An Introduction to Quantum Statistics

1. The Occupancy of States and Distributions:

In statistical physics two particular distributions occupy attention: the Bose-Einstein, and the Fermi-Dirac. For our purposes we will spend much more time on the latter but it is useful to see how both enter the picture.

In the case of the Bose-Einstein formalism we are concerned with bosons which have integral spin values (i.e. an orbital may be populated by any number of bosons so the Pauli Exclusion principle doesn’t apply).

Consider the distribution function for a system of non-interacting bosons in which the system is in

thermal and diffusive contact with a reservoir.  Thus we may allow a circumstance such as this:

Fig. 1: Illustrating systems in thermal and diffusive contact.

Thus heat can be exchanged between the systems A and B, as well as particles.  In the formalism treatment we let e denote the energy of a single orbital when occupied by one particle. Hence, when n particles occupy the orbital then the energy must be n e.  To simplify, we treat one orbital as a system and ignore all others.  The grand sum taken over one orbital would then be:

å^¥ _{n = 0}   å _e   exp ( Nm - e)/ t) =  å^¥ _{n = 0}    lⁿ e ^{- n e/t}

=  å ^¥ _{n = 0}    ( l e ^{- n e/t}   )   ⁿ

Now, let x =  l e ^{- n e/t}     and sum the series in closed form so the grand sum is now:

Ž   =  å ^¥ _{n = 0}    (x) ⁿ  =  1/ 1 – x = 1/ [   1 - l e ^{- n e/t} ]

Provided:  l e ^{- n e/t}       <  1

 

The ensemble average of the number of particles in the orbital is by definition of the average value:

[ n(e )]  =   å ^¥ _{n = 0}    (nx) ⁿ  /  å ^¥ _{n = 0}    (x) ⁿ  = 

x (d/dx)  å ^¥ _{n = 0}   (x) ⁿ / å ^¥ _{n = 0}   (x) ⁿ  = 

x (d/dx) (1 – x) / (1 – x) ^-1

After some calculus and messy algebra we find:

[ n(e )]  =  1/ exp ( e -  m)/ t

Which is the Bose- Einstein distribution.

(N.B. Square brackets used above in place of  R-L 'angle' brackets to prevent problems with blogger)

For the Fermi-Dirac distribution we consider a simplified system represented by a cubical box of volume V = L ³

The number of electrons (Fermi particles) within occurs between energies e and e + de.  The energies occur in a quantized way and are depicted schematically below:

Fig 2. Box with Fermi occupancy and energy levels.

 

The diagram shows each level corresponding to two orbitals, one for spin up the other for spin down. 

The Fermi energy:

e _F = ħ²  ( p n _F / L) ² /2m

Is determined by the requirement that the system in the ground state holds N electrons – with each orbital filled with one electron if the energy of the orbital is less than:

 e _F.

Since for a 3D box the quantized energy is:

ħ²/2m [p n² / L]

where n² =  n _x ²  +  n _y ²    + n _z ² 

Then the number of states possessed by a 3D box is:

1/8 [4p n³ / 3]

For a sphere, the number of orbitals in some radius is:

n=  g  [4p n³ / 3]  (1/8) =  g p n³  / 6

For electrons,  g   = 2 so:   n=  p n³  / 3

Hence, if the box holds N electrons the orbitals must be filled up to the quantum number n _F.    So the total number of electrons is:

N =    p n _F³  / 3  so     n _F  = (3N/p) ^1/3

The Fermi energy would then be:

e _F = ħ²  ( p / L) ²  n _F ²/ 2m   =

ħ²  ( p / L) ²  (3N/p) ^1/3/ 2m  

Since  V = L ³ we can write:

e _F = ħ²  / 2m   [3p² N/ v] ^2/3

This yields good results for monovalent atoms because they have only one electron per atom.

2. Treating the Nucleus as a Fermi Gas

One of the useful applications of the Fermi quantum statistics is to the atomic nucleus, such as that for the deuteron.  The key to the application is assuming an independent particle model so each nucleon moves in a smooth potential hole as opposed to being subject to the actions of (A- 1) other nucleons.  In this sense, the model would exhibit similar properties to the “electron gas” which we examined being confined to a 3D box in the previous section.  In this case, the nucleons will be constrained to move in a spherical potential hole of radius:

R = r _o A ^1/3

(I.e. analogous to the case of electrons)

We focus then on all momentum states being filled up to the Fermi momentum,   p _F.

