Download 1 Quantization of the Electromagnetic Field

1
Quantization of the Electromagnetic Field
1.
The electric and magnetic field are given by the vector potential:
!
E
!
B
!
1 ∂A
c ∂t
!
= ∇×A
= −
The electromagnetic energy reads:
H=
!
 $

%2
&
'2
!
1
1
∂
A
! 

d3!r
+ ∇×A
8π c2 ∂t
The time dependence of the vector potential can be restored by applying the rule:
!
â(!k, λ) → â(!k, λ)e−iE(k)t/!
* *
* *
where E(!k) = !c *!k *. The time dependent vector potential is given by:
! r, t) =
A(!
!
√
.
d3!k 4π!c , !
!
!
!
!
+
!%(k, λ) â(!k, λ)eik·!r−iE(k)t/! + â(!k, λ)† e−ik·!r+iE(k)t/!
(2π)3 2E(!k) λ=1,2
NOTE: There are a few typos in the given expression. Consult the Lecture Notes, page 115.
To simplify the derivation a bit we are going to introduce the notation:
√
.
, !
!
! !k, t) = + 4π!c
A(
!%(−!k, λ)â(−!k, λ)e−iE(k)t/! + !%(!k, λ)â(!k, λ)† eiE(k)t/!
2E(!k) λ=1,2
!
d3!k ! !
!
! r, t) =
A(!
A(k, t)e−ik·!r
3/2
(2π)
It is not difficult to show that after carrying out the Fourier transform the energy will take the form:
/
!
' &
'0
&
1
1 ∂ ! !
3! 1 ∂ ! !
!
!
!
!
!
!
H=
d k
A(k, t) ·
A(−k, t) + k × A(k, t) · k × A(−k, t)
8π
c ∂t
c ∂t
This is obtained by just performing the spacial integration and eliminating the δ(!k + !k " ) which will emerge.
For the magnetic field term we can do the algebra:
&
' &
'
&&
'
'
!k × A(
! !k, t) · !k × A(−
! !k, t)
! !k, t) × A(−
! !k, t)
= −!k · !k × A(
&
&
' &
' '
! !k, t) !k · A(−
! !k, t) − A(
! !k, t) · A(−
! !k, t) !k
= −!k · A(
&
'
! !k, t) · A(−
! !k, t)
= !k 2 A(
1
(1)
(2)
&
'
&
'
! !k, t)
In the first line we used the identity !a · !b × !c = −!b · (!a × !c) and not the !a · !b × !c = !b · (!c × !a) because A(
! !k, t) are operators and we cannot permute them. Accordingly in the second line we used the identity
and
& A(−
'
!b × !c × !a = !c(!a · !b) − (!a · !c)!b. In the third line the gauge condition !k · A(
! !k, t) = !0 was used to eliminate the first
term.
To proceed we to put in the table that the polarization vectors satisfy (Lect. Notes, page 109):
%(!k, λ) · %(!k, λ" ) = δλλ!
%(!k, λ) · %(−!k, λ" ) = (−1)λ δλλ!
This means that apart from the fact that %(!k, λ) for λ = 1, 2 being orthonormal, that %(−!k, 1) ↑↑ %(!k, 1) and
%(−!k, 1) ↑↓ %(!k, 1).
Now that everything is set up lets evaluate the electric field term:
1 ∂ ! !
1 ∂ ! !
A(k, t) ·
A(−k, t) =
c ∂t
c ∂t
.
4π!2 i2 E 2 (!k) , !k, λ)â(−!k, λ)e−iE(!k)t/! + !%(!k, λ)â(!k, λ)† eiE(!k)t/! ·
−!
%
(−
2E(!k) !2
λ=1,2
.
, !
!
"
!
−!%(k, λ )â(!k, λ" )e−iE(k)t/! + !%(−!k, λ" )â(−!k, λ" )† eiE(k)t/!
λ! =1,2
= −2πE(!k)
+2πE(!k)
,
λ
.
!
!
(−1)λ â(−!k, λ)â(!k, λ)e−2iE(k)t/! + â(!k, λ)† â(−!k, λ" )† e−2iE(k)t/!
,λ
.
â(−!k, λ)â(−!k, λ)† + â(!k, λ)† â(!k, λ)
!2
! !k, t) · A(−
! !k, t) term is the same. The difference is that all terms will appear with a positive
Calculating the k A(
sign, because there is no i2 term
* *2 and no minus sign due to exponent in the first term of each vector potential. Also
4π!2 c2 *! *
the prefactor will be
*k * = 2πE(!k) which is the same. In other words we will have the same expression
2E(!
k)
but in the first term we will have a plus sign instead and the two time dependent terms (from the electric and the
magnetic field) will cancel. The expression for the energy is going to be:
!
