## Dipole and Quadropole electrostatic potential moments and coefficents

<a href=”https://peeterjoot.com/archives/math2016//momentCoeffiecients.pdf”>[Click here for a PDF of this post with nicer formatting]</a>

In class Thursday we calculated the $$q_{1,1}$$ coefficient of the electrostatic moment, as covered in [1] chapter 4. Let’s verify the rest, as well as the tensor sum formula for the quadropole moment, and the spherical harmonic sum that yields the dipole moment potential.

The quadropole term of the potential was stated to be

\label{eqn:momentCoeffiecients:120}
\inv{4 \pi \epsilon_0} \frac{4 \pi}{5 r^3} \sum_{m=-2}^2 \int (r’)^2 \rho(\Bx’) Y_{lm}^\conj(\theta’, \phi’) Y_{lm}(\theta, \phi)
=
\inv{2} \sum_{ij} Q_{ij} \frac{x_i x_j}{r^5},

where

\label{eqn:momentCoeffiecients:140}
Q_{i,j} = \int \lr{ 3 x_i’ x_j’ – \delta_{ij} (r’)^2 } \rho(\Bx’) d^3 x’.

Let’s verify this. First note that

\label{eqn:momentCoeffiecients:160}
Y_{l,m} = \sqrt{\frac{2 l + 1}{4 \pi} \frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta) e^{i m \phi},

and
\label{eqn:momentCoeffiecients:180}
P_l^{-m}(x) =
(-1)^m \frac{(l-m)!}{(l+m)!} P_l^m(x),

so
\label{eqn:momentCoeffiecients:200}
\begin{aligned}
Y_{l,-m}
&= \sqrt{\frac{2 l + 1}{4 \pi} \frac{(l+m)!}{(l-m)!} }
P_l^{-m}(\cos\theta)
e^{-i m \phi} \\
&=
(-1)^m
\sqrt{\frac{2 l + 1}{4 \pi} \frac{(l-m)!}{(l+m)!} }
P_l^m(x)
e^{-i m \phi} \\
&=
(-1)^m Y_{l,m}^\conj.
\end{aligned}

That means

\label{eqn:momentCoeffiecients:220}
\begin{aligned}
q_{l,-m}
&=
\int (r’)^l \rho(\Bx’)
Y^\conj_{l,-m}(\theta’, \phi’)
d^3 x’ \\
&=
(-1)^m
\int (r’)^l \rho(\Bx’)
Y_{l,m}(\theta’, \phi’)
d^3 x’ \\
&=
(-1)^m q_{lm}^\conj.
\end{aligned}

In particular, for $$m \ne 0$$

\label{eqn:momentCoeffiecients:320}
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, -m}^\conj (\theta’, \phi’) r^l Y_{l, -m}(\theta, \phi)
=
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, m} (\theta’, \phi’) r^l Y_{l, m}^\conj(\theta, \phi) ,

or
\label{eqn:momentCoeffiecients:340}
(r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi)
+ (r’)^l Y_{l, -m}^\conj (\theta’, \phi’) r^l Y_{l, -m}(\theta, \phi)
=
2 \textrm{Re} \lr{ (r’)^l Y_{l, m}^\conj (\theta’, \phi’) r^l Y_{l, m}(\theta, \phi) }.

To verify the quadropole expansion formula in a compact way it is helpful to compute some intermediate results.

\label{eqn:momentCoeffiecients:360}
\begin{aligned}
r Y_{1, 1}
&= -r \sqrt{\frac{3}{8 \pi}} \sin\theta e^{i\phi} \\
&= -\sqrt{\frac{3}{8 \pi}} (x + i y),
\end{aligned}

\label{eqn:momentCoeffiecients:380}
\begin{aligned}
r Y_{1, 0}
&= r \sqrt{\frac{3}{4 \pi}} \cos\theta \\
&= \sqrt{\frac{3}{4 \pi}} z,
\end{aligned}

