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Introduction to Quantum Chemistry
part I
Maciej Bobrowski
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History of the Revolution
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in 19th century
... physicists were satisfied
• they had two kingdoms: particles and electromagnetic waves.
• particles were described by sir Isaac Newton’s laws.
• waves were described by Jamess Clerk Maxwel’s equations.
• Max Planck were advised to abandon the thinking about physical studiesbecause everything was already discovered.
• ... just few details were unclear yet: the blackbody radiation, strangephotoelectric effect and atomic spectra.
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Blackbody radiation
An idealized object that absorbs all electromagnetic radiation that falls on it.No electromagnetic radiation passes through it and none is reflected.
Ultraviolet catastrophe.Ultraviolet : 10 - 400 nm, energy: 3 eV to 124 eV
Blackbody radiation spectrum and comparison with the classical theory of
Rayleigh-Jeans
Rayleigh-Jeans, Wien and Planck’s laws for a body at 8 mK tempe-
rature
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Electromagnetic spectrum
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Electromagnetic spectrum
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Blackbody radiation
1. Max Karl Ernst Ludwig Planck (1858-1947). Nobel Prize „in recognition of the services he rendered to theadvancement of Physics by his discovery of energy quanta”
Planck postulated that electromagnetic energy did not follow the classicaldescription, but could only oscillate or be emitted in discrete packets ofenergy proportional to the frequency
Planck introduced the justification of his equation on 14th of Decemeber1900. This date is seemed to be the beginning of quantum mechanics.
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Blackbody radiation - 1905
Planck’s density of energy of the blackbody radiation
u(ν;T ) =8πν2
c3hν
ehνkT − 1
For large frequencies it converts into the Wien’s displacement law
u(ν;T ) =8πhν3
c3e−
hνkT
and for small frequencies it converts into Rayleigh-Jeans law
u(ν;T ) =8πν2
c3kT
But, the two last equations are generally not valid. Using theRayleigh-Jeans’ equation the energy per a volume equals
U(T ) =
∫ ∞
0
u(ν;T )dν = ∞
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Photoelectric effect - 1905
Einstein’s equation: hν = 12mv
2 +W
Traditional explanation - the energy of the emitted electrons does notdepend on the intensity of the incoming light, but only on the energy of theindividual photons. Here it was a partcile nature of the light (wave-particleduality).
modern explanaition - Fermi’s golden rule based upon quantummechanics. This method does not actually treat light as a particlephenomenon, but rather as an electromagnetic wave passing from oneeigenstate to another.
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Niels Bohr (1885 - 1962)
Niels Bohr Bohr and Einstein at Paul Ehrenfest’s home in Leiden
(December 1925).
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Hydrogen atom model - Bohr model - 1913
The model’s assumptions
• Energy (a foton i.e. light quantum) can be emitted only when an electronchanges its state, not constantly. The energy equals ∆E = hν.
• The angular momentum equals L = n~
• The model is similar in structure to the solar system, but with electrostaticforces providing attraction, rather than gravity.
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Bohr model - 1913
If electrons travel with velocity v in circular orbits (radius r) around thenucleus with charge Ze, then the angular momentum equals
L = mvr
The centripetal force is equal to the Coulomb force.
mv2
r=
Ze2
4πǫ0r2
and the total energy of the electron (kinetic and potential) equals
E =1
2mv2 − Ze2
4πǫ0r
Collecting everything together we obtain energy of the n-th level determinedby the radius
En =Z2me4
8n2ǫ20h2
=E1
n2= −13.6eV
n2
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The Broglie waves - 1923
2. Louis-Victor-Pierre-Raymond, 7th duc de Broglie, (15 August 1892 - 19 March 1987). Nobel Prize for: „hisdiscovery of the wave nature of electrons”.
any moving particle or object has an associated wave.
λ =h
p
h = Planck’s constantp = momentum.
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The Compton Effect - 1923
Arthur Holly Compton (Sep-
tember 10, 1892 - March 15,
1962). Nobel Prize „for his di-
scovery of the effect named
after him” (half of the prize, the
rest for Charles Thomson and
Rees Wilson).
A photon of wavelength λ comes in from the left, collides
with a target at rest, and a new photon of wavelength λ′
emerges at an angle θ (between 0◦ and 180◦ )
The collision of electron and photon can be described in the same way likeordinary particles (conservation of energy and conservation of momentum)
λ′ − λ =h
mec(1 − cos θ).