.
,1
H=
â(−!k, λ)â(−!k, λ)† + â(!k, λ)† â(!k, λ)
d3!k4πE(!k)
8π
λ
where in the second line we remembered the commutation relation which implies that â(−!k, λ)â(−!k, λ)† =
1 + â(−!k, λ)† â(−!k, λ). The constant will give the vacuum energy which in this calculation diverges (there is a really
really long story about how to deal with this). In the second term we perform the !k → −!k transformation in the
summation and exploit the fact that E(−!k) = E(!k). The energy finally reads:
!
,
H = d3!k
E(!k)n̂(!k, λ)
λ
where n̂(!k, λ) = â(!k, λ)† â(!k, λ) is the number operator. Intuitively the energy is just a sum of the energies E(!k) of
all the occupied modes.
Using the same notation we will now evaluate the momentum operator:
!
'
! &
1
3 ∂A
!
d
!
r
×
∇
×
A
P =−
4πc2
∂t
2
As before we will use equation 2 to perform the spacial integration:
P
!
'
! !k, t) &
∂ A(
! !k, t)
× i!k × A(−
∂t
1$
2
%
!
&
'
! !k, t)
∂
A(
i
∂
!k · A(
! !k, t)
! !k, t) !k −
! !k, t) A(−
d3!k
= −
· A(−
4πc2
∂t
∂t
= −
1
4πc2
d3!k
The second term is zero due to choice of gauge. The first term, after doing the same kind of algebra as we did
for the energy will become:
1 ∂ ! !
! !k, t) =
A(k, t) · A(−
c2 ∂t
.
4π!2 iE(!k) , !
!
−!%(−!k, λ)â(−!k, λ)e−iE(k)t/! + !%(!k, λ)â(!k, λ)† eiE(k)t/! ·
2E(!k) ! λ=1,2
.
, !
!
!%(!k, λ" )â(!k, λ" )e−iE(k)t/! + !%(−!k, λ" )â(−!k, λ" )† eiE(k)t/!
λ! =1,2
= 2πi!
,
λ
+2πi!
.
!
!
(−1)λ −â(−!k, λ)â(!k, λ)e−2iE(k)t/! + â(!k, λ)† â(−!k, λ" )† e−2iE(k)t/!
,λ
.
−â(−!k, λ)â(−!k, λ)† + â(!k, λ)† â(!k, λ)
All of this is multiplied by a !k in the integral. The term â(−!k, λ)â(!k, λ)!k changes sign under a !k → −!k because
the two operators commute. Therefore in the integral this term will contribute zero. Same for the a† a† term. For
the second line we also perform a !k → −!k for the first term to get:
!
.
,i
!kâ(!k, λ)â(!k, λ)† + !kâ(!k, λ)† â(!k, λ)
P = − 2πic! d3!k
4π
λ
!
.
,1
3!
!
k + 2!kâ(!k, λ)† â(!k, λ)
c! d k
=
2
λ
!
,
=
d3!k
!c!kn̂(!k, λ)
λ
In the second line we used the commutation relation and in the third line the fact that
total momentum is the sum of the momenta !!k of the occupied modes.
3
!kd3!k = 0. Again the
2.
By commuting the just terms that correspond to the same mode one can easily show that:
[Ai (!r, t), Aj (!r" , t)] = 0
[Ei (!r, t), Ej (!r" , t)] = 0
Both the electric field and the vector potential at the same time and at different points commute. For the
magnetic field we don’t need to evaluate anything: the magnetic field is just the spacial derivative of the vector
3
potential. If we realise that the derivative is nothing but the limit of the difference of the vector potential between
neighboring points and at the same time, we can argue that the magnetic field is the limit of a linear combination of
vector potentials at different points. Since the vector potential between any points commutes, so does the magnetic
field:
[Bi (!r, t), Bj (!r" , t)] = 0
For the commutator of the electric field and the vector potential we need to do a bit more work. The electric
field is give by:
√
!
.
,
!
d3!k 4π!c
!
!
!
!
! !k, t) = − 1 ∂ A = i
+
E(
E(!k)
!%(!k, λ) â(!k, λ)eik·!r−iE(k)t/! − â(!k, λ)† e−ik·!r+iE(k)t/!
(3)
c ∂t
!c
λ=1,2
(2π)3 2E(!k)
Using this expression along with the commutation relations will give:
!
-.
.
.
,
i 4π!2 c2
3!
