Recall from last time that we worked out that the ket pointing in the
\(\theta\) and \(\phi\) direction is given by the rotation
of the state from the north-pole state via \[\left|\theta,\phi\right\rangle=e^{-i\phi\frac{\hat{L}_z}{\hbar}}e^{-i\theta\frac{\hat{L}_y}{\hbar}}\left|\theta{=}0,\phi{=}0\right\rangle.\]
The spherical harmonic is then defined to be the overlap of this angular
position state \(\left\langle\theta,\phi\right|\) with the
angular momentum state that has definite total and \(z\)-component of angular momentum \(\left|lm\right\rangle\). We have, \[\begin{aligned}
Y_{lm}(\theta,\phi)=& \
\left\langle\theta,\phi\middle|lm\right\rangle \\
=&\langle{\theta{=}0,\phi{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}e^{i\phi\frac{\hat{L}_z}{\hbar}}}|{lm}\rangle.
\end{aligned}\] But \(\hat{L}_z\left|lm\right\rangle=m\hbar\left|lm\right\rangle\),
so \[Y_{lm}(\theta,\phi)=\langle{\theta{=}0,\phi{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}e^{im\phi}}|{lm}\rangle.\]
Inserting the identity operator within the angular momentum multiplet
with total angular momentum \(l\) just
to the right of \(\left\langle\theta{=}0,\phi{=}0\right|\)
gives us \[Y_{lm}(\theta,\phi)=e^{im\phi}\sum_{m'}\left\langle\theta{=}0,\phi{=}0\middle|lm'\right\rangle\langle{lm'}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle.\]
The matrix element \(d^{(l)}_{mm'}\equiv\langle{lm'}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle\)
is called the rotation matrix. It is a continuous matrix
representation of the rotation group with \((2l+1)\times(2l+1)\) matrices. Note that we
computed \(d^{(\frac{1}{2})}_{mm'}\) when we
worked with the Pauli matrices in Lecture 1.
Fortunately, we do not need the whole matrix.
Note that \(\left\langle\theta{=}0,\phi\right|=\left\langle\theta{=}0,\phi{=}0\right|e^{i\phi\frac{\hat{L}_z}{\hbar}}=\left\langle\theta{=}0,\phi{=}0\right|\).
Think about what this is physically: It is a state pointing along the
north pole. If I rotate about the north pole, I do nothing to the
state.
In other words, the state for \(\theta=0\) is the same for all \(\phi\). But, we already saw that \[\begin{aligned}
\left\langle\theta{=}0,\phi\middle|lm\right\rangle=& \
\langle{\theta{=}0,\phi{=}0}|{e^{i\phi\frac{\hat{L}_z}{\hbar}}}|{lm}\rangle
\\
=&
e^{im\phi}\left\langle\theta{=}0,\phi{=}0\middle|lm\right\rangle.
\end{aligned}\] Hence, \[\left\langle\theta{=}0,\phi{=}0\middle|lm\right\rangle=\begin{cases}0
& m\ne 0 \\
\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle &
m{=}0\end{cases}.\] Note that this also implies that \(l\) is an integer since \(m{=}0\) only occurs when \(l\in\mathbb{Z}\). Hence, we have \[Y_{lm}(\theta,\phi)=e^{im\phi}\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle\langle{l,m{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle,\]
that is, only the \(m{=}0\) term
survives in the summation.
To calculate \(\langle{l,m{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle\),
we use exponential disentangling. We have \[\langle{l,m{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle=\langle{l,m{=}0}|e^{-\tan\left
(\frac{\theta}{2}\right )\frac{\hat{L}_-}{\hbar}}e^{\ln\cos^2\left
(\frac{\theta}{2}\right )\frac{\hat{L}_z}{\hbar}}e^{\tan\left
(\frac{\theta}{2}\right )\frac{\hat{L}_+}{\hbar}}|{lm}\rangle.\]
Now, recall that \[\begin{aligned}
\frac{\hat{L}_+}{\hbar}\left|lm\right\rangle=& \
\sqrt{(l-m)(l+m+1)}\left|l,m+1\right\rangle \\
\frac{\hat{L}_-}{\hbar}\left|lm\right\rangle=& \
\sqrt{(l+m)(l-m+1)}\left|l,m-1\right\rangle.
