Lecture 03

Lecture Lecture 3: Identica Identicall particles, quantum statistics again (9/27/2005) Today, oday, we start with Bloch’s theory theory that we did not finish last time. And then we will have a look at the Bose-Einstein and Fermi-Dirac distributions from a different (combinatorial) perspective than the quick approach in the first class. class. Our discussion discussion will include the Bose-Einstein Bose-Einstein condensation condensation and Planck’s Planck’s blackbody formula. formula. Sadly, Sadly, I must must ask you to invest invest your time and read the part of the notes that we won’t finish today later (or in the class if  you can do multitasking).

Additiona Additionall reading reading if you wish: wish: Griffiths Griffiths ch. 5.4.

General principles of statistical physics Each distinct state with the same energy must have the same probability. However, the number of states depends on the statistics of your particles.

An example: example: 3 particles particles in an infinite infinite well well This example example is similar similar to the first homework homework problem, but it’s not quite the same thing. Imagine Imagine three three non-intera non-interacting cting particles particles in the one-dimension one-dimensional al infinite potential well. The total energy is ¯2 2 π2h  = (n1 + n22 + n23 ). E  = 2 2ma Imagine that the total energy is ¯2 π2h  = E  = 2ma2

 × 243. 243.

There There are several several ways ways how can we get 243 by summing summing three squares. squares. The triplets triplets of positive positive integer integerss (n1 , n2 , n3 ) are (9, (9, 9, 9) (3, (3, 3, 15), 15), (5, (5, 7, 13), 13),

(5, (5, 13, 13, 7), 7),

(3, (3, 15, 15, 3), 3),

(13, (13, 5, 7), 7),

(15, (15, 3, 3)

(13, (13, 7, 5), 5), 1

(7, (7, 5, 13), 13),

(7, (7, 13, 13, 5)

If the particles are distinguishable, there is one state from the first line, three different states from the second line, and six different states from the third line. If they are indistinguishable bosons, there is one state coming from each line because the permutations do not matter. If they are indistinguishable fermions, the Pauli principle dictates that there are no states coming from the first two lines, and only one state from the last line (5, 7, 13). You can see that statistics matters, indeed.

General case Imagine that the particles can be in states whose energies are E 1 , E 2 , E 3 , . . . and whose degeneracies are d1 , d2 , d3 , . . . and we label the numbers of particles in these states as N 1 , N 2 , N 3 , . . . You can see that the total number of particles and the total energy are

 N  = N  ,

 E  = N  E  .

i

i

i

i

i

How many ways there are to arrange the particles to these “boxes” with fixed values of  N i ? For distinct particles, the total number of arrangements is Qdistinct

N i i

 d  = N ! i

N i !

This is because every permutation of the N  particles gives a new state (thus N !) except for the permutations of the particles inside the groups – which reduces it by the product i (N i !) – and every particle gives an extra factor di  coming from the degeneracy of its state whose energy is  E i . For identical fermions we only get a nonzero number of possibilities if  the degeneracies di  are enough to host the particles, di N i , and the total



 ≥

2

number of possibilities is the product of the combination numbers (di  choose N i ): di ! Qfermions  = N i )! i N i !(di



−

For bosons, we need a similar but slightly different product. What are the factors? We need the combinatorial number that counts the ways in which N i  identical particles can be separated into di  boxes. This problem can be solved exactly. Imagine that you describe the problem by N i +di 1 slots, and each slot can be either filled with the word “particle” (or an actual particle) or the words “new box” (or a separator). Moreover, you require that the number of “particles” is N i  and the number of the words “new box” – that act as separators between the different  d i  boxes – must be d i 1. Obviously, such arrangements of the words are in one-to-one correspondence with the configurations of bosons, and therefore the total number of the states for bosons is the product of the combinatorial numbers (N i  + di 1 choose N i  – that determine which of the slots are particles (and which of them are separators) – which means

−

−

−

Qbosons  =

 (N  + d − 1)! . i

i

N i !(di

i

− 1)!

Adding the Lagrange multipliers We want to find such numbers N i  in the case of each statistics that maximize the number Q  of different multiparticle states. But we only want to do so given the assumption that the total number of particles and total energy are fixed: N  = N i , E  = N i E i .





i

i

How can one find such a constrained maximum? First of all, a maximum of  Q occurs at the same place as the maximum of ln Q (log is an increasing function) and we will choose the latter because it is convenient. We define

      G = ln Q + α N  − N  + β  E  − N  E  i

i

i

i

and require that ∂G = 0, ∂N i

∂G = 0, ∂α 3

∂G = 0. ∂β 

i

The last two equations impose the previous conditions for the total number of particles and total energy, and because α, β   can be arbitrary, the first equation (actually a set of equations) only imposes the “constrained” maximality of  Q. For large degeneracies d i , we may simplify and approximate the formulae for Q using the Stirling formula. For large x, the factorial can be approximated by x x 2πx x! e which means that the limit of the ratio of LHS/RHS goes to one as  x . This can be proved from the Euler representation of the factorial

≈

√ 



  x! =

∞

→∞

dt e t tx −

0

by identifying the maximum of the integrand near t = x   and using the saddle point approximation (of the integrand by a Gaussian). We also have the Stirling formula for the logarithms ln(N !)

√ 

≈ N (ln N  − 1)

where even the 2πN  is negligible and similar formulae for N i ! etc. Combining these insights, the previous formula for Q distinct  reduces to

[N  ln d − ln(N !)]

ln Qdistinct  = ln(N !) +

i

i

i

i

and thus

[N  ln d − N  ln N  + N  ] +       α N  − N  + β  E  − N  E  .