"
!k, λ)%j (!k, λ) â(!k, λ), â(!k, λ)† ei!k·(!r−!r! ) − â(!k, λ)† , â(!k, λ) e−i!k·(!r−!r! )
[Ei (!r, t), Aj (!r , t)] =
%
(
k
d
i
!c (2π)3 2
λ=1,2
where %i = êi · % with i = 1, 2, 3 is the ith component of !%. First we will perform the λ summation. Lets fix
some !k and with no loss of generality choose a system of coordinates such that %1 → (1, 0, 0) and %2 → (0, 1, 0) (the
z direction is the the direction of !k). Also let e!i and e!j be two arbitrary vectors represented by column vectors.
Then:
 
 

1
0
,
8
9
8
9
 0  · 1 0 0 +  1  · 0 1 0  · !ej
%i (!k, λ)%j (!k, λ) = !etr
i ·
0
0
λ=1,2


1 0 0
 0 1 0  · !ej = !ei · !ej − (!ei · k̂)(!ej · k̂)
= = !etr
i ·
0 0 0
where k̂ is the direction of !k. If e!i and e!j are orthonormal we can write:
,
λ=1,2
%i (!k, λ)%j (!k, λ) = δij −
ki kj
!k 2
Substituting gives:
:
;
!
.
2π!c
ki kj - i!k·(!r−!r! )
3!
−i!
k·(!
r −!
r! )
[Ei (!r, t), Aj (!r , t)] = i
d
k
δ
−
e
+
e
ij
!k 2
(2π)3
;
:
!
3!
!
ki kj
d k
!
eik·(!r−!r )
δ
−
= i4π!c
ij
3
!k 2
(2π)
"
In the last line we just set !k → −!k in the second term. This integral is called transverse commutator.
The magnetic field is obtained by Bm = εajm ∂a" Aj and the electric-magnetic field commutator is given by:
[Ei (!r, t), Bm (!r , t)] = i4π!c
"
!
d3!k
(2π)3
4
:
ki kj
δij −
!k 2
;
!
!
(−i)eik·(!r−!r ) εajm ka
&
'
but we get εajm ka kj = !k × !k
m
= 0. The commutator is then:
[Ei (!r, t), Bm (!r , t)] = i4π!c
"
!
.
d3!k - i!k·(!r−!r! )
−i!
k·(!
r −!
r! )
i
−e
+
e
εaim ka = 4π!cεaim i∂a" δ(!r − !r" )
(2π)3
3
First we will need the commutators:
.
â(!k " , λ" ), â(!k, λ)† â(!k, λ)
.
â(!k " , λ" )† , â(!k, λ)† â(!k, λ)
= â(!k, λ)δλλ! δ(!k − !k " )
= −â(!k, λ)† δλλ! δ(!k − !k " )
Integrating over !k and summing over λ gives:
.
!
!
â(!k, λ)e−iE(k)t/! , N̂
= â(!k, λ)e−iE(k)t/! =
-
!
â(!k, λ)† eiE(k)t/! , N̂
.
i
!
∂t â(!k, λ)e−iE(k)t/!
!
E(k)
!
= −â(!k, λ)† eiE(k)t/! =
i
!
∂t â(!k, λ)† eiE(k)t/!
!
E(k)
In either case taking the commutator with the number operator is like taking the time derivative and dividing
by the energy of the mode. If a commutator vanishes it vanishes identically for each mode separately. Clearly
this cannot happen by just taking a time derivative so neither the electric or the magnetic field commute with the
number operator. The physical interpretation for this is that the number of photons is not conserved when the
electromagnetic field is coupled to matter.
4.
To verify those equations it suffices to consider one mode. In the same way that we took the commutator with
the number operator at part (3) we can take the commutator with the total energy. The only difference is that we
multiply by E(!k) before integrating over !k and summing over λ. The result is:
.
!
!
!
â(!k, λ)e−iE(k)t/! , Ĥ
= E(!k)â(!k, λ)e−iE(k)t/! = i∂t â(!k, λ)e−iE(k)t/!
.
!
!
!
â(!k, λ)† eiE(k)t/! , Ĥ
= −E(!k)â(!k, λ)† eiE(k)t/! = i∂t â(!k, λ)† eiE(k)t/!
In both cases, commuting with the Hamiltonian is like taking the derivative. Same applies to any operator that
can be expressed as a linear combination of a and a† like the electric field, the magnetic field, the vector potential
etc. There are two Maxwell equations that involve the time derivatives:
!
!
1 ∂B
! ×E
! = −1∇
! × ∂A
⇒∇
c ∂t
c
∂t
&
'
!
!
1 ∂E
1 ∂2A
! ∇
! ·A
! +∇
! 2A
!
