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LECTURE 14
Quantized Energy Levels We know from quantum mechanics that the solutions to Schroedinger’s equation
Hψ = Eψ (1)
have quantized energy levels. For example, a particle of mass m in a box with infinitely high walls, has energy levels given by
En =
n^2 π^2 ¯h^2 2 ma^2
where n = 1, 2 , 3 , .... A harmonic oscillator is another example. The energy eigenvalues are
En = (n +
)¯hω (3)
where n = 0, 1 , 2 , 3 , .... Notice once again that the energy levels are quantized. In this case they are evenly spaced by an amount ∆E = ¯hω. Electromagnetic radiation is also quantized. Light can be described as waves or as particles called photons. A photon has energy hν where ν is the frequency of the electromagnetic wave. Recall that ω = 2πν and that ν = c/λ where c is the speed of light. Often one speaks in terms of the wavenumber k = 2π/λ. If we make it a vector
quantity ~k, then we call it a wavevector. This is related to the momentum by ~p = ¯h~k and to the frequency by ω = ck. So if the electromagnetic wave has a short wavelength, it has a high frequency and the photon carries a lot of energy. Lots of wiggles means lots of energy. Photons are massless and they travel at the speed of light. Periodic Boundary Conditions:Counting States So energy is quantized into discrete energy levels. Each energy level is associated with a mode or eigenfunction. We have seen that it is often useful to be able to count the number of modes in a box that have energies between E and E + dE. Suppose we have a 3 dimensional box whose walls are parallel to the x, y, and z axes with lengths Lx, Ly, and Lz. Thus the volume is V = LxLyLz. We can solve this as a particle in a box problem. Inside the box the potential is zero. The eigenmodes are waves. However, let’s choose boundary conditions such that the solution of Schroedinger’s equation are wavefunctions that are plane waves:
Ψ = A exp[i(~k · ~r − ωt)] = ψ(~r) exp(−iωt) (4)
This is a propagating wave that is never reflected. So our box can’t have hard walls. Rather let’s imagine that our box is embedded in an infinite set of similar boxes in each of which the physical situation is exactly the same. In other words, each of these boxes is a repeat of the original box.
To describe this situation, we use periodic boundary conditions which we can write as
ψ(x + Lx, y, z) = ψ(x, y, z) ψ(x, y + Ly, z) = ψ(x, y, z) ψ(x, y, z + Lz ) = ψ(x, y, z)
If we require our traveling wave solution
ψ(~r) = exp(i~k · ~r) = exp[i(kxx + kyy + kz z)] (5)
to satisfy these boundary conditions, then we must require that
kx(x + Lx) = kxx + 2πnx (6)
where nx is an integer. We can rewrite this as
kx =
2 π Lx
nx (7)
Similarly,
ky =
2 π Ly
ny
kz =
2 π Lz
nz
Here the numbers nx, ny, and nz are any set of integers: positive, negative, or zero. We can use p = ¯hk and E = p^2 / 2 m to deduce that
E(nx, ny, nz ) =
¯h^2 2 m
(k x^2 + k y^2 + k^2 z ) =
2 π^2 ¯h^2 m
( n^2 x L^2 x
n^2 y L^2 y
n^2 z L^2 z
) (8)
The factor of 2 comes from the fact that there are 2 photon polarizations. The polariza- tion refers to the direction of the electric field vector E~ in the electromagnetic radiation. Since E~ must be perpendicular to ~k, there are 2 polarization directions. We will use (14) in deriving blackbody radiation. Sometimes the term “density of states” for photons is used to refer to the number of states per unit volume per unit energy:
N (ω) =
2 π^2 c^3
ω^2 =
π^2 c^3
ω^2 (15)
The density of states is very useful for converting sums into integrals as we shall see. Recap So let’s recap where we are and what we’ve found. If a system with lots of particles has many–particle states R with energy ER, then the average of some quantity A is given by
A =
Z
∑
R
ARe−βER^ (16)
where the partition function Z Z =
∑
R
e−βER^ (17)
This is for the canonical ensemble with fixed temperature T and fixed particle number N. We have seen that ln Z is very useful in finding other quantities. For example,
F = −kB T ln Z (18)
S = kB (ln Z + βE) (19)
E = −
∂ ln Z ∂β
CV =
∂E
∂T
∣∣ ∣∣ ∣V^ (21)
p =
β
∂ ln Z ∂V
But the problem is that it is very difficult to solve Schroedinger’s equation to get ER:
HψR = ERψR (23)
It is much easier to solve Schroedinger’s equation to get single particle energies. So we consider systems (gases) of noninteracting particles. If we know how many particles are in each single particle state, then we just sum over all the particles to get the appropriate average, e.g., the mean energy of the whole system. If we have to treat the particles quantum mechanically because their wavefunctions overlap or because the temperature is low, then we need to pay attention to whether the particles are fermions or bosons.