Since:   p _F ²/ 2m  =  ħ²  / 2m   [3p² N/ v] ^2/3

the number of states up to  p _F  per unit volume is:

N/ v  =    4p/ 3  ( p _F ³/ h³ )

Or the volume in momentum space divided by the Planck constant. Let’s now apply this to the case of the deuteron, for which we have a neutron and proton in each state and each can have spin up or spin down. Then we get a total of four states so multiply  the expression by 4 to get:

4N =  4[4p/ 3 ( c h³ ) (4p R ³/ (2 p h³ ) 3]

Simplifying:

p _F r _o / h =    Ö (9 p/ 8)  =   1.52

This prescribes:  the maximum momentum present in the nucleus in terms of the nuclear radius  r _o .

p _F  =    1.52 (h/ r _o)

Taking r _o =   1.1 fm (e.g. 1.1 x 10 ^-15 m)

We learn the maximum kinetic energies in the nucleus are as large as:

K _F  =    p _F ²/ 2m  =  39 MeV

If the binding energy of the ‘last nucleon’, e.g. at the top of the Fermi distribution) is » 8 MeV then this result shows that the nucleus looks to each nucleon (say to the neutron or proton  in Deuterium) like a potential hole of 39 MeV depth.

 Suggested Problems:

1) Find the Fermi sphere parameters: e _F ,  v _F  and   T _F for He3 at absolute zero, viewed as a gas of non-interactive fermions. (The density of the liquid is 81 kg/ m ³).

2) a) Show that (- ¶f / ¶ e)  evaluated at the Fermi level (e  = m) has the value (4 k_B T) ^-1. Thus, the lower the temperature, then the steeper the slope of the Fermi-Dirac function.

Hint: Use f(e)  =  1/ [exp  (e- m )/ t + 1]

b) Make a careful plot of   (- ¶f / ¶ e) vs.  e/ k_B  for  the specific cases:  

m/ k_B   =  5 x 10 ⁴ K  and  5 x 10 ² K .

3) Let e =  m +  d,  show that: f(d  ) = 1 – f( -d )

Hint: Let f(d  ) =  1/ [exp (m - e)/ t + 1]

 

4) Treating the Tritium nucleus as a nucleonic electron gas, assume an independent particle model (based on Fermi energy levels)  so each nucleon moves in a smooth potential hole.

Thereby obtain the maximum momentum present in the nucleus in terms of the nuclear radius  r _o .

p _F  =    1.52 (h/ r _o)

(Take r _o =   1.1  fm )

Use this to estimate the depth of the potential hole, i.e. the magnitude of the well in MeV for how the nucleus looks to each nucleon (say to the neutrons or proton in Tritium) .

Give a general idea of how errors enter your estimate and explain why it can’t be taken too literally.

An Introduction to Nuclear Physics (1)

1.The Liquid Drop model

Nuclear models occur under a set of hypotheses, each of which explains some aspect of nuclear behavior. The most basic is perhaps the “liquid drop model” which is used to account for nuclear fission, radioactivity. This is illustrated in Fig, 1 below:

Fig. 1: The Liquid Drop model showing U235 Fission

As the name implies, the nucleus is depicted in terms of a “liquid drop”, which analogous to a liquid macroscopic drop, exhibits surface tension and an excitation energy when disturbed.

This model is premised on the exceedingly short range of nuclear forces which requires that nearest neighbor attractions predominate. This is somewhat similar to the type of attraction between the molecules of a liquid which leads to the property of its surface tension.  Thus, in a liquid drop nucleus each nucleon shares its total binding energy with every nucleon.