⇒ 2 2 = −∇
c ∂t
c ∂t
! ×E
!
∇
= −
! ×B
!
∇
=
Those correspond to the cases that there is no source current.
5
! =∇
! ×A
! implies that E
! = −1 ∂ A
! − ∇φ
! but in the absence of charges the second
The first one, along with B
c ∂t
term vanishes. In momentum space the second equation will become:
! !k, t)
∂ 2 A(
! !k, t)
= c2!k 2 A(
∂t2
!
which is automatically satisfied because the time dependence of each mode is e±iE(k)t/! with E(!k) = !c|!k|.
5.
! ·A
! = 0 in the given gauge. One can see
! ·B
! is identically zero. Also ∇
! ·E
! = −1 ∂ ∇
This is a tricky question. First ∇
c ∂t
! !k, t) = 0 because the polarization vectors are perpendicular to !k. The results
that formally in Fourier space: !k · E(
mean that there are no sources of electric or magnetic field (free space). Therefore all commutators are identically
zero.
6.
The normalized wave functions are:
|!
p, R|L) = a† (!k, R|L) |0)
(4)
Consulting the Lect.. Notes page 118, we get that:
a† (!k, 1) =
a† (!k, 2) =
a† (!k, R) + a† (!k, L)
√
2
† !
a (k, R) − a† (!k, )
√
i 2
It is straightforward to show that in the R, L basis the operators satisfy the same commutation relations. Using
this fact we can show that:
* =
* = < *
< *
p − p!" )
p − p!" ) + a† (!
p" , µ" )a(!
p, µ)* 0 = δµµ! δ(!
*!
p, µ|!
p" , µ" ) = 0 *a(!
p, µ)a† (!
p" , µ" )* 0 = 0 *δµµ! δ(!
where µ = R, L. Therefore the above states are correctly normalized.
Now we will obtain expressions of the Hamiltonian, momentum and angular momentum using the L, R polarization vectors.
The number operators will become:
n̂(!k, 1) =
n̂(!k, 2) =
a† (!k, R) + a† (!k, L) a(!k, R) + a(!k, L)
n̂(!k, R) + n̂(!k, L) + a† (!k, L)a(!k, R) + a† (!k, R)a(!k, L)
√
√
=
2
2
2
† !
† !
†
a (k, R) − a (k, L) a(!k, R) − a(!k, L)
n̂(!k, R) + n̂(!k, L) − a (!k, L)a(!k, R) − a† (!k, R)a(!k, L)
√
√
=
2
i 2
−i 2
From this we get:
,
n̂(!k, λ) =
,
µ=L,R
λ=1,2
6
n̂(!k, µ)
Using this we can quickly get the same expressions for the momentum and the Hamiltonian, only this time the
polarization summation is over R and L:
!
,
H =
d3!k
E(!k)n̂(!k, µ)
P!
=
!
µ=R,L
d3!k
,
!!kn̂(!k, µ)
µ=R,L
The algebra for obtaining the angular momentum is already in the Lect.. Notes in pages 118-119:
!
&
'
!
L = d3!k!k̂ n̂(!k, R) − n̂(!k, R)
! have the same general form:
All the operators H, P! and L
!
,
G(!k, µ)n̂(!k, µ)
Q = d3!k
µ=R,L
Before applying all those operators in the states 4 we need one more little results:
p − p!" )â(!
p, µ)† |0)
p − p!" ) |0) + â(!
p" , µ" )† a† (!
p, µ)â(!
p" , µ" ) |0) = δµµ! δ(!
n̂(!
p" , µ" )a† (!
p, µ) |0) = â(!
p" , µ" )† δµµ! δ(!
Applying the general Q operator will give:
!
,
p − p!" )â(!
p, µ)† |0) = G(!
p, µ)
G(!
p" , µ" )δµµ! δ(!
Q |!
p, µ) = d3 p!"
µ! =R,L
Therefore applying Q on the single photon state just picks the corresponding term. Therefore the states 4 have
energy E(!k), momentum !!k and angular momentum +! in the R polarization and −! in the L polarization.
2
Scattering of Light
1.
The incoming state has N photons all with the same momentum p! and polarization ν. The outgoing photon state
has one photon removed from the (!
p, ν) mode and one single photon in the mode (!
p" , ν " ). The correctly normalized
photon states can be constructed with the creation operators for photons.
>
3
(2π) a†N (!
p, ν)
√
|0)
V
N!
>
3
(2π) a†N −1 (!
p, ν)a† (!
p" , ν " )
?
|Fphoton ) = |N − 1, 1) =
|0)
V
(N − 1)!
|Iphoton ) = |N, 0) =
The first slot corresponds to the (!
p, ν) mode and the second to the (!
p" , ν " ) mode. This way we avoid the rather
"
"
confusing N (!
p , ν ) and N (!
p, ν) notation.