\label{eqn:momentCoeffiecients:400}
\begin{aligned}
r^2 Y_{2, 2}
&= -r^2 \sqrt{\frac{15}{32 \pi}} \sin^2\theta e^{2 i\phi} \\
&= – \sqrt{\frac{15}{32 \pi}} (x + i y)^2,
\end{aligned}

\label{eqn:momentCoeffiecients:420}
\begin{aligned}
r^2 Y_{2, 1}
&= r^2 \sqrt{\frac{15}{8 \pi}} \sin\theta \cos\theta e^{i\phi} \\
&= \sqrt{\frac{15}{8 \pi}} z ( x + i y ),
\end{aligned}

\label{eqn:momentCoeffiecients:440}
\begin{aligned}
r^2 Y_{2, 0}
&= r^2 \sqrt{\frac{5}{16 \pi}} \lr{ 3 \cos^2\theta – 1 } \\
&= \sqrt{\frac{5}{16 \pi}} \lr{ 3 z^2 – r^2 }.
\end{aligned}

Given primed coordinates and integrating the conjugate of each of these with $$\rho(\Bx’) dV’$$, we obtain the $$q_{lm}$$ moment coeffients. Those are

\label{eqn:momentCoeffiecients:460}
q_{11}
= -\sqrt{\frac{3}{8 \pi}} \int d^3 x’ \rho(\Bx’) (x – i y),

\label{eqn:momentCoeffiecients:480}
q_{1, 0}
= \sqrt{\frac{3}{4 \pi}} \int d^3 x’ \rho(\Bx’) z’,

\label{eqn:momentCoeffiecients:500}
q_{2, 2}
= – \sqrt{\frac{15}{32 \pi}} \int d^3 x’ \rho(\Bx’) (x’ – i y’)^2,

\label{eqn:momentCoeffiecients:520}
q_{2, 1}
= \sqrt{\frac{15}{8 \pi}} \int d^3 x’ \rho(\Bx’) z’ ( x’ – i y’ ),

\label{eqn:momentCoeffiecients:540}
q_{2, 0}
= \sqrt{\frac{5}{16 \pi}} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 }.

For the potential we are interested in

\label{eqn:momentCoeffiecients:560}
\begin{aligned}
2 \textrm{Re} q_{11} r^2 Y_{11}(\theta, \phi)
&= 2 \frac{3}{8 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{ (x’ – i y’)( x + i y) } \\
&= \frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) \lr{ x x’ + y y’ },
\end{aligned}

\label{eqn:momentCoeffiecients:580}
q_{1, 0} r Y_{1,0}(\theta, \phi)
= \frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) z’ z,

\label{eqn:momentCoeffiecients:600}
\begin{aligned}
2 \textrm{Re} q_{22} r^2 Y_{22}(\theta, \phi)
&= 2 \frac{15}{32 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{
(x’ – i y’)^2
(x + i y)^2
} \\
&= \frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \textrm{Re} \lr{
((x’)^2 – 2 i x’ y’ -(y’)^2)
(x^2 + 2 i x y -y^2)
} \\
&= \frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
},
\end{aligned}

\label{eqn:momentCoeffiecients:620}
\begin{aligned}
2 \textrm{Re} q_{21} r^2 Y_{21}(\theta, \phi)
&= 2 \frac{15}{8 \pi} \int d^3 x’ \rho(\Bx’) z \textrm{Re} \lr{ ( x’ – i y’ ) (x + i y) } \\
&= \frac{15}{4 \pi} \int d^3 x’ \rho(\Bx’) z \lr{ x x’ + y y’ },
\end{aligned}

and
\label{eqn:momentCoeffiecients:640}
q_{2, 0} r^2 Y_{20}(\theta, \phi)
= \frac{5}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }.

