- #1

Ratzinger

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thank you

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- Thread starter Ratzinger
- Start date

- #1

Ratzinger

- 287

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thank you

- #2

Edgardo

- 705

- 15

Do you have the second edition?

- #3

Norman

- 896

- 4

Cheers,

Ryan

- #4

Doc Al

Mentor

- 45,424

- 1,870

Why don't you scan in the page so we can have a look.

- #5

Ratzinger

- 287

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it's really only a kids question for you guys

- #6

snooper007

- 33

- 1

They are evolution equations, when t->0, (5.1.10) really stands for

a delta function.

a delta function.

- #7

Galileo

Science Advisor

Homework Helper

- 1,994

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(5.1.10)

[tex]U(x,t;x')\equiv \langle x|U(t)|x' \rangle = \int_{-\infty}^{\infty}\langle x|p\rangle \langle p|x' \rangle e^{-ip^2t/2m\hbar}dp

=\frac{1}{2\pi \hbar}\int_{-\infty}^\infty e^{ip(x-x')/\hbar} \cdot e^{-ip^2t/2m\hbar}dp=\left(\frac{m}{2\pi \hbar it}\right)^{1/2}e^{im(x-x')^2/2\hbar t[/tex]

(5.1.11)

[tex]\psi(x,t)=\int U(x,t;x')\psi(x',0)dx'[/tex]

- #8

Ratzinger

- 287

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Galileo said:

(5.1.10)

[tex]U(x,t;x')\equiv \langle x|U(t)|x' \rangle = \int_{-\infty}^{\infty}\langle x|p\rangle \langle p|x' \rangle e^{-ip^2t/2m\hbar}dp

=\frac{1}{2\pi \hbar}\int_{-\infty}^\infty e^{ip(x-x')/\hbar} \cdot e^{-ip^2t/2m\hbar}dp=\left(\frac{m}{2\pi \hbar it}\right)^{1/2}e^{im(x-x')^2/2\hbar t[/tex]

(5.1.11)

[tex]\psi(x,t)=\int U(x,t;x')\psi(x',0)dx'[/tex]

These are the equations. Thanks Galileo, thanks snooper007, thanks everbody else. I will learn Latex now.

My trouble was that I thought wave(x,t)=propagator acts on wave(x,0), without any integral.

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