Multi Mode Quant

8/12/2019 Multi Mode Quant

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31.05.07

Quantum Optics. Multimode Field Quantization

Dr. B.I.Lembrikov

I. Electromagnetic Radiation in the Coulomb Gauge

We start with the Maxwell equations in the free space without external electric

current density jand charge density :

j = 0; = 0 (1)

They have the form

curlE =@B@t

(2)

curlH = @D

@t (3)

divB = 0 (4)

divD = 0 (5)

where the electric induction D, the magnetic inductionB, and the electric eldE, the magnetic eld Hare related by

D ="0E;B =0H (6)

Taking into account equation (4) we introduce now the vector potential A asfollows

B =curlA (7)

sincedivcurlA 0

Substitute (6) and (7) into (2) and (3). We obtain

curlE =@

@t (curlA) (8)

and

curlcurlA =0"0@E

@t (9)

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From (8) we get

curlE + @A@t = 0 (10)which gives

E +@A

@t =grad (11)

where is a scalar potential, and we used the identity

curl (grad)0 (12)Then

E =@A@t

grad (13)Substitute (13) into (9). Using the identity

curlcurlA =graddivA r2A (14)we obtain

graddivA r2A =0"0@

@t

@A

@t grad

(15)

and

r2A+1c2

@2A

@t2 +grad

divA+

1

c2@

@t

= 0 (16)

where

0"0 = 1

c2 (17)

Any vector function A is dened by curlA anddivA. We now use the so-calledCoulomb gauge, or the transverse gauge in order to dene the vector potentialA.

divA = 0; = 0 (18)

The advantage of the Coulomb gauge is that the radiation eld is completelydescribed by the vector potential A:

E =@A@t

;B =curlA ! H = 10

curlA (19)

The free space can be modeled as a cubic cavity of side Lwith perfectly reectingwalls, where L is very large compared with dimensions of atoms. After allcalculations the cavity dimension is excluded from the results by taking thelimit L! 1. The periodic boundary conditions on the faces of the cavity

give:exp ikxx= exp [ikx(x+L)]!kx= 2

Ll ; l= 0; 1; 2; ::: (20)

Similarly,

ky =2

Lm; m= 0; 1; 2; :::; kz =2

Ln; n= 0; 1; 2; ::: (21)

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where k = (kx; ky; kz)is the wave vector. In the Coulomb gauge eq. (16) reducesto the wave equation

r2A+1c2

@2A@t2

= 0 (22)

and

r2A =

@2

@x2+

@2

@y2+

@2

@z2

A (23)

We expand the solution in the Fourier series with respect to the plane waves.

A =Xk;s

ek;s

Ak;s(t)exp(ikr) +Ak;s(t)exp(ikr)

(24)

where ek;s is the polarization vector. For the transverse wave dened by (18)there are two independent orthogonal polarizations s such that

(ek;sek;s0) = ss0= 1 s= s00 s6=s0 (25)and

(ek;sks) = 0 (26)Substitute (24) into (22). Taking into account that

r2 (exp(ikr)) =k2 exp(ikr) (27)we get the following equation for the amplitudes Ak;s(t).

@2Ak;s

@t2 +!2kAk;s= 0 (28)

where!k=

kc (29)

The solution of (28) is well known. It has the form

Ak;s= Ak;s(0) exp (i!kt) (30)Then (24) takes the form

A =Xk;s

ek;s

Ak;s(0) exp (i (kr !kt)) +Ak;s(0) exp (i (kr !kt)) (31)Using (19) we obtain

E =@A@t

=iXk;s

ek;s!k

Ak;s(0) exp (i (kr !kt)) Ak;s(0) exp (i (kr !kt)) (32)and

B =curlA =iXk;s

[k ek;s]

Ak;s(0) exp (i (kr !kt)) Ak;s(0) exp (i (kr !kt)) (33)3


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II. Classical Expression of the Electromagnetic Field En-

ergy

The energy of the electromagnetic eld in a classical case has the form

Hem=1

2

ZV

dV

"0E

2 + 1

0B2

(34)

Evaluate (34). We start with the electric eld energy.

