a charged particle in an electromagnetic field

A Charged Particle in an Electromagnetic Field

2006 Quantum Mechanics Prof. Y. F. Chen




We will apply quantum mechanics to treat the motion of a charged particle

in an external electromagnetic field.

The electromagnetic field is assumed to be produced by charges and

currents other than the one that we are considering; the field produced by

the charge that we are studying is also neglected.

Although the motion of the charged particle will be quantum mechanically

given, the treatment is inherently a semiclassical approach because the

external electromagnetic field is treated classically.




The classical theory of the electromagnetic field, including Maxwell’s

equations, the formalism of the scalar and vector potentials, and the gague

transformation:

The vacuum Maxwell’s equations are given by

The continuity equation is given by

A Brief Review of Electrodynamics

tc

tcc

∂∂

+=×∇=⋅∇

∂∂

−=×∇=⋅∇

EBB

BEE

14 0

1 4

cjπ

πρ

0=⋅∇+∂

∂cjt

cρ



The Maxwell’s equations reveal that the most general electric and magnetic

fields can be expressed in terms of the scalar and vector potential, φ and A,

in the following way:

These equations are not unique in defining electric and magnetic fields.

If an arbitrary vector, , is added to the vector potential, the magnetic field

is unchanged because of .

If the quantity is added to the scalar potential, the electric field

is also not changed.


AA×∇=

∂∂

−−∇= BE 1tc

ϕ

f∇

0=∇×∇ f

))(/1( tfc ∂∂−



The electric and magnetic fields remain invariant under the following

transformation of the potentials:

This transformation is called the gauge transformation.

A common choice is to make: (Coulomb gauge )

suppose that an arbitrary set of potentials is given by φ and A, the gauge

transformation is used to obtain a new set of potentialsφ’and A’.

In order to obtain , the following condition needs to be satisfied :


tf

cf

∂∂

−=′∇+=′1 ϕϕAA

0=⋅∇ A

0=′⋅∇ A

0=∇+⋅∇ f2A



is just Poisson’s equation defining f in terms of the

specified function

The function f is given by

can be always obtained by a gauge transformation.

In empty space the choice brings about .


A⋅∇

0=∇+⋅∇ f2A

∫∫∫ ′′′′

⋅∇= zdydxdf

|r-r|A

π41

0=′⋅∇ A

0=⋅∇ A 0=ϕ

012 =∂∂⋅∇−−∇=⋅∇

tcAϕE



The fact leads to that is simply Laplace’s equation.

It is well known that the only solution of this equation that is regular over all

of space is .

In empty space the following expressions for the fields can be obtained:

subject to the condition that


0=⋅∇ A 02 =∇ ϕ

0=ϕ

AA×∇=

∂∂

−= BE 1tc

0=⋅∇ A



Considering a particle of charge q and mass m and in terms of φand A, the

classical Hamiltonian function is given by

where is the part of the potential energy which is of nonelectromagnetic

origin.

The Hamiltonian for a Charged Particle in an Electromagnetic Field

ϕqVcq

mH ++⎟

⎠⎞

⎜⎝⎛ −= )(

21 2

rAp

)(rV

Appcq

−→

ii

ii q

HppHq

∂∂

−=∂∂

= && ; ⎟⎠⎞

⎜⎝⎛ −= Apr

cq

m1

&



When there is a vector potential, the canonical momentum, p, is no longer

equal to its conventional value of .

It represents exactly the classical equations of motion.


r&m

( ) ( )[ ]

( ) ( ) ( )[ ]

BE ×++−∇=

⋅∇+×∇×−−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

−∇−+∇−⋅∇=

⋅∇+×∇×−−∂∂

−∇−∇−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−∇=

⎥⎦

⎤⎢⎣

⎡∇⋅+

∂∂

−−∇=⎟⎠⎞

⎜⎝⎛ −=

vr

AvAvArAv

AvAvArAp

AvAApr

cqqV

cq

tcqV

cq

cq

tcqqV

cq

m

tcqH

cq

tdd

tddm

)(

1)(

)(21

)(

2

2

2

ϕ

ϕ



The quantum-mechanical Hamiltonian operator can be obtained by

replacing p by the operator :

It is useful to derive the expression of the probability current density for a

charged particle under the influence of a static magnetic field.


