1

1

Brief review of quantum mechanics. Quantum-mechanical

perturbation theory for the nonlinear optical susceptibility.

1st, 2nd and 3rd order susceptibilities. Susceptibility

resonances.

Lecture 4

1

2

QUANTUM MECHANICS FOR PEDESTRIANS

𝑖ℏ�̇� = &𝐻𝜓 the reduced Planck’s constant

𝜓- wave function (is the state vector of the quantum system). Contains all information about the system

&𝐻 - Hamiltonian operator (~ total energy)

The most general form is the time-dependent Schrödinger equation

or

𝑖ℏ𝑑𝜓𝑑𝑡 =

&𝐻𝜓

To apply the Schrödinger equation, write down the Hamiltonian &𝐻 for the system, accounting for the kinetic and potential energies of the particles constituting the system, then insert it into the Schrödinger equation. The resulting partial differential equation is solved for the wave function 𝜓.

=6.626×10−34/2π = 1.0546×10−34 J.s

For example, the probability of finding a particle at the position r is ~ |ψ(r)|2 =ψ(r)ψ(r)*

(4.1)

*ψψ∗𝑑,𝑟 = 1 – i.e. a particle should be somewhere

2

2

3

Quantum mechanics for pedestriansIn the coordinate representation, quantum mechanical operators are represented by :

ordinary numbers for positions :/𝑥 -> x

by differential operators for momenta:�̂� -> 𝑖ℏ 3

34

If U does not depend on time, the Schrödinger equation allows stationary solutions in the form of energy eigenstates 𝜓5 with time evolution given by a simple exponential phase factor:

Then (4.1)

𝜓5 𝑟, 𝑡 = 𝑢5 𝑟 𝑒9:;<=/ℏ

&𝐻𝑢5 = 𝐸5𝑢5 (4.2)

This is where quantization comes from !

For a free particle the Hamiltonian &𝐻 becomes:

Here, the form of the Hamiltonian operator comes from classical mechanics, where the Hamiltonian function is the sum of the kinetic and potential energies (Similar to p2/2m+U(x) expression in mechanics)

&𝐻 = −ℏA

2𝑚𝑑A𝜓𝑑𝑥A + 𝑈(𝑥)

kinetic energy potential energy

𝑢5 – spatially varying part of the wavefunction

time-independent Schrödinger equation𝑖ℏ𝑑𝜓𝑑𝑡 =

&𝐻𝜓 becomes :

3

4

The particle-in-a-box problemA particle that can move in only one dimension:

Quantum well: U = 0 between 0 < x < a, and U = ∞ outside 0 <x < aExample: semiconductor quantum well: a thin (few nm) layer of material with low bandgap (Eg) is sandwitched between two layers with high Eg

x=0 x=a

U=0

The Hamiltonian &𝐻 : &𝐻 = −ℏA

2𝑚𝑑A𝜓𝑑𝑥A + 𝑈(𝑥)

Solve (4.2): &𝐻𝑢5 = 𝐸5𝑢5

Rewrite this:

−ℏA

2𝑚𝑑A𝜓𝑑𝑥A

+ 0 = 𝐸5𝜓 inside the well-> ℏI

AJ3IK34I + 𝐸5𝜓 =0

and 𝜓=0 outside the well3IK34I + 𝜅

A𝜓 =0, 𝜅A=𝐸5 / ( ℏI

AJ)

E3

E2

E1

Solution: 𝜓 = 𝐶𝑜𝑛𝑠𝑡 Q sin(𝜅𝑥), boundary condition : 𝜅a = π𝑛, 𝑛 = 1,2,3 . .

-> 𝜅A=𝐸5 / ( ℏI

AJ)=(π𝑛/𝑎)A, -> 𝐸5= ℏI

AJ(Z[)A𝑛A, n=1,2,3 ...

