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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
$$ U(t) = \sum_E E\rangle \langle Ee^{iEt/\hbar}.$$
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 Ee^{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.
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.
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