∇− hi

22

22

2

2

2)(

2)(

2

)(21

Acm

qcmqiqV

m

qVcqi

mH

+⋅∇+∇⋅+++∇−

=

++⎟⎠⎞

⎜⎝⎛ −∇−=

AAr

rA

hh

h

ϕ

ϕ



Using the time-dependent Schrödinger equation , we

obtain

the probability current density is given by


ψψ Hti =∂∂ )(h

( ) )(2

)(2

)(2

22)(

22

2222

22

22

ψψψψψψ

ψψψψ

ψψψψψψψψψψρ

∗∗∗

∗∗

∗∗∗

∗∗

⋅∇+∇−∇⋅∇−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+∇⋅+⋅∇+∇−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+∇⋅+⋅∇−∇−−=

⎟⎠⎞

⎜⎝⎛ −∇−+⎟

⎠⎞

⎜⎝⎛ −∇−=

∂∂

+∂∂

=∂∂

=∂∂

A

AAAA

AA

cmq

mi

Acq

cqi

miA

cq

cqi

mi

cqi

micqi

mitttt

h

hh

h

hh

h

hh

hh

0=⋅∇+∂∂ j

tρ

( ) )(2

ψψψψψψ ∗∗∗ −∇−∇= Acm

qmihj



The complete Schrödinger equation is expressed as

Making a gauge transformation, the Schrödinger equation becomes

After some algebra

in terms of the new potentials,φ’and A’ , a new solution is given by


( )ψϕψψ qVcqi

mti ++⎟

⎠⎞

⎜⎝⎛ −∇−=

∂∂ )(

21 2

rAhh

( ) ψψϕψψtf

cqqVf

cq

cqi

mti

∂∂

+′++⎟⎠⎞

⎜⎝⎛ ∇+′−∇−=

∂∂ )(

21 2

rAhh

( ) ( ) ( )( )ψϕψψ cfiqcfiqcfiq eqVecqi

me

ti hhh hh //

2/ )(

21 ′++⎟

⎠⎞

⎜⎝⎛ ′−∇−=

∂∂ rA

ψψ cfiqe h/=′



Imagine a particle constrained to move in a circle of radius b. There is a

solenoid of radius a<b along the center of the circle, carrying a steady

electric current I. If the solenoid is extremely long, the magnetic filed inside it

is uniform, and the field outside is zero. But the vector potential outside the

solenoid is not zero. Adopting the convention gauge condition ,

the vector potential is given by

where is the magnetic flux through the solenoid.

The Aharonov-Bohm Effect

0=⋅∇ A

φπarA ˆ

2)(

21

2

2

rra BΦ

=×= B

B2aB π=Φ



Consider a charged particle that is moving through a region where B is zero

( ), but A itself is not.

solve the Schrödinger equation without the vector potential A, and find the

eigenfunction to be

the eigenfuction for the presence of a vector potential is given by

moving a path through a region where the field is zero, but not the vector

potential, the wave function of the charged particle acquires an additional

phase.


0=×∇ A

ψ

ψψ δie=′ ∫ ⋅=s

dc

q lAh

δ

where s denotes the path of integration



Aharonov and Bohm proposed an experiment in which a beam of electrons

is split in two, and passed either side of a long solenoid, before being

recombined. The phase difference accruing from traveling the different

paths is given by

The combination of path 1 and back by path 2 makes a closed loop. With

the Stoke’s theorem


[ ] ∫∫ ∫ ⋅=⋅−⋅=− lll dc

qddc

q AAAhh 1 212 δδ

Bddd Φ=⋅=⋅×∇=⋅ ∫∫∫∫∫ SSAA B)(l



This phase shift leads to measurable interference, which has been

confirmed experimentally by Chambers (Phys. Rev. Lett. 5, 3, 1960) and

others.


BcqΦ=

hδ



Considering a charged particle moving in a constant magnetic field, the

classical trajectory is a helical path that is uniform translational motion in the

direction parallel to the magnetic field and uniform rotational motion in the

plane perpendicular to .