Energy is quantized (!) and scales as n2

n=3

n=2

n=1

4

3

5

Susceptibilities – derived via perturbation solution to Schrödinger’s equation

In optics: we have a free atom with a Hamiltonian \𝐻] and an external EM field, which we regard as a perturbation:&𝐻 = \𝐻] + �̂�(t)

(after Boyd book)

perturbation

Now introduce 𝜆 (0 < 𝜆 < 1) a ‘tuning’ parameter (strength of the interaction): 𝜆 = 0 field is off; 𝜆 = 1 field is on &𝐻 = \𝐻] + 𝜆 �̂�(t)

Seek a solution to Schrödinger’s equation in the form of a power series in λ

𝜓 𝑟, 𝑡 = 𝜓 ] 𝑟, 𝑡 + 𝜆𝜓 b 𝑟, 𝑡 + 𝜆A𝜓 A 𝑟, 𝑡 + …

Perturbation technique in mathematics – start from the exact solution of a related, simpler problem and then solve the main problem by adding correction terms.

Plug this 𝜓 𝑟, 𝑡 into the main Eq. (4.1): 𝑖ℏ 3K3== &𝐻𝜓 and require that terms proportional to λN satisfy the equality separately:

𝑖ℏ(3Kd

3=+ 𝜆 3K

e

3=+ 𝜆A 3K

I

3=+. . ) = \𝐻]𝜓 ] +\𝐻]𝜆𝜓 b +\𝐻]𝜆A𝜓 A + 𝜆�̂�𝜓 ] + 𝜆A�̂�𝜓 b + 𝜆,�̂�𝜓 A +..

(0 -order approxim.)

(1-st -order)

(2-nd -order).......

(N-th -order)

𝑖ℏ 3Kd

3== \𝐻]𝜓 ]

𝑖ℏ 3Ke

3== \𝐻]𝜓 b + �̂�𝜓 ]

𝑖ℏ 3KI

3== \𝐻]𝜓 A + �̂�𝜓 b

– simply Schrödinger’s equation for the atom in the absenceof its interaction with the applied field

(4.3)

(4.4)

(4.5)𝑖ℏ 3Kf

3== \𝐻]𝜓 g + �̂�𝜓 g9b

......

5

6

QM perturbation solution

3

2

1

Initially, atom is in the state 1 (ground state) so that the solution to the 0-order equation is:

𝜓(]) 𝑟, 𝑡 = 𝑢b 𝑟 𝑒9:;e=/ℏ = 𝑢b 𝑟 𝑒9:he=, 𝑤𝑖𝑡ℎ 𝜔b=𝐸b/ℏ

Now, expand 𝜓(g) (N=1,2,3...) as a sum of energy eigenfunctions of an unperturbed system:

𝑖ℏ∑m(�̇�mg − 𝑖𝜔m𝑎m

g ) 𝑢m 𝑒9:hn== ∑m \𝐻]𝑎mg 𝑢m𝑒9:hn= + ∑m �̂�𝑎m

g9b 𝑢m𝑒9:hn=

𝜓(g) 𝑟, 𝑡 =om𝑎mg (𝑡) 𝑢m 𝑟 𝑒9:hn= 𝑤𝑖𝑡ℎ 𝜔m=𝐸m/ℏ

probability amplitude

(4.6)

time independent energy eigenfunction

its exponential phase factor

cancelcancelNow plug (4.6) into (4.5) :

𝑖ℏ∑m �̇�mg 𝑢m 𝑒9:hn== ∑m �̂�𝑎m

g9b 𝑢m𝑒9:hn=

multiply each side from the left by 𝑢J∗ and integrate over all space (take into account orthogonality for 𝑚 ≠ 𝑙 )

𝑖ℏ�̇�Jg 𝑒9:hr== ∑m 𝑉Jm𝑎m

g9b 𝑒9:hn=

�̇�Jg (𝑡) =(𝑖ℏ)9b ∑m 𝑎m

(g9b)𝑉J5𝑒:hrn=

where we have introduced the matrix elements of the perturbing Hamiltonian

(4.7)