E2 = (E E) =Xk;s

Xk0;s0

(ek;sek0;s0) !k!k0

[Ak;s(0) Ak0;s0(0) exp [i (k + k0) r i (!k+!k0) t]

+Ak;s(0) Ak0;s0(0) exp [i (k + k0) r +i (!k+!k0) t]Ak;s(0) Ak0;s0(0) exp [i (k k0) r i (!k !k0) t]

Ak0;s0(0) Ak0;s0(0) exp [i (k k0) r +i (!k !k0) t]] (35)Substitute (35) and evaluate integrals. First of all we use the condition (25).Then we have

ZV

dV "0E2 ="0

Xk;s

Xk0s0

(ek;sek0;s0)LZ0

dx

LZ0

dy

LZ0

dz

[Ak;s(0) Ak0;s0(0) exp [i (k + k0) r i (!k+!k0) t]

+Ak;s(0) A

k0;s0(0) exp [i (k + k

0

) r +i (!k+!k0) t]Ak;s(0) Ak0;s0(0) exp [i (k k0) r i (!k !k0) t]

Ak0;s0(0) Ak0;s0(0) exp [i (k k0) r +i (!k !k0) t]] (36)

Consider the rst integral in the right hand side (RHS) of (36). Taking intoaccount (20), (21) we obtain

LZ0

dx

LZ0

dy

LZ0

dzexp (i (k + k0) r) =

LZ0

dx exp(i (kx+k0x) x)

L

Z0

dy exp i ky+ k0y yL

Z0

dzexp (i (kz

+ k0z

) z)

LZ0

dx exp(i (kx+k0x) x) =

L k0x=kx

0 k0x6=kx =Lk0

x(kx)

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and nally

LZ0

dxLZ0

dyLZ0

dzexp (i (k + k0) r) = L3k0x(kx)k0y(ky)k0z(kz)

=V k0(k) (37)

SimilarlyLZ0

dx

LZ0

dy

LZ0

dzexp (i (k + k0) r) = V k0(k) (38)

LZ0

dx

LZ0

dy

LZ0

dzexp [i (k k0) r] = V kk0 (39)

Substituting (37)-(39) into (36) and taking into account that according to (29)

k =k0 !!k = !k0 (40)we obtainZ

V

dV "0E2 ="0V

Xk;s

Xk0;s0

2kk0(ek;sek0;s0) !k!k0

Ak;s0(0) Ak0;s0(0)

"0VXk;s

Xk0;s0

k(k0)(ek;sek0;s0) !k!k0

Ak;s(0) Ak0;s0(0) +Ak;s(0) A

k0;s0(0)

= 2"0VXk;s!2kAk;s(0) A

k;s(0)

"0VXk;s;s0

!2k(ek;sek;s0)

Ak;s(0) Ak;s0(0) +Ak;s(0) A

k;s0(0)

(41)

where we used the condition (25), and as a result the summation overs0 vanishes.

Now evaluate the magnetic eld contribution to the energy (34).ZV

dV 1

0B2 =

ZV

dV 1

0

Xk;s

Xk0;s0

([k ek;s] [k0 ek0;s0 ])

[Ak;s(0) Ak0;s0(0) exp [i (k + k0) r i (!k+!k0) t]+Ak;s(0) A

k0;s0(0) exp [i (k + k0) r +i (!k+!k0) t]

Ak;s(0) Ak0;s0(0) exp [i (k k0) r i (!k !k0) t]A

k0;s0(0) Ak0;s0(0) exp [i (k k0) r +i (!k !k0) t]] (42)We rst use the identity

([k ek;s] [k0 ek0;s0]) =k k0 (ek;sek0;s0)

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(k ek0;s0) (k0 ek;s) (43)

Then we once more evaluate the integrals similarly to the previous case. Thenwe obtain ZV

dV 1

0B2 = V

0

Xk;s

Xk0;s0

k k0 (ek;sek0;s0)

k(k0)

Ak;s(0) Ak0;s0(0) +Ak;s(0) A

k0;s0(0)

+

V

0

Xk;s

Xk0;s0

k k0 (ek;sek0;s0) 2kk0Ak;s(0) Ak0;s0(0)

