interaction of electrons with photonswimd/524-13em2.pdf• use radiation gauge • vector potential...

E&M

Interaction of electrons with photons• Complete Hamiltonian includes interaction of charges and their

coupling to the electromagnetic field• Use radiation gauge

• Vector potential in minimal substitution

• Hamiltonian for Z electrons in an atom plus radiation field

Hem =X

k�

~�ka†k�ak�

H =ZX

i=1

�p

i

+ e

c

A(xi

, t)�2

2m�

ZX

i=1

Ze2

|xi

| +ZX

i<j

e2

|xi

� x

j

| + Hem

= Helectrons

+Hint

+ Hem

E&M

Electron and interaction Hamiltonian

Electrons

Coupling

with

No spin yet. Add by hand

as before from

Helectrons

=ZX

i=1

p

2i

2m�

ZX

i=1

Ze2

|xi

| +ZX

i<j

e2

|xi

� x

j

|

Hint =ZX

i

e

2mc(pi ·A(xi, t) +A(xi, t) · pi) +

e2

2mc2A(xi, t) ·A(xi, t)

�

A(xi, t) =X

k�

✓

2�~c2⇥kV

◆1/2n

ak�ek�e

i(k·xi�⇥kt) + a†k�ek�e

�i(k·xi�⇥kt)o

Hspinint =

e

mc

ZX

i=1

si · [r�A(x, t)]x=xi

E = �µ ·B

E&M

Towards transitions between atomic levels

• Solve electron Hamiltonian (approximately)• Hartree-Fock method for example

• Ground state: occupy lowest HF orbits

• Treat atoms in IPM with e.g. in second quantization

• Free electromagnetic field solved

• Transitions between states --> coupling

• Usually emission or absorption of one photon

• Use second quantization for electrons as well

Helectrons =X

n�m�ms

�n� a†n�m�msan�m�ms

|atom� |photons�

E&M

Second quantizedUsing transversalityRemember

then

and

neglect term with vector potential squared

pi ·A(xi, t) = A(xi, t) · pi

A(xi, t) =X

k�

✓

2�~c2⇥kV

◆1/2n

ak�ek�e

i(k·xi�⇥kt) + a†k�ek�e

�i(k·xi�⇥kt)o

Hspinint =

e

m

X

⇥⇤

X

k�

✓

2⇤~⌅kV

◆1/2

(ik � ek�)

·n

⇥�| ei(k·x�⌅kt) s |⇥⇤ a†⇥a⇤ak� + ⇥�| e�i(k·x�⌅kt) s |⇥⇤ a†⇥a⇤a†k�

o

Hint =e

m

X

⇥⇤

X

k�

✓

2⇤~⌅kV

◆1/2

ek� ·

n

��| ei(k·x�⌅kt) p |⇥⇥ a†⇥a⇤ak�

+ ��| e�i(k·x�⌅kt) p |⇥⇥ a†⇥a⇤a†k�

o

E&M

Next step• Use standard time-dependent perturbation theory for transitions

of the type

• Do only lowest order (otherwise squared term must be included)

• Validity– present results for “slow” particles

– not good for interaction with modes (--> pair creation)

– can be eliminated by cut-off: sum only with

– should still be large with respect to transition frequency of particles so

– Hydrogen:

–

~� � mc2

|k| � kc ~ckc = ~�c ⌧ mc2

� =e2

~c ' 1

137

|A⇤ |nk↵⇤ � |A;nk↵⇤ ⇥ |B;nk↵ ± 1⇤

~⇥0 ⇠ �2mc2 ' 1H

~⇥c ⇠ � mc2

mc2 ⇠ 0.5 MeV

~�0 ⌧ ~�c ⌧ mc2

E&M

Apply time-dependent perturbation theory• Results from TDPT• Constant potential

– Transition rate from Fermi’s Golden Rule

– No change when Fock space formulation is used

– Except: “potential” now includes or

– So instead of --> for removing a photon (absorption)

– --> for adding a photon (emission)

