Change of variables in a propagator

[Jacob Nie]

Homework Statement

I'm reading Shankar's quantum mechanics. He states that the equation for the propagator $U(t)$ (such that $|\psi(t)\rangle = U(t)|\psi(0)\rangle$) is
$$ U(t) = \sum_E |E\rangle \langle E|e^{-iEt/\hbar}.$$
Later, when deriving the propagator for a free particle, he changes variable from $E$ to $p$ (for each $E$ there are two possible $p$ with that energy: $\pm (2mE)^{1/2}$):
$$U(t) = \int_{-\infty}^{\infty} |p\rangle \langle p | e^{-iE(p)t/\hbar} \ dp.$$
My question: why is it $dp$ and not $dE?$ Shouldn't turning $\sum_E |E\rangle \langle E|e^{-iEt/\hbar}$ into an infinite dimensional version translate to
$$U(t) = \int_{-\infty}^{\infty} |p\rangle \langle p | e^{-iE(p)t/\hbar}\ dE\ ?$$
Then, to properly change to $dp,$ wouldn't one use the proper substitution method using $dE = \dfrac{dE}{dp}dp$?

Relevant Equations

$|E\rangle$ are the eigenkets of the relevant Hamiltonian $H.$

Some other relevant equations: $|E,+\rangle = |p=(2mE)^{1/2}\rangle$ and $|E,-\rangle = |p = -(2mE)^{1/2}\rangle.$

I'm guessing that there must be some nuance that I do not quite understand in the difference between $|p\rangle$ and $|E\rangle$?

Like, later in the book even $dk$ is used as a variable of integration, where $k = p/\hbar.$ Surely this has effects on the value of the integral - wouldn't it change it by a numerical factor? In the case of $p$ and $E,$ it would change by more than just a numerical factor, because they're not linearly related.

 

[anuttarasammyak]

U(0)=1=\int_{-\infty}^{+\infty} |p>dp<p|= \int_0^{\infty} |E>dE<E|
seems all right.

<E'|E>=\delta(E'-E)=\delta(p'^2/2m - p^2/2m)=2m\delta(p'^2-p^2)=m/|p|\{\delta(p'-p)+\delta(p'+p)\}=m/2|p|\{<p'|p>+<-p'|p>+<p'|-p>+<-p'|-p>\}

|E>=\sqrt{\frac{m}{2|p|}}\{|p>+|-p>\}

dE=d(p^2/2m)=\frac{p}{m}dp

\int_0^\infty |E>dE<E|=\int_{0}^{+\infty} \{|p>+|-p>\}dp\{<p|+<-p|\}=\int_{-\infty}^{+\infty}|p>dp<p|+ \int_{-\infty}^{+\infty}|-p>dp<p|

The first RHS term is $1$ and the second term is operator of reversing momentum.

How should we exclude the second term to meet the anticipation ? Your advise is appreciated.