Permutation Symmetry

and Identical Particles

Until now, our focus has largely been on the study of quantum mechanics of individual

particles.

However, most physical systems involve interaction of many (about 1023!) particles, e.g.

electrons in a solid, atoms in a gas, etc.

In classical mechanics, particles are always distinguishable

In quantum mechanics, particles can be identical and indistinguishable, e.g. electrons

in an atom or a metal.

The intrinsic uncertainty in position and momentum therefore demands separate consideration

of distinguishable and indistinguishable quantum particles.

Here we define the quantum mechanics of many-particle systems, and address a few implica-

tions of particle indistinguishability.

Two particle systems

A general two-particle wave function with a particular energy can be written as

/

2121 ),(),,( iEterrtrr

The Hamiltonian for whole system is therefore,

),,(22

21

2

2

2

22

1

1

2

trrVmm

H

The normalized two-particle wavefunction has probability

of finding particle 1 in the volume and particle 2 in the volume 2

3

1

32

21 |),,(| rdrdtrr

1

3rd

2

3rd

1|),,(| 2

3

1

32

21 rdrdtrr

is the normalization condition

If particle 1 is in the one particle state and particle 2 in the state then )(ra

)(rb

)()(),( 2121 rrrr ba

However, this is possible only if we can distinguish particle 1 and 2. In quantum mechanics

two identical particles are indistinguishable so it is impossible to assign each particle to a

particular state. Instead, there are two ways of writing the wavefunction, not assigning each

particle to a particular state, namely:

)]()()()([),( 212121 rrrrArr abba

The theory admits two kinds of identical particles: bosons, for which we use the plus sign

and fermions, for which we use the minus sign. It turns out that

All particles with integer spin are bosons, and

All particles with half-integer spin are fermions

The two-particle wavefunction makes sense only if

),(),(|),(| |),(| 1221

2

12

2

21 rrerrrrrr i

If we introduce exchange operator ),(),(ˆ1221 rrrrP

Since ,

It follows that the eigenvalues of P are ±1. Hence

or 012 ie

Particles with half-integer spin are fermions and their wavefunction must be antisymmetric

under particle exchange e.g. electron, positron, neutron, proton, quarks, muons, etc.

Particles with integer spin (including zero) are bosons and their wavefunction must be

symmetric under particle exchange e.g. pion, kaon, photon, gluon, etc.

Symmetrization requirement ),(),( 1221 rrrr

baabba rrrrArr if ,0)]()()()([),( 212121

Observe that for fermions

In other words, two identical fermions cannot occupy the same state. This is the famous

Pauli exclusion principle

Example:

Consider two non-interacting particles in an infinite potential well. As we know, the one

particle states are:

2

222

2 where, ),sin(

2)(

maKKnE

a

xn

ax nn

Ignoring the spins, find the ground state for (a) two distinguishable particles

(b) two identical bosons (c) two identical fermions

KKEEa

x

a

x

aG 2)11( and sinsin

2 22

1121

11

a)

b) For two identical bosons KKEEa

x

a

x

aG 2)11( and sinsin

2 22

1121

11

c) For two identical fermions there is no state with energy E11

)sin()sin()sin()sin(2

2

1)()()()(

2

12121 2

2

2

21122211 a

x

a

xx

a

x

axxxx

)sin()sin()sin()sin(2

2121 2

2

2

a

x

a

xx

a

x

a

KKEEG 5)21( 22

12

Now consider noninteracting two spin-1/2 particles. What is their ground state and energy?

For a general state, total wavefunction for two spin-1/2 particles is of the form:

),(),(),;,( 21212211 ssxxsxsx

As we saw in the last lecture, the total spin of two spin-1/2 particles can be either the

singlet or triplet state, with

is antisymmetric under particle interchange.

they are symmetric under particle interchange.

For two electrons, total wavefunction, , must be antisymmetric under particle exchange.

i.e. spin singlet state must have symmetric spatial wavefunction; spin triplet states have

antisymmetric spatial wavefunction.

For the ground state, the symmetric spin triplet states are excluded because the required

antisymmetric spatial state is identically zero. The ground state of two spin-1/2 fermions

is therefore

singlet

What about the excited states?

The triplet state has the form

121122211 )()()()(2

1 xxxx nnnn

T

, where χ1 is defined as before

So T could be any of the three configurations in 1 with energy E12 = E1 + E2 = 5K

The singlet state is

021122211 )()()()( xxxx nnnn

S

, where χ0 is defined as before

with energy E12 = 5K

symmetric antisymmetric

antisymmetric symmetric

To summarize the previous examples; complete quantum state vector then has the form

For bosons

For fermions

The two parts must have the same exchange symmetry,

or else the full eigenstate would not be symmetric.