– Corresponding Golden Rule becomes

– With no longer including time dependence

wi![f ] =2�

~ ⇥(Ef ) |�f |V |i⇥|2

e�i�t ei�t

Ef = Ei Ef = Ei + ~�

Ef = Ei � ~�

wi![f ] =2�

~

��f | H 0int |i⇥

��2⇥f

H 0int

E&M

Emission of a photon• We want to describe transitions of the kind

• So we need a transition rate of the kind

• Density of states --> # of allowed states in interval for photon emitted into solid angle

• First evaluate (note )

|A;nk↵ = 0⇥ � |B;nk↵ = 1⇥

wd� =2�

~

��B;nk� = 1| H 0int |A;nk� = 0⇥

��2⇥~⇥,d�

~� + d(~�), ~�d⌦

(# of states � ~⇥) =X

|k|��/c

⇥Z

dnx

Zdny

Zdnz with |k| � ⇥/c

=V

(2�)3

Zd3k =

V

(2�)3

Zdk k2 d�

dnx

=L

2�dk

x

E&M

Density of states• Required density of states is then obtained from

• Therefore

• Initial state

• with

• Final state

• with

• such that

⇥~�,d� =V

(2�)3⇤2

~c3 d�

(# of states ~(⇤ + d⇤)� (# of states ~⇤) = ⇥~�,d� d~⇤

=V

(2�)3

Z (�+d�)/c

0dk k2 d�� V

(2�)3

Z �/c

0dk k2 d�

=V

(2�)3d�

Z �+d�

�

d⇤0

c

✓⇤0

c

◆2

=V

(2�)3d�

1

3

✓⇤0

c

◆3��

�+d�

�

) V

(2�)3⇤2

~c3 d~⇤ d�

|A� |0� = |i�

|B� |1k�� = |f�

⇣H

electrons

+ Hem

⌘|i� = E

A

|i�

⇣H

electrons

+ Hem

⌘|f� = (E

B

+ ~�k

) |f�

EA = EB + ~�k

E&M

Corresponding rate• Insert density of states

• Keep in mind

• Only second term contributes

• Single-particle matrix elements require evaluation

• Also matrix element connecting initial and final atomic state plus photon involving

wd� =2�

~V

(2�)3⇥2k

~c3 d��f | H 0

int |i⇥��2

H ⇥int =

e

m

X

⇥⇤

X

k

0�

✓

2⇤~⌅⇥kV

◆1/2

ek

0� ·n

��| eik0·x p |⇥⇥ a†⇥a⇤ak0�

+ ��| e�ik0·x p |⇥⇥ a†⇥a⇤a†k

0�

o

a†�a�a†k0↵

E&M

Rate continued• So right now

• Typical transition: optical ~ eV --> (green --> 2 eV)

