Discrepancies In Clebsch–Gordan Calculations (Dipole Transitions)

  • Thread starter Thread starter flyusx
  • Start date Start date
flyusx
Messages
63
Reaction score
9
Homework Statement
Calculate the dipole transitions $$\langle n',l',m'|\textbf{r}|n,l,m\rangle$$
Relevant Equations
$$\langle l',m'|Y_{1q}|l,m\rangle=\sqrt{\frac{3(2l+1)}{4\pi(2l'+1)}}\langle l,1;0,0|l',0\rangle\langle l,1;m,q|l',m'\rangle$$
A table of Clebsch–Gordan relations is hyperlinked below (table B.2)
This is a solved problem from Zettili Chapter 7 Problem 7.8(b) but I am having troubles reproducing some of the quantities he produces. Zettili approaches this problem by describing ##\textbf{r}## using a spherical basis: a product between a radial and angular part $$r_{q}=\sqrt{\frac{4\pi}{3}}rY_{1q}(\theta,\phi)\quad\quad q=1,0,-1$$. The dipole term can be rewritten as $$\sqrt{\frac{4\pi}{3}}\langle n',l'|r_{q}|n,l\rangle\langle l',m'|Y_{1q}|l,m\rangle$$ where the radial part can be calculated from an integral (I have no problem with this) and the angular part can be resolved as follows. Since ##m'=m+q## then ##m'=m,m-1,m+1## and since ##\vert l_{1}-l_{2}\vert\leq l'\leq l_{1}+l_{2}## then ##l'=l,l-1,l+1##. The parity selection rule erases all ##l'=l## so there are just six angular terms to focus on. $$l'=l+1,m'=m+1$$ $$l'=l-1,m'=m+1$$ $$l'=l+1,m'=m$$ $$l'=l-1,m'=m$$ $$l'=l+1,m'=m-1$$ $$l'=l-1,m'=m-1$$
He refers to the 'relevant Clebsch–Gordan coefficients from standard tables' which from an internet search, I have found Table B.2 of this pdf. It concerns Clebsch–Gordan coefficients for ##j_{2}=1## and ##m_{2}=1,0,-1##.

Out of the six terms, my work differs from Zettili's solution on two of them. For instance, I take
$$\langle l-1,m+1|Y_{11}|l,m\rangle=\sqrt{\frac{3(2l+1)}{4\pi(2l'+1)}}\langle l,1;0,0|l-1,0\rangle\langle l,1;m,1|l-1,m+1\rangle$$
The square root coefficient is $$\sqrt{\frac{3(2l+1)}{4\pi(2(l-1)+1)}}=\sqrt{\frac{3(2l+1)}{4\pi(2l-1)}}$$
The centre term (the first CG coefficient) is calculated in Table B.2 within the cell bordered by ##m_{2}=0## and ##j-j_{1}=-1## which states
$$\langle j_1,1;m-m_2,m_2|j,m\rangle=-\sqrt{\frac{(j_1-m)(j_1+m)}{j_1(2j+1)}}$$
Where plugging in ##j_1=l##, ##m=0## and ##j=l-1## gives $$\langle l,1;0,0|l-1,0\rangle=\langle l,1;0,0|l-1,0\rangle=-\sqrt{\frac{l^2}{l(2l+1)}}$$
The last term is found in Table B.2 cell ##j-j_1=-1## and ##m_2=1## where $$\langle j_1,1;m-m_2,m_2|j,m\rangle=\sqrt{\frac{(j_1-m+1)(j_1-m)}{2j_1(2j_1+1)}}$$
So the associated CG term is
$$\langle l,1;m,1|l-1,m+1\rangle=\sqrt{\frac{(l-m)(l-m-1)}{2l(2l+1)}}$$
Therefore $$\langle l-1,m+1|Y_{11}|l,m\rangle=-\sqrt{\frac{3(2l+1)}{4\pi(2l-1)}}\sqrt{\frac{l^2}{l(2l+1)}}\sqrt{\frac{(l-m-1)(l-m)}{2l(2l+1)}}=-\sqrt{\frac{3(l-m)(l-m-1)}{8\pi(2l+1)(2l-1)}}$$
However, Zettili replaces ##2l-1## in the denominator with ##2l+3## and loses the negative sign.

I have a similar issue for ##\langle l-1,m-1|Y_{1-1}|l,m\rangle## where ##l'=l-1## and ##m'=m-1## where ##q=-1##. The square root is $$\sqrt{\frac{3(2l+1)}{4\pi(2l'+1)}}=\sqrt{\frac{3(2l+1)}{4\pi(2l-1)}}$$
The first CG term is in cell ##j-j_1=-1## and ##m_2=0## which gives $$\langle l,1;0,0|l-1,0\rangle=-\sqrt{\frac{l^{2}}{l(2l+1)}}$$
The other CG term is in cell ##j-j_1=-1## and ##m_2=-1## which gives $$\langle l,1;m,-1|l-1,m-1\rangle=\sqrt{\frac{(l+m)(l+m-1)}{2l(2l+1)}}$$
Multiplying these terms together, I get $$\langle l-1,m-1|Y_{1-1}|l,m\rangle=-\sqrt{\frac{3(l+m)(l+m-1)}{8\pi(2l-1)(2l+1)}}$$
Which differs from Zettili's solution by exactly a negative sign.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top