# Trace of density of states

1. Jul 19, 2010

### john0909

regarding the density of states:
how I GET THE FOLLOWING EQUALITY?
$$\langle E_n\mid \delta(E-\widehat{H}) \mid E_n \rangle = \sum_n \delta(E-E_n)$$

Last edited: Jul 19, 2010
2. Jul 19, 2010

### jostpuur

If

$$H|E_n\rangle = E_n|E_n\rangle$$

and if $f:\mathbb{R}\to\mathbb{R}$ is some function, then the operator $f(H)$ is defined by using the eigenbasis of $H$, like this:

$$f(H)|E_n\rangle = f(E_n)|E_n\rangle$$

Then, if you think that the delta function is like any function, you can do this:

$$\delta(E - H)|E_n\rangle = \delta(E-E_n)|E_n\rangle$$

In order to understand better what's going on, you should take a closer look at how you got the $H$ inside the delta function in the first place.

3. Jul 19, 2010

### john0909

yes but then you get:

$$\sum_n \langle E_n\mid \delta(E-E_n) \mid E_n \rangle.$$

So how do you eliminate the bra and kets? $$\langle E_n| , |E_n\rangle$$

4. Jul 19, 2010

### jostpuur

If you think that the delta function is like any function, then $\delta(E - E_n)$ is a number, and it can be taken out from between the brackets, by bilinearity of the inner product.

$$\langle E_n|\delta(E - E_n)| E_n\rangle = \delta(E - E_n)\langle E_n| E_n\rangle$$

5. Jul 19, 2010

### john0909

But you didn't answer my question:

let me explain you my problem:

The density of states n(E) is defined as the trace of the spectral operator
$$\delta(E-\hat{H}), \newline n(E)\equiv Tr \delta(E-\hat{H}).$$

this expression is equal $$= \sum_n \langle E_n|\delta(E- \hat{H})| E_n\rangle.$$

My question is how do I get the final expression:$$\sum_n \delta(E-E_n)?$$
According to what you said above I get: $$\sum_n \delta(E - E_n) \langle E_n| E_n\rangle$$
BUT HOW DO I ELIMINATE THE BRA AND KETS?
Because finally I need to get $$\sum_n \delta(E-E_n)$$.

Last edited: Jul 19, 2010
6. Jul 19, 2010

### jostpuur

I just decided that I'm in a nasty mood, and I refuse to answer your final question, even though I know the answer. BUHAHAHahahahahh.....!!!! :rofl:

7. Jul 19, 2010

### Demystifier

What is $$\langle \psi | \psi \rangle$$ for any conventionally normalized state $$| \psi \rangle$$?