The dipole term of the potential is

\label{eqn:momentCoeffiecients:660}
\begin{aligned}
\inv{ 4 \pi \epsilon_0 } \frac{4 \pi}{3 r^3}
\lr{
\frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) \lr{ x x’ + y y’ }
+
\frac{3}{4 \pi} \int d^3 x’ \rho(\Bx’) z’ z
} \\
&=
\inv{ 4 \pi \epsilon_0 r^3}
\Bx \cdot \int d^3 x’ \rho(\Bx’) \Bx’ \\
&=
\frac{\Bx \cdot \Bp}{ 4 \pi \epsilon_0 r^3},
\end{aligned}

as obtained directly when a strict dipole approximation was used.

Summing all the terms for the quadrople gives

\label{eqn:momentCoeffiecients:680}
\begin{aligned}
\inv{ 4 \pi \epsilon r^5 } \frac{ 4 \pi }{5}
\Bigl(
&\frac{15}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
&+
\frac{15}{4 \pi} \int d^3 x’ \rho(\Bx’) z z’ \lr{ x x’ + y y’ } \\
&+
\frac{5}{16 \pi} \int d^3 x’ \rho(\Bx’) \lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }
\Bigr) \\
=
\inv{ 4 \pi \epsilon r^5 }
\int d^3 x’ \rho(\Bx’)
\inv{4}
\Bigl(
&3
\lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
&+
12
z z’ \lr{ x x’ + y y’ } \\
&+
\lr{ 3 (z’)^2 – (r’)^2 } \lr{ 3 z^2 – r^2 }
\Bigr).
\end{aligned}

The portion in brackets is

\label{eqn:momentCoeffiecients:700}
\begin{aligned}
3
&\lr{
((x’)^2 -(y’)^2) (x^2 -y^2)
+ 4 x x’ y y’
} \\
+
12
& z z’ \lr{ x x’ + y y’ } \\
+
&\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2} \lr{ 2 z^2 – x^2 -y^2 } \\
=
x^2 &\lr{
3 (x’)^2 – 3(y’)^2

\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2}
} \\
+
y^2 &\lr{
-3 (x’)^2 + 3 (y’)^2

\lr{ 2 (z’)^2 – (x’)^2 – (y’)^2}
} \\
+
2 z^2 &\lr{
2 (z’)^2 – (x’)^2 – (y’)^2
} \\
+
&12{ x x’ y y’ + x x’ z z’ + y y’ z z’ } \\
=
2 x^2 &\lr{
2 (x’)^2 – (y’)^2 – (z’)^2
} \\
+
2 y^2 &\lr{
2 (y’)^2 – (x’)^2 – (z’)^2
} \\
+
2 z^2 &\lr{
2 (z’)^2 – (x’)^2 – (y’)^2
} \\
+
&12{ x x’ y y’ + x x’ z z’ + y y’ z z’ }.
\end{aligned}

The quadopole sum can now be written as
\label{eqn:momentCoeffiecients:720}
\inv{2}
\inv{ 4 \pi \epsilon r^5 }
\int d^3 x’ \rho(\Bx’)
\biglr{
x^2 \lr{ 3 (x’)^2 – (r’)^2 }
+y^2 \lr{ 3 (y’)^2 – (r’)^2 }
+z^2 \lr{ 3 (z’)^2 – (r’)^2 }
+
3 \lr{
x y x’ y’
+y x y’ x’
+x z x’ z’
+z x z’ x’
+y z y’ z’
+z y z’ y’
}
},

which is precisely \ref{eqn:momentCoeffiecients:120}, the quadropole potential stated in the text and class notes.

<h1>References</h1>

[1] JD Jackson. <em>Classical Electrodynamics</em>. John Wiley and Sons, 2nd edition, 1975.

## Totally asymmetric potential

December 16, 2015 phy1520 , , ,

### Q: [1] pr 4.11

(a) Given a time reversal invariant Hamiltonian, show that for any energy eigenket

\label{eqn:totallyAsymmetricPotential:20}
\expectation{\BL} = 0.

(b) If the wave function of such a state is expanded as

\label{eqn:totallyAsymmetricPotential:40}
\sum_{l,m} F_{l m} Y_{l m}(\theta, \phi),

what are the phase restrictions on $$F_{lm}$$?