Fermions can have at most one particle in a state while bosons can have umpteen particles in a state. So now the mean energy is given by
E =
∑ s
εsns (24)
where the mean number of particles in state s is given by
ns =
eβ(εs−μ)^ ± 1
- is for fermions and − is for bosons. It’s usually not easy to do sums, so it would be nice if we could convert the sum into an integral. That’s why we calculated density of single particle states. Then we can do the conversion: ∑ s
∫ ρ(ε)dε (26)
or (^) ∑
s
∫ ρ(ω)dω (27)
So the mean energy becomes
E =
∫ dερ(ε)n(ε)ε (28)
where
n(ε) =
eβ(ε−μ)^ ± 1
Applications Now let’s go do some examples of this strategy. We will cover the following examples:
- Monatomic Ideal Gas
- Black Body Radiation
- Electron Gas (electrons in a metal)
- Bose–Einstein Condensation (if time permits) Monatomic Ideal Gas Let’s start with our tried and true example of a monatomic ideal gas in the classical limit of low density or high temperature. We want to calculate the partition function. We found earlier that
Z =
ζN N!
where ζ is the partition function for one particle.
ζ =
∑ r
e−βεr
∑
kx,ky ,kz
exp
[ −
βh¯^2 2 m
( k^2 x + k^2 y + k z^2
)] (31)
on it. If its temperature is kept constant, then the amount of power it radiates must equal the amount of power it absorbs. Otherwise it would heat up or cool off. We can imagine the black body being kept inside some kind of closed container which is at the same temperature T. The radiation field inside this enclosure is in equilibrium. In other words there is a gas of photons in thermal equilibrium inside the enclosure. By thermal equilibrium, we mean that the average occupation number ns of the single particle states is given by the Planck distribution that we talked about in lecture 13.
ns =
eβεs^ − 1
One can imagine making a histogram by counting the photon energy density in each frequency range from ω to ω + dω. It turns out that this distribution of the energy density of blackbody radiation is a universal curve that depends only on the temperature T. In other words if one plots the distribution of the photon energy density (counting both directions of polarization) as a function of photon (angular) frequency ω, the shape of the curve is universal and the position of the peak is a function only of the temperature. When we say that the curve is universal, we mean that it doesn’t depend on the size or shape of the box, or what the walls are made of. All that matters is the temperature. Blackbody radiation is historically important in physics for two reasons. The first is that the measurement of the spectral distribution in the late 1800’s led Planck to come up with the idea of energy quantization. He couldn’t explain the distribution unless he postulated that E = hν. This marked the birth of quantum mechanics. The second reason that blackbody radiation is important is that 3 K black body radiation pervades the universe and is the remnant of the Big Bang. This radiation is in the microwave region. Let’s calculate the distribution of the mean energy density of blackbody radiation. Since the size and shape of the box don’t matter, let’s imagine a rectangular box of volume V filled with a gas of photons that are in thermal equilibrium. The box has edges with lengths Lx, Ly, and Lz such that each of these lengths is much larger than the longest wavelength of significance. There are 2 factors that determine the energy density at a given frequency. The first is the average energy in each state s which is given by
nsεs =
εs eβεs^ − 1
If we set εs = ¯hω and ns = n(¯hω), we can rewrite this to give:
n(¯hω)¯hω =
¯hω eβ¯hω^ − 1
The second factor is the number of states per unit volume whose frequency lies in the range between ω and ω + dω. This is given by (15)
N (ω)dω =
π^2 c^3
ω^2 dω (42)
So at temperature T the mean energy density u(ω, T )dω contained in the photon gas by photons whose frequencies are between ω and ω + dω is given by the product of the average energy in each single photon state and the density of states which lie in this frequency range:
u(ω, T )dω = n(¯hω)¯hωN (ω)dω =
ω^2 π^2 c^3
¯hω (eβ¯hω^ − 1)
dω (43)
We can rewrite this to give:
u(ω, T )dω =
¯hω^3 π^2 c^3
dω (eβ¯hω^ − 1)
This is Planck’s law for the blackbody spectrum.