Gdistinct   = ln(N !) +

i

i

i

i

i

i

+

i

i

i

i

i

Its derivatives must vanish: ∂G =0 ∂N i

⇒

ln di

− ln N  − α − βE   = 0 i

i

We see that N i  = d i e

−

α−βE i

.

The normalization e α is determined from the total number of particles and the coefficient β   is identified with 1/kB T . That’s the usual MaxwellBoltzmann result. −

4

Fermi-Dirac distribution For fermions, some details in G fermions  must be modified. We obtain Gfermions =

 {ln(d !) − ln(N !) − ln(d − N )!} +       α N  − N  + β  E  − N  E  [ln(d !) − N  ln(N ) + N  − (d − N )(ln(d − N ) − 1)]       α N  − N  + β  E  − N  E  i

i

i

i

i

+

i

i

i

≈ +

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

i

The derivatives must vanish again: 0=

∂G fermions = ∂N i

− ln(N  ) + ln(d − N  ) − α − βE  i

i

i

i

which means, for identical fermions, that N i  =

di eα+βE  + 1 i

.

Note that unlike the Maxwell-Boltzmann case, α affects not only the overall normalization. You should notice that we were not cheating: N i  was large even in the case of fermions because these  N i  fermions have di  (degeneracy) available slots.

Bose-Einstein distribution: the result You are positively invited to repeat the very same calculation for the bosons, using Q bosons . What you get at the end only differs by a sign from the FermiDirac distribution above: N i  =

di eα+βE 

i

− 1.

For every physical system, we must then determine  α, β   from the conditions that the total number and total energy are what they should be, namely N, E . Let’s finish this task for the three-dimensional box – or, equivalently, the three-dimensional fermion gas – in all three cases of different statistics. 5

Requiring the right total number and the total energy For distinct particles, we have α−βE i

d e N  =

−

i

i

(2s + 1)V  = e 2π2

−

α

∞

 

e

0

or N  = (2s + 1)V e

−

α



β ¯ h2 k2 /2m 2

k dk

−

m 2πβ h ¯2

3/2



where we inserted the spin degeneracy (2s + 1) and replaced the summation by the integration using the approximate formula for the degeneracy: (2s + 1)V  4πk 2 dk (2s + 1)V k 2 dk = dk = 8 2π 2 π3 /V  This means that (2s + 1)V  eα = N 



m 2πβ h ¯2

3/2



A similar constraint is derived from the total energy condition by inserting an extra factor of  E i :

 d E  e E  =

−

i

i

i

α−βE i

(2s + 1)V  = e 2π2

−

h2 α  ¯ 2m

 

∞

0

or (2s + 1)3V  E  = e 2β 

−

Substitute our previous result for e E  =

−

α

α



m 2πβ h ¯2

e

β ¯ h2 k2 /2m 4

−

k dk

3/2



.

to see that

3N   , 2β 

3 E  = . N  2β 

This nicely agrees with the usual definition β  = 1/kB T   because of the equipartition theorem that implies E  3 = kB T. N  2

6

In the case of distinguishable particles,  α only determines the normalization. But in the cases of all statistics, we usually write down the exponent in a new form µ(T ) α = kB T  and interpret µ(T ) as the chemical potential: it is the energy needed to add a particle for constant S, V . Note that in the case of all three statistics, N i was proportional to di , so if we evaluate the total number of particles per one (non-degenerate) state, we simply divide the result by  d i  to obtain

−

n() =

1 e(

−

µ)/kB T 

−σ

where σ   = 1 for bosons, 1 for fermions, and 0 for distinguishable particles. As long as the occupation numbers n are low, the distibutions coincide because the exponential dominates over σ in the denominator.

 −

Temperatures near zero: fermions What happens with the Fermi-Dirac distribution for tiny T  see that 1,  < µ n() 0,  > µ

→   0? You can



→

which simply means that µ(T  = 0) may be identified with the Fermi energy E F  and the fermions occupy all states up to E F .

Temperatures near zero: bosons For bosons, note that n() must be positive which really means that µ 0

  T   = N 

3/2

T 0

where the normalization was obtained from N  = N  at T  = T 0   while the rest N E =0 = N  N E>0 = N [1 (T /T 0 )3/2 ] of the particles is found in the condensate. You can actually try to use these equations. For example, for Helium 4 you may use ρHe = 0.15 g/cm3 . This will give you T 0 = 3.1 kelvins, very close to the experimentally determined temperature  T 0 = 2.17 kelvins where Helium 4 becomes a superfluid. Recently, the Bose-Einstein condensation was observed on many real gases with low  ρ  and correspondingly low T 0 .

 −

−

8

Planck’s blackbody formula Photons are a special case of bosons. However, we must

•  set 2S +1 → 2 even though the spin is S  = 1 because photons only have

two (transverse) polarizations and the third, longitudinal polarization is unphysical

•  we must set µ = 0 because the number of photons may change – ther-

malization may involve emission and absorption of photons and we should have avoided the Lagrange multiplier α  i.e. µ completely

•  realize that E  = ¯hω for photons, which allows us to choose the angular frequency  ω

•   appreciate that E  = c| p|  i.e. ω = c|k| The energy is conserved and  β  enters as usual. The occupation numbers are N ω =

dk e¯hω/k

B

where

T 

−1

2from spinV  2 V ω2 dk = k dk = 2 3 dω 2π2 π c The contribution of a frequency interval (ω, ω + dω) to the energy density can be expressed in two related ways: N ω ¯hω/V  = ρ(ω)dω Finally we obtain, using more deductive approach than Planck himself in 1900, his formula ¯hω 3 ρ(ω) = 2 3 ¯hω/k T  . 1] π c [e B

−

Unlike Max Planck 105 years ago, you should actually try to believe that this formula and the derivation are true.

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