Dirak notation

|𝑎mb |2 is the probabilty of

being at a given energy state

𝑢m 𝑟 – constitute a complete set of orthogonal basis functions in the sense ∫𝑢J∗ 𝑢5 𝑑,𝑟 = 1 𝑖𝑓 𝑚 = 𝑛, and = 0 𝑖𝑓 𝑚 ≠ 𝑛

n

à

6

4

7

1st -order perturbation solution

�̇�Jb (𝑡) =(𝑖ℏ)9b 𝑉Jb𝑒:hre=

In QM, the interaction Hamiltonian of the atom with the electromagnetic field is the form:

where w𝝁 = −𝑒/𝒓 , is the electric dipole moment operator and −e is the charge of the electron.

Calculate 1st -order correction – induced linear polarization P (1)

From (4.7) and N=1, we get

𝑉Jm = −𝜇Jm 𝐸 𝑡 where the matrix element is the electric dipole transition moment. we get:

With an external EM wave electric field in the form: 𝐸 𝑡 = 𝐸𝑒9:h=

(4.8)

𝑎Jb (𝑡) =∫9{

= (𝑖ℏ)9b 𝑉Jb𝑒:hre=| 𝑑𝑡}=∫9{

= (𝑖ℏ)9b (−𝜇Jb)𝐸𝑒9:h=|𝑒:hre=|𝑑𝑡}By integrating (4.8) we get

(4.9)

(4.10)

=:ℏ ∫9{

= 𝜇Jm𝐸𝑒:(hre9h)=|𝑑𝑡} = ~re

ℏ;��(�re��)�

(hre9h)

𝜓(b) = ∑J𝑎Jb (𝑡) 𝑢J 𝑟 𝑒9:hr==∑J

~reℏ

;���(�e��)�

(hre9h)𝑢J

Finally,

7

8

1st -order perturbation solution𝜓 𝑟, 𝑡 = 𝜓 ] 𝑟, 𝑡 + 𝜓 b 𝑟, 𝑡Our 1-st order-corrected time-dependent wave function is:

Polarization induced per atom is: 𝑝(b) = 𝜓 ] + 𝜓 b |w𝝁| 𝜓 ] + 𝜓 b

only this combination gives polarization proportional to E𝜓 ] |w𝝁| 𝜓 b + 𝜓 b |w𝝁| 𝜓 ]

𝜓 ] |w𝝁| 𝜓 b = 𝑢b𝑒9:he= |w𝝁| ∑J~reℏ

;���(�e��)�

(hre9h)𝑢J =∑J

~er~reℏ

;�����

(hre9h)

𝑝(b) = ∑J~er~re

ℏ;�����

(hre9h)+ 𝑐. 𝑐.Thus This is an electric dipole moment of a

single atom induced by the field E at frequency 𝜔

– this is the linear (1st -order) susceptibility

The linear polarization (dipole moment per unit volume) is: 𝑃(b) = 𝑁𝑝(b)and since 𝑃(b) = 𝜖]𝜒(b)𝐸, and 𝐸 = Re 𝐸𝑒9:h= , we get ∶

(N is the density of atoms)

𝜒(b) =𝑁𝜖]ℏ

oJ

|𝜇bJ|A

(𝜔Jb−𝜔)(4.11)

+its complex conjugate

8

5

9

1st -order perturbation solution

3

2

1

Let us simplify (11) - take a system with just 3 levels (”1” is the ground state) and leave only resonant terms

𝜔𝜔,b

𝜔Ab

𝜒(b) =g�dℏ

∑J|~er|

I

(hre9h)

𝑁𝜖]ℏ

|𝜇bA|A

(𝜔Ab−𝜔)+

𝑁𝜖]ℏ

|𝜇b,|A

(𝜔,b−𝜔)

This formula makes sense: susceptibility 𝜒, and refractive index

𝑛 = 1 + 𝜒– grow with frequency between the two poles

𝜔𝜔Ab 𝜔,b

𝜒

(4.12)