For a positive charge q, the circular motion is in the counterclockwise

direction when viewed from the leaving direction of translational motion.

The Coulomb gauge leads to . Therefore,

the quantum Hamiltonian is given by

Quantum Eigenstates of a Charged Particle in a Constant Magnetic Field

B

0=⋅∇ A ( )ψψ ∇⋅=⋅∇ AA)(

22

22

2

22A

cmq

cmqi

mH +∇⋅+∇

−= Ahh



One of the convenient expressions for the vector potential of a uniform

constant magnetic field is given by

As a result, we find that

In addition, we use equation mentioned above to obtain


BrA ×−=21

LBprBrBBrA ⋅−=×⋅−=∇×⋅=∇⋅×−=∇⋅hh ii 2

1)(21)(

21)(

21

[ ]22222 )(41)(

41 BrBr ×−=×= BrA



After some algebra, the Hamiltonian for a charged particle in a constant

magnetic field is given by

Assuming to be in the z-direction, the Eq. can be expressed as


[ ]2222

22

2

)(822

BrLB ×−+⋅−∇−

= Brcm

qcm

qm

H h

B

)(21

22222

2

yxmLm

H LzL ++−∇−

= ωωh



where is the Lamor precession frequency. The

corresponding Schrödinger equation in cylindrical coordinates is given by

The Schrödinger equation is separable in this case and its solution can be

expressed as


),,(),,(2111

222

2

2

2

2

22

22

zrEzrrmizrrrrm LL θψθψω

θω

θ=⎥

⎦

⎤⎢⎣

⎡+

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂−

hh

zikli zeerRzr θθψ )(),,( =

)2/( mcqBL =ω



This solution is an eigenfunction of both the and z kinetic energy

terms which together have contributions to the total eigenenergy given by

and , respectively. The remaining 2D problem is exactly

the same as the 2D isotropic harmonic oscillator in polar coordinates:

where is related to the total energy as


zL̂

Ll ωh− )2/(22 mkzh

)()(211

222

2

2

2

22

rErrmrl

rdd

rrdd

m L ψψω ′=⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

− h

E ′

mklEE z

L 2

22hh +−′= ω



In terms of the dimensionless variable , where

can be reduced to

with the dimensionless eigenvalue and .


wr /2=ξ )/(2 Lmw ωh=

)()(211

222

2

2

2

22

rErrmrl

rdd

rrdd

m L ψψω ′=⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

− h

0)(~1 22

2

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−++ ξψξ

ξε

ξξξl

dd

dd

)/(2 LE ωε h′= )()(~ rψξψ =



Since the behavior of at infinity is important the existence of

normalizable solutions, the Eqs. above can be examined for large ξ:

,

which has approximation solution, .

On the other hand, the behavior near the origin is guaranteed to be the form

. Therefore, the eigenfunction can be expressed as


)(~ ξψ

0)(~22

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛− ξψξ

ξdd

2/2

)(~ ξξψ −∝ e

|| lξ )(~ ξψ

)()(~ 2/|| 2

ξξξψ ξ gel −=



Substituting Eq. above into , we

have


0)(~1 22

2

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−++ ξψξ

ξε

ξξξl

dd

dd

0)()2||2()(21||2)(

)(~1

2

2

22

2

2

2

=−−+⎟⎟⎠

⎞⎜⎜⎝

⎛−

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−++

ξεξξξ

ξξξ

ξψξξ

εξξξ

gldgdl

dgd

ldd

dd



Here we have used the following identities

and


[ ] )(11||)(12

2/||2/|| 22

ξξξξ

ξξξξξ

ξξ gddlege

dd ll

⎟⎟⎠

⎞⎜⎜⎝

⎛+−= −−

[ ]

