Identical particles

Bosons and fermions

Two-particle systems

( ) ( ) ( )

1 2 1 2

2 22 21 2 1 2 1 2

1 1

1 2 1 2 1 2

1 2 1 1 2 2

221 1

1

( , , ) depends on coordinates and spin variables of both particles

( , , , )2 2

If particles do not interact ( , , , ) ( ) ( ),

( )2

t

H V tm m

V t V V

Vm

σ σ

σ σ

σ σψ ψ ψ

Ψ

= − ∇ −− ∇ +

= +

=

⎡− ∇ +⎣

r r

r r

r r r rr r r r

r ( ) ( )

( ) ( )

1 1

222 2 2 2

1

( )2

n n n

n n n

E

V Em

ψ ψ

ψ ψ

⎤=⎢ ⎥

⎦⎡ ⎤− ∇ + =⎢ ⎥⎣ ⎦

r r

r r r

Identical particlesIf, however, particles are indistinguishable then interchanging particles should not produce any observable difference. This means that probability distribution corresponding to any two-particle state must remain the same when we exchange the particles. Mathematically speaking we can say that

1 2 2 11 2 2 1( , ) ( , )P Pσ σ σ σ=r r r r

Using connection between the probability density and the wave functions we can obtain

( ) ( )1 2 2 11 2 2 1, ,σ σ σ σψ ψ= ±r r r r

It turns out that the choice of + or – sign in this relation depends on the spin of particles. Particles with integer spin, such as photons, W and Z0 bosons require symmetric wave functions, while particles with half-integer spin such as electrons, protons, neutrons, and others require an antisymmetric wave functions.

Coordinate and spin states of identical particles

Symmetrization or antysimmetrization requirements apply to entire wave functions including their spin dependence. If spin and coordinate variables can be separated we can analyze consequences of these requirements for each of the variables separately.

( ) ( ) ( )1 2 1 2 1 2 1 2, ,σ σψ ϕ χ σ σ=r r r r

In the case of fermions the antisymmetric wave function can be obtained by two different ways:

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 2 2 11 2 2 1

1 2 2 1 1 2 2 1

1 2 2 1 1 2 2 1

, , if

1. , , and

2. , , and

σ σ σ σψ ψ

ϕ ϕ χ σ σ χ σ σ

ϕ ϕ χ σ σ χ σ σ

= −

= − =

= = −

r r r r

r r r r

r r r r

Basis of single-particle statesLet us assume that we know all single particle states of our system. These are stationary states a Schrödinger equation with omitted interaction term. They form a basis in the space containing coordinates of each particle.

( )1 1, nψ σr

( ) ( )1 1 2 2, ,n mψ σ ψ σr r

basis in the space of the first particle

( )2 2, basis in the space of the second particlenψ σrWe need to construct a basis in the space containing two-particle wave functions. Each of single particle spaces contain 2N states (including spin). Thus the two-particle state must have dimension 2Nx2N. We can form functions (spinors)

That form a set of 2Nx2N linearly independent orthonormal spinors, and can, therefore, we considered as a basis for the two-particle state. If, however, the particles are undistinguishable this basis spinors have to be properly symmetrizied. For fermions we should have

( ) ( ) ( ) ( )1 1 2 2 1 1 2 2, , , ,n m m nψ σ ψ σ ψ σ ψ σ−r r r r

Non-interacting distinguishable particles. Example.

In the case of particles in the infinite square one particle states are

20

2( ) sin ;n nnx x E E n

a aπψ ⎛ ⎞= =⎜ ⎟⎝ ⎠

If particles are distinguishable, the two particle states are

( )2 21 2 0

1 2

2( ) sin sin ;

2( ) sin sin

nm n

nm

n mx x x E E n ma a a

n mx x xa a a

π πψ

π πψ

⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Ground state corresponds to n=m=1 02E E=

The first excited state is double degenerate with n=1, m=2, and n=2, m=1 giving the same energy

Non-interacting fermions in an infinite square well.

Now the two-particle functions have to be made antisymmetric (we are ignoring spin for now:

( )1 2 1 2

2 20

2 1( ) sin sin sin sin ;2nm

nm

n m m nx x x x xa a a a a

E E n m

π π π πψ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

= +

States with n=m=1 do not exist in this system. The lowest energy state will correspond to n=1, m=2

12 1 2 1 2

0

2 1 2 2( ) sin sin sin sin2

5

x x x x xa a a a a

E E

π π π πψ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠=

Non-interacting bosons in an infinite square well.

Now the two-particle functions have to be made symmetric

( )1 2 1 2

2 20

2 1( ) sin sin sin sin ;2nm

nm

n m m nx x x x xa a a a a

E E n m

π π π πψ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

= +

Ground state is the same as for distinguishable particles, but the next state is no longer degenerate because n=1, m=2 and n=2, m=1 give the same wave function

12 1 2 1 2

0

2 2 2( ) sin sin sin sin

5

x x x x xa a a a a

E E

π π π πψ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠=

Singlet statesNow, let us take spin into account. Assuming again that coordinate and spin variables can be separated we have for non-symmetric basis functions

( ) ( )1 21 2 (1) (2)n m σ σϕ ϕ χ χr r

In the case of fermions we have two possibilities. First, the coordinate wave function is symmetric, and spinor is antisymmetric:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2 1 2

1 2 1 2

1 2 1 2

1 (1) (2) (1) (2)21 002

12

n m m n

n m m n

n m m n

ϕ ϕ ϕ ϕ χ χ χ χ

ϕ ϕ ϕ ϕ

ϕ ϕ ϕ ϕ

↑ ↓ ↑ ↓+ − =⎡ ⎤⎣ ⎦

+ =⎡ ⎤⎣ ⎦

+⎡ ⎤⎣ ⎦

r r r r

r r r r

r r r r This corresponds to a singlet state of two spins, in which total spin of two fermions is zero.