9

10

2nd -order nonlinearity – perturbation solutionNow, let us calculate the 2nd -order correction – nonlinear polarization P(2)

𝜓 b = 𝑎bb 𝑢b𝑒9:he= + 𝑎A

b 𝑢A𝑒9:hI= + 𝑎,b 𝑢,𝑒9:h�= = ∑: {

~IeAℏ( ;�������

(hIe9h��)+ ;�∗�����

(hIe h��))𝑢A +

~�eAℏ( ;�������

(h�e9h��)+ ;�∗�����

(h�e h��))𝑢,}

�̇�JA (𝑡) = (𝑖ℏ)9b ∫ 𝑑,𝑟 𝑢J∗ �̂� ∑:

~IeAℏ( ;�������

(hIe9h��)+ ;�∗�����

(hIe h��))𝑢A +

~�eAℏ( ;�������

(h�e9h��)+ ;�∗�����

(h�e h��))𝑢, 𝑒:hr= =

From the previous result (1st -order approx.)

Assume we have two external EM fields witℎ 𝜔¢b and witℎ 𝜔¢A: 𝐸 𝑡 = 𝑅𝑒(𝐸b𝑒9:h�e= + 𝐸A𝑒9:h�I=)=∑:bA(𝐸:𝑒9:h��= + 𝐸:𝑒:h��=)

𝑖 = 1.2

=0 because 𝜇bb = 0 (𝑛𝑜 𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑛𝑡 𝑑𝑖𝑝𝑜𝑙𝑒)

from (4.7) stating that : �̇�Jg (𝑡) =(𝑖ℏ)9b ∑m 𝑎m

(g9b)𝑉J5𝑒:hrn=

it follows (for N=2)

Assume a system with just 3 levels

10

6

11

2nd -order nonlinearity – perturbation solutionThe 2nd-order contribution to the induced dipole moment per atom is :

< 𝑝(A) > = < 𝜓 ] + 𝜓 b + 𝜓 A |w𝝁| 𝜓 ] + 𝜓 b + 𝜓 A >

since only this combination gives polarization proportional to E2

we have from previous:𝜓(]) 𝑟, 𝑡 = 𝑢b 𝑟 𝑒9:;e=/ℏ = 𝑢b𝑒9:he=, 𝜔b=𝐸b/ℏ

𝜓(b) 𝑟, 𝑡 = bAℏ∑J,¢ 𝐸(𝜔¢)

~re(hre9h�)

𝑢J𝑒9:(he h�)=m = 2,3𝜔¢ = ±𝜔¢b; ±𝜔¢A

𝜓(A) 𝑟, 𝑡 = ∑5 𝑎5A (𝑡) 𝑢5𝑒9:h<==

b¦ℏI

∑J,5,§,¢~<r~re;(h�); h¨

(h<e9h�9h¨)(hre9h�)𝑢5𝑒9:(he h� h¨)=

< 𝑝(A) > =...

< 𝑢b 𝑒9:he=|𝜇|b¦ℏI

∑J,5,§,¢~<r~re;(h�); h¨

(h<e9h�9h¨)(hre9h�)𝑢5𝑒9:(he h� h¨)= >= b

¦ℏI∑J,5,§,¢

~e<~<r~re;(h�); h¨(h<e9h�9h¨)(hre9h�)

𝑒9:(h� h¨)=

n,m = 2,3

< bAℏ∑J,¢ 𝐸(𝜔¢)

~re(hre9h�)

𝑢J𝑒9:(he h�)= |𝜇| bAℏ∑J,¢ 𝐸(𝜔¢)

~re(hre9h�)

𝑢J𝑒9:(he h�)=>= b¦ℏI

∑J,5,§,¢~e<~<r~re;(h�); h¨(h<e9h¨)(hre9h�)

𝑒9:(h�9h¨)=

c.c.

