# Where does the Density of States come from?

There is an existing question here, which asks about the propagator for a free particle and the difference in its form when expressed as an integral over $$p$$ or over $$E$$. The accepted answer points to a function called the Density of States. I am wondering where this Density of States comes from and why it is necessary?

In particular, I am confused as to why the following discrete sum is always true for an orthonormal eigenbasis $$|E\rangle$$:

$$\sum_{E} \lvert E\rangle\langle E\rvert=I\tag{1}$$

However, the corresponding integral for a continuous $$E$$ is, in general, not always equal to $$I$$ (because the Density of States function ($$\rho(E)$$) is required:

$$\int_{-\infty}^{\infty} \lvert E\rangle\langle E\rvert dE \neq I \tag{2}$$

Presumably, it is because of the addition of the $$dE$$ term in the integral, which in itself requires normalization (even though $$|E\rangle$$ is already orthonormal)?

The answer lies in the tricky nature of the continuum. As you probably know, when we normalize our basis and the Hilbert space is compact, then we can choose $$\langle n | m \rangle = \delta_{m,n}$$ and $$|n\rangle$$ is dimensionless. However, when during class one introduces a basis which is a continuum of states (let's say the position states) then one normalizes $$\langle x | x' \rangle = \delta(x-x')$$ which is very different than $$\delta_{x,x'}$$! One difference is that $$|x\rangle$$ now has dimensions. Let's say that now we want to perform a 'squeeze' transformation on our space such that $$|y\rangle = |a x\rangle$$. The new states maintain $$\langle y | y' \rangle = \delta(ax-ax') = \delta(x-x')/a$$. What happened here? The $$a$$ factor came our naturally, because the new states are concentrated in a new density with measure $$a$$ over the original states.
One way to stay always on a safe ground, and for me also to gain intuition, is to start from finite space and then gradually to take the continuum limit. In finite space of length $$L$$, with periodic boundary conditions, the momentum values are spaced with $$2\pi/L$$. So we have $$I = \sum_n |p=2\pi n / L \rangle \langle p=2\pi n /L|$$. And we can write the wave function explicitly $$\psi_p(x) = e^{ipx}/\sqrt{L}$$. Note that $$\sqrt{L}$$ is here to make sure that the wave-functions are normalized. Without it, you will not get the identity!
Now we want to take $$L\to \infty$$. Two things happen - one is that the sum for $$I$$ becomes 'denser' as the $$p$$ become closer, but also $$\psi_p(x)\to 0$$. This is not good as we want a finite wave-function. So we choose $$\psi_p(x) = e^{ipx}$$, (which is the usual choice for plane waves in an infinite space), and get that $$I = 1/L \sum_p |p\rangle \langle p|$$. Now we can take the continuum limit, and the sum becomes an integral. What is $$dp$$? It is the spacing between consecutive values of $$p$$ i.e. $$2\pi/L$$. So we get $$I = \frac{1}{2\pi}\int\! dp |p\rangle \langle p|$$ and voila - the $$1/2\pi$$ is exactly the density-of-states of our $$p$$ eigenbasis. Now we can do a convenient choice, and incorporate it into our definition of the wave-function, $$\psi_p(x) = e^{ipx}/\sqrt{2\pi}$$. The d.o.s. is now $$1$$, and indeed $$I=\int\! dp |p\rangle \langle p$$. Note that both choices are ok. They just represent different normalizations of our eigenbasis. In the first choice $$\langle p | p' \rangle = 2\pi \delta(p-p')$$ and in the second one $$\langle p | p' \rangle = \delta(p-p')$$. In both cases it is derived from a limit of the integral $$\int_{-L/2}^{L/2}\! dx e^{i(p-p')x}$$ when we took $$L\to\infty$$, where in the second case we just put $$2\pi$$ by hand.
• Thanks very much again. This is a great answer and very helpful. One slight quibble though: shouldn't the p values for this example be $2\pi\hbar n/L$ (since $p = \hbar k$)? Although, perhaps I am getting confused somewhere. Commented Jan 28, 2020 at 21:15
• no you're not missing anything, I'm just using units in which $\hbar=1$ :)
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