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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]

??

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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