• Atomic dimension: ~

• So from

• therefore

• Electric dipole (E1) approximation

• Photon matrix element

• Consider alkali atom in IPM

wd� =2⇤

~V

(2⇤)3⌅2k

~c3e2

m2

✓

2⇤~V

◆

�

�

�

�

�

�

X

⇥⇤

X

k

0�

1p

⌅⇥k

ek

0� ·n

��| e�ik0·x p |⇥⇥ �B1k⌅| a†⇥a⇤a

†k

0� |A⇥o

�

�

�

�

�

�

2

~!k

10�10 m

~�k ) kr =~�k

~c r ' eV 10�10 m

1.24⇥ 10�6 eV m⇠ 10�4

e�ik0·x ⇥ 1� ik⇥ · x ⇥ 1

�1k�| a†k0↵ |0⇥ = �kk0��↵

E&M

Atom• Alkali atom: one particle outside closed shell

• Transition to final state

• Evaluate

• not unexpected…

• So we also need

|A� = |n�m�ms;�0� = a†n�m�ms|�0�

|B� = |n��m��m

�s;�0� = a†n0�0m0

�m0s|�0�

⇥B| a†�a⇥ |A⇤ = ⇥�0| an0⇤0m0�m

0sa†�a⇥a

†n⇤m�ms

|�0⇤

= ⇥�0|⇣�n0� � a†�an0⇤0m0

�m0s

⌘⇣�n⇥ � a†n⇤m�ms

a⇥⌘|�0⇤

= �n0��n⇥

ek� · �n0�0m0⇥m

0s|p |n�m⇥ms⇥

E&M

Dipole matrix element• Use central field from

• to evaluate

• First note that

• So

• Replace

ek� · �n0�0m0⇥m

0s|p |n�m⇥ms⇥

Helectrons =X

n�m�ms

�n� a†n�m�msan�m�ms � H0 |n⇥m�ms⇥ = �n� |n⇥m�ms⇥

⇥p

2,x⇤= �2i~p

p

2

2m,x

�= �i~ p

mp

2

2m+ Vnucleus + Vcentral,x

�= �i~ p

m

[H0,x] = �i~ p

m

p = im

~ [H0,x]

ek,1 E&M

Matrix element• Then (hence dipole approximation) & note change in parity

• Insert all ingredients

Sum over polarization & integrate over all directions

ek� · ⇥n0⌅0m0⇥m

0s|p |n⌅m⇥ms⇤ = ek� · ⇥n0⌅0m0

⇥m0s| i

m

~ [H0,x] |n⌅m⇥ms⇤

= im

~ ek� · ⇥n0⌅0m0⇥|x |n⌅m⇥⇤ �msm0

s(⇤n0⇥0 � ⇤n⇥)

= �im⇥kek� · ⇥n0⌅0m0⇥|x |n⌅m⇥⇤ �msm0

s

x

k

ek,2

⇥2

⇥1

✓

�

wd� =e2

2�

⇥3k

~c3 d⇥ |ek� · �n0⇤0m⇥0 |x |n⇤m⇥⇥|2

=e2

2�

⇥3k

~c3 d⇥ cos2 �� |�n0⇤0m⇥0 |x |n⇤m⇥⇥|2

cos�1 = sin ✓ cos�

cos�2 = sin ✓ sin�X

�

) sin2 ✓

Zd� sin2 � = 2⇥

Zd(cos �) sin2 � =

8⇥

3

E&M

Rate• Therefore• Note

• Need

• So that

w =4e2�3

k

3~c3 |�n0⇥0m�0 |x |n⇥m�⇥|2

|⇤f |x |i⌅|2 = |⇤f |x |i⌅|2 + |⇤f | y |i⌅|2 + |⇤f | z |i⌅|2

=

�

�

�

�

⇤f |� 1⇧2(x+ iy) |i⌅

�

�

�

�

2

+

�

�

�

�

⇤f | 1⇧2(x� iy) |i⌅

�

�

�

�

2

+ |⇤f | z |i⌅|2

=4�

3

n

|⇤f | rY11 |i⌅|2 + |⇤f | rY1�1 |i⌅|2 + |⇤f | rY10 |i⌅|2o

⇥ 4�

3

X

µ

|⇤n0⇥0m0�| rY1µ |n⇥m�⌅|

2

⇥n0⇥0m0�| rY1µ |n⇥m�⇤ =

(⇥ m� 1 µ | ⇥0 m0� )⌅

2⇥+ 1⇥n0⇥0||rY1||n⇥⇤

=

Zdr r2

⇢Zd� Y ⇤

�0m0`(�)Y1µ(�)Y�m`(�)

�Rn0�0(r)rRn�(r)

=

Zdr r3Rn0�0(r)Rn�(r)�

r3

4�

r2⇥+ 1

2⇥0 + 1(⇥ 0 1 0 | ⇥0 0 ) (⇥ m� 1 µ | ⇥0 m0

� )

⇤n0⇥0||rY1||n⇥⌅ =

Zdr r3Rn0�0(r)Rn�(r)�

r3

4�

2⇥+ 1⇧2⇥0 + 1

(⇥ 0 1 0 | ⇥0 0 )

⇥ In0�0n�

r3

4�

2⇥+ 1⇧2⇥0 + 1

(⇥ 0 1 0 | ⇥0 0 )

e��At = e�t/⇥A

E&M

Experimental conditions• Sum also over all projections of final state

• So

• Lifetime: exponential decay:

m0`

X

m0`

|�..|x |..⇥|2 =X

m0`

4�

3

X

µ

|�n0⇥0m0�| rY1µ |n⇥m�⇥|

2

X

m0`

wn�!n0�0 =e2

~c4�3

k

3c2I2n0�0n� (⇥ 0 1 0 | ⇥0 0 )