As the simplified model diagram (for U 235 fission) shows, we expect  a spherical configuration in the minimum energy or unexcited state.  The unstable (compound) nucleus then yields two “daughter nuclei” : La 139 and Mo 95, with two additional neutrons released

    The total binding energy initially holding all the nucleons together in the liquid drop nucleus model is comprised of a sum of separate energies, including:  surface or E(s), volume or E(v), and Coulombic repulsion or E(c) – the latter due to the electrostatic repulsion between protons. Thus, the binding energy:

 E(B) = E(s) + (E(v) + E(c)

     A reduction in any one or all of these constituent energies reduces the overall binding energy and hence the bonding energy between any pair of nucleons. To account for energy release in the fission of heavy nuclei we need only represent the nucleus as a liquid drop and its transition from the unexcited to the excited state as shown in Fig. 1.

 Thus, between the spherical and compound nucleus phase, the “drop” may be said to undergo oscillations which lead it to overshoot sphericity in two directions – vertically (converting to an elliptical configuration with semi-major axis in the vertical sense, and horizontally, with semi-major axis in the horizontal sense. It finally reaches a point – the compound nucleus phase – at which the surface tension restoring force is no longer able to contain the long –range Coulombic repulsion force and the inertia of nuclear matter.

    At this point, the drop reaches a stage of distortion in which the nucleus splits into two fission fragments, denoted by [A1, Z1] and [A2, Z2]. In order to further assess the nuclear changes, we use what is called a “semi-empirical mass formula” which sets out terms as follows:


(1) f_o(Z, A) = 1.008142Z + 1.008982 (A – Z)

     Where the Z coefficient is the mass of the hydrogen atom in atomic mass units and the (A – Z) coefficient is the mass of the neutron in amu. The remaining terms listed correct for several effects which contribute to the total nuclear binding energy:

(2) f₁ (Z, A) = -a₁ A

      This term accounts for the binding energy and is essentially proportional to the nuclear mass or the volume of the nucleus.

(3) f₂ (Z, A) = +a₂ A^2/3

     This term is a positive correction to the surface area of the nucleus, i.e.  the effect of surface tension energy.

(4) f₃ (Z, A) = + a₃ (Z²/ A^1/3 )

     This term accounts for the positive Coulomb energy of the charged nucleus, assumed to be a sphere of radius proportional to  A^1/3.


(5) f₄ (Z, A) = + a₄ (Z  -  A/2)²/ A

     This term and the last (6) introduce properties specific to the nucleus. This term is zero for the case of Z = (A – Z) or 2Z = A.

(6) f₅ (Z, A) = 0, - f(A), +f(A)

    Where  0 is the result when ‘Z even’ pairs with (A – Z) odd, or ‘Z odd’ pairs with (A – Z) even; and –f(A) is the result when Z even pairs with (A – Z) even, and f(A) is the results when Z odd is paired to (A – Z) odd.  The form of f(A) is determined by fitting the data. It is found that the best fit for a simple power law relation is:

f (A) = a₅ A ^{- 1/2}

Combining all the terms we get:

M(Z, A) = 1.008142Z + 1.008982 (A – Z) -a₁ A+ a₂ A^2/3

+ a₃ (Z²/ A^1/3 ) + a₄ (Z  -  A/2)²/ A +  (-1, 0 , 1) a₅ A ^{– 1/2}

    The parameter values designated are:  a₁ = 0.01692,  a₂ = 0.01912,  a₃ = 0.000763,  a₄  = 0.10178, and  a₅  = 0.012.

2. The  Nuclear Shell Model:

The shell model of the nucleus treats a different aspect, namely specific energy levels within the nucleus.  To achieve this the nucleons are treated as independent particles, each of which moves in its own spherically symmetrical potential well about 50 MeV deep with rounded edges. 

Fig. 2: The potential well for the Shell model


The Schrodinger Equation  is then set up for this potential and solved. The solution yields stationary quantum states, somewhat like the states which are associated with electrons in the outer shells of the atom. 

Like the electrons occupying atomic shells, the nucleons are fermions, i.e. particles each with spin ½  and therefore obeying the Pauli Exclusion Principle so that no more than two nucleons of the same type can occupy the same energy level (e.g. one with spin +½, the other with spin -½.). This leads to particles pairing up in the nucleus: spin up (+ ½) protons with spin down (-½) protons, and spin up neutrons with spin down neutrons. 