)(2||21||2||

)(||

)(

2

22

2

22/||

2/||

2/||2

2

2

2

2

ξξξ

ξξ

ξξ

ξ

ξξ

ξξ

ξξ

ξξξ

ξ

ξ

ξ

gdd

ddlllle

gddle

dd

gedd

l

l

l

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛−++−−

−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

−

−

−



Next, a change of variable, , leads Eq. to be


2ξη =

( ) 0)(4

)2||2()(1||)(

0)()2||2()(221||2)(2)(4

0)()2||2()(21||2)(

2

2

2

22

2

2

=−−

+−++⇒

=−−+⎟⎟⎠

⎞⎜⎜⎝

⎛−

++⎟⎟

⎠

⎞⎜⎜⎝

⎛+⇒

=−−+⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

ηεηηη

ηηη

ηεηηξξ

ξηη

ηηξ

ξεξξξ

ξξξ

Gld

dGldGd

Gld

dGld

dGdGd

gldgdl

dgd



Here we have used

The Eq. from previous page is the differential equation for Laguerre

polynomials, and, consequently the solution is given by with the

dimensionless eigenvalue


)()( ηξ Gg =

ηηξ

ξξ

ddG

dgd )(2)(

=

ηη

ηηξ

ξξ

ddG

dGd

dgd )(2)(4)(

2

22

2

2

+=

)()( ηη lpLG =

2||244

)2||2(++=⇒=

−− lppl εε



where n. To sum up, the eigenfunctions and eigenvalues

can be given by

and

where are the normalization constants to be determined later.


2...... 1, ,0=p

ziklilp

l

lp

ziklilp

llp

z

z

eew

rLw

rwrC

eeLeCzr

θ

θη ηηθψ

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

= −

2||2

,

2/||2/,

22exp

)(),,(

mkllpE

lpElp

zLLklp

L

z 2)1||2(

)1||2(2||24

22

,,h

hh

h

+−++=⇒

++=′⇒++=

ωω

ωε

lpC ,



The Laguerre polynomials, , are defined by the formula

for arbitrary real .

From above, the first few Laguerre polynomials are

Definition and Generating Function of the Laguerre Polynomials

)(xLlp

.....2,1,0 ,)(!

)( == +−−

pxexd

dpxexL lpx

p

plxl

p

1−>l

])2(2)2)(1[(21)(

,1)(,1)(

22

10

xxlllxL

xlxLxL

l

ll

++−++=

−+==



In general, Leibniz’z formula enable us to have

where for all k < p the ratio of gamma functions can be replaced by the

product .


∑= −

−++Γ++Γ

=p

k

klp kpk

xlklpxL

0 )!(!)(

)1()1()(

)1()1)(( ++−++ kllplp L



The generating function for Laguerre polynomials is given by

With the generating function, the following identity can be easily verified:

After algebra, we find that

which gives

when the coefficient of is set equal to zero.


1||,)()1(),(0

)1(/1 <=−= ∑∞

=

−−−− ttxLettxW p

p

lp

ttxl

0)]1)(1([)1( 2 =+−−+∂∂

− Wltxt

Wt

0)()]1)(1([)()1(0

1

0

2 =+−−+− ∑∑∞

=

−∞

=

p

p

lp

p

p

lp txLltxtxLpt

0)()()()12()()1( 11 =++−−−++ −+ xLlpxLplxxLp lp

lp

lp

pt




Similarly, substituting

into the identity

we obtain

which indicates

1||,)()1(),(0

)1(/1 <=−= ∑∞

=

−−−− ttxLettxW p

p

lp

ttxl

0)1( =+∂∂

− Wtx

Wt

0)()(

)1( 1

00

=+− +∞

=

∞

=∑∑ p

p

lp

p

p

lp txLt

dxxdL

t

0)()()(

11 =+− −− xL

dxxdL

dxxdL l

p

lp

lp



From we

obtain

Replacing p by p-1 and using equation from previous page to eliminate ,

then we get


0)()()()12()()1( 11 =++−−−++ −+ xLlpxLplxxLp lp

lp

lp

0)()1(

)()22()(

)1()(

)1(

1

1

=+−

−+++++−−

+

+

xLp

xLxlpdx

xdLp

dxxdL

px

lp

lp

lp

lp

dxxdL lp /)(1−

)()()()(

1 xLlpxpLdx

xdLx l

plp

lp

−+−=



The last equation indicates that the derivative of a Laguerre polynomial can

be in terms of another Laguerre polynomial. Using the recurrence relations

Eqs. before, we can derive a differential equation satisfied by the Laguerre

polynomials.