### A: part (a)

For a time reversal invariant Hamiltonian $$H$$ we have

\label{eqn:totallyAsymmetricPotential:60}
H \Theta = \Theta H.

If $$\ket{\psi}$$ is an energy eigenstate with eigenvalue $$E$$, we have

\label{eqn:totallyAsymmetricPotential:80}
\begin{aligned}
H \Theta \ket{\psi}
&= \Theta H \ket{\psi} \\
&= \lambda \Theta \ket{\psi},
\end{aligned}

so $$\Theta \ket{\psi}$$ is also an eigenvalue of $$H$$, so can only differ from $$\ket{\psi}$$ by a phase factor. That is

\label{eqn:totallyAsymmetricPotential:100}
\begin{aligned}
\ket{\psi’}
&=
\Theta \ket{\psi} \\
&= e^{i\delta} \ket{\psi}.
\end{aligned}

Now consider the expectation of $$\BL$$ with respect to a time reversed state

\label{eqn:totallyAsymmetricPotential:120}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\bra{ \psi} \Theta^{-1} \BL \Theta \ket{\psi} \\
&=
\bra{ \psi} (-\BL) \ket{\psi},
\end{aligned}

however, we also have

\label{eqn:totallyAsymmetricPotential:140}
\begin{aligned}
\bra{ \psi’} \BL \ket{\psi’}
&=
\lr{ \bra{ \psi} e^{-i\delta} } \BL \lr{ e^{i\delta} \ket{\psi} } \\
&=
\bra{\psi} \BL \ket{\psi},
\end{aligned}

so we have $$\bra{\psi} \BL \ket{\psi} = -\bra{\psi} \BL \ket{\psi}$$ which is only possible if $$\expectation{\BL} = \bra{\psi} \BL \ket{\psi} = 0$$.

### A: part (b)

Consider the expansion of the wave function of a time reversed energy eigenstate

\label{eqn:totallyAsymmetricPotential:160}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} e^{i\delta} \ket{\psi} \\
&=
e^{i\delta} \braket{\Bx}{\psi},
\end{aligned}

and then consider the same state expanded in the position basis

\label{eqn:totallyAsymmetricPotential:180}
\begin{aligned}
\bra{\Bx} \Theta \ket{\psi}
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \ket{\Bx’}\bra{\Bx’} } \ket{\psi} \\
&=
\bra{\Bx} \Theta \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} } \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \Theta \ket{\Bx’} \\
&=
\bra{\Bx} \int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \ket{\Bx’} \\
&=
\int d^3 \Bx’ \lr{ \braket{\Bx’}{\psi} }^\conj \braket{\Bx}{\Bx’} \\
&=
\int d^3 \Bx’ \braket{\psi}{\Bx’} \delta(\Bx- \Bx’) \\
&=
\braket{\psi}{\Bx}.
\end{aligned}

This demonstrates a relationship between the wave function and its complex conjugate

\label{eqn:totallyAsymmetricPotential:200}
\braket{\Bx}{\psi} = e^{-i\delta} \braket{\psi}{\Bx}.

Now expand the wave function in the spherical harmonic basis

\label{eqn:totallyAsymmetricPotential:220}
\begin{aligned}
\braket{\Bx}{\psi}
&=
\int d\Omega \braket{\Bx}{\ncap}\braket{\ncap}{\psi} \\
&=
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) \\
&=
e^{-i\delta}
\lr{
\sum_{lm} F_{lm}(r) Y_{lm}(\theta, \phi) }^\conj \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj Y_{lm}^\conj(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{lm}(r)}^\conj (-1)^m Y_{l,-m}(\theta, \phi) \\
&=
e^{-i\delta}
\sum_{lm} \lr{ F_{l,-m}(r)}^\conj (-1)^m Y_{l,m}(\theta, \phi),
\end{aligned}

so the $$F_{lm}$$ functions are constrained by

\label{eqn:totallyAsymmetricPotential:240}
F_{lm}(r) = e^{-i\delta} \lr{ F_{l,-m}(r)}^\conj (-1)^m.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.