Lecture 27 PHYS 416 Tuesday November 30 Fall 2021

1. Reminder: Lect 26 was the online Higgs Boson lecture; expect a Final exam question on it.

2. Reminder: details on final exam time & place etc…3. Review Quiz 9 solutions4. HW 7.2 Photons & phonons are bosons5. Review 7.6.2 Blackbody radiation6. 7.6.3 Bose condensation7. Postpone to Thursday: 7.7 Metals & the Fermi gas

Note: e-µ>>kT happens at high T, which may seem odd at first. As T gets large, however, µ gets large and negative at a faster rate. This is due to more states available for occupation at higher T; see footnote 20 (Ch 7).

ψ = 2L

⎛⎝⎜

⎞⎠⎟

3 2

cos2πn1Lx

⎛⎝⎜

⎞⎠⎟cos

2πn2L

⎛⎝⎜

⎞⎠⎟cos

2πn3L

⎛⎝⎜

⎞⎠⎟

7.6 Black-body radiation7.6.1 Free particles in a box

We talk about particles in a box, because photons in a cavity behave like particles (which of course also behave like waves….)

Assume particles are identical, non-interacting, and free (no potentials). Box has volume V=L3, and for simplicity apply periodic boundary conditions.

Redundancy: there are 8 combinations of +/- ni that give the same energy (ni

2).

density of plane waves in k - space =V 8π 3

ψk = 1 L( )3 2 exp ik ⋅ r( )We can describe the same states by using complex exponentials and vector notation:

k = 2π / L( ) n1,n2 ,n3( )

Viewed in reciprocal space (k-space), each node is 2p/L from the next one, so the volume per point is (2p/L)3. Invert this to get the density of points:

g ω( )dω = 4π k 2( ) d kdω dω⎛

⎝⎜

⎞

⎠⎟

2V

2π( )3⎛

⎝⎜⎜

⎞

⎠⎟⎟

7.6.2 Black-body radiation

Each quantized mode of electromagnetic radiation (a photon) becomes one quantum harmonic oscillator. (Each wave vector has 2 modes, one for each polarization.)

The equipartition theorem says that in thermodynamic equilibrium, each mode gets 1/2kT of energy. This means higher frequencies yield more modes yields higher energy density, which leads to the “ultraviolet catastrophe.” Let’s set up the problem…

Recall w=ck. The number of states within a small range dw is:

Here we are modelling the volume element dw as a spherical shell of width dk in k-space.

#of photons ( )dω =g ω( )

e!ω kBT −1dω

g ω( ) = Vω2

π 2c3k 2 =ω 2 / c2 , d k / dω = 1/ cUsing: We get:

Because the number of photons is not fixed (they can be created and destroyed), the chemical potential must be zero. This now gives us the number of photons in a frequency range dw:

That is, here we are applying boson statistics to photons.

This tells us how many photons can occupy a given mode, which differs greatly from classical statistics:

nBE

= 1e!ω /kT −1

Vu ω( )dω =!ωg ω( )e!ω kBT −1

dω

= V!π 2c3

ω 3dωe!ω kBT −1

VuRJ ω( )dω =VkBTπ 2c3

⎛⎝⎜

⎞⎠⎟ω 2dω = kBTg ω( )

The photon energy spectrum is thus:

At low frequencies, expand the exponential to get:

Each mode does have 1/2kT, consistent with the equipartion theorem.(Why does equipartition break down?)

What is quantum statistics doing?

Classically, a mode with a given energy can have any (small) amount of energy added to it:

!ω → ! ω +δ( )Quantum-mechanically, any additional energy must be quantized:

!ω → !ω + !ω

This introduces an energy gap, which changes everything…

7.6.2 Bose condensation

Consider now a constant number of particles N (no creating or destroying). Assume these are non-interacting, and they are free particles:

k 2 =ω 2 / c2 , d k / dω = 1/ cPhotons:

Particles: E= KE= 12mv2 = p2

2m

p= hλ=!k p2 = 2mε, p= 2mε

dε dp= 1m

2mε = 2ε / m

g ω( ) = Vω2

π 2c3Photon result:

g ε( )dε = 4π p2( ) d pdε dε⎛

⎝⎜

⎞

⎠⎟

V

2π!( )3

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= 4π 2mε( )( ) m2εdε

⎛

⎝⎜

⎞

⎠⎟

V

2π!( )3

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= Vm3 2

2π 2!3ε dε

Particle result:

NOTE: This is a continuum approximation to a discrete distribution.

N µ( ) = g ε( )e ε−µ( ) kBT −10

∞

∫ dε

We have a density of states (g(e)), and an occupation function (fBE(e,T)). The number of particles is their product, added up (integrated over) all allowed states.

Typically the number N is fixed. For a given temperature, this integral determines the value of µ, the chemical potential. (𝜇 ≤ 𝜀! = 0)

In experiments, the temperature is varied. This changes the thermal deBroglie wavelengths of the particles, moving from classical to quantum behavior at low enough T. (Why?)