The step is to differentiate with

respect to x and then to eliminate and .


)()()()(

1 xLlpxpLdx

xdLx l

plp

lp

−+−=

dxxdL lp /)(1− )(1 xL l

p−



Consequently, we obtain

,

which indicates that is a particular solution of the second-order

linear differential equation

By making changes of variables, it can be derived other differential equations

whose integrals can be expressed in terms of Laguerre polynomials.


0)()(

)1()(

2

2

=+−++ xpLdx

xdLxl

dxxLd

x lp

lp

lp

)(xLu lp=

0)1( =+′−++′′ puuxlux



For instance, it can be shown that the differential equations

have the particular solutions


0)(42

1)21( =⎥⎦⎤

⎢⎣⎡ −

+−+

++′−++′′ ux

lxlpulux ννν

)()( 2/ xLexxu lp

x−= ν



Next we demonstrate one of the most important properties of the Laguerre

polynomials, i.e., their orthogonality with weight on the interval

.

Setting

and recalling eqution from previous page, we find that and

satisfy the differential equations


2/2/ xl ex −

∞<≤ x0

)()( 2/2/ xLexxu lp

xlp

−=

)(xu p )(xu p

0442

1)(2

=⎥⎦

⎤⎢⎣

⎡−−

+++′′ pp u

xlxlpux

0442

1)(2

=⎥⎦

⎤⎢⎣

⎡−−

+++′′ qq u

xlxlqux



After subtracting, and integrating from 0 to ∞, we obtain

For the first term of

vanishes at both limits, and therefore

if

In other words,

if


0)()(00

=−+′−′ ∫∞∞

dxuuqpuuuux pqpqqp

[ ] )(11||)(12

2/||2/|| 22

ξξξξ

ξξξξξ

ξξ gddlege

dd ll

⎟⎟⎠

⎞⎜⎜⎝

⎛+−= −−

1−>l

00

=∫∞

dxuu pqqp ≠

0)()(0

=∫∞ − dxxLxLex l

qlp

xl 1, −>≠ lqp



Now we evaluate the value of the integral for . Recall the recurrence

relation

We can replace the index p by p-1 in the Eq. above and multiply the result by

. Then from this equation we subtract above Eq. multiplied by ,

obtaining


qp =

0)()()()12()()1( 11 =++−−−++ −+ xLlpxLplxxLp lp

lp

lp

)(xL lp )(1 xL l

p−

[ ] [ ]...3,2,0)()()1()()(2

)()()1()()()(

21

112

12

==−+++

+−+−

−−

−+−

pxLxLlpxLxL

xLxLpxLlpxLplp

lp

lp

lp

lp

lp

lp

lp



Multiplying this equation by , integrating from 0 to ∞, using the

orthogonality property ,

we obtain

Repeated application of this formula gives


xl ex −

0)()()()12()()1( 11 =++−−−++ −+ xLlpxLplxxLp lp

lp

lp

[ ] [ ]∫∫∞

−−∞ − =+=

0

210

2 ...3,2,)()()( pdxxLexlpdxxLexp lp

xllp

xl

[ ] [ ]

...3,2,!

)1(

)(23)1(

)2()1)(()(0

210

2

=++Γ

=

⋅−+−++

= ∫∫∞ −∞ −

pp

lp

dxxLexpp

llplpdxxLex lxllp

xl

L

L



It follows by direct substitution that this formula is also valid for ,

and hence

As a consequence, the functions

form an orthonormal system on the interval .


1,0=p

[ ] ...3,2,1,0,1,!

