way to get there.
Thank you! I will try the tensor products.

blue_leaf77
Homework Helper
What are the tensor products doing there?
I believe this is a multiparticle system where each particle obeys the ordinary harmonic oscillator Hamiltonian. A basis in the whole system is then a tensor product between the single particle state ##|n_1\rangle \otimes |n_2\rangle \otimes \ldots \otimes |n_N\rangle## and the corresponding symmetry operator must be a Kronecker product from each particle's symmetry operator. Moreover, the formal form of a symmetry generator of the system, such as the Hamiltonian, is
$$H = H_1\otimes I_2 \otimes \ldots \otimes I_N + I_1\otimes H_2 \otimes \ldots \otimes I_N + \ldots + I_1\otimes I_2 \otimes \ldots \otimes H_N$$
where ##H_i = \hbar\omega(a_i^{\dagger}a_i + 1/2)##. Using a hand-waving notation omitting those jumbled identity operators, the total Hamiltonian is rewritten as
$$H = \sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)$$
As far as I can see, at the end of #3 you already have the correct result.
Yes, but the OP was trying to prove that.

This is great, thank you very much! You are completely right with your assumption about the many particle system.
Now the last part remains:

When evaluating the geometric series, I get a density of states which looks like
$$\Omega(\epsilon)=\frac{1}{2\pi}\int dk\exp\left(N\left[ik\epsilon-\ln \left(2i\sin(k\hbar\omega/2)\right)\right]\right)$$
with ##\epsilon=E/N##

I have to prove, that the average entropy per oscillator looks like
$$s(\epsilon)=\lim_{N\rightarrow\infty}\frac{\ln\Omega(\epsilon)}{N}=\left(\frac{\epsilon}{\hbar\omega}+1/2\right)\ln\left(\frac{\epsilon}{\hbar\omega}+1/2\right)-\left(\frac{\epsilon}{\hbar\omega}-1/2\right)\ln\left(\frac{\epsilon}{\hbar\omega}-1/2\right)$$
by using the saddle point approximation in the limit ##N\rightarrow\infty##.
I'm familiar with the saddle point approximation but don't know how to evaluate the ln, how to find the ##x_0## and how to get the final result.

I hope you have any advice. Thank you!