7
2.
In case the volume is finite the k−integration is replaced by a summation using the rule:
!
(2π)3 ,
d3!k →
V
k
Also the creation/annihilation operators are renormalized by a factor
! r, t) =
A(!
,
!
k
+
V
(2π)3 .
The vector potential reads:
√
.
4π!c , !
!
!
!
!
+
!%(k, λ) â(!k, λ)eik·!r−iE(k)t/! + â(!k, λ)† e−ik·!r+iE(k)t/!
V 2E(!k) λ=1,2
The matrix element we need is
@
(5)
*
*
A
* !2 *
Fphoton *A
(!r)* Iphoton
For the element to be non zero, in the initial state exactly one a† (!
p" , ν " ) and one a(!
p, ν) should be applied
to give the correct final state. Those can appear in any order and we have two possibilities: a† (!
p" , ν " )a(!
p, ν) or
"
†
"
"
† "
"
p − p! ) + a (!
p , ν )a(!
p, ν). Because the initial and final states have different occupations the
a(!
p, ν)a (!
p , ν ) = δνν ! δ(!
p − p!" ) term between them will vanish, so both probabilities will give the same result. This gives us a factor
δνν ! δ(!
of 2. If we also put all the other factors that accompany the creation/annihilation operators in the expression of
the vector potential we will get:
*
*
=
! <
4π!2 c2
!%(!
p" , ν " ) · !%(!
p, ν)ei!r·(p!−!p ) Fphoton *a† (!
p" , ν " )a(!
p, ν)* Iphoton
2 ?
"
V 4E(!
p)E(!
p)
To proceed we will apply the operators (conjugated) to the left, that is the Final photon state:
√ √
a(!
p, ν)† a(!
p" , ν " ) |N − 1, 1) = N 1 |N, 0)
We are getting for the matrix element the expression:
3.
√
!
4π!c2 N!%(!
p" , ν " ) · !%(!
p, ν)ei!r·(p!−!p )
?
V ω(!
p)ω(!
p" )
The density operator affects the degrees of freedom of the atom but not the the electromagnetic field. Therefore
we can perform the factorization:
!
*
*
*
*
@
@
A
A
r0
* !2 *
* (2) *
*n| d3!rρ (!r) Fphoton *A
(!r)* Iphoton |n = 0)
M = n; N − 1, 1 *Hint * 0; N, 0
=
2
√
!
!
2r0 π!c2 N!%(!
p" , ν " ) · !%(!
p, ν)
?
=
*n| d3!rρ (!r) ei!r·(p!−!p ) |n = 0)
"
V ω(!
p)ω(!
p)
The integral gives nothing by the Fourier transform of the density ρ (!
p − p!" ).
8
4.
The differential cross section is given by the transition rate (Fermi’s Golden Rule) divided by the incoming photon
flux:
dσ0→n
V 2π
2
=
|M | D(E " )
dΩ
cN !
where D(E " ) is the density of states of the outgoing photon with energy E " = E(!
p" ) = En − E0 + E(!
p). The
constraints are the the photon should move in a infinitesimal solid angle dΩ.
The phonon density of states is given by:
V 3
V
V
d p! = 3 p2 dpdΩ = 3 3 E 2 dEdΩ = D(E)dEdΩ
3
h
h
h c
(NOTE: h is not !).
Then the differential cross section becomes:
dσ0→n
dΩ
V 2π
2 V
|M | 3 3 E "2 =
cN !
h c
"
2
2
"
"
2ω
p , ν ) · !%(!
p, ν)| |*n |ρ (!
p − p!" )| 0)|
= r0 |!%(!
ω
=
5.
If !r" is the position of the electron, the density operator is just δ(!r − !r" ). We then have:
!
!
!
!
!
!
!
!
!
1
d3!r" e−iq!n ·!r ei!r ·(p!−!p ) eiq!0 ·!r = δp!+q!0 ,!p! +q!n
*n| d3!rδ(!r − !r" )ei!r·(p!−!p ) |0) = *n| ei!r ·(p!−!p ) |0) =
V
The δ is a Kronecker delta and not a delta function and it enforces the momentum conservation between the
initial and final states. The differential cross section in this case is:
dσ0→n
dΩ
= r02
ω"
2
|!%(!
p" , ν " ) · !%(!
p, ν)| δp!+q!0 ,!p! +q!n
ω
with the additional constraint that the energy is conserved in the process:
c |!
p| +
3
!2 !q02
!2 !qn2
= c |!
p" | +
2m
2m
Spontaneous Emission
1.