)1( )(0

2=−>

++Γ=∫

∞ − plp

lpdxxLex lp

xl

L,2,1,0,)()1(

!)( 2/2/2/1

, =⎥⎦

⎤⎢⎣

⎡++Γ

= − pxLexlp

px lp

xllpψ

∞<≤ x0



The θ

integral is conveniently done using , hence

Writing , and further change the variable to

, the right side of the above equation becomes


1sin||sin),(),(2

0

22

0,*, ==Ω ∫∫∫ −−

ππθφθθφθφθ l

llll CdddYY

θcos=x

1)1(||21

1

22 =−∫− dxxC lπ

)1)(1(1 2 xxx +−=−

12 −= tx

∫∫ −=− + 1

0

1221

0

2 )1(2||2)]22(2[||2 dtttCdtttC llll ππ



From the property of the Beta function

With the last three equations, we obtain


)()()()1(),(

1

0

11

qpqpdtttqpB qp

+ΓΓΓ

=−= ∫ −−

π

ππ

4)!12(

!21||

1)!12(

)!(2||4)22(

)1()1(2||22

22122

+=⇒

+=

+Γ+Γ+Γ+

ll

C

llC

lllC

l

ll



The commonly used convention is not to have an additional phase factor,

hence

From Eqs. mentioned before, we have


φθπ

φθ lillll ell

Y −−

+= sin

4)!12(

!21),(,

1,)1)((,ˆ +++−=+ mlmlmlmlL



Eq. of last page indicates that all can be obtained by successively

acting on .

With the equation mentioned before

if ,

the spherical harmonic can be expressed as


),(, φθmlY

+L̂ ),(, φθllY −

00

=∫∞

dxuu pq qp ≠

),(, φθmlY

,

,

,

1 1 1 1( , )( 1)( ) ( 2)( 1) ( 1)(2) ( )(1)

cot ( , )

( - )! = cot ( , )(2 )! ( )!

l m

l m

il l

l m

il l

Yl m l m l m l m l l l l

i e i Y

l m e i Yl l m

φ

φ

θ φ

θ θ φθ φ

θ θ φθ φ

+

−

+

−

=− + + − + + − + − +

⎡ ⎤⎛ ⎞∂ ∂× − +⎢ ⎥⎜ ⎟∂ ∂⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞∂ ∂+⎢ ⎥⎜ ⎟+ ∂ ∂⎝ ⎠⎣ ⎦

Lh h h h

h



The φ derivative always give eigenvalues. But each time acts, there is

a factor of and makes the eigenvalue of increase one by one.

As a result, the last equation can be expressed as


+L̂ie φ i φ− ∂ ∂

,1 (2 1)! ( - )!( , ) ( 1)cot

2 ! 4 (2 )! ( )!

( 2)cot ( 1)cot ( ) cot sin

iml m l

l

l l m dY e ml l l m d

d d dm l ld d d

φθ φ θπ θ

θ θ θ θθ θ θ

+ ⎛ ⎞= − −⎜ ⎟+ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎛ ⎞× − − − − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

L



With the property that ,

equation from last page can be written as

where


1cot sinsin

nn

d dnd d

θ θθ θ θ+ =

( 1), ( 1)

( 2) 1( 2) 1

1 (2 1) ( - )! 1( , ) sin2 ! 4 ( )! sin

1 1 1 sin sin sin sinsin sin sin

1 (2 1) ( - )! 2 ! 4 ( )!

im ml m l m

m l l lm l l

il

l l m dY el l m d

d d dd d d

l l m el l m

φθ φ θπ θ θ

θ θ θ θθ θ θ θ θ θ

π

− −− −

− − −− − −

+ ⎛ ⎞= ⎜ ⎟+ ⎝ ⎠

⎛ ⎞ ⎛ ⎞⎛ ⎞×⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

+=

+

L

2

2 / 2 2

2 / 2 2

1sin sinsin

1 (2 1) ( - )! (1 cos ) (1 cos )2 ! 4 ( )! cos

( 1) (2 1) ( - )! (1 ) (1 )2 ! 4 ( )!

l mm m l

l mim m l

l

l ml mim m l

l

dd

l l m del l m d

l l m de x xl l m dx

φ

φ

φ

θ θθ θ

θ θπ θ

π

+

+

++

⎛ ⎞⎜ ⎟⎝ ⎠

+ ⎛ ⎞= − − −⎜ ⎟+ ⎝ ⎠

− + ⎛ ⎞= − −⎜ ⎟+ ⎝ ⎠

cosx θ=



It is of significant importance to visualize the distribution patterns of spherical

harmonics in a pedagogical way. Before implementing the visualization of

spherical harmonics, an explicit polynomial expression for the associated

Legendre functions is essential.

