Quantum Mechanics of Many-Particle Systems: Fermions and Bosons, Lecture notes of Physics

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Lecture 11

Identical particles

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 (ca.

23

!) particles, e.g. electrons in a solid, atoms in a gas, etc.

In classical mechanics, particles are always distinguishable – at least

formally, “trajectories” through phase space can be traced.

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 (just) a few implications of particle indistinguishability.

Quantum statistics: preliminaries

But which sign should we choose?

ψ(x 1

, x 2

) = ψ(x 2

, x 1

) bosons

ψ(x 1

, x 2

) = −ψ(x 2

, x 1

) fermions

All elementary particles are classified as

fermions or bosons:

(^1) 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.

2 Particles with integer spin (including zero) are bosons and their

wavefunction must be symmetric under particle exchange.

e.g. pion, kaon, photon, gluon, etc.

Quantum statistics: remarks

Within non-relativistic quantum mechanics, correlation between spin

and statistics can be seen as an empirical law.

However, the spin-statistics relation emerges naturally from the

unification of quantum mechanics and special relativity.

The rule that fermions have half-integer spin and

bosons have integer spin is internally consistent:

e.g. Two identical nuclei, composed of n nucleons

(fermions), would have integer or half-integer spin

and would transform as a “composite” fermion or

boson according to whether n is even or odd.

Quantum statistics: fermions

We could achieve antisymmetrization for particles 1 and 2 by

subtracting the same product with 1 and 2 interchanged,

ψ a

(1)ψ b

(2)ψ c

(3) $→ [ψ a

(1)ψ b

(2) − ψ a

(2)ψ b

(1)] ψ c

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:

ψ abc

ψ a

(1) ψ b

(1) ψ c

ψ a

(2) ψ b

(2) ψ c

ψ a

(3) ψ b

(3) ψ c

and can be generalized to N, ψ i 1 ,i 2 ,···iN

(1, 2 , · · · N) = det(ψ i

(n))

Quantum statistics: fermions

ψ abc

ψ a

(1) ψ b

(1) ψ c

ψ a

(2) ψ b

(2) ψ c

ψ a

(3) ψ b

(3) ψ c

Antisymmetry of wavefunction under particle exchange follows from

antisymmetry of Slater determinant, ψ abc

(1, 2 , 3) = −ψ abc

Moreover, 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.

Wavefunction is exact for non-interacting fermions, and provides a

useful platform to study weakly interacting systems from a

perturbative scheme.

Space and spin wavefunctions

When Hamiltonian is spin-independent, wavefunction can be

factorized into spin and spatial components.

For two electrons (fermions), there are four basis states in spin

space: the (antisymmetric) spin S = 0 singlet state,

|χ S

1

2

1

2

and the three (symmetric) spin S = 1 triplet states,

|χ

1

T

1

2

〉, |χ

0

T

1

2

1

2

〉) , |χ

− 1

T

1

2

Space and spin wavefunctions

For a general state, total wavefunction for two electrons:

Ψ(r 1

, s 1

; r 2

, s 2

) = ψ(r 1

, r 2

)χ(s 1

, s 2

where χ(s 1

, s 2

) = 〈s 1

, s 2

|χ〉.

For two electrons, total wavefunction, Ψ, must be antisymmetric

under exchange.

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

spin triplet states have antisymmetric spatial wavefunction.

For three electron wavefunctions, situation becomes challenging...

see notes.

The conditions on wavefunction antisymmetry imply spin-dependent

correlations even where the Hamiltonian is spin-independent, and

leads to numerous physical manifestations...

Example II: Excited states spectrum of Helium

Although, after hydrogen, helium is simplest

atom with two protons (Z = 2), two neutrons,

and two bound electrons, the Schr¨odinger

equation is analytically intractable.

In absence of electron-electron interaction, electron Hamiltonian

H

(0)

=

2 ∑

n=

[

pˆ

2

n

2 m

• V (r n

]

, V (r ) = −

4 π& 0

Ze

2

r

is separable and states can be expressed through eigenstates, ψ n"m

of hydrogen-like Hamiltonian.

Example II: Excited states spectrum of Helium

H

(0)

=

2 ∑

n=

[

pˆ

2

n

2 m

• V (r n

]

In this approximation, ground state wavefunction involves both

electrons in 1s state! antisymmetric spin singlet wavefunction,

g.s.

〉 = (| 100 〉 ⊕ | 100 〉)|χ S

Previously, we have used perturbative theory to determine how

ground state energy is perturbed by electron-electron interaction,

H

(1)

=

4 π& 0

e

2

|r 1

− r 2

What are implications of particle statistics on spectrum of lowest

excited states?

Example II: Excited states spectrum of Helium

|ψ p,o

(| 100 〉 ⊗ | 2 %m〉 ±| 2 %m〉 ⊗ | 100 〉) |χ

ms

S,T

Despite degeneracy, since off-diagonal matrix elements between

different m, % values vanish, we can invoke first order perturbation

theory to determine energy shift for ortho- and parahelium,

∆E

p,o

n"

= 〈ψ p,o

H

(1)

|ψ p,o

e

2

4 π& 0

d

3

r 1

d

3

r 2

|ψ 100

(r 1

)ψ n" 0

(r 2

) ± ψ n" 0

(r 1

)ψ 100

(r 2

2

|r 1

− r 2

(+) parahelium and (-) orthohelium.

N.B. since matrix element is independent of m, m = 0 value

considered here applies to all values of m.

Example II: Excited states spectrum of Helium

∆E

p,o

n"

e

2

4 π& 0

d

3

r 1

d

3

r 2

|ψ 100

(r 1

)ψ n" 0

(r 2

) ± ψ n" 0

(r 1

)ψ 100

(r 2

2

|r 1

− r 2

Rearranging this expression, we obtain

∆E

p,o

n"

= J

n"

± K

n"

where diagonal and cross-terms given by

J

n"

e

2

4 π& 0

d

3

r 1

d

3

r 2

|ψ 100

(r 1

2

|ψ n" 0

(r 2

2

|r 1

− r 2

K

n"

e

2

4 π& 0

d

3

r 1

d

3

r 2

ψ

∗

100

(r 1

)ψ

∗

n" 0

(r 2

)ψ 100

(r 2

)ψ n" 0

(r 1

|r 1

− r 2

Example II: Excited states spectrum of Helium

|ψ p,o

(| 100 〉 ⊗ |n%m〉 ±| n%m〉 ⊗ | 100 〉) |χ

ms

S,T

∆E

p,o

n"

= J

n"

± K

n"

Example II: Excited states spectrum of Helium

Finally, noting that, with S = S 1

+ S

2

2

2 S

1

· S

2

2

[

(S

1

+ S

2

2

− S

2

1

− S

2

2

]

= S(S + 1) − 2 × 1 /2(1/2 + 1) =

1 / 2 triplet

− 3 / 2 singlet

the energy shift can be written as

∆E

p,o

n"

= J

n"

2

S

1

· S

2

K

n"

From this result, we can conclude that electron-electron interaction

leads to effective ferromagnetic interaction between spins.

Similar phenomenology finds manifestation in metallic systems as

Stoner ferromagnetism.