In this problem we are going to allow both the proton and the electron to move and interact with the electro
magnetic field. Therefore there is one more degree of freedom compared to the problem solved in Shankar.
The states of the system are
|CM ) ⊕ |nlm) ⊕ |photon)
The first part describes the center of mass motion of the neutral atom. The second term describes the internal
states of the atom and the third term describes the photon field. The photon part is described by a list of occupation
number of the various photon modes (momentum and polarization). The internal states wave function expressed
9
in the position basis will depend only on the difference !r = !re − !rp . Finally in the position basis the CM wave
!
function is just a plane wave: eiP ·(!re +!rp )/! where P! is the total momentum. The initial and final states of the
system atom-photons are:
*
A
*
|I) = *P! = 0 ⊕ |211) ⊕ |0)
* A
* A
*
*
|F ) = *P!f ⊕ |100) ⊕ *!kν
where the polarization of the photon can by R or L. Originally the atom will be at rest, but when the photon
is emitted there is going to be a recoil and the center of mass momentum will be non zero. Intuitively the photon
momentum will be opposite the center of mass momentum of the atom, but we will show that through the algebra.
Lets revise the Lect. Notes page 123. Because of the fact that the proton has a different charge than then electron,
the Hamiltonian of the system of two particles is:
H=
! re , t)/c)2
! rp , t)/c)2
(!
pe + eA(!
(!
pp − eA(!
+
2mp
2me
where external potentials have been ignored. From this equation we can get the interaction Hamiltonian by a
!
simple Taylor expansion for small A:
Hint
= −
!j(!k) =
e
c
/
!
! x, t)
d3 !x!j(!x) · A(!
p!p
p!e
1 p!p −i!k·!rp
p!e −i!k·!re
!
!
e
+ e−ik·!rp
−
e
− e−ik·!re
2 mp
mp
me
me
0
The current is expressed through its Fourier transform. Note that the terms in the current that correspond
! = mp!rp + me!re and
to the electron come with a different sign due to the opposite charge. Now lets define M R
m
m
p
!+
! − e !r. For the transformation to be canonical
!r = !re − !rp with M = me + mp so that !re = R
r and !rp = R
M !
M
!
!
p
!
p
!
p
!
p
!
p
e
we also have to set P! = p!e + p!p and
=
−
which means that p!e = P + p! and p = P − p! . The
µ
me
mp
me
M
me
mp
M
mp
capital operators commute with all the small operators (the center of mass degrees of freedom are irrelevant to the
internal degrees of freedom). We can express everything in terms of the new operator and the current will become:
Replacing the will give the current:
$
$
%
%
P!
p!
P!
p!
! i!
!
−i!
k·R
k·!
r me /M
i!
k·!
r me /M −i!
k·R
!
!
2j(k) =
−
e
e
+e
e
−
M
mp
M
mp
$
%
$
%
P!
p!
p!
P!
! −i!
!
−i!
k·R
k·!
re mp /M
−i!
k·!
re mp /M −i!
k·R
−
+
e
e
−e
e
+
M
me
M
me
%
$
'
! & !
P! −i!k·R!
! ! P
!
e
+ e−ik·R
eik·!rme /M − e−ik·!rmp /M
=
M
M
:
;
p! i!k·!rme /M
p!
p! −i!k·!rmp /M
p!
! !
!
!
−e−ik·R
e
+ eik·!rme /M
+
e
+ e−ik·!rmp /M
mp
mp
me
me
Now we need to evaluate the transition matrix element:
*F |Hint | I)
10
* A
@ *
*!
*
The first step is to evaluate the transition element for the photons !kν *A(!
x, t)* 0 . We can just follow exactly
the algebra in the Lect. Notes to get the result of page 126:
B
C 2π!e2 @ *
* A
@
A
C
& ' 0 **!j(!k)** n · !%(!k, ν)
0; !kλ |Hint | 0; 0 = −D
V ω !k
here I use the notation |0) and |n) to represent the final and initial state of the atom, including both the internal
and the center of mass degrees of freedom. Now we have to evaluate this matrix element. The job is not as hard as
it seems. We start by integrating out the CM degrees of freedom:
* F
E *
* P!
'
! **
P
P!f @ ! ** −i!k·R! ** ! A P!f & !
!
!
!
!
*
−i
k·
R
−i
k·
R
Pf *e
+e
δ Pf − !!k
P!f * e
*0 =
* !0 =
*M
M*
M
M
This means that after integrating out the CM degrees of freedom we get an effective correct involving only the
internal degrees of freedom:
&
'
2!jef f (!k)δ P!f − !!k
=
'
'&
P!f & !
