Shankar (5.1.10)

  • Thread starter Ratzinger
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  • #1
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Main Question or Discussion Point

Would someone mind opening his or her Shankar page 153 and tell me what (5.1.10) and (5.1.11) exactly mean? Does (5.1.10) stand for a delta function?

thank you
 

Answers and Replies

  • #2
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which Shankar book are you talking about? The "Principles of quantum mechanics"? I have one (1st edition) here, but equation (5.1.10) and (5.1.11) are on page 161 and deal with propagators.

Do you have the second edition?
 
  • #3
887
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Why don't you write the equations out for us in LaTeX and we can then figure out what they are, since most people probably don't have this book. You will get a lot more help.
Cheers,
Ryan
 
  • #4
Doc Al
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Why don't you scan in the page so we can have a look.
 
  • #5
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it's the second edition of "Principles of Quantum Mechanics" and unfortunately I haven't learnt the Latex thing yet and neither do I have a scanner...it's the free particle propagator represented in position space which multiplied with a wave function at t=0 under an integral over space gives the wave function at a later time

it's really only a kids question for you guys
 
  • #6
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They are evolution equations, when t->0, (5.1.10) really stands for
a delta function.
 
  • #7
Galileo
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Are these the equations?

(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
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Galileo said:
Are these the equations?

(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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