0e+00 2e+15 4e+15 6e+15 8e+15 1e+ angular frequency (radians/sec)
0.0e+
5.0e+
1.0e+
1.5e+
u(
ω
, T) (K−sec/m
Black Body Radiation
3000 K
4000 K
5000 K
We can take the high temperature limit to get the classical limit of this spectrum. In the high temperature limit, β is small so we can expand the exponential in the denominator:
eβ¯hω^ − 1 ≈ (1 + β¯hω) − 1 = β¯hω (45)
So the high temperature limit of (44) is
u(ω, T )dω ≈
¯hω^3 π^2 c^3 (β¯hω)
dω (46)
So if at temperature T 1 the maximum occurs at frequency ω 1 ,max, then at some other temperature T 2 the maximum occurs at ω 2 ,max. This is because
ηmax =
hω¯ 1 ,max kB T 1
¯hω 2 ,max kB T 2
or ω 1 ,max T 1
ω 2 ,max T 2
This is called the Wien displacement law. It says that
ωmax ∝ T (52)
This was initially an empirical relation that was deduced from the experimental data. We see that it also follows from Planck’s law. It is often useful in physics to express things in terms of dimensionless parameters. The Wien displacement law is an example of useful scaling relations that can result from this. We can also calculate the total energy density uo(T ) contained in the photon gas at temperature T by integrating (44) over frequency:
uo(T ) =
∫ (^) ∞
0
u(ω, T )dω (53)
Using (49), we can rewrite this as
uo(T ) =
h¯ π^2 c^3
( kB T ¯h
) 4 ∫ (^) ∞
0
η^3 dη eη^ − 1
One can evaluate the integral exactly. The answer is
∫ (^) ∞
0
η^3 dη eη^ − 1
π^4 15
Using this, one finds
uo(T ) =
π^2 15
(kB T )^4 (¯hc)^3
This is known as the Stefan–Boltzmann law. The important point is that the total energy density goes as the fourth power of the temperature:
uo(T ) ∝ T 4 (57)
Finally the mean pressure p exerted on the walls of the enclosure by the radiation is simply related to the total energy density:
p =
∑ s
ns
( −
∂εs ∂V
)
T
uo(T ) (58)
To see where this comes from, recall that
εs = ¯hωs ω = ck (59)
Now use
kx =
2 π Lx
nx ky =
2 π Ly
ny kz =
2 π Lz
nz (60)
to obtain
εs = ¯hck = ¯hc
√ k^2 x + k^2 y + k z^2
= ¯hc
( (^2) π
L
) √ n^2 x + n^2 y + n^2 z
= 2 π¯hcV −^1 /^3
√ n^2 x + n^2 y + n^2 z (61)
where V is the volume. So the pressure associated with state s is
ps = −
∂εs ∂V
2 π¯hcV −^4 /^3
√ n^2 x + n^2 y + n^2 z =
εs 3 V
So the average pressure for the system is
p =
∑ s
psns =
3 V
∑ s
nsεs =
E
3 V
or
p =
uo (64)
(The pressure can also be written as (^1) β
( (^) ∂ ln Z ∂V
) T
= − ∂F ∂V
∣∣ ∣ T .) The “3” in the denominator
reflects the fact that the box is 3 dimensional. Radiation pressure is quite small, but it is what gives comets their tails. Solar radiation is what pushes tiny bits of dust and ice that come from the ice ball away from the sun and produces the tail. The comet tail always points away from the sun. The power emitted ∼ flux ∼ cuo(T ). Principle of Detailed Balance If an object is sitting in a cavity filled with radiation (photons) and is in equilibrium at temperature T , then the
- power radiated by body = power absorbed by body
If this were not true, the body would be losing or gaining energy and would get cooler or would heat up. As a result, its temperature would no longer be the same as the ambient photons at temperature T ; it would no longer be in equilibrium. So it must absorb the same amount of power as it emits in order to stay in equilibrium. We can make an even stronger statement. Namely, that in equilibrium the power radiated and absorbed by the body must be equal for any particular element of area of the body, for any particular direction of polarization, and for any frequency range.
