Quantum Mechanics: Particle in a Central Potential and Generalities of Angular Momentum Operator, Lecture notes of Quantum Mechanics

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Quantum Mechanics

Luis A. Anchordoqui

Department of Physics and Astronomy

Lehman College, City University of New York

Lesson IX

April 9, 2019

Table of Contents

(^1) Particle in a central potential

Generalities of angular momentum operator

Schr ¨odinger in 3D

Internal states of the hydrogen atom

Particle in a central potential Generalities of angular momentum operator

Complete fundamental commutation relations

of coordinate and momentum operators are:

[ xˆ, ˆpx ] = [ yˆ, ˆpy] = [ zˆ, ˆpz] = iℏ

and

[ xˆ, ˆpy] = [ xˆ, ˆpz] = · · · = [ zˆ, ˆpy] = 0

It will be convenient to use following notation

x ˆ 1

= xˆ, xˆ 2

= yˆ, xˆ 3

= zˆ and pˆ 1

= pˆ x

, pˆ 2

= pˆ y

, pˆ 3

= pˆ z

Summary of fundamental commutation relations

[ xˆ i

, ˆp j

] = iℏ δ ij

Kronecker symbol:

δ ij

1 if i = j

0 if i 6 = j

Particle in a central potential Generalities of angular momentum operator

Commutation relations for components of angular momentum operator

Convenient to get at first commutation relations with xˆ i

and pˆ i

Using fundamental commutation relations

ˆ L x

= yˆ pˆ z

− zˆ pˆ y

• [ xˆ, ˆL x

] = 0

ˆ Ly = zˆ pˆx − xˆ pˆz + [ xˆ, ˆLy] = [ xˆ, ˆz pˆx ] − [ xˆ, ˆx pˆz] = iℏ zˆ

similarly + [ xˆ, ˆL z

] = −iℏ yˆ

We can ummarize the nine commutation relations

[ xˆ i

, ˆL

j

] = iℏ e ijk

xˆ k

and summation over the repeated index k is implied

Levi-Civita tensor

e ijk

1 if (ijk) = (1, 2, 3) or (2, 3, 1) or(3, 1, 2)

− 1 if (ijk) = (1, 3, 2) or (3, 2, 1) or (2, 1, 3)

0 if i = j or i = k or j = k

Particle in a central potential Generalities of angular momentum operator

Construct operator that commutes with all components of

L

L

2

=

L

2

x

L

2

y

L

2

z

It follows that + [

L

x

, ˆL

2 ] = [

L

x

, ˆL

2

x

L

2

y

L

2

z

] = [

L

x

, ˆL

2

y

] + [

L

x

, ˆL

2

z

]

There is simple technique to evaluate commutator like [

L

x

, ˆL

2

y

]

write down explicitly known commutator

[

ˆ L x

, ˆL y

] =

ˆ L x

ˆ L y

−

ˆ L y

ˆ L x

= iℏ

ˆ L z

multiply on left by

ˆ Ly

ˆ L y

ˆ L x

ˆ L y

−

ˆ L

2

y

ˆ L x

= iℏ

ˆ L y

ˆ L z

multiply on right by

ˆ Ly

ˆ Lx

ˆ L

2

y

−

ˆ Ly

ˆ Lx

ˆ Ly = iℏ

ˆ Lz

ˆ Ly

Add these commutation relations to get

ˆ Lx

ˆ L

2

y

−

ˆ L

2

y

ˆ Lx = iℏ(

ˆ Ly

ˆ Lz +

ˆ Lz

ˆ Ly)

Similarly +

Lx

L

2

z

L

2

z

Lx = −iℏ(

Ly

Lz +

Lz

Ly)

All in all + [

L

x

, ˆL

2 ] = 0 and likewise [

L

y

, ˆL

2 ] = [

L

z

, ˆL

2 ] = 0

Particle in a central potential Generalities of angular momentum operator

Summary of angular momentum operator

L =

~r ×

~p = −iℏ

~r ×

in cartesian coordinates

L

x

= yˆ pˆ z

− pˆ y

zˆ = −iℏ

y

∂ z

∂ y

z

Ly = zˆ pˆx − pˆz xˆ = −iℏ

z

∂ x

∂ z

x

Lz = xˆ pˆy − pˆx yˆ = −iℏ

x

∂ y

∂ x

y

commutation relations

[

L

i

, ˆL

j

] = iℏ ε ijk

L

k

and [

L

2

, ˆLx] = [

L

2

, ˆLy] = [

L

2

, ˆLz] = 0 (3)

L

2

=

L

2

x

L

2

y

L

2

z

Particle in a central potential Schr ¨odinger in 3D

Continuity equation

We are often concerned with moving particles

e.g. collision of particles

Must calculate density flux of particle beam~ 

From conservation of probability

rate of decrease of number of particles in a given volume

is equal to total flux of particles out of that volume

∂ t

∫

V

ρ dV =

∫

S

~  · n dSˆ =

∫

V

∇ ·~  dV (7)

(last equality is Gauss’ theorem)

Probability and flux densities are related by continuity equation

∂ρ

∂ t

Particle in a central potential Schr ¨odinger in 3D

Flux

To determine flux...

First form ∂ρ / ∂ t by substracting wave equation multiplied by −i ψ

∗

from the complex conjugate equation multiplied by −i ψ

∂ρ

∂ t

2 m

( ψ

∗

∇

2

ψ − ψ ∇

2

ψ

∗

) = 0 (9)

Comparing this with continuity equation + probability flux density

iℏ

2 m

( ψ

∗

∇ ψ − ψ ∇ ψ

∗

) (10)

Example + free particle of energy E and momentum ~p

ψ = Ne

i~p·~x−iEt

(11)

has + ρ = |N|

2 and~  = |N

2 | ~p/m

Particle in a central potential Schr ¨odinger in 3D

To look for solutions...

