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Propagator for a free particle / schrodinger equation

  • Thread starter boboYO
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http://img18.imageshack.us/img18/4295/eqn.png [Broken]


here is the text preceding the exercise:

http://yfrog.com/5mch5p


in the exercise, where does the factor [tex]\frac{m}{(2mE)^{1/2}}[/tex] come from? Comparing that equation with 5.19 (bottom right of link), why can't we just replace |p> with |E,+> and |E,->?




if I start from scratch:

[tex]| \psi \rangle = \sum_{\alpha=\pm} \int_0^{\infty}\! |E,\alpha\rangle\langle E,\alpha|\psi \rangle \,\,dE [/tex]


do [tex]\left( i \hbar \frac{\partial}{\partial t} - H \right)[/tex] to both sides:

[tex] \left( i \hbar \frac{\partial}{\partial t} - H \right) |\psi \rangle = 0 = \sum_{\alpha=\pm} \int_0^{\infty}\!\left[i \hbar \langle E,\alpha|\dot{\psi}\rangle -E\langle E\langleE,\alpha|\psi\rangle \right] |E, \alpha \rangle \,\, dE [/tex]

[tex]\implies i \hbar \langle E,\alpha|\dot{\psi}\rangle -E\langle E\langleE,\alpha|\psi\rangle = 0 [/tex]

[tex] \implies \langle E,\alpha| \psi \rangle = \langle E,\alpha|\psi_0\rangle e^{-iEt/\hbar}[/tex]


[tex]\therefore | \psi \rangle = \sum_{\alpha=\pm} \int_0^{\infty}\! |E,\alpha\rangle \langle E,\alpha|\psi_0\rangle e^{-iEt/\hbar} \,\,dE [/tex]


and so the propagator is


[tex] U(t) = \sum_{\alpha=\pm} \int_0^{\infty}\! |E,\alpha\rangle \langle E,\alpha| e^{-iEt/\hbar} \,\,dE [/tex]

??
 
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Answers and Replies

  • #2
vela
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Split (5.1.9) into two integrals, one for positive p and one for negative p, and when you change variables from p to E, remember that you have to express dp in terms of dE.
 
  • #3
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Thanks. I mistakenly assumed dp=dE because |p>=|E>


but I still can't find the mistake in my derivation? Is the first line correct?
 
  • #4
vela
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Just a guess, but maybe it's a just difference in normalization.
 
  • #5
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It can not be a just difference in normalization, since [tex]\frac{m}{(2mE)^{1/2}}[/tex] is the factor of the integrand. I am facing the same problem, any help is appreciated.
 
  • #6
fzero
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That factor is necessary as a measure to properly count the density of states. The point is that the number of states in the interval [tex](E,E+dE)[/tex] must be the same as the number of states in the interval [tex](p,p+dp)[/tex] when [tex]E = p^2/(2m)[/tex].

For a free particle, the momentum states are parameterized as a linear function of wavenumbers, so it's obvious that

[tex]\sum_p \rightarrow \int dp[/tex]

without any factors. If we compute

[tex]\int_{\sqrt{2mE}}^{\sqrt{4mE}} = (\sqrt{2}-1) \sqrt{2mE} \neq \int_E^{2E} dE.[/tex]

However

[tex] \int_E^{2E} dE \frac{m}{\sqrt{2mE}} = (\sqrt{2}-1) \sqrt{2mE}.[/tex]

We therefore conclude that

[tex]\sum_E \rightarrow \int dE ~\frac{m}{\sqrt{2mE}}.[/tex]
 
  • #7
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I do not understand why the number of states in the two interval must be same, through it leads to the desired result. But the point is, what is wrong with the derivation above? Isn't that the normal form of propagator?
 
  • #8
fzero
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I do not understand why the number of states in the two interval must be same, through it leads to the desired result. But the point is, what is wrong with the derivation above? Isn't that the normal form of propagator?
It is, but you still have to change the momentum integration to an energy integration, which introduces the same factor. Start with

[tex] U(t) = \int_{-\infty}^\infty dp | p \rangle \langle p | e^{-iEt/\hbar}[/tex]

We can introduce complete sets of energy eigenstates:

[tex] U(t) = \int_{0}^\infty dp \int_0^\infty dE' \int_0^\infty dE'' |E',+\rangle\langle E'',+ | \langle E'',+ | p \rangle \langle p | E',+ \rangle e^{-iEt/\hbar}
+ (+\rightarrow -).[/tex]

Now we know that

[tex]\langle p | E',+\rangle = \langle E=p^2/2m, + | E',+\rangle = \delta(E-E'),[/tex]

but to use that we'll need to rewrite the integral over momentum in terms of an integral over energy:

[tex]\int_0^\infty dp = \int_0^\infty \frac{m}{\sqrt{2mE}} dE.[/tex]
 
  • #9
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Thank you for your reply. I am puzzled with what you "start with". We are dealing with the free particle, so [tex]H=\frac{m}{(2mE)^{1/2}} [/tex]. If a state is an eigenstate of [tex]H[/tex], it should be an eigenstate of [tex]P^2[/tex], and thus an eigenstate of [tex]P[/tex]. (Why? We know that any eigenstate of [tex]P[/tex] is also eigenstate of [tex]P ^2[/tex], but vice versa?) As the result, an eigenstate of energy can be labeled as |p>.

The normal form of propagator is [tex]
U(t) = \int_{0}^\infty dE | E \rangle \langle E | e^{-iEt/\hbar}
[/tex].
Then replace |E> with |p>, it seems that the propagator should be
[tex]
U(t) = \int_{0}^\infty dE | p \rangle \langle p | e^{-iEt/\hbar}
[/tex]. What's wrong?

I'm reading shankar. If you have a book at hand, you may see p.146 and p.152. Thanks.
 
  • #10
fzero
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Thank you for your reply. I am puzzled with what you "start with". We are dealing with the free particle, so [tex]H=\frac{m}{(2mE)^{1/2}} [/tex].
That's not the free particle Hamiltonian. Among other things, that expression has the wrong units!.

If a state is an eigenstate of [tex]H[/tex], it should be an eigenstate of [tex]P^2[/tex], and thus an eigenstate of [tex]P[/tex]. (Why? We know that any eigenstate of [tex]P[/tex] is also eigenstate of [tex]P ^2[/tex], but vice versa?)
The vice versa comes from the theorem that if two observables commute, they have simultaneous eigenstates.

As the result, an eigenstate of energy can be labeled as |p>.

The normal form of propagator is [tex]
U(t) = \int_{0}^\infty dE | E \rangle \langle E | e^{-iEt/\hbar}
[/tex].
Then replace |E> with |p>, it seems that the propagator should be
[tex]
U(t) = \int_{0}^\infty dE | p \rangle \langle p | e^{-iEt/\hbar}
[/tex]. What's wrong?

I'm reading shankar. If you have a book at hand, you may see p.146 and p.152. Thanks.
This thread was about a particular definition of the propagator that is equation 5.1.9 in the link in the OP. It is more common to define the propagator as a sum over energy eigenstates. However, if you rewrite your last expression above as an integral over momentum, you'll find an extra factor of [tex]p/m[/tex]. So the two definitions of propagator are not perfectly in agreement, but I believe that the qualitative physics will be the same.
 

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