Radiation Emitted by a Body Let us now apply the principle of detailed balance to a body at temperature T in equilibrium with radiation (photon gas) inside an enclosure at this temperature. Let Pi(k, α) be the incident radiation power on a unit area of this body per unit frequency and solid angle range about the vector k with polarization α. Let a(k, α) be the fraction of incident power absorbed, the rest being reflected. By the principle of detailed balance, the power absorbed must equal the power emitted Pe(−k, α) in the opposite direction −k: Pe(−k, α) = a(k, α)Pi(k, α) (69)
For a blackbody, a(k, α) = 1; a good absorber is a good emitter and vice-versa. If we integrate over all directions ˆk and polarizations α, the total power emitted per unit area into the frequency range between ω and ω + dω is
Pe(ω)dω = a(ω)Pi(ω)dω (70)
In the appendix to these notes, we show that
Pe(ω)dω = a(ω)
[ 1 4
cu(ω)dω
] (71)
It makes sense to see cu(ω) for the flux. The factor of 1/4 comes from geometric consid- erations. Using Eq. (44), we can write this as
Pe(ω)dω = a(ω)
h¯ 4 π^2 c^2
ω^3 dω eβ¯hω^ − 1
The total power Petot emitted per unit area of the body is obtained by integrating Eq. (72) over frequency as we did in Eqs. (54)-(56) to obtain
Petot = a
( 1 4
cuo
) = a
( σT 4
) (73)
where uo is given by Eq. (56):
uo(T ) =
π^2 15
(kB T )^4 (¯hc)^3
Eq. (73) is another form of the Stefan-Boltzmann law. The Stefan-Boltzmann constant σ is
σ ≡
π^2 60
k^4 B c^2 ¯h^3
= (5. 6697 ± 0 .0029) × 10 −^5 erg/(sec cm^2 deg^4 ) (75)
For a perfect blackbody, a = 1. For something shiny like gold, a ≈ 0 .01. Appendix: Calculation for Radiation Emitted by a Body Let Pi(k, α) be the incident radiation power on a unit area of this body per unit frequency and solid angle range about the vector k with polarization α. Let a(k, α) be the fraction of incident power absorbed, the rest being reflected. By the principle of
detailed balance, the power absorbed must equal the power emitted Pe(−k, α) in the opposite direction −k: Pe(−k, α) = a(k, α)Pi(k, α) (76)
For a blackbody, a(k, α) = 1; a good absorber is a good emitter and vice-versa. Let us now calculate explicitly the power Pi(k, α) incident per unit area of a body in an enclosure at temperature T. This is the incident energy flux. Let f (k)d^3 k be the mean number of photons per unit volume with a given polarization whose wavevector lies between k and k + dk. So (c dt cos θ)f (k)d^3 k photons of a given frequency and polarization strike a unit area of the body in a time dt. Since each photon carries energy ¯hω, one obtains Pi(k, α)dωdΩ = (¯hω)(c cos θf (k)d^3 k) (77)
k Area dA
z
c dt
Converting d^3 k to spherical coordinates and using k = ω/c, we obtain
d^3 k = k^2 dkdΩ =
ω^2 c^3
dωdΩ (78)
and
Pi(k, α)dωdΩ =
¯hω^3 c^2
f (k)dωdΩ cos θ (79)
If the body absorbs isotropically, then the fraction of incident radiation absorbed is a(k, α) = a(ω), i.e., a is independent of the direction k. We are also assuming a is independent of the polarization direction. So the power emitted in the direction k′^ = −k is Pe(k′, α) = a(ω)Pi(k, α) (80) Now let us find the total power Pe(ω)dω emitted per unit area into the frequency range between ω and dω for both polarization directions by integrating over the solid angle. Using dΩ = sin θdθdφ and multiplying by 2 for both polarizations, we write
Pe(ω)dω = 2
∫
Ω
Pe(k′, α)dωdΩ