𝑖𝑡𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒+

+

11

12

2nd -order nonlinearity – perturbation solution

Finally, the 2nd-order contribution to the induced dipole moment per unit volume:

𝑃(A) = 𝑁𝑝(A)

3

2

1

Restrict ouservesto a 3-level system

𝜒(A) = g�dℏI

∑J,5,§,¢ {~e<~<r~re

(h<e9h�9h¨)(hre9h�)+ ~e<~<r~re

(hre h� h¨)(h<e h¨)+ ~e<~<r~re

(h<e h¨)(hre9h�)}

The nonlinear succeptibility is :

(we omitted some details of derivation - see Stegeman book)

(4.13)

𝜔¢ and𝜔§ run through ±𝜔¢b; ±𝜔¢A 𝑚= 2,3,… 𝑛 = 2,3,…

Even for a 3-level system, there are 2x2x4x4=64 elements in this sum !

Sum over all possible pairs of states. Sum over p and q are over all of the incident fields.

12

7

13

2nd -order perturbation solution

3

2

1

𝜔,b

𝜔Ab 𝜔𝑝

𝜔,A 𝜔𝑞𝜔𝑝 + 𝜔𝑞

For the sum frequency generation 𝜔, = 𝜔b + 𝜔A, let us now select only resonant terms in (4.13)

𝜒(A) --> g�dℏI

{ ~e�~�I~Ie(h�e9h�9h¨)(hIe9h�)

+ ~e�~�I~Ie(h�e9h�9h¨)(hIe9h¨)

+

+ ~e�~�I~Ie(h�e9h¨)(hIe9h�)

+ ~e�~�I~Ie(h�e9h�)(hIe9h¨)

} + ....

Strictly speaking, transition frequencies 𝜔Ab , 𝜔,b need to be complex quantities 𝜔 → 𝜔+𝑖𝛾, to incorporate damping phenomena into the theory. This allows to avoid infinities at exact resonances.

double resonance: at 𝜔,b and 𝜔Ab𝜔¢ , 𝜔¢ - two pump freqencies

𝜔b → 𝜔¢, 𝜔A → 𝜔§, 𝜔, → 𝜔¢ + 𝜔§

13

14

3nd -order nonlinearity – perturbation solution

The QM microscopic expression for 3rd order nonlinear succeptibility 𝜒(,) looks even more complicated !

see Stegeman or Boyd book

14

8

15

Model system for optical 𝜒(A) nonlinearities: semiconductor quantum well

E3

E2

E1

infinite-barrier quantum well (QW)

𝜒(A) --> g�dℏI

{ ~eI~I�~�e(h�e9h�9h¨)(hIe9h�)

+ ~eI~I�~�e(h�e9h�9h¨)(hIe9h¨)

+...}

Symmetric semiconductor quantum well

From previous page:

z àa

z=0

Note that in a symmetric quantum well𝜇bA≠0, 𝜇A,≠0, but 𝜇,b =0.

Recall that 𝜇,b=−𝑒 ∫𝑢,∗ 𝑧 𝑢b𝑑𝑧 =0;

Hence 𝜇b,𝜇,A𝜇Ab=0

This is quite clear, because there should be some asymmetry to achieve 𝜒(A)

even functioneven functionodd function

~ 5-10 nm

𝜔¢ , 𝜔¢ - two pump freqencies

15

16

2nd -order nonlinearity – perturbation solution

(𝜓b + 𝜓A)2

(𝜓A + 𝜓,)2

(𝜓b + 𝜓,)2

𝜓b

𝜓A

𝜓,

16

9

17

Model system for optical nonlinearities: semiconductor quantum well

Asymmetric quantum well : now all 𝜇bA≠0, 𝜇b,≠0, and 𝜇,b ≠0.

3

2

1

Morover, SHG (𝜔 + 𝜔 = 2𝜔) with double resonance : 𝜔~𝐸12, 2𝜔~𝐸13

17