The two parts cannot have the same exchange symmetry,

or else the full eigenstate would not be antisymmetric.

To construct wavefunctions for three or more fermions, let us suppose that they do not interact,

and are confined by a spin-independent potential,

i

iiS rpHH nHamiltonia particle single theis H where),,( S)(

2

2

rVm

pH S

Eigenfunctions of Schrodinger equation involve products of states of single-particle

Hamiltonian, HS

However, simple products do not have required

antisymmetry under exchange of any two particles.

We could achieve antisymmetrization for particles 1 and 2 by subtracting the same product

with 1 and 2 interchanged,

However, wavefunction must be antisymmetrized under all possible exchanges. So, for 3

particles, we must add together all 3! permutations of 1, 2, 3 in the state a, b, c with factor

-1 for each particle exchange.

Such a sum is known as a Slater determinant:

and can be generalized to N particles as:

Antisymmetry of wavefunction under particle exchange follows from antisymmetry of Slater

determinant,

The determinant is non-vanishing only if all three states a, b,c are different – manifestation

of Pauli’s exclusion principle: two identical fermions can not occupy the same state.

The conditions on wavefunction antisymmetry imply spin-dependent correlations even where

the Hamiltonian is spin-independent, and leads to numerous physical manifestations...

Example : Hydrogen H2 gas

With two spin-1/2 proton degrees of freedom, H2 can adopt a spin singlet (parahydrogen)

or spin triplet (orthohydrogen) wavefunction

The space part is described by rotational modes with orbital quantum numbers l = 0, 1, 2, . . .

• Orthohydrogen: Nuclear spin part is symmetric (spin triplet) and

nuclear space part is antisymmetric (odd l).

• Parahydrogen: Nuclear spin part is antisymmetric (spin singlet)

and nuclear space part is symmetric (even l).

H2 gas at equilibrium contains 75% orthohydrogen and 25%

parahydrogen. The 3:1 ratio is a reflection of the nuclear spin

degeneracy.

These two forms have different sets of rotational energies and hence different specific heats

Moment of inertia of molecule

Another implication of the symmetrization of the wavefunctions is the separation between

the particles.

If we ignore spin, and consider just the spatial wavefunction with for bosons, and

for fermions, we can calculate the expectation value of the square of the separation between

two particles.

Case I: Distinguishable Particles

Similarly and

where we write to indicate that we just have the x2 expectation value w.r.t. the state .

. Our distinguishable particles are then, on average,

apart

Case II: Indistinguishable particles

Here, we have

The relevant expectation values are

and provided then orthonormality gives

and the above reduces to

since there is no distinguishing between the particles.

But

where the last term is just the obvious

If we compare with the distinguishable case we see that the difference resides in the

final term

So for the bosonic (upper signs) case, the particles are slightly closer together than distin

guishable particles, and for fermions (lower signs) case, they are slightly further apart.

Again, we see that the cross term vanishes if the wave functions do not significantly overlap,

another indication that localization (i.e. a classical interpretation) works at some level.

The “force” is fictitious in a classical sense, but we can think of a bosonic attractive force

pulling two particles closer together, and a fermionic repulsive force tending to push two

particles apart. This is the so-called “exchange” force.

Total wave function has to be symmetric or antisymmetric, we have to put together complete

two-electron state including the spin:

Thus, a symmetric spin state (e.g., a triplet) must be associated with an antisymmetric

spatial wave function (an antibonding combination), while an antisymmetric spin state

(e.g., a singlet) must be associated with a symmetric spatial w.f. (a bonding combination).

Sure enough, the chemists tell us that the covalent bonding requires the two electrons to

occupy the singlet state, with total spin zero.

Schematic picture of the covalent bond: (a) symmetric combination produces attractive

force; (b) antisymmetric combination produces repulsive force.

For a neutral atom,

For fermions, the spatial part of the wave function must satisfy and the

complete wave function

must be antisymmetric with respect to interchange of any two particles.

For hydrogen, Z = 1, so there is no contribution from the electron-electron interaction term.

is then simply a one-electron wave function.

Electron-electron interaction Atoms

Helium atom

For helium, Z = 2, leading to a single electron-electron interaction term. This is enough to turn

the proper solution of the Schrodinger equation into a true many-body wave function.

The Hamiltonian now reads

If we ignore the e-e interaction for the moment, then the spatial

part of the wavefunction can be approximated as the product of

two one electron hydrogen wave functions:

where the Bohr radius is half as large as for hydrogen and

Specifically, the ground state is

E0 = 8(-13.6) = -109 eV

Because 0 is a symmetric function, the spin wave function must be antisymmetric. Thus, the

ground state of He is a singlet configuration, wherein the spins are aligned oppositely.