The form or expression for the energy of a nucleus is very closely approximated by  the energy associated with a “square well” potential. The Fermi energy in more explicit form (which we will not elaborate upon too much beyond this) is:

E_F = 3 ^2/3 p ^4/3 ħ² r ^2/3/ 2M

 

Where M is the nucleon mass, ħ = h/2p is the adjusted Planck number (h divided by 2p) and r is the nucleon density. Performing the operation with known or estimated values, the result is E_F =  30 MeV.   If we assume a spherical nucleus then the radius is just:

R = r _o A ^1/3

This is also the dimension on either side of the central symmetry line of Fig. 2. The evidence for the shell model includes the following:

i)Particularly stable nuclei are borne out by “closed shells”, based on our observations.

ii) The model predicts (correctly) that the even N (or A – Z) nuclei will be most stable and the odd Z, odd N nuclei the least stable. Experiments bear this out: there are 160 stable nuclides (with even Z, N) and only 4 with odd Z, N.

iii) The model predicts that for even N, Z nuclides the total angular momentum J = 0, and that for odd nuclides it is half-integral. (Borne out by measurements of nuclear magnetic moments.)

Problem: Find the diameter of the oxygen (O ¹⁶) nucleus.

Solution:

We apply:  R = r _o A ^1/3

Where:  r _o  =  1.2 x 10 ^-15 m  is the Bohr radius, and A = 16.  Then:

R = (1.2 x 10 ^-15 m)  (16)  ^1/3= 3.0238 x 10 ^-15 m

 

And the diameter D = 2R = 2(3.0238 x 10 ^-15 m)

D = 6.05 x 10 ^-15 m

The preceding result can also be generalized to a situation in which the ratio of radii (between two nuclei) is sought. In this case, we can write:

R1/ R2  =  [r _o A1 ^1/3] /  [r _o A2  ^1/3]

Or:

R1/ R2 = (A1/ A2) ^1/3

Example: How much larger is a copper nucleus than an oxygen nucleus?

 

Let R1 denote  the radius of the copper nucleus, and R2, oxygen’s. Then: A1 = 64, and A2 = 16. So:

R1/ R2 = (64/16) ^1/3 =   4/ 2.51 = 1.587

Thus, the copper nucleus is about 1.587 times the size of the oxygen nucleus.

Problems:

 1) Evaluate the terms of the semi-empirical mass formula for the U 238 nucleus,  i.e. if A = 238 and Z = 92. Use this information to find the total mass in atomic mass units (u) and compare it to the standard mass expression. Thence or otherwise, obtain the mass defect and the binding energy as well as the binding energy per nucleon.

2)The sketch graph shown below plots the mass number A vs. the binding energy per nucleon (BE/nucleon) on the vertical axis.

Using any models or other explanations, account for the region of greatest stability indicated. 

3) A quantum square well potential is  schematically defined according to: 

It is found that the Schrodinger equation to solve becomes:

-ħ²/ 2m ( dy²/dx² ) = Ey

Thence:

dy²/dx² + 2m/ ħ² (Ey) = 0

And the quantized energy is found to be:

E _n = (h²/ 8m L²) n²

This can be modified to yield a simplified Fermi energy in one dimension by using instead:

E _n = (h²  p²/ 8m L²) (N/2)²

 Where all the Fermi shell energy levels are presumed occupied up to N/2.  Use this to obtain a simplified estimate of the energy associated with the oxygen nucleus if we assume its shells filled up to N =4 and use an estimate for L as:

  R = r _o A ^1/3

(Take m = 1.7 x 10 ^-27 kg)

4)(a) Compare the value you obtain in (3) to the actual Fermi (shell) energy for oxygen, if the nuclear density r » 2.3 x 10¹⁷ kgm^-3.

(b) Write out at least one technique that might be used to find the density of an atomic nucleus. State clearly any assumptions made and how you quantify them.

5)(a) Find the ratio of the helium nucleus radius to that of the uranium 238 nucleus.

(b) Estimate, using any technique you can think of, the ratio of the nuclear densities for part (a).

6) An element has mass number A = 202 and atomic number Z = 80.

a) Find the diameter of the nucleus and how many times it is greater than that of hydrogen.

b) Find the mass defect D M for this nucleus

c) From the result in (b) obtain the binding energy and the binding energy per nucleon, E_B/ A.  Using the graph on the opposite page, mark the location of E_B/ A.