- #1
maximus123
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Hello,
I have this matrix element
[itex]\langle n', l', m'| x |n, l, m \rangle[/itex]
where [itex]x=rsin\theta cos\phi[/itex] in spherical polars
Which I write explicitly as:
[itex]=\int \int \int R_{n', l'} R_{n, l} A_{l', m'} A_{l, m} P^{m'}_{l'} P^{m}_{l} e^{im\phi} e^{-im\phi} rsin\theta cos\phi r^{2} sin \theta d \theta d \phi dr[/itex][itex]=A_{l', m'} A_{l, m} \int dr R_{n', l'} r^{3} R_{n, l} \bullet \int d \theta sin^{2} \theta P^{m'}_{l'} P^{m}_{l} \bullet \int^{2\pi}_{0} e^{i\phi(m-m')} cos \phi d \phi[/itex]
The question just wants me to do the integral for the [itex]d \phi[/itex] bit then show that it is only non zero for [itex]m=m' \pm 1[/itex] hence [itex](m-m')= \pm 1[/itex]
So I let [itex]i(m-m')=a[/itex]
[itex]I=\int^{2\pi}_{0} e^{a\phi} cos \phi[/itex]
[itex]u=e^{a\phi}[/itex]
[itex]du=a e^{a\phi}[/itex]
[itex]dv=cos \phi[/itex]
[itex]v=sin \phi[/itex]
[itex]I=sin \phi e^{a\phi}-a \int^{2\pi}_{0} sin \phi e^{a\phi}[/itex]
for the integral here I do the same again
[itex]u=e^{a\phi}[/itex]
[itex]du=a e^{a\phi}[/itex]
[itex]dv=sin \phi[/itex]
[itex]v=-cos \phi[/itex]
[itex]I_{2}=-cos \phi e^{a\phi} +a \int^{2\pi}_{0} cos \phi e^{a\phi}[/itex]
So in total
[itex]\int^{2\pi}_{0} e^{a\phi} cos \phi=sin \phi e^{a\phi}+acos \phi e^{a\phi}-a^{2}\int^{2\pi}_{0} e^{a\phi} cos\phi[/itex]
Then I take the integral on the RHS over to the left, factor out the integral and divide through by [itex]1+a^{2}[/itex] giving:
[itex]\int^{2\pi}_{0} e^{a\phi}cos \phi=\frac{sin\phi e^{a\phi}+acos\phi e^{a\phi}}{1+a^{2}}[/itex]
[itex]= [\frac{sin\phi e^{i(m-m')\phi}+i(m-m')cos\phi e^{i(m-m')\phi}}{1-(m-m')^{2}}]^{2\pi}_{0}[/itex]
Is this integration ok? I kind of made it up as I went along as I wasn't sure how to integrate this function.
Assuming it is could someone explain to me how it follows that it is only non zero if [itex]m=m' \pm 1[/itex]
I have this matrix element
[itex]\langle n', l', m'| x |n, l, m \rangle[/itex]
where [itex]x=rsin\theta cos\phi[/itex] in spherical polars
Which I write explicitly as:
[itex]=\int \int \int R_{n', l'} R_{n, l} A_{l', m'} A_{l, m} P^{m'}_{l'} P^{m}_{l} e^{im\phi} e^{-im\phi} rsin\theta cos\phi r^{2} sin \theta d \theta d \phi dr[/itex][itex]=A_{l', m'} A_{l, m} \int dr R_{n', l'} r^{3} R_{n, l} \bullet \int d \theta sin^{2} \theta P^{m'}_{l'} P^{m}_{l} \bullet \int^{2\pi}_{0} e^{i\phi(m-m')} cos \phi d \phi[/itex]
The question just wants me to do the integral for the [itex]d \phi[/itex] bit then show that it is only non zero for [itex]m=m' \pm 1[/itex] hence [itex](m-m')= \pm 1[/itex]
So I let [itex]i(m-m')=a[/itex]
[itex]I=\int^{2\pi}_{0} e^{a\phi} cos \phi[/itex]
[itex]u=e^{a\phi}[/itex]
[itex]du=a e^{a\phi}[/itex]
[itex]dv=cos \phi[/itex]
[itex]v=sin \phi[/itex]
[itex]I=sin \phi e^{a\phi}-a \int^{2\pi}_{0} sin \phi e^{a\phi}[/itex]
for the integral here I do the same again
[itex]u=e^{a\phi}[/itex]
[itex]du=a e^{a\phi}[/itex]
[itex]dv=sin \phi[/itex]
[itex]v=-cos \phi[/itex]
[itex]I_{2}=-cos \phi e^{a\phi} +a \int^{2\pi}_{0} cos \phi e^{a\phi}[/itex]
So in total
[itex]\int^{2\pi}_{0} e^{a\phi} cos \phi=sin \phi e^{a\phi}+acos \phi e^{a\phi}-a^{2}\int^{2\pi}_{0} e^{a\phi} cos\phi[/itex]
Then I take the integral on the RHS over to the left, factor out the integral and divide through by [itex]1+a^{2}[/itex] giving:
[itex]\int^{2\pi}_{0} e^{a\phi}cos \phi=\frac{sin\phi e^{a\phi}+acos\phi e^{a\phi}}{1+a^{2}}[/itex]
[itex]= [\frac{sin\phi e^{i(m-m')\phi}+i(m-m')cos\phi e^{i(m-m')\phi}}{1-(m-m')^{2}}]^{2\pi}_{0}[/itex]
Is this integration ok? I kind of made it up as I went along as I wasn't sure how to integrate this function.
Assuming it is could someone explain to me how it follows that it is only non zero if [itex]m=m' \pm 1[/itex]
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