2

(` 0 1 0 | `0 0 )

2 ⇥ `+ 1

2`+ 1

for `0 = `+ 1

⇥ `

2`+ 1

for `0 = `� 1

1

⇥A=

X

f

wA)Bf = �A

=4�

3

X

m0`µ

(⇥ m� 1 µ | ⇥0 m0� )2

2⇥+ 1I2n0�0n�

3

4�

(2⇥+ 1)2

2⇥0 + 1(⇥ 0 1 0 | ⇥0 0 )

2

=X

m0`µ

2⇥0 + 1

2⇥+ 1

(⇥0 �m0� 1 µ | ⇥ �m� )2

2⇥+ 1I2n0�0n�

(2⇥+ 1)2

2⇥0 + 1(⇥ 0 1 0 | ⇥0 0 )

2

= I2n0�0n� (⇥ 0 1 0 | ⇥0 0 )2

E&M

Explicit example• Hydrogen atom transition• Radial wave functions

• Lifetime

• using

• in agreement with experiment ...

2p ) 1s

2p ) 1p24a30

r

a0e�r/2a0

1s ) 2

a30e�r/a0

�(2p ! 1s) = 3~ce2

3c2

4⇥3k

I�2 = 1.6⇥ 10�9s

I2 = a20

"4!p6

✓2

3

◆5#2

E&M

General issues related to absorption (emission)• Absorption

– Initial state: assume only one type of photons

– Atom absorbs 1 photon• initial state

• final state

– contribution with so

– “Before” TDPT -->

– Can obtain equivalent classical result by taking classical vector potential

– for large; then do minimal substitution

k� ) nk↵

|nk↵� |A�|nk↵ � 1⇥ |B⇥

Hint ak↵ ak↵ |nk↵⇥ =⇤nk↵ |nk↵ � 1⇥

nk↵

A

abs(x, t) = A

abs0 ei(k·x�⇥kt)

A

abs0 = c

✓2�~n

k�

⇥kV

◆1/2

e

k�

�B| Hint |A⇥ = e

mc

✓2⇤~n

k�

⌅kV

◆1/2

cX

⇥⇤

ek� · ��| ei(k·x�⌅kt) p |⇥⇥ �B| a†⇥a⇤ |A⇥

E&M

Absorption rate in dipole approximation• TDPT

Absorption cross section• Defined --> Energy per unit time absorbed by atom A --> B

energy flux of radiation field

wi�[f ] =2⌅

~e2

m2

2⌅~nk�

⇧kV

��

X

⇥⇤

ek� · ⇥�|p |⇥⇤ ⇥B| a†⇥a⇤ |A⇤

��

2

⇤(EB � EA � ~⇧k)

⇧abs(⌃) =~⌃ wA�B

nk�~⌃c/V

=4⌅2e2

m2⌃c

��

X

⇥⇤

ek� · ⇥�|p |⇥⇤ ⇥B| a†⇥a⇤ |A⇤

��

2

⇤(EB � EA � ~⌃)

E&M

Example• Take photon momentum along z-axis and polarized light --> x-axis• As before use

• Initial state: ground state of closed shell atom

• Final state: excited state

• Simple particle-hole state

• Evaluate• So absorption cross section

• IPM

• and

• thus

px

=m

i~ [x,H0]

|A� = |�0�

|B� = a†n>�0m0�m

0san<�m�ms |�0�

�B| a†�a⇥ |A⇥ = ��0| a†n<an>a

†�a⇥ |�0⇥ = �n>��n<⇥

⇤abs(⌅) =4⇥2e2

~⌅~c |⇥n>⇧0m0

�m0s| [x,H0] |n<⇧m�ms⇤|

2�(EB � EA � ~⌅)