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Invention of the spin - 1925
Wolfgang Ernst Pauli (25.04.1900 -
15.12.1958). Nobel Prize „for the di-
scovery of the Exclusion Principle,
also called the Pauli Principle”
George Uhlenbeck, Hendrik Kramers, and Samuel
Goudsmit around 1928. Uhlenbeck and Goudsmit pro-
posed the idea of electron spins 3 years earlier
In 1924 Wolfgang Pauli introduced what he called a „two-valued quantum degree of freedom” associatedwith the electron in the outermost shell. This allowed him to formulate the Pauli exclusion principle, statingthat no two electrons can share the same quantum state at the same time (Pauli’s Principle).
In the fall of 1925, two Duch physicists - George Uhlenbeck and Samuel Goudsmit - proposed an
explanaition of an experiment in which a beam of silver atoms split into two beams under a magnetic field.
They suggested, that electrons have additional degree of freedom.
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More events ...
• 1925 (Werner Heisenberg) - matrix mechanics,• 1926 (Erwin Schrödinger) - wave equation,• 1926 (Max Born) - statistical interpretation of the wave function,• 1927 (Werner Heisenberg) - uncerteinty principle,• 1927 (Clinton Davisson, Lester Germer, George Thomson) - wave nature
of electrons - deflection of electrons beam through a crystalline grid,• 1927 (Walter Heitler and Fritz Wolfgang London) - they explained why two
neutral atoms attract each other like charged ions. The day June 30, 1927is a date of born of „quantum chemistry”,
• 1928 (Paul Dirac) - relativistic equation for electron and positron.• 1929 (Werner Heisenberg and Wolfgang Pauli) - an electron, photon or
any other particle is an excited state of appropriate field (electron,electromagnetic and other),
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More events ...
• 1932 (Carl Anderson) - experimental prove on existance of antimatter(here positron),
• 1948 (Richard Feynman, Julian Schwinger, Shinichiro Tomonaga) -quantum electrodynamics,
• 1964 (John Bell) - Bell’s inequality,• 1982 (Alain Aspect) - photons do not obey the Bells inequality (the World
is nonlocal or not real :) ),• 1997 (Anton Zeilinger) - teleportation of photon.• ... and many other important events.
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Axioms
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1st axiom - Quantum state
Quantum mechanical state of any system consisting of N particles (each ithparticle is indicated by a vector ri = (xi, yi, zi)) can be described by a wavefunction Ψ(r1, r2, . . . , rN , t) which square of its module is proportional to thedensity of probability of finding of the system in the time t.
Generally, wave functions values are complex.
Ψ - real values,Ψ∗ - complex values.
according to Max Born’s statistical interpretation of a wave function
p(r1, r2, . . . , rN , t) =∫
r1
∫
r2
. . .
∫
rN
Ψ∗(r1, r2, . . . , rN , t)Ψ(r1, r2, . . . , rN , t)d3r1d
3r2 . . . d3rN
where p is a probability that in the time t particle 1 is in the volume V1,particle 2 is in the volume V2, etc.
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The probability
mass = density x volume (m = ρv).
ρ = density of probability = Ψ∗Ψ.
If the p(r1, r2, . . . , rN , t) is a probability, the wave function must obey
∫
r1
∫
r2
. . .
∫
rN
|Ψ(r1, r2, . . . , rN , t)|2d3r1d3r2 . . . d
3rN = 1
which is called the normalization condition.
Example:
Let’s say that p = A, 0 < A 6= 1 and we have only one dimension - x.Then, we can choose: Ψ = NΨ with the N that 1 =
∫ +∞−∞ Ψ ∗ (x, t)Ψ(x, t)dx
= N ∗N∫ +∞−∞ Ψ∗(x, t)Ψ(x, t)dx = A|N |2. Then |N | = 1√
A.
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Spherical coordinates
zr
= cosθ z = r cos θ r =√
x2 + y2 + z2
OAr
= sinθ OA = r sin θ θ = arc cos zr
xOA
= cosφ x = OA cosφ = r sin θ cosφ φ = arc tg yx
yOA
= sinφ y = OA sinφ = r sin θ sinφ
Volume element dτ = dxdydz = dφ sin θdθr2dr
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Normalization - example
Normalize the function e−ar, where r =√
x2 + y2 + z2.