!
!
δ Pf − !!k eik·!rme /M − e−ik·!rmp /M
M
;
&
' : p!
p!
p!
p! −i!k·!rmp /M
!
!
!
−δ P!f − !!k
eik·!rme /M + eik·!rme /M
+
e
+ e−ik·!rmp /M
mp
mp
me
me
This way we got the momentum conservation: the phonon momentum is equal to the atom momentum as we
expected. Also inside a box this is not a delta function but a Kronecker delta, so when we square the matrix element
it will give a clean Kronecker delta.
There is this identity:
e−i!a·!r p!ei!a·!r = p! + !!a
which is similar as the one found in the Lect. Notes page 134. With the help of this we can write the current as:
'
P!f & i!k·!rme /M
!
e
− e−ik·!rmp /M
M$
1
1
2
2
%
1
me !!k i!k·!rme /M
mp !!k −i!k·!rmp /M
1
2!
p−
2!
p+
−
e
+
e
mp
M
me
M
;
:
1 −i!k·!rmp /M
1 i!k·!rme /M
!
e
−
e
= Pf
mp
me
:
;
1 i!k·!rme /M
1 −i!k·!rmp /M
−2!
p
e
+
e
mp
me
2!jef f (!k) =
In the second line I used P!f = !!k. Now it is time for some approximations.
Fermi’s Golden Rule will give:
$
%
* * *@
*
*
*2
'
A
, ! V d3!k , 4π 2 e2 &
Pf2
*
*
*
*!
!
!
!k)** 211 · !%(!k, ν)**
!
&
'
Γ211→100 =
δ
P
−
!
k
δ
E
−
E
−
−
!c
k
100
j
(
*
*
*
*
f
2
1
ef
f
3
2M
(2π) ! V ω !k
ν
Pf
11
where the energy conservation delta function takes into account both the energy of the recoil and the fact that
the photon and atom momentum is the same. The summation extends over all possible final states which are defined
by the value of P!f and !k. The P!f summation can go first:
:
; *@
*
*
*2
A
,!
e2 2
!2 k 2
*
*
*
*
dkdΩ
k δ E2 − E1 −
− !ck * 100 *!jef f (!k)* 211 · !%(!k, ν)*
Γ211→100 =
2πck
2M
ν
The angular integration is not so easy without the matrix element but the dk integration is trivial. Solving the
equation in the delta function will give one positive solution:
$G
%
:
;
2∆E
∆E
∆E
∆E
∆E M c2
1+
−1 =
f
≈
k0 =
!c ∆E
M c2
!c
M c2
!c
where we ignored with not much thought the ratio of the transition energy (few eV ) and the relativistic mass
of the atom (1GeV ).
0)
Then we use the identity δ(f (x)) = δ(x−x
f ! (x0 ) and f (x0 ) = 0 to get:
:
;
!2 k 2
1
1
1
δ E2 − E 1 −
− !ck = δ(k − k0 ) !2 k
≈ δ(k − k0 )
= δ(k − k0 ) +
2M
!c
!c 1 + 2∆E
M + !c
M c2
We also need to simplify the current jef f . The photon wave length is going to be negligible. More specifically
lets compare it with the Bohr radius a0 :
1
∆E
k0 a0 =
a0 ≈
!c
1726
So the atom is of the order of a thousand times smaller than the wavelength of the photon k0 a0 - 1. Multiplying
this on both sides by either me /M or mp /M only enhances the inequality, so both exponentials can be set to one
with great accuracy:
;
:
1
p!
1
!jef f (!k) = − 1 P!f
−
−
2
me
mp
µ
Because the transition is between different states the P!f term will not contribute anything. So the transition
rate will become:
Then:
*
*H
*2
*
I
!
* p!a0 *
*
*
2e2 !k0 , 1
!
*
*
*
Γ211→100 = 2 2 2
dΩ * 100 *
211 · !%(k, ν)**
µ c a0 ν 4π
! *
One can proceed by evaluating the integral explicitly using the known analytical expressions for the hydrogen
atom wave functions. There is a different way. We can use the Wigner-Eckart theorem (check Sakurai, page 239).
Applied here it would give:
CGC(1q, lm; l" m" )
√
*n" l" m" |pq | nlm) =
Φ(n" l" ; nl)
2j + 1
where CGC(1q, lm; l" m" ) is the Clebsh-Gordan coefficient for adding the angular momenta 1 and l and Φ(n" l" ; nl)
that does not depend on m, m" or q. The vector p! is treated as an object of angular momentum one:
p0
p1
p−1
= pz
px + ipy
√
=
2
px − ipy
√
=
2
12
For the particular quantum numbers the only allowed value for q = −1. Otherwise the matrix element will be
zero. To simplify notation lets define:
*
*
H
I
* p!a0 *
* 211
*
!a = 100 *
! *
Back to the transition rates: the summation over the polarizations will give (see problem 1 part 2):
/
!