The actual ground state is indeed a singlet, but the energy is only -79 eV. Such poor agreement

is expected since we neglected the positive (repulsive) contribution of e-e interaction

The excited states of He consist of one electron in the hydrogenic ground state and the other in

an excited state:

The spatial portion of this wave function can be constructed either symmetrically or

antisymmetrically

leading to the possibility of either antisymmetric or symmetric spin portions, respectively,

known as parahelium and orthohelium.

If you try to put both electrons in excited states, one immediately drops into the ground state,

releasing enough energy to knock the other electron into the continuum and yielding a He ion

(He+). This process is known as an Auger transition.

Energy level diagram for He (relative to He+, -54.4 eV).

Note that parahelium (antisymmetric spin) energies are

uniformly higher than orthohelium counterparts.

Quantum Mechanical Statistics (brief)

So far, we have been dealing with the ground state of systems. Statistical mechanics deals

with the occupation of states when a system is excited. The fundamental assumption of

statistical mechanics is that, in thermal equilibrium, every distinct state with the same

total energy is occupied with equal probability. Temperature is simply a measure of the total

energy of a system in thermal equilibrium.

The only change from classical statistical mechanics occasioned by quantum mechanics

has to do with how we count distinct states, which depends on whether the particles

involved are distinguishable, identical fermions, or identical bosons.

Simple example: Consider a system of 2 identical particles for which 3 single-particle states are

available. If, as in classical mechanics, the particles are considered distinguishable, they may be

labeled as A and B. On the other hand, both particles must have the same label, say A, if they

are considered indistinguishable. The tables below enumerate all distinct states for distinguisha

ble classical particles or for indistinguishable fermions and bosons.

Classically, 32 = 9 distinct microstates are

available, but only 3 are available to fermions

or 6 to bosons.

So, the results vary dramatically depending on the fundamental nature of the particles.

Quantum statistical mechanics

( )x

One particle Two particles

1 2( , )x x

N particles

1 2 3( , , ..... )Nx x x x

Quantum statistical mechanics

N particles

Thermal equilibrium, T

Quantization of the energy

for individual particles

1 2 3( , , ,...)E E E

Total energy: 1 1 2 2 3 3 ...E N E N E N E

1 2 3 ...N N N N

Quantum statistical mechanics

3 kind of particles

• Distinguishable particle

• Identical fermions

• Identical bosons

Antisymmetric state

Symmetric state

For each type of particles:

• List all the possible configurations

• Determine the number of combinations of each configuration

• Determine the probability of each energy for a given particle

Generally for an arbitrary potential, for which the one-particle energies are E1, E2, E3…with

degeneracies d1, d2, d3… If we put N particles into this potential, the number of distinct states

corresponding to the configuration (N1, N2, N3,…) is Q, which depends strongly on whether

the particles are distinguishable, identical fermions, or identical bosons.

• Distinguishable particle 1 2 3

1

, , ,.. !!

nN

ndis

n n

dQ N N N N

N

• Identical fermions 1 2 3

1

!, , ,..

! !

nfermions

n n n n

dQ N N N

N d N

1 2 3

1

1 !, , ,..

! 1 !

n n

bosons

n n n

N dQ N N N

N d

• Identical bosons

In thermal equilibrium, the most probable configuration is that one which can be attained

in the largest number of different ways. We want to find this subject to two constraints:

This is best handled using Lagrange multipliers. It is also useful to maximize ln (Q) rather

than Q, which turns the products of factorials into sums. Let

where and b are the Lagrange multipliers. We now seek the values of Nn which satisfy

The most probable occupation numbers are:

Distinguishable particles:

Identical fermions:

Identical bosons:

Physical significance of and b

Insight into the physical meaning of and b can be gained by substituting one of the

above equations for Nn into the sums which yield N and E, but that requires assuming

a specific model for V. Let's use a simple V, a three-dimensional infinite square well:

z

z

y

y

x

x

kl

n

l

n

l

nkk

mE

,, where,

2

22

For distinguishable particles,

The second equation leads us to define T by

is customarily expressed in term of a chemical potential:

We can now write expressions for the most probable number of particles in a particular

one-particle state with energy e as:

Distinguishable particles (Maxwell-Boltzmann):

Identical fermions (Fermi-Dirac):

Identical bosons (Bose-Einstein):

The Fermi-Dirac distribution for T = 0 and for T somewhat above zero.