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[tex] \langle \mathbf{x} \vert \dfrac{1}{E- H_{0} \pm \varepsilon} \vert \mathbf{x'} \rangle = [/tex]
[tex]\int d^{3}p' \int d^{3}p'' \langle \mathbf{x} \vert \mathbf{p'} \rangle \mathbf{p'} \vert \dfrac{1}{E- H_{0} \pm i\varepsilon} \vert \mathbf{p''} \rangle \langle \mathbf{p''} \vert \mathbf{x'} \rangle [/tex]
Operator, acts to the left in this case.
[tex]H_0 = \dfrac{\mathbf{p}}{2m}[/tex]
Evaluating the parts in the integral:
[tex]\langle \mathbf{p'} \vert \dfrac{1}{E- H_{0} \pm \varepsilon} \vert \mathbf{p''} \rangle = [/tex]
[tex]\dfrac{\delta ^{(3)} (\mathbf{p'} - \mathbf{p''} )}{E- \frac{\mathbf{p'}}{2m} \pm i\varepsilon} [/tex]
[tex]\langle \mathbf{x} \vert \mathbf{p'} \rangle = \dfrac{e^{i\mathbf{x}\mathbf{p'}}}{(2 \pi \hbar)^{3/2}} [/tex]
[tex]\langle \mathbf{p''} \vert \mathbf{x'} \rangle = \dfrac{e^{-i\mathbf{x'}\mathbf{p''}}}{(2 \pi \hbar)^{3/2}} [/tex] (1)
Now this last line is wrong (?), it should be:
[tex]\langle \mathbf{p''} \vert \mathbf{x'} \rangle = \dfrac{e^{-i\mathbf{x'}\mathbf{p'}}}{(2 \pi \hbar)^{3/2}} [/tex] (2)
According to Sakurai p381, eq (7.1.14)
the integral should become this one when integrating with respect to p''
[tex]\int d^{3}p' \dfrac{e^{i \mathbf{p'}( \mathbf{x}- \mathbf{x'})}}{E- \frac{ \mathbf{p'}}{2m} \pm i\varepsilon}[/tex] (3)
If would have continue with my expression for <p''|x'> (1)
[tex] \int d^{3}p'' \delta ^{(3)}( \mathbf{p'} - \mathbf{p''}) e^{-i\mathbf{x'}\mathbf{p''}}
= e^{-i \mathbf{x'} \mathbf{p'}} [/tex] (4)
Which yields the same result?
Can someone please give some Ideas on this one.
I am unsure if my expression for <p''|x'> is right, and if it is right, if I get the final result (3), and i (4) is right too.
[tex]\int d^{3}p' \int d^{3}p'' \langle \mathbf{x} \vert \mathbf{p'} \rangle \mathbf{p'} \vert \dfrac{1}{E- H_{0} \pm i\varepsilon} \vert \mathbf{p''} \rangle \langle \mathbf{p''} \vert \mathbf{x'} \rangle [/tex]
Operator, acts to the left in this case.
[tex]H_0 = \dfrac{\mathbf{p}}{2m}[/tex]
Evaluating the parts in the integral:
[tex]\langle \mathbf{p'} \vert \dfrac{1}{E- H_{0} \pm \varepsilon} \vert \mathbf{p''} \rangle = [/tex]
[tex]\dfrac{\delta ^{(3)} (\mathbf{p'} - \mathbf{p''} )}{E- \frac{\mathbf{p'}}{2m} \pm i\varepsilon} [/tex]
[tex]\langle \mathbf{x} \vert \mathbf{p'} \rangle = \dfrac{e^{i\mathbf{x}\mathbf{p'}}}{(2 \pi \hbar)^{3/2}} [/tex]
[tex]\langle \mathbf{p''} \vert \mathbf{x'} \rangle = \dfrac{e^{-i\mathbf{x'}\mathbf{p''}}}{(2 \pi \hbar)^{3/2}} [/tex] (1)
Now this last line is wrong (?), it should be:
[tex]\langle \mathbf{p''} \vert \mathbf{x'} \rangle = \dfrac{e^{-i\mathbf{x'}\mathbf{p'}}}{(2 \pi \hbar)^{3/2}} [/tex] (2)
According to Sakurai p381, eq (7.1.14)
the integral should become this one when integrating with respect to p''
[tex]\int d^{3}p' \dfrac{e^{i \mathbf{p'}( \mathbf{x}- \mathbf{x'})}}{E- \frac{ \mathbf{p'}}{2m} \pm i\varepsilon}[/tex] (3)
If would have continue with my expression for <p''|x'> (1)
[tex] \int d^{3}p'' \delta ^{(3)}( \mathbf{p'} - \mathbf{p''}) e^{-i\mathbf{x'}\mathbf{p''}}
= e^{-i \mathbf{x'} \mathbf{p'}} [/tex] (4)
Which yields the same result?
Can someone please give some Ideas on this one.
I am unsure if my expression for <p''|x'> is right, and if it is right, if I get the final result (3), and i (4) is right too.
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