# Selection rule for electric dipole transitions in x direction

1. Dec 4, 2013

### maximus123

Hello,
I have this matrix element

$\langle n', l', m'| x |n, l, m \rangle$

where $x=rsin\theta cos\phi$ in spherical polars

Which I write explicitly as:

$=\int \int \int R_{n', l'} R_{n, l} A_{l', m'} A_{l, m} P^{m'}_{l'} P^{m}_{l} e^{im\phi} e^{-im\phi} rsin\theta cos\phi r^{2} sin \theta d \theta d \phi dr$

$=A_{l', m'} A_{l, m} \int dr R_{n', l'} r^{3} R_{n, l} \bullet \int d \theta sin^{2} \theta P^{m'}_{l'} P^{m}_{l} \bullet \int^{2\pi}_{0} e^{i\phi(m-m')} cos \phi d \phi$

The question just wants me to do the integral for the $d \phi$ bit then show that it is only non zero for $m=m' \pm 1$ hence $(m-m')= \pm 1$

So I let $i(m-m')=a$

$I=\int^{2\pi}_{0} e^{a\phi} cos \phi$

$u=e^{a\phi}$
$du=a e^{a\phi}$
$dv=cos \phi$
$v=sin \phi$

$I=sin \phi e^{a\phi}-a \int^{2\pi}_{0} sin \phi e^{a\phi}$

for the integral here I do the same again

$u=e^{a\phi}$
$du=a e^{a\phi}$
$dv=sin \phi$
$v=-cos \phi$

$I_{2}=-cos \phi e^{a\phi} +a \int^{2\pi}_{0} cos \phi e^{a\phi}$

So in total

$\int^{2\pi}_{0} e^{a\phi} cos \phi=sin \phi e^{a\phi}+acos \phi e^{a\phi}-a^{2}\int^{2\pi}_{0} e^{a\phi} cos\phi$

Then I take the integral on the RHS over to the left, factor out the integral and divide through by $1+a^{2}$ giving:

$\int^{2\pi}_{0} e^{a\phi}cos \phi=\frac{sin\phi e^{a\phi}+acos\phi e^{a\phi}}{1+a^{2}}$

$= [\frac{sin\phi e^{i(m-m')\phi}+i(m-m')cos\phi e^{i(m-m')\phi}}{1-(m-m')^{2}}]^{2\pi}_{0}$

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 $m=m' \pm 1$

Last edited: Dec 4, 2013
2. Dec 4, 2013

### TSny

OK, except when you substituted for $a$ in the denominator in the last step.

3. Dec 4, 2013

### maximus123

Thanks, I've fixed that in the original post

4. Dec 4, 2013

### TSny

So, what do you get when m ≠ m' ±1 and what do you get when m = m' ±1?

5. Dec 4, 2013

### maximus123

I'm having trouble getting passed the fact that for $m=m'\pm 1$ the denominator will be zero
e.g for $m=m'+1$

$= [\frac{sin\phi e^{i\phi}+icos\phi e^{i\phi}}{1-1}]^{2\pi}_{0}$

6. Dec 4, 2013

### TSny

Right. For this case you will need to go back and substitute $m=m'\pm 1$ before evaluating the integral.

7. Dec 4, 2013

### maximus123

OK, I think I get this now, so for example for $(m-m')=+1$

$\int^{2\pi}_{0} e^{i(m-m')\phi}cos\phi=\int^{2\pi}_{0} e^{i \phi}cos\phi$

$=\int^{2\pi}_{0} e^{i \phi} \frac{e^{i\phi}+e^{-i\phi}}{2}$

$=\frac{1}{2}\int^{2\pi}_{0}e^{2i\phi}+1$

$=\frac{1}{2}[\frac{e^{2i\phi}}{2i}+\phi]^{2\pi}_{0}$

$=\frac{1}{2}[\frac{cos(2\phi)+isin(2\phi)}{2i}+\phi]^{2\pi}_{0}$

$=\frac{1}{2}[\frac{1}{2i}+2\pi-\frac{1}{2i}-0]$

$=\pi$

Similarly for the case when $(m-m')=-1$

$\int^{2\pi}_{0} e^{i(m-m')\phi}cos\phi=\int^{2\pi}_{0} e^{-i \phi}cos\phi$

$=\int^{2\pi}_{0} e^{-i \phi} \frac{e^{i\phi}+e^{-i\phi}}{2}$

$=\frac{1}{2}\int^{2\pi}_{0}e^{-2i\phi}+1$

$=\frac{1}{2}[-\frac{e^{-2i\phi}}{2i}+\phi]^{2\pi}_{0}$

$=\frac{1}{2}[\phi -\frac{cos(2\phi)-isin(2\phi)}{2i}]^{2\pi}_{0}$

$=\frac{1}{2}[2\pi-\frac{1}{2i}+\frac{1}{2i}]$

$=\pi$

When $m=m'\pm 1$ one of the exponential terms multiplies with another to make 1. This means after integration there is a non trigonometric term which will become zero when zero is substituted in to it. However when $m\neq m'\pm 1$ both terms have an exponential in after the integral is evaluated so the two terms you get after subbing in the $2\pi$ will equal the two terms you get after subbing in the zero. Does that sound right? Thanks for the help.

8. Dec 4, 2013

Sounds good.