- #1
karlsson
- 3
- 0
I hope this is the correct place for my question. I posted it here, because it`s from Peskin & Schroeder:
"Consider the amplitude for a free particle to propagate from [tex] \mathbf{x}_{0} [/tex] to [tex] \mathbf{x} [/tex] :
[tex]U(t)=\left\langle \mathbf{x}\right|e^{-iHt}\left|\mathbf{x_{0}}\right\rangle[/tex]
In nonrelativistic quantum mechanics we have E=p^2/2m, so
[tex] U(t)&=&\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{x}_{0}\right\rangle [/tex]
[tex]
=\int\frac{d^{3}p}{(2\pi)^{3}}\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right.\left|\mathbf{x_{0}}\right\rangle [/tex]
[tex]
=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i(\mathbf{p}^{2}/2m)t}\cdot e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_{0})}
[/tex]
[tex]
=\left(\frac{m}{2\pi it}\right)^{3/2}\, e^{im(\mathbf{x}-\mathbf{x}_{0})/2t}}
[/tex]
."
I don't understand the last equation.
Why I can't use the fourier-transformation:
[tex]
=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i\mathbf{p}\cdot(\mathbf{x}_{0}-\mathbf{x})}\widetilde{f}(\mathbf{p})
[/tex]
[tex]
=f(\mathbf{p})
[/tex]
[tex]
=e^{-i(\mathbf{x}-\mathbf{x}_{0})^{2}/2m)t}
[/tex]
"Consider the amplitude for a free particle to propagate from [tex] \mathbf{x}_{0} [/tex] to [tex] \mathbf{x} [/tex] :
[tex]U(t)=\left\langle \mathbf{x}\right|e^{-iHt}\left|\mathbf{x_{0}}\right\rangle[/tex]
In nonrelativistic quantum mechanics we have E=p^2/2m, so
[tex] U(t)&=&\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{x}_{0}\right\rangle [/tex]
[tex]
=\int\frac{d^{3}p}{(2\pi)^{3}}\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right.\left|\mathbf{x_{0}}\right\rangle [/tex]
[tex]
=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i(\mathbf{p}^{2}/2m)t}\cdot e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_{0})}
[/tex]
[tex]
=\left(\frac{m}{2\pi it}\right)^{3/2}\, e^{im(\mathbf{x}-\mathbf{x}_{0})/2t}}
[/tex]
."
I don't understand the last equation.
Why I can't use the fourier-transformation:
[tex]
=\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i\mathbf{p}\cdot(\mathbf{x}_{0}-\mathbf{x})}\widetilde{f}(\mathbf{p})
[/tex]
[tex]
=f(\mathbf{p})
[/tex]
[tex]
=e^{-i(\mathbf{x}-\mathbf{x}_{0})^{2}/2m)t}
[/tex]