\end{aligned}\] We can expand each exponential factor in a power
series. \[e^{\tan{\frac{\theta}{2}}\frac{\hat{L}_+}{\hbar}}=\sum_{n=0}^\infty\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\frac{\hat{L}_+}{\hbar}\right)^n.\]
So we have that \[e^{\tan{\frac{\theta}{2}}\frac{\hat{L}_+}{\hbar}}\left|lm\right\rangle=\sum_{n=0}^\infty\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\frac{\hat{L}_+}{\hbar}\right)^n\left|lm\right\rangle.\]
But \((\hat{L}_+)^n\left|lm\right\rangle=0\) if
\(n>l-m\) since \(\hat{L}_+\left|l,m=l\right\rangle=0\). So,
the summation truncates after a finite number of terms, and we have
\[\begin{aligned}
e^{\tan{\left (\frac{\theta}{2}\right
)}\frac{\hat{L}_+}{\hbar}}\left|lm\right\rangle=& \
\sum_{n=0}^{l-m}\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\frac{\hat{L}_+}{\hbar}\right)^n\left|lm\right\rangle
\\
=& \
\sum_{n=0}^{l-m}\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\prod_{r=1}^{n}\sqrt{(l-m-r+1)(l+m+r)}\right)\left|l,m+n\right\rangle.
\end{aligned}\] Now, we operate the next factor \(e^{\ln\cos^2\left (\frac{\theta}{2}\right
)\frac{\hat{L}_z}{\hbar}}\) onto this. It is easy since \(\frac{\hat{L}_z}{\hbar}\left|lm\right\rangle=m\left|lm\right\rangle\).
\[e^{\ln\cos^2{\left (\frac{\theta}{2}\right
)\frac{\hat{L}_z}{\hbar}}} e^{\tan{\left (\frac{\theta}{2}\right
)}\frac{\hat{L}_+}{\hbar}}\left|lm\right\rangle=\sum_{n=0}^{l-m}\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\prod_{r=1}^{n}\sqrt{(l-m-r+1)(l+m+r)}\right)\left(\cos{\frac{\theta}{2}}\right)^{2m+2n}\left|l,m+n\right\rangle.\]
Finally, we operate \[e^{-\tan\left
(\frac{\theta}{2}\right
)\frac{\hat{L}_-}{\hbar}}=\sum_{n'=0}^\infty\frac{(-\tan{\frac{\theta}{2}})^{n'}}{n'!}\left(\frac{\hat{L}_-}{\hbar}\right)^{n'}\]
onto the state. But since we multiply by \(\left\langle l,m{=}0\right|\) on the left
in the end, we only need to include the term where \(n'=n+m\). Since \(n'\ge 0\), \(n\) must be at least \(-m\) for a nonzero result. So \(\langle{l,m{=}0}|{e^{i\theta\frac{\hat{L}_y}{\hbar}}}|{lm}\rangle\)
becomes \[\begin{gathered}
\sum_{n=\max(0,-m)}^{l-m}=\frac{(-\tan{\frac{\theta}{2}})^{m+n}}{(m+n)!}\frac{(\tan{\frac{\theta}{2}})^n}{n!}\left(\prod_{r=1}^{n}\sqrt{(l-m-r+1)(l+m+r)}\right)\left(\cos{\frac{\theta}{2}}\right)^{2m+2n}
\\ \times \prod_{s=1}^{m+n}\sqrt{(l+m+n-s+1)(l-m-n+s)}
\end{gathered}\] because \(\left\langle
l,m{=}0\middle|l,m{=}0\right\rangle=1\).
Now, we need to simplify. Note that \[\begin{aligned}
\left(-\tan{\frac{\theta}{2}}\right)^{m+n}\left(\tan{\frac{\theta}{2}}\right)^n\left(\cos{\frac{\theta}{2}}\right)^{2m+2n}=&
\
(-1)^{m+n}\left(\sin{\frac{\theta}{2}}\right)^{2n+m}\left(\cos{\frac{\theta}{2}}\right)^m
\\
=& \
(-1)^{m+n}\left(\frac{1}{2}\right)^m(\sin{\theta})^m\left(\sin^2{\frac{\theta}{2}}\right)^n
\\
=& \
\left(\frac{1}{2}\right)^{m+n}(\sin{\theta})^m(1-\cos{\theta})^n,
\end{aligned}\] where we made use of the half-angle identities
\(\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}=\frac{\sin{\theta}}{2}\)
and \(\cos{\theta}=1-2\sin^2{\frac{\theta}{2}}\).
The two product factors become \[\begin{gathered}
\prod_{r=1}^n\sqrt{(l-m-r+1)(l+m+r)}\prod_{s=1}^{m+n}\sqrt{(l+m+n-s+1)(l-m-n+s)}\\=
\sqrt{\frac{(l-m)!}{(l-m-n)!}\frac{(l+m+n)!}{(l+m)!}\frac{(l+m+n)!}{l!}\frac{l!}{(l-m-n)!}}.