*
*2 0
1
2e2 !k0
*
*
2
dΩ *!a · k̂ *
Γ211→100 = 2 2 2 |!a| −
µ c a0
4π
*
*2
*
*
dΩ *!a · k̂ * . Because we are integrating on the whole sphere we
2
23
a| .
can use spherical coordinates with the z direction along !a. Then the integral will become |!a| dΩ cos2 θ = 4π
3 |!
The vector !a = *100 |!
p| 211) has complex coordinates ax , ay and az if we want to perform the rotation to a0 , a1
2
2
2
2
and a−1 . The norm will be just |!a| = |a0 | + |a−1 | + |a1 | . In this case only a−1 is non zero and the transition
rate will become:
*H
*
*
I*2
* a0 (p̂x − ip̂y ) *
*
4e2 !k0 **
*
*
*
211
100
Γ211→100 =
2
*
*
*
*
3µ2 c2 a0
!
Now we need to evaluate something of the form:
3
where the number in the brackets is dimensionless.
2 2
!
If we replace e2 = αc!, a0 = µcα
and En = − µc2na2 , where α = 1/137 is the fine structure constant, the prefactor
can be written as:
4e2 !k0
α5 µc2
4 α3 ∆E
=
=
2
2
2
3µ c a0
3 !
2!
which fortunately has exactly the right units.
If we want to maintain the term
∆E
3 µ 2
α
x=
=
M c2
4M
all we need to do is multiply the above with:
g(x) =
9
1 8√
2
3
5
35
1
√
=
= 1 − x + x2 − x3 + O(x4 )
1+x−1 √
x
2
2
8
1+x
1+x+ 1+x
In this case we would get:
Γ211→100
*
*H
*
I*2
* a0 (p̂x + ip̂y ) *
*
*
α5 µc2
2
*
*
*
√
=
211 *
100 **
*
*
2! 1 + x + 1 + x
!
Notice that I swapped the wave function in the integral and I took the hermitian conjugate of the operator.
Now we will try to evaluate the remaining matrix element. Since the only characteristic scale is the Bohr radius
and everything else is dimensionless for this calculation we can set a0 = ! = 1. The relevant wave functions are:
ψ100
=
ψ211
=
1
√ e−r
π
1
1
√
re−r/2 = √ re−r/2 Y11
64π
2 6
13
Applying the momentum operator on the ground state is easy:
−e−r xi
= −i √
π r
G
:
;
−r
e
x + iy
e−r
2 −r 1
iφ
(p̂x + ip̂y ) ψ100 = i √
= i √ sin θe = i2
e Y1
r
3
π
π
* *
The angular integration is trivial it is just *Y11 * = 1. The radial integration will be:
pi ψ100
H
*
*
I !
* a0 (p̂x + ip̂y ) *
* 100 =
*
211 *
*
!
0
∞
2
G
drr i2
i
2 −r 1
√ re−r/2 =
e
3
3
2 6
!
∞
r3 e−3r/2 dr = i
0
32
81
And now the final result:
Γ211→100 =
29 α5 µc2
2
√
≈ 2 × 108 s−1
38 ! 1 + x + 1 + x
µ 2
α ≈ 3 × 10−8 . The life time is the inverse of the transition rate and is about 5ns and this is the
where x = 34 M
correct order of magnitude for transitions.
2.
Anchoring the proton means taking its mass to infinity so that M → ∞ and µ → me . You are asked to compare
the result that you got in the first part for these two cases:
√
:
;:
;
Γanchored
me 1 + x + 1 + x
me
3 ∆E
=
≈ 1+
1+
Γf ree
µ
2
mp
2 M c2
The anchored proton has a larger transition rate. The percentage difference is approximately 100/1836.0%=0.05%
which is negligible.
3.
p −ip
If the photon leaves from the ẑ-axis its polarization vector will be on the x̂ − ŷ plane. Also p! · !%(!k, ν) = x√2 y and
we know beforehand that the outgoing photon is going to be right handed and it will have angular momentum +!.
We see that angular momentum is conserved because +! was also the original angular momentum of the atom.
If the proton is anchored the total momentum is not conserved, the photon momentum is given by whatever
holds the proton in place. If the proton is free the total momentum is conserved as the delta function δ(P!f − !!k)
that we derived implies.
14