# Density of states with delta function

• I
• Arnd Obert
In summary, the author is trying to figure out how to calculate the density of states for an unknown quantum mechanical Hamiltonian. He uses the Dirac delta function and the property that eikE1eikE1e^{ikE1} is diagonal to get the trace. He then uses the Trace to get the density of states. He is stuck on how to calculate the ln and x_0.

#### Arnd Obert

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

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.

Arnd Obert said:
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
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.

vanhees71 said:
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
$$H = \sum_{j=1}^{N}\hbar\omega(a_j^{\dagger}a_j + 1/2)$$
vanhees71 said:
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!

## What is the "density of states with delta function"?

The density of states with delta function is a mathematical concept used in physics and materials science to describe the energy states of a system. It is a measure of the number of allowed energy states per unit energy interval. The delta function, also known as the Dirac delta function, is a mathematical function that represents an idealized point mass or point charge. When combined with the density of states, it allows for a more precise description of the energy states of a system.

## How is the density of states with delta function used in research?

The density of states with delta function is commonly used in research to model the energy states of materials such as semiconductors and metals. It is an important tool for understanding the electronic properties of these materials and can be used to predict their behavior under different conditions. It is also used in theoretical studies of quantum mechanics and statistical mechanics.

## What is the difference between the density of states with delta function and the density of states without delta function?

The density of states with delta function takes into account the contribution of discrete energy states, while the density of states without delta function only considers the contribution of continuous energy states. In other words, the delta function accounts for the energy levels of individual particles or atoms, while the non-delta function accounts for the overall distribution of energy levels in a material.

## What are some advantages of using the density of states with delta function?

One advantage of using the density of states with delta function is that it allows for a more precise and detailed description of the energy states of a material. It can also simplify calculations in certain situations, such as when dealing with localized energy states. Additionally, it is often easier to interpret and visualize the results obtained using the density of states with delta function compared to other methods.

## Are there any limitations to using the density of states with delta function?

While the density of states with delta function is a useful tool, it is not always applicable in all situations. It assumes that the energy states are well-defined and discrete, which may not be the case in some materials or under certain conditions. Additionally, it may not accurately describe the behavior of materials with highly correlated or interacting particles. In these cases, other methods may be more appropriate.