Triplet statesSecond possibility is antisymmetric coordinate function and symmetric spin function. It is immediately seen that there three possibilities to form symmetric spin functions:

( )1; ;2

↑↑ ↑↓ + ↓↑ ↓↓

Which we immediately recognize as three states of a triplet configuration of two spins. Thus we can present the total wave function as

( ) ( ) ( ) ( )1 2 1 21 1 ; 1,0,12( , ) 0 Pauli exclusion principle

n m m n s sm m

P

ϕ ϕ ϕ ϕ− = −⎡ ⎤⎣ ⎦

=

r r r r

r r

Exchange correlationsThe symmetry requirements in the case of fermions and bosons result in specific quantum mechanical correlation between the two particles even in the absence of any direct interaction between them. Consider probability to find two electrons within the same region of space for the spatially symmetric state (in spatially antisymmetric state this probability is zero):

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

*1 2 1 2 1 2 1 2 1 2

†

1 2 2 1 1 2 2 1

1 2 2 12 2 1 1 1 2

1 2 2 1

0 0 0 0

2 2

1( , )4

( , )

2

s n m m n n m m n

s n m

n m

P

P

ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

ϕ ϕ

ϕ ϕ

= + + ×⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤↑ ↓ − ↑ ↓ ↑ ↓ − ↑ ↓⎣ ⎦ ⎣ ⎦⎡ ⎤↑ ↑ ↓ ↓ + ↑ ↑ ↓ ↓ −⎢ ⎥⎢ ⎥= =⎢ ⎥↑ ↓ ↓ ↑ − ↑ ↓ ↓ ↑⎢ ⎥⎣ ⎦

r r r r r r r r r r

r r r r

r r This probability is twice larger than in the case of distinguishable particles.

Exchange correlations and interaction energy

Consider two interacting electrons. Their classical interaction energy is a positive quantity (Coulomb repulsion). Quantum mechanical energy would involve integration of the Coulomb term with the square of the respective two-electron wave function, which we, of course, do not know. Qualitatively, however, we understand, now that this wave function would depend on the spin state of the electrons. In the triplet configuration exchange correlations cause electrons to avoid each other: wave functions goes to zero, when electrons approach each other. In the singlet configuration exchange correlations favor electrons being closer to each other. Thus we can say, that the exchange correlations imitate an attractive potential for singlet thus lowering the energy of the system, and they imitate an additional repulsive potential for triplet state increasing the energy. We can conclude, therefore, that the energy of two interacting spins depends on their spin state with lower energy corresponding to the singlet state.

Exchange correlations and chemical bonds

Consider now two hydrogen atoms in their ground state. When they are far away from each other the electron wave functions do not overlap and exchange contribution to energy is zero.

When they approach each other we have to distinguish between electrons in triplet and singlet states.

In the singlet state (a) electrons tend to be closer to each other thus pushing protons toward each other and keeping them together. This is how covalent bond arises. In one of the triplet states electrons avoid each other thus pushing protons away from each other resulting in an anti-bonding state.

Atoms

2 2 22

1 0 0

1 1 12 4 2 4

Z Z

jj j kj j k

Ze eHm r r rπε πε= ≠

⎧ ⎫⎪ ⎪= − ∇ − +⎨ ⎬−⎪ ⎪⎩ ⎭

∑ ∑

Wave functions of electrons in a many-electron atom are stationary states of the Hamiltonian below.

Because of the interaction term find these wave functions exactly is not possible. In the first approximation we can neglect the interaction terms and find single electron states, and then use them to calculate average energy of the total Hamiltonian, hoping that it would give at least some idea about the energy of the respective states. The multi-electron states are formed as appropriately symmetrized products of single electron states. They are characterized by single electron quantum numbers n,l,m and by the spin state ms . One particle states are called orbitals. Orbitals with the same n form a shell. Horizontal rows in Periodic Table correspond to filling out each shell.

Z=1 through 10

2 1

means 0; means 1; means 2; means 3

Capital , , , refer to the total orbital moments of all electrons

S here means total spin, designates grand

total momentum ( )

SJ

s l p ld l f l

S P D F

LJ

L S

+

= == =

+

( ) ( ) ( )2 2 2100 200 210 21 1 2111 2 2 2 2 2( )

2 electrons with 1 2,1,04 electrons with 0, spin state of 2 electrons can be either 1, or 0

can be 3, 2,1,0. Actual state corresponds to 1, 0,

s s p a b cl LS S S

J L J S

ψ ψ ψ ψ ψ−→ + + + +

= ⇒ == = =

= = =1

Z=11 through 30

Hund’s rules:

1. The state with the highest total Shave the lowest energy

2. For a given spin configuration, the state with highest Lconsistent with overall antisymmetrization has lowest energy

3. If a subshell (n,l) is no more than half-filed, then J=L-S, otherwise, it is J=L+S