+V

0

Xk;s

Xk0;s0

(k ek0;s0) (k0 ek;s)

k(k0) Ak;s(0) Ak0;s0(0) +A

k;s(0) A

k0;s0(0)

2V0

Xk;s

Xk0;s0

(k ek0;s0) (k0 ek;s) kk0Ak;s(0) Ak;s(0)

= V

0

Xk;s;s0

k2 (ek;sek;s0)

Ak;s(0) Ak;s0(0) +Ak;s(0) A

k;s0(0)

+2V

0

Xk;s

k2Ak;s(0) Ak;s(0) (44)

where(k ek0;s0) (k0 ek;s) k(k0) =(k ek;s0) (k ek;s) = 0 (45)

and

(k ek0;s0) (k0 ek;s) kk0 = (k ek;s0) (k ek;s) = 0 (46)according to the transversality condition (26). Substitute (41) and (46) into(34).

Hem=1

22"0V

Xk;s

!2kAk;s(0) Ak;s(0)

12

"0VXk;s;s0

!2k(ek;sek;s0)

Ak;s(0) Ak;s0(0) +Ak;s(0) A

k;s0(0)

+1

2

2V

0

Xk;s

k2Ak;s(0) Ak;s(0)

+

1

2

V

0 Xk;s;s0

k2

(ek;sek;s0) Ak;s(0) Ak;s0(0) +Ak;s(0) Ak;s0(0) (47)Taking into account that

k2

0=

!2kc20

=!2k"00

0=!2k"0 (48)

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and substituting (48) into the third and fourth terms of (47) we obtain

Hem= 2"0VXk;s

!2kAk;s(0) Ak;s(0)

12

"0VXk;s;s0

!2k(ek;sek;s0)

Ak;s(0) Ak;s0(0) +Ak;s(0) A

k;s0(0)

+1

2"0V

Xk;s;s0

!2k(ek;sek;s0)

Ak;s(0) Ak;s0(0) +Ak;s(0) A

k;s0(0)

= 2"0V

Xk;s

!2kAk;s(0) Ak;s(0) (49)

III. Quantization of the Multi Mode Field

In order to quantize the classical Hamiltonian (49) we introduce the classicalcanonical variables having the meaning of the coordinate qks and momentum

pks.

Ak;s= 1

2!kp

"0V[!kqks+ipks] (50)

Ak;s= 1

2!kp

"0V[!kqks ipks] (51)

Substitute (50) and (51) into (49). We obtain.

Hem= 2"0VXk;s

!2kAk;s(0) Ak;s(0) = 2"0V

1

4"0V

Xk;s

!2k1

!2k

!2kq

2ks+p

2ks

=12Xk;s

!2kq2ks+p2ks (52)which corresponds to the sum of unit mass harmonic oscillator energies

!2kq

2ks+p

2ks

.

Now we replace the classical variables with operatorsbqks andbpks satisfying thecommutation relations

[bqks;bqk0s0] = 0; [bpks;bpk0s0] ; [bqks;bpk0s0] = ihkk0ss0 (53)These relations dier from the one particle operators relations with functionsof Kroneker. This dierence simply means that the operators of the dierentoscillators are independent. The Hamiltonian operator takes the form.

bHem=1

2 Xk;s !2kbq

2ks+bp

2ks (54)

Now we introduce the annihilation and creation operators for each oscillatorsimilarly to the single mode eld.

baks = 1p2h!k

[!kbqks+ibpks] (55)7


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bayks =

1p2h!k

[!k

bqks i

bpks] (56)

Using the commutation relations (53) we once more obtain similarly to the singlemode eld quantization the following commutation relations for the photonoperatorsbaks andbayks.