H0 |B⇥ = EB |B⇥ = H0a†n>�0m0

�m0san<�m�ms |�0⇥ = (�n>�0 � �n<� + E�0) |B⇥

H0 |A� = EA |A� = H0 |�0� = E�0 |A� H0 |n>⇥0m0

�m0s� = �n>�0 |n>⇥

0m0�m

0s�

H0 |n<⇥m�ms� = �n<� |n<⇥m�ms�

⌅abs(⇧) =4⇤2e2

~c (⌃n>�0 � ⌃n<�) |⇥n>⌥0m0

�m0s|x |n<⌥m�ms⇤|

2⇥(⌃n>�0 � ⌃n<� � ~⇧)⇥m0

sms

= 4⇤2� ⇧n>n< |⇥n>⌥0m0

�m0s|x |n<⌥m�ms⇤|

2⇥(⇧n>n< � ⇧)⇥m0

sms

E&M

Thomas-Reiche-Kuhn sum rule• Simple model of absorption cross section: delta spike at every

allowed combination of • Dipole matrix element: see before

• Consider integral over all possible absorption contributions

• More general expression

n>�0 ! n<�

Zd⌅ ⇤abs(⌅) = 4⇥2�

X

n>n<

⌅n>n< |�n>⇧0m0

�m0s|x |n<⇧m�ms⇥|

2

Zd⌅ ⇤abs(⌅) =

4⇥2�

~X

B

(EB � EA)

��⇥B|ZX

i=1

xi |A⇤

��

2

=4⇥2�

~X

B

(EB � EA) ⇥A|XZ |B⇤ ⇥B|XZ |A⇤

=4⇥2�

~ {⇥A|XZHXZ |A⇤ � 12 ⇥A|HXZXZ |A⇤ � 1

2 ⇥A|XZXZH |A⇤}

=4⇥2�

~ ⇥A| 12 [XZ , [H,XZ ]] |A⇤

E&M

Evaluate double commutator• Only kinetic contribution of Hamiltonian survives

• and therefore

• Planck’s constant has disappeared --> classical result (Jackson)

Zd⌅ ⇤abs(⌅) =

4⇥2�

~ �A| 12 [XZ , [H,XZ ]] |A⇥

=4⇥2e2

~c~Z~22m

= Z2⇥2c

✓e2

mc2

◆

[XZ

, [H,XZ

]] =1

2m

2

4X

i

xi

,

2

4X

j

p2xj,X

k

xk

3

5

3

5

=1

2m

X

ijk

hxi

,hp2xj, x

k

ii

=1

2m

X

ijk

⇥xi

, (�2i~)pxj�jk

⇤

= � i~m

X

ij

⇥xi

, pxj

⇤= � i~

m

X

ij

i~�ij

=Z~2m

E&M

Absorption cross sections in nature• Atoms

E&M

More• Big molecules

E&M

and more• Silicon

E&M

for nuclei• 197Au nucleus

E&M

and finally• Proton

E&M

Photoelectric effect (--> beginning 1905)• Absorb high-energy photon (energy still much less than electron

rest mass)• Must overcome binding of electron

• Close to threshold Coulomb cannot be neglected for outgoing electron

• At higher energy approximate final electron by plane wave

• Use absorption cross section but replace delta function by appropriate density of final states

• But don’t make dipole approximation!• Initial state

• Final state

• Evaluate density of states for plane wave

|A� = |�0�

|B� = a†kfmsan<�m�ms |�0�

E&M

Density of states• Wave function• As usual etc.

• Energy

• So

• Cross section

�x|kf ⇥ =1⇤Veikf ·x

kxf =

2�

Lnx

Ef =~2k2f2m

(# of states Ef + dEf )� (# of states Ef ) = ⇥d� dEf

=V

(2�)3

Z kf+dkf

0dk k2 d�� V

(2�)3

Z kf

0dk k2 d�

=V

(2�)3d�

Z Ef+dEf

Ef

dEmkf~2 =

V

(2�)3mkf~2 d� dEf

d⌅abs(⇧)