From

,

the expression of the associated Legendre functions for is given by

Graphical Visualization of Spherical Harmonics

),,(),,(2111

222

2

2

2

2

22

22

zrEzrrmizrrrrm LL θψθψω

θω

θ=⎥

⎦

⎤⎢⎣

⎡+

∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

+∂∂−

hh

( ) lml

mlm

l

lm

l xdxdx

lxP )1(1

!2)1()( 22/2 −−

−= +

+



To obtain an explicit expression, we expand

Although the sum extends for , the term in the first square

parentheses vanishes for and the term in the second square

parentheses vanishes for . Therefore, the sum is taken only

for .


⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

+−=−

−+

−++

=

+

+

+

+

∑ lkml

kmll

k

kml

k

llml

mll

ml

ml

xdxdx

dxd

kml

xxdxdx

dxd

)1()1(

)1()1()1(

0

2

mlk +≤≤0

lk >

lkml >−+

lkm ≤≤



Then


( ) ( )

( )

( )

( )

( )l

ml

ml

m

m

lsml

smll

s

sml

sm

m

msslsml

sm

m

msslsmml

sm

msslsmml

sm

msslsmml

sm

ssmlsmml

s

mkklkl

mk

lkml

kmll

k

kml

k

lml

ml

xdxd

mlml

x

xdxdx

dxd

sml

mlml

x

xsm

lxsll

ssmlml

mlml

x

xslx

smll

smslml

x

xslx

smll

smslml

xx

slx

smll

smml

x

xslx

smll

smml

xmk

lxkl

lkml

xdxdx

dxd

kml

xdxd

)1()!()!(

1)1(

)1()1()!()!(

1)1(

)1()!(

!)1()!(!)1(

!)!()!(

)!()!(

1)1(

)1(!!)1(

)!(!)1(

)!()!()!(

11

)1(!!)1(

)!(!)1(

)!()!()!(

11)1(

!!)1(

)!(!)1(

11

)1(!!)1(

)!(!)1()1(

)!(!)1(

)!(!)1(

)1()1()1(

2

2

02

02

02

02

02

0

0

2

−−+

−

−=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−+

−

−=

⎥⎦

⎤⎢⎣

⎡+

+⎥⎦

⎤⎢⎣

⎡−

−−

−−−

−+

−

−=

⎥⎦⎤

⎢⎣⎡ +⎥

⎦

⎤⎢⎣

⎡−

−−−

+−+

−=

⎥⎦⎤

⎢⎣⎡ +⎥

⎦

⎤⎢⎣

⎡−

−−−

+−+

−=⎥⎦

⎤⎢⎣⎡ +⎥

⎦

⎤⎢⎣

⎡−

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

⎥⎦⎤

⎢⎣⎡ +⎥

⎦

⎤⎢⎣

⎡−

−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=⎥⎦

⎤⎢⎣

⎡+

−⎥⎦

⎤⎢⎣

⎡−

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=−

−

−

−−

−−−

=

+−−

=

+−+−

=

+−+−

=

+−+−

=

−−+−

=

−−

=

−+

−++

=+

+

∑

∑

∑

∑∑

∑∑

∑



With previous page, the alternative definition of the associated Legendre

functions is given by

With above equation, the alternative expression for the spherical harmonics

with is given by


( )

( ) lml

mlm

l

ml

lml

mlm

l

lm

l

xdxdx

mlml

l

xdxdx

lxP

)1(1|)!|(|)!|(

!2)1(

)1(1!2)1()(

2||

||2/||2

2||

||2/||2

−−−+−

=

−−−

=

−

−−

+

+

+

0≥m

lml

mlmmi

l

l

lml

mlmmi

l

ml

ml

mimml

dde

mlmll

l

dde

mlmll

l

PemlmllY

)cos1()(cos

sin)!()!(

412

!2)1(

)cos1()(cos

sin)!()!(

412

!2)1(

)(cos)!()!(

412)1(),(

2

2

θθ

θπ

θθ

θπ

θπ

φθ

φ

φ

φ

−−++−

=

−+−+−

=

+−+

−=

−

−−

+

++

a charged particle in an electromagnetic field

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