- #1
flyusx
- 63
- 10
- 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.
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.
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