Nmaxcont =

g ε( )eε kBT −1∫ dε

=Vm3 2

2π 2!3dε εeε kBT −10

∞

∫

=V2πmkBTh

⎛

⎝⎜⎜

⎞

⎠⎟⎟

3

2

πz

ez −10

∞

∫ dz

=Vλ 3

⎛⎝⎜

⎞⎠⎟ζ 3 2( )

OK, let’s do the integral, in the continuum limit. Also let’s set µ=0, which maximizes N. (For negative µ there are fewer particles. Positive values are not allowed.)

N. b. ζ(3 / 2)≈ 2.612

λ=h

2πmkBT

(Why should there even be a maximum number of particles?)

Nmaxcont

V=ζ 3 2( )λ 3 = 2.612particles

deBroglie volume

kBTcBEC = h2

2πmN

Vζ 3 2( )⎛

⎝⎜

⎞

⎠⎟

2 3

OK, so here’s the (supposedly) maximum number of particles that you can stuff into the volume V, expressed as the density:

Put in the definition of l and solve for the critical temperature:

What happens below this temperature? If you have the maximum N above Tc, then you will exceed the maximum allowed for lower T. What does that mean???

It means that every single “extra” particle must end up in the very lowest energy (e=0) allowed state. Every single one of them. Even if there are 1022 of them.

This is Bose-Einstein condensation.

How did this happen?

One way of looking at it is that the separation between e0 and µ (as µ increased) got smaller than the gap between e1 and e0 (which is only ~10-11 K!). This, however, is not much of an explanation….

It turns out that the problem is the continuum approximation. In this situation, we can’t ignore the ”quantization” of the allowed particle states.

Let’s follow the approach in Kittel:

The number of particles N is obtained by summing over the occupation of all states:

N = fn = N0(τ )+ Nexcited states(τ )n∑

= Nε=0(τ )+ dεD(ε ) fBE (ε ,τ )0

∞

∫N.b. D(e)=g(e)=density of states;Set e=0. The state N0 was left out of the continuum model.

Let’s examine some behaviors as e, T goes to zero:

f (0,τ ) = 1e−µ/τ −1

limτ→0

fBE (0,τ ) = N ≈ limτ→0

1e−µ/τ −1

= 1

1− (µτ)−1

= −τ / µ

Wait! limτ→0e−µ/τ = 1− µ / τ ???

Shouldn’t this exponential blow up? (Remember, µ<0.) But then N could not be large. This means that µ(t) must go to zero faster than t does.

limτ→0

fBE (0,τ ) = N ⇒ N ! −τ / µ, or µ = −τ / N

λ ≡ eµ/τ ! 1+ µ / τ = 1− 1N

1e(ε−µ )/τ −1

⇒ 1e−µ/τ −1

= 1λ −1 −1

= N0

N = N0(0,τ )+ Nexc(ε ,τ )

= −τ / µ + 1.306V4

2mτπ!2

⎛⎝⎜

⎞⎠⎟

32

= −τ / µ + 2.612nQV

with nQ ≡mτ

2π!2

⎛⎝⎜

⎞⎠⎟

3/2

= "quantum concentration"

We just compute Nexc, then subtract from N to get N0, the occupation of the ground state. NOTE: this is a macroscopic number!!

NexcN

= ττ E

⎛

⎝⎜⎞

⎠⎟

3/2

Carl Wieman’s group experiment:

Observation of BEC in rubidium by the JILA group. The upper left sequence of pictures shows the shadow created by absorption in the expanding atomic cloud released from the trap (Rb87).

http://nobelprize.org/nobel_prizes/physics/laureates/2001/phyadv.pdf

M. H. Anderson, J. R. Ensher, M. R. Matthews, C. E. Wieman and E. A. Cornell, Science 269, 198 (1995)

~170 nK

• The BEC phase transition occurs at a temperature of about 600 nK. The last plot displays the spatial distribution of a condensate with 105 atoms and no discernible non-condensed fraction.

• http://www.mpq.mpg.de/qdynamics/projects/bec/

1200 nK310 nK

170 nK

Rempe’s group (Max-Planck Institute)

OK, so this 1995 work got the Nobel prize for Bose-Einstein condensation. But what about the superfluid transition in liquid Helium. Isn’t that Bose-Einstein condensation too?

Yes, it is! Specific heat anomaly discovered in 1927 (Keesom), superfluidity seen in 1937 (Kapitsa; Allen & Misener).

Superfluid Helium versus Rubidium gas:

• Our theory is for a non-interacting gas. The Rb experiment comes very close to this.

• Helium is a liquid, indicating stronger interactions than expected in a gas.• Helium tends to be somewhat like a gas, however, because it has a very

low density due to zero-point motion (quantum harmonic oscillator effect). (Factor of 3.1 change in density!)

• Non-interacting theory predicts BE transition for helium at 3.1 K; superfluid transition is at 2.17 K. Not bad!

• Note: superfluidity is a result of interactions; it is not a property of all BE condensates.

• Compare to superconductivity = a “gas” of spin-zero electron pairs.