Change of variables in a propagator

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SUMMARY

The discussion focuses on the nuances between the quantum states ##|p\rangle## and ##|E\rangle##, particularly in the context of variable changes in integrals. The relationship between momentum ##p## and energy ##E## is highlighted, emphasizing that they are not linearly related, which affects the value of integrals. The integration of states is examined, revealing that the transformation from momentum to energy introduces complexities that require careful consideration of factors such as the Jacobian in the change of variables. The participants seek clarification on excluding terms related to momentum reversal to align with expected outcomes.

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  • Understanding of quantum mechanics, specifically state representations.
  • Familiarity with the concepts of momentum ##p## and energy ##E## in quantum systems.
  • Knowledge of integration techniques in quantum mechanics, including variable changes.
  • Basic grasp of Dirac notation and delta functions in quantum physics.
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Jacob Nie
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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
$$ U(t) = \sum_E |E\rangle \langle E|e^{-iEt/\hbar}.$$
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.
 
Last edited:
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[tex]U(0)=1=\int_{-\infty}^{+\infty} |p>dp<p|= \int_0^{\infty} |E>dE<E|[/tex]
seems all right.

[tex]<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>\}[/tex]

[tex]|E>=\sqrt{\frac{m}{2|p|}}\{|p>+|-p>\}[/tex]

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

[tex]\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|[/tex]

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.
 
Last edited:

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