Change of variables in a propagator

  • Thread starter Jacob Nie
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  • #1
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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:

Answers and Replies

  • #2
anuttarasammyak
Gold Member
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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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