Use separation of variable methods + ψ (r, ϑ , ϕ ) = R(r)Y( ϑ , ϕ )

−

ℏ

2

2 m

[

Y

r

2

d

dr

(

r

2

dR

dr

)

R

r

2 sin ϑ

∂

∂ϑ

(

sin ϑ

∂ Y

∂ϑ

)

R

r

2 sin

2 ϑ

∂

2 Y

∂ϕ

2

]

• V(r)RY = ERY

Divide by RY/r

2 and rearrange terms

−

ℏ

2

2 m

[

1

R

d

dr

(

r

2

dR

dr

)]

• r

2 (V − E) =

ℏ

2

2 mY

[

1

sin ϑ

∂

∂ϑ

(

sin ϑ

∂ Y

∂ϑ

)

1

sin

2 ϑ

∂

2 Y

∂ϕ

2

]

Each side must be independently equal to a constant + κ = −

ℏ

2

2 m

l(l + 1 )

Obtain two equations

1

R

d

dr

(

r

2

dR

dr

)

−

2 mr

2

ℏ

2

(V − E) = l(l + 1 ) (15)

1

sin ϑ

∂

∂ϑ

(

sin ϑ

∂ Y

∂ϑ

)

1

sin

2 ϑ

∂

2 Y

∂ϕ

2

= −l(l + 1 )Y (16)

What is the meaning of operator in angular equation?

Particle in a central potential Schr ¨odinger in 3D

Choose polar axis along cartesian z direction

After some tedious calculation + angular momentum components

L

x

= iℏ

sin ϕ

∂θ

• cot ϕ cos ϕ

∂ϕ

Ly = −iℏ

cos ϕ

∂ϑ

− cot ϑ sin ϕ

∂ϕ

L

z

= −iℏ

∂ϕ

Form of

L

2 should be familiar

L

2

= −ℏ

2

[

sin ϑ

∂ϑ

sin ϑ

∂ϑ

sin

2 ϑ

2

∂ϕ

2

]

Eigenvalue equations for

L

2 and

L

z

operators:

L

2 Y( ϑ , ϕ ) = ℏ

2 l(l + 1 )Y( ϑ , ϕ ) and

L

z

Y( ϑ , ϕ ) = ℏmY( ϑ , ϕ )

Particle in a central potential Schr ¨odinger in 3D

Solution of angular equation

sin ϑ

∂ϑ

sin ϑ

∂ Y

m

l

( ϑ , ϕ )

∂ϑ

sin

2 ϑ

2 Y

m

l

( ϑ , ϕ )

∂ϕ

2

= −l(l + 1 )Y

m

l

( ϑ , ϕ )

Use separation of variables + Y( ϑ , ϕ ) = Θ( ϑ )Φ( ϕ )

By multiplying both sides of the equation by sin

2 ϑ /Y( ϑ , ϕ )

Θ( ϑ )

[

sin ϑ

d

d ϑ

sin ϑ

dΘ

d ϑ

)]

• l(l + 1 ) sin

2 ϑ = −

Φ( ϕ )

d

2 Φ

d ϕ

2

2 equations in different variables + introduce constant m

2 :

d

2 Φ

d ϕ

2

= −m

2

Φ( ϕ ) (20)

sin ϑ

d

d ϑ

sin ϑ

dΘ

d ϑ

= [m

2 − l(l + 1 ) sin

2 ϑ ]Θ( ϑ ) (21)

Particle in a central potential Schr ¨odinger in 3D

Solution of angular equation

First equation is easily solved to give + Φ( ϕ ) = e

im ϕ

Imposing periodicity Φ( ϕ + 2 π ) = Φ( ϕ ) + m = 0, ±1, ±2, · · ·

Solutions to the second equation + Θ( ϑ ) = AP

m

l

(cos ϑ )

P

m

l

• associated Legendre polynomials

Normalized angular eigenfunctions

Y

m

l

( ϑ , ϕ ) =

( 2 l + 1 )

4 π

(l − m)!

(l + m)!

P

m

l

(cos ϑ )e

im ϕ

(22)

Spherical harmonics are orthogonal:

∫ π

0

∫ 2 π

0

Y

m

l

∗

( ϑ , ϕ ) Y

m

′

l

′ sin^ ϑ d ϑ d ϕ^ =^ δ ll′^ δ mm′^ ,^ (23)

Particle in a central potential Schr ¨odinger in 3D

Solution of radial equation

d

dr

r

2

dR(r)

dr

2 mr

2

2

(V − E) = l(l + 1 )R(r) (24)

to simplify solution + u(r) = rR(r)

2

2 m

d

2 u

dr

2

[

V +

2 m

l(l + 1 )

r

2

]

u(r) = Eu(r) (25)

define an effective potential

V

′ (r) = V(r) +

2

2 m

l(l + 1 )

r

2

(25) is very similar to the one-dimensional Schr ¨odinger equation

Wave function + need 3 quantum numbers (n, l, m)

ψ n,l,m

(r, ϑ , ϕ ) = R n,l

(r)Y

m

l

( ϑ , ϕ ) (27)

Particle in a central potential Internal states of the hydrogen atom

Internal states of the hydrogen atom

We start with the equation for

the relative motion of electron and proton

We use the spherical symmetry of this

equation

and change to spherical polar coordinates

From now on, we drop the subscript r in the

operator

2

2

2

H

V U E U

P

ª º

 ’  « »

¬ ¼

r

r r r

=

2

’