...
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2nd axiom - Correspondence rules
CORRESPONDENCE RULES of replacing physical quantities withoperators.
To every physical quantity (observable) can be assigned a linear operator
Linear operator A:
A(c1Ψ1 + c2Ψ2) = c1AΨ1 + c2AΨ2
Examples of linear operators: ·x,∫
, ∂∂y
, ...
Examples of nonlinear operators: √, ()2, sin(), ...
Examples of linear operators in quantum mechanics
1 - elementary operator 1Ψ = Ψ
x - operator of x component of location xΨ = x · Ψpx - operator of x component of the momentum pxΨ = −i~ ∂
∂xΨ
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Operators
Having the 1, x and px we can construct next operators
• angular momentum L = r × p
Lx = ypz − zpy ⇒ Lx = ~
i
(
y ∂∂z
− z ∂∂y
)
• Energy of interaction of electron with nucleus.
V = −Ze2
r2⇒ V = −Ze2
r2
• Kinetic energy of a particle
T = p2
2m =p2x+p2y+p2z
2m ⇒ T = − ~2
2m
(∂2
∂x2 + ∂2
∂y2 + ∂2
∂z2
)
= − ~2
2m∆
• Total energy of a particle (Hamiltonian operator)
H = T + V
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Hermitian operators
The operator is hermitian if for any two functions of Q class ψ and φ
∫ ∞
−∞ψ∗(x)Aφ(x)dx =
∫ ∞
−∞[Aψ(x)]∗φ(x)dx
Dirac’s notation
ψ(r, t) − wave function → |ψ〉︸︷︷︸
ket
−vector
∫ψ∗
1(r, t)ψ2(r, t)d3r → 〈ψ1|
︸︷︷︸
bra
ψ2〉
Now it is formally simpler
〈ψ|φ〉 = 〈φ|Ψ〉∗〈ψ|A|φ〉 = 〈Aψ|φ〉
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Hermitian operators - an example
Check if the operator A = x is hermitian.
Solution. The condition is now
∫ ∞
−∞ψ∗(x)xφ(x)dx =
∫ ∞
−∞[xψ(x)]∗φ(x)dx
Because
∀x∈R[xψ(x)]∗ = xψ∗(x)
then both sides of the equation are equal and x is a hermitian operator.
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Hermitian operators - another example
Check if the operator A = ∂∂x
is hermitian.
∫ ∞
−∞ψ∗(x)
d
dxφ(x)dx =
∫ ∞
−∞
[d
dxψ(x)
]∗
φ(x)dx
Left side
∫ ∞
−∞ψ∗(x)
d
dxφ(x)dx =
{
f(x) = ψ∗(x), f ′(x) =[dψ(x)dx
]∗
g′(x) = dφ(x)dx
, g(x) = φ(x)
}∞
−∞
=
= [ψ∗(x)φ(x)]∞−∞
︸ ︷︷ ︸
0
−∫ ∞
−∞
[dψ(x)
dx
]∗
φ(x)dx =
= −∫ ∞
−∞
[dψ(x)
dx
]∗
φ(x)dx
what means that the ∂∂x
is not hermitian operator. It is easy to check now,
that the B = i ∂∂x
is a hermitian operator.
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3rd axiom - Wavefunction evolution in time
Wavefunction evolution is described by the Schrödinger equation with thetime variable t
i~∂ψ(r1, r2, . . . , rN , t)
∂t= Hψ(r1, r2, . . . , rN , t)
3. Erwin Rudolf Josef Alexander Schrödinger (12 August 1887 - 4 January 1961). Nobel Prize in 1933 „for thediscovery of new productive forms of atomic theory” (together with Paul Dirac).