d�=

4⇤2e2

m2⇧c

��

X

⇥⇤

ek� · ��| eik·xp |⇥⇥ �B| a†⇥a⇤ |A⇥

��

2

V

(2⇤)3mkf~2

E&M

General issues related to emission• Emission

– Initial state: assume only one type of photons

– Atom emits 1 photon• initial state

• final state

– contribution with so

– Induced emission

– Can obtain equivalent classical result by taking classical vector potential

– for large; then do minimal substitution

– QM: induced and spontaneous emission on the same footing

k� ) nk↵

|nk↵� |A�

Hint

nk↵

|nk↵ + 1� |B�a†k↵ a†k↵ |nk↵� =

⇥nk↵ + 1 |nk↵ + 1�

�B| H ⇥int |A⇥ = e

m

✓2⇤~(n

k� + 1)

⌅kV

◆1/2 X

⇥⇤

ek� · ��| e�ik·x p |⇥⇥ �B| a†⇥a⇤ |A⇥

Aemis0 = c

✓2�~(nk� + 1)

⇥kV

◆1/2

ek�

E&M

Other applications• Remember• Photon scattering can also be handled with this Hamiltonian

• Squared vector potential term contributes directly

• Linear terms in vector potential should be considered simultaneously in second order

Hint =ZX

i

e

2mc(pi ·A(xi, t) +A(xi, t) · pi) +

e2

2mc2A(xi, t) ·A(xi, t)

�

�B;nk0�0 = 1|H 0int |A;nk� = 1⇥

E&M

Towards Planck’s radiation law --> 1900• Consider atoms and radiation field that exchange energy by a

reversible process such that thermal equilibrium is established

• population of higher level

• population of lower level

• Equilibrium

• and also

• emission

• absorption

A , � +B

N(A)

N(B)

N(B) wB!Aabs = N(A) wA!B

emis

�B| H ⇥int |A⇥ = e

m

✓2⇤~(n

k� + 1)

⌅kV

◆1/2 X

⇥⇤

ek� · ��| e�ik·x p |⇥⇥ �B| a†⇥a⇤ |A⇥

�A| H �int |B⇥ = e

m

✓2⇤~n

k�

⌅kV

◆1/2 X

⇥⇤

ek� · ��| eik·x p |⇥⇥ �A| a†⇥a⇤ |B⇥

N(A)

N(B)=

e�EA/kBT

e�EB/kBT= e�~�k/kBT ~�k = EA � EB

E&M

Thermal occupation of modes• Ratio of rates

• and therefore

• So thermal occupation

• Familiar?

• Onward to Planck!

wA⇥Bemis

wB⇥Aabs

=(n

k� + 1)��P

⇥⇤ ek� · ��| e�ik·x p |⇥⇥ �B| a†⇥a⇤ |A⇥��2

nk�

��P

⇥⇤ ek� · ��| eik·x p |⇥⇥ �B| a†⇥a⇤ |A⇥��2 =

nk� + 1

nk�

N(B)

N(A)=

wA!Bemis

wB!Aabs

= e~⇥k/kBT =nk� + 1

nk�

nk�(T ) =1

e~⇥k/kBT � 1

E&M

Derivation of Planck• Consider radiation in a black box / cavity• Made of atoms that emit and absorb all types of radiation

• Use previous results to determine energy density of radiation field in angular frequency interval

• Familiar calculation: count contribution of all states in interval

• Before

• Now all angles and polarizations --> 2

• Multiply with energy X population per volume

� + d�,�

(# of states (⇤ + d⇤)� (# of states ⇤) = ⇥! d⇤

) V

(2�)3⇤2

c3d⇤ d�

d� ) 4�

U(⇥) =1

e~�/kBT � 1

~⇥V

4� 2V

(2�)3⇥2

c3=

~⇥3

�2c31

e~�/kBT � 1

E&M

Planck --> 1900 where it all began• Switch to frequency distribution

• Planck’s famous radiation law!

U(�) = U(⇤)d⇤

d�=

~⇤3

⇥2c31

e~⇥/kBT � 12⇥

=8⇥h�3

c31

eh�/kBT � 1

The end! Thanks for an enjoyable year!

interaction of electrons with photonswimd/524-13em2.pdf• use radiation gauge • vector potential...

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