Quantization of Normal Modes
2
2.1
Wave equation
Want to review modes of electromagnetic radiation in cavity. Start with
Maxwell’s equations in free space(SI units)
∇·E
∇·B
∂B
∇×E+
∂t
∂E
∇ × B − µ0 ² 0
∂t
= 0
= 0
(1)
(2)
= 0
(3)
= 0
(4)
∂
∂
∂
+ ĵ
+ k̂
∂x
∂y
∂z
(5)
where ∇ is gradient operator
∇ = î
Will also need subsidiary condition, charge conservation:
∂ρ
+∇·j=0
∂t
(6)
F = q(E + v × B)
(7)
and Lorentz force:
Now take curl of Eq. (3), and use identity ∇×(∇×E = ∇(∇·E)−∇2E:
0 = ∇ × (∇ × E +
1
∂B
)
∂t
(8)
∂
∇×B
∂t | {z }
µ0²0 ∂E
∂t
2
∂ E
= −∇2E + µ0²0 2
∂t
= −∇2E +
(9)
(10)
(11)
i.e. a wave equation for each component of the electric field E for waves
√
travelling with speed c ≡ µ0²0. Why is it a wave eqn.? Consider soln.
E = E0f (x − ct), with f any function. Check! Compare field “profile”
f (x) at time t = 0 with profile time t later: simply translated to right
distance ct =⇒ “pulse” f (x) moving to right with velocity c (vel. of light!)
Check that
1. E ⊥ î (direction of propagation)
2. B field ⊥ î, E
2.2
Normal modes of cavity
Example: transverse waves on string.
displacement of string δ(x, t), obeys
2
∂ 2δ
2∂ δ
= cs 2
∂t2
∂x
wave equation
2
(12)
where cs ≡ velocity of sound. Boundary conditions: suppose
δ(x = 0, t) = 0,
δ(x = L, t) = 0
(13)
(14)
Normal modes of vibration: solutions of form
δ = f (x)cos(ωt + φ)
(15)
consistent with boundary conditions. Note these are standing waves.
Substitute in (12)–solution only if
2
−ω f (x) =
2
2d f
cs 2 .
dx
(16)
(16) called “Eigenvalue” problem (German: proper value) since eqn has
solutions consistent with B.C. only for special values (eigenvalues) ω.
General soln. to (16):
f (x) = A sin
ωx
ωx
+ B cos
cs
cs
(17)
B.C. δ(x = 0) = 0 =⇒ B = 0, δ(x = L) = 0 =⇒ A = 0 or
ω = nπcs/L,
n = 1, 2, 3...
(18)
N.B. This purely classical result has connection with quantization as used
in SHO problem.
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Modes in cavity: wave eqn. for electric field. Assume walls perfect
conductors, so B.C. are Ek = 0, B⊥ = 0 at walls. Search for solns of
Maxwell eqns. of form
E = cos(ωt + φ)f (r)
(19)
consistent with these B.C.
2.3
Periodic boundary conditions
We can find solutions with fixed (standing wave) boundary conditions, but
since it is sometimes convenient to use other B.C., let’s look at periodic
B.C., i.e. assume cavity is periodically repeated with period L along x, y,
z — x, y, z is same pt. as x + L, y, z, etc. This is ok since the shape of
cavity, exact pos. of walls can’t matter if we are calculating something
which depends only on the density of modes in freq. range. such that
L À sizes of interest.
Consider solutions
E = Re E0ei(k·r−ωt)
4
(20)
= |E0| cos(k · r − ωt)
(21)
E0 = const. φ = const
Re=“real part of”
Remarks:
• Plug into wave eqn. =⇒ ω 2 = k 2c2
• ∇ · E = 0 =⇒ k · E0 = 0 transverse wave
• Periodic B.C. =⇒
kxL = 2πnx, nx = 0, ±1, ±2...
ky L = 2πny , ny = 0, ±1, ±2...
kz L = 2πnz , nz = 0, ±1, ±2...
(22)
• Note allowed k’s separated by 2× larger interval compared to fixed
b.c., but nx,y,z can be ± =⇒ same # per dν.
• note that for, e.g., nx = ±1, + and - give independent solutions –
check! This was not true for fixed B.C.
Allowed frequencies are therefore
5
ω = kc =
2πc 2
(nx + n2y + n2z )1/2.
L
(23)
Maxwell’s eqns linear—lin. comb. of solns. is soln. General:
E = Re
X
X
p=1,2 nx ,ny ,nz
E0(n, p)ei(k·r)−ωt
(24)
here
p is polarization of wave: 2 lin. ind. directions of E for given k
n = nx , n y , n z
k = 2πn/L
Exercise: derive form for B from Eq. (24) and James Clerk’s equations...
2.4
Counting modes
Need # of solutions ∼ ei(k·r−ωt) with freqs in range ν to ν + δν. Look at
“k-space”:
6
• Allowed k form cubic lattice in k-space, spacing 2π/L
• “volume” occupied by each allowed k is (2π/L)3
• volume in k-space between k and k + δk is 4πk 2δk
• therefore number of k’s in this shell is
δN = 4πk 2δk · (L/2π)3
(25)
Now relate number of k’s in (k, k + δk) to number of allowed solutions
with ν’s in (ν, ν + δν)
(recall ω = 2πν = kc !) :
δN = |{z}
2
2 polarizations
2.5
·
4π
|{z}
·
L3
8π 3
·
θ, φ integration
= 8πL3ν 2δν/c
2π ¶3 2
ν δν
c
{z
}
|
k = 2πν/c
µ
(26)
Planck Law
Planck idea (1900): each mode of radiation in cavity acts like SHO —
suppose it contributes mean energy hν/(exphν/kT −1). Mean energy in
(ν, ν + δν) is then
8πL3hν 3δν
δE = 3 hν/kT
c (e
− 1)
(27)
Energy per unit volume and bandwidth:
1 δE
8πhν 3
uν ≡ 3
= 3 hν/kT
L δν
c (e
− 1)
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(28)
(• Aside: at high temperatures, hν/kT ¿ 1, expand exponential to find
uν = 8πkT ν 2/c3, Rayleigh-Jeans law. Note there’s no h–classical result.)
Net energy density now
u=
Z ∞
0
8πh Z ∞
ν 3dν
uν dν = 3 0
c
(ehν/kT − 1)
(29)
Make integral dimensionless, put α = hν/kT , dν = kT dα/h:
8πh µ kT ¶4 Z ∞ x3dx
u= 3
c
h | 0 e{zx − 1}
(30)
π 4/15
So Planck also recovered :
u = aT 4 Stephan-Boltzmann law
and determined
8π 5 k 4
a =
15 (hc)3
(31)
(32)
But if had let h → 0 as originally intended, → nonsense! Although
made no sense classically, P. brave enough to allow h to stay. Stephan’s
const. pretty well known expt’lly in 1900 =⇒ Planck determined h =
6.55 × 10−27 erg − sec, close to modern h = 6.625 × 10−27 erg − sec!
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