\end{gathered}\] Simplifying, we find \[=\frac{(l+m+n)!}{(l-m-n)!}\sqrt{\frac{(l-m)!}{(l+m)!}}.\]
So the spherical harmonic becomes (deep breath) \[\begin{gathered}
Y_{lm}(\theta,\phi)=\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle
e^{im\phi}\sum_{n=\max(0,-m)}^{l-m}\left(\frac{1}{2}\right)^{m+n}(\sin{\theta})^m(1-\cos{\theta})^n
\\
\times\frac{(l+m+n)!}{(l-m-n)!}\sqrt{\frac{(l-m)!}{(l+m)!}}\frac{1}{n!}\frac{1}{(m+n)!}.
\end{gathered}\] Pulling out the \(n\)-independent factors, \[\begin{gathered}
Y_{lm}(\theta,\phi)=\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle
e^{im\phi}(\sin{\theta})^m\left(-\frac{1}{2}\right)^m\sqrt{\frac{(l-m)!}{(l+m)!}}\\
\times\sum_{n=\max(0,-m)}^{l-m}\left(-\frac{1}{2}\right)^n(1-\cos{\theta})^n\frac{(l+m+n)!}{(l-m-n)!}\frac{1}{n!(m+n)!}
\end{gathered}\] In essence, we are finished. But it is customary
to re-express this in terms of functions defined by dead French and
German mathematicians. To do this, recall the associated Legendre
polynomial, which is given by \[P_l^m(\cos{\theta})=\frac{1}{2^m}(\sin{\theta})^m\sum_{n=0}^{l-m}(-1)^n\frac{(l+m+n)!}{(l-m-n)!(m+n)!n!}\left(\frac{1-\cos{\theta}}{2}\right)^n.\]
In spite of its name, it is not a polynomial of its argument. So, for
\(m\ge 0\), we have \[\boxed{Y_{lm}(\theta,\phi)=e^{im\phi}(-1)^m\sqrt{\frac{(l-m)!}{(l+m)!}}P_l^m(\cos{\theta})\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle}\]
In my personal opinion, this is a much clearer way to find the spherical
harmonics than using differential equations. It is a classic "French
cooking" exercise. Every step is rather simple, but there are many of
them. One needs to work carefully and cautiously to get to the right
final answer.
What about \(m<0\)? We use the fact
that \(Y_{l,-|m|}(\theta,\phi)=(-1)^{|m|}(Y_{l,|m|}(\theta,\phi))^*\)
to find it. One can also compute it directly using a similar scheme to
what we just did.
We still need \(\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle\) which comes from normalization. One can do this most easily for \(m=l\): \[P_l^l(\cos{\theta})=\frac{1}{2^l}(\sin{\theta})^l\frac{(2l)!}{l!}\] So, \[Y_{ll}(\theta,\phi)=\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle e^{il\phi}\left(-\frac{1}{2}\right)^l(\sin{\theta})^l\frac{\sqrt{(2l)!}}{l!}.\] Integrate to find the normalization: \[\begin{aligned} 1=& \ \int_0^\pi d\theta \ (\sin{\theta})^{2l+1}\frac{1}{4^l}\frac{(2l)!}{(l!)^2}|\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle|^2\int_0^{2\pi}d\phi \\ =& \ \frac{2\pi}{4^l}\frac{(2l)!}{(l!)^2}|\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle|^2\int_0^\pi d\theta \ (\sin{\theta})^{2l+1}. \end{aligned}\] By Wolfram Alpha, I find \[\int_0^\pi d\theta \ (\sin{\theta})^{2l+1}=\frac{\sqrt{\pi} \ \Gamma(l+1)}{\Gamma\left(l+\frac{3}{2}\right)}=\frac{2^{2l+1}(l!)^2}{(2l+1)!}\] Therefore, \[1=|\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle|^2\frac{4\pi}{l+1}\implies\left\langle\theta{=}0,\phi{=}0\middle|l,m{=}0\right\rangle=\sqrt{\frac{2l+1}{4\pi}}.\] So, in conclusion, for \(m\ge 0\) \[\boxed{Y_{lm}(\theta,\phi)=\sqrt{\frac{(2l+1)(l-m)!}{4\pi(l+m)!}}(-1)^mP_l^m(\cos{\theta})e^{im\phi}}\] and for \(m<0\), \[\boxed{Y_{l,-|m|}(\theta,\phi)=(-1)^{|m|}Y^*_{l,|m|}(\theta,\phi)}.\]
I encourage you to think about the other way you solved this problem with differential equations and think about which method felt more concrete and clear to you about what you calculated.