[baks;bak0s0] = 0; hbayks;bayk0s0i= 0 (57)hbaks;bayk0s0i= kk0ss0 (58)Evidently, the photon annihilation and creation operators of the same oscilla-tor with the same k, s have the same commutation relations as in the singlemode eld, while the operators of dierent oscillators are commutative. Fromexpressions (55) and (56) we have

bqks= p2h!k2!k

hbaks+bayksi=r

h

2!k

hbaks+bayksi (59)and

bpks=p

2h!k2i

hbaksbayksi= ir

h!k2

hbayksbaksi (60)

Substitute (59), (60) into (54). We obtain

bHem= 12

Xk;s

!2k

h

2!k

baks+bayks2 h!k2bay

ksbaks2

=

1

2 Xk;s

h!k2 ba2ks+ bayks

2

+ 2baksbayksbayks2

ba2ks+ 2bayksbaks=

1

2

Xk;s

h!khbay

ksbaks+baksbayksi (61)According to condition (58) we havehbaks;bayksi= 1!baksbayks= 1 +bayksbaks (62)Substituting (62) into (61) we nally obtain

bHem=

1

2 Xk;sh!k

hb

ayks

baks+ 1 +

bayks

baks

i

=Xk;s

h!k

bayksbaks+ 12

(63)

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Expression (63) shows that the multimode electromagnetic eld can be presentedas a number of harmonic oscillators with dierent frequencies. We dene the

photon number operatorbnks as followsbnks =bayksbaks (64)

and bHem= Xk;s

h!k

bnks+ 12

(65)

Each of the modes characterized by the wave vector k and polarization s isindependent from other modes. It has an associated set of number eigenstatesjnksi. The eigenstates of the j th mode can be dened asjnjiwherebnkjsjbnj .Then the Hamiltonian takes the form.

bHem= Xj

h!j bnj+

1

2 (66)

The multimode photon number state is a product of the number states of allmodes which has the form

jn1i jn2i jn3i ::: jn1; n2; n3;:::i=jfnjgi (67)This is an eigenstate of the HamiltonianbHem such thatbHem jfnjgi= Ejfnjgi (68)where

E=Xj

h!j

nj+

1

2

(69)

These number states are orthogonal:

hn1; n2; n3;:::jn01; n02; n03;:::i= n1n01n2n02n3n03 ::: (70)The action of annihilation and creation operators on thej th mode is

bajjn1; n2; n3;:::nj ; ::i=pnjjn1; n2; n3; :::nj 1; ::i (71)bayjjn1; n2; n3; :::nj ; ::i= pnj+ 1 jn1; n2; n3;:::nj+ 1; ::i (72)

The multimode vacuum state is dened as follows.

jf0gi=j01; 02; 03;:::0j ; ::i (73)and

bajjf0gi= 0 (74)

for all j. All number states can be generated from the vacuum state similarlyto the single mode eld.

jfnjgi=Yj

bayjnjp

njjf0gi (75)

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IV. Operators of the Field

Comparing (50), (51) and (59), (60) we obtain the vector potential amplitudeoperators. Namely

Ak;s= 1

2!kp

"0V[!kqks+ipks] =

= 1

2!kp

"0V

p2h!kbaks =

r h

2!k"0Vbaks (76)

and

Ak;s=

1

2!kp

"0V[!kqks ipks]

= 1

2!kp"0Vp2h!kbayks

= r h

2!k"0Vbayks

(77)

The quantized vector potential takes the form.

bA = Xk;s

ek;s

r h

2!k"0V

hbaksexp (i (kr !kt)) +bayksexp (i (kr !kt))i (78)

The electric eld operator has the form

bE (r; t) = i

Xk;s

ek;s

rh!k2"0V

hbaksexp (i (kr !kt)) bayksexp (i (kr !kt))i (79)

and the magnetic eld operator is given by

bB =iXk;s

[k ek;s]r

h

2!k"0V

hbaksexp (i (kr !kt)) bayksexp (i (kr !kt))i (80)

The electric eld operator (79) can be written as follows.

bE (r; t) =bE(+) (r; t) +bE

() (r; t) (81)

wherebE(+) (r; t) is called the positive frequency part and has the formbE(+) (r; t) = iX

k;s

ek;s

rh!k2"0V

baksexp (i (kr !kt)) (82)

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The term (82) is a collective annihilation operator. The term

bE() (r; t)is called

the negative frequency part and it is given by

bE() (r; t) =iXk;s

ek;s

rh!k2"0V

bayksexp (i (kr !kt)) (83)

The term (83) is a collective creation operator.

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