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Stationary states
The states where density of probability doesn’t depend on time
If the Hamiltonian doesn’t depend on time (because the V doesn’t dependon time) we can write for example
Ψ(r, t) = ψ(r) · f(t) = ψ(r)e−ig(t), g(t) ∈ R
ρ(r, t) = Ψ∗(r, t)Ψ(r, t) = ψ∗(r)eig(t)ψ(r)e−ig(t) = |ψ(r)|2 · 1 = ρ(r)
~ψ(r)dg(t)
dt= Hψ(r)
If the Hamiltonian doesn’t depend on time then the right side of the equationdoesn’t depend on time. So, the left as well. It is fulfilled when
g(t) = at+ b, a, b ∈ R
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Stationary states
the b can be derived from boundary condition, where
Ψ(r, 0) = ψ(r) ⇒ b = 0
Now, the equation takes form
Hψ(r) = a~ψ(r)
Inserting a~ = E the general form of wave function of N particles looks like
Ψ(r1, r2, . . . , rN , t) = ψ(r1, r2, . . . , rN )ei~Et
and the Schrödinger equation independent on time
HΨ(r1, r2, . . . , rN ) = EΨ(r1, r2, . . . , rN )
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4th axiom - Eigenvalues and eigenstates
If a given operator A which represents a given physical quantity togetherwith a function of Q class satisfy the following equation
A|ψn〉 = an|ψn〉then the |ψn〉 function is called an eigenfunction and the an is called aneigenvalue.
If to a given eigenvalue correspond K linearly independent eigenfunctions,
A|ψn,i〉 = an|ψn,i〉, i ∈ (1,K) ∩ N
then the an is K-times degenerated.
If the |ψn〉 is an eigenfunction of the operator A then the functionc|ψn〉, c ∈ R is also an eigenfunction of the A operator and moreover withthe same eigenvalue an.
A|ψn〉 = an|ψn〉 ⇒ A(c|ψn〉) = cA|ψn〉 = can|ψn〉 = anc|ψn〉
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Properties of quantum-mechanical operators
1. The eigenvalues in quantum mechanics are real
Let the A operator be hermitian one, i.e.
〈ψn|A|φn〉 = 〈Aψn|φn〉Then, let’s take the eigen problem
A|ψn〉 = an|ψn〉and conjugated eigen problem
A∗〈ψn| = a∗n〈ψn|
〈ψn|anφn〉 = 〈anψn|φn〉 ⇒ an = a∗n ⇒ an ∈ R
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Properties of quantummechanical operators
2. eigenfunctions of hermitian operators with different eigenvalues areorthogonal
〈ψ|φ〉 = 0
Let’s take two different eigenfunctions of the same hermitian operator
A|ψ1〉 = a1|ψ1〉
A|ψ2〉 = a2|ψ2〉Multiplying the first equation by 〈ψ2| and conjugated (to second equation)
A ∗ 〈ψ2| = a2〈ψ2|by |ψ1〉, recollecting that A is hermitian we get
〈ψ2|A|ψ1〉 − A ∗ 〈ψ2|ψ1〉 = (a1 − a2)〈ψ2|ψ1〉 = 0
and because a1 6= a2 thus 〈ψ2|ψ1〉 = 0.
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Properties of quantummechanical operators
3. The eigenvalues of hermitian operator form complete configuration, i.e.any function from this configuration can be expressed as linearconfiguration of other (from this set) functions.
|φ〉 =
M∑
l=1
cl|ψl〉
and
A|ψl〉 = al|ψl〉and the coefficients we can assign from scalar multiplication of this linearcombination by 〈ψk|. We got
ck = 〈ψk|φ〉
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Properties of quantummechanical operators
4. If two hermitian operators commute (i.e. they are commutative
(alternate), i.e. [A, B] = 0) then these two operators have common set ofeigenvalues
Let’s say that all eigenvalues of both operators are not degenerated. Ifadditionaly
A|ψn〉 = an|ψn〉and recollecting that B is linear
BA|ψn〉 = Ban|ψn〉 = anB|ψn〉From the commutation we get
AB|ψn〉 = BA|ψn〉AB|ψn〉 = Ban|ψn〉AB|ψn〉 = anB|ψn〉
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Continuation ...
This means that eigenfunction B|ψn〉 (function |ψn〉 modified by operator Bis also an eigenfunction of A operator with eigenvalue an.
Moreover, because we assumed that eigenvalue an is not degenerated,then the functions |ψn〉 and B|ψn〉 differs at best by a constant, and
B|ψn〉 = bn|ψn〉
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Properties of quantummechanical operators
5. If two hermitian operators have the same set of eigenfunctions thenboth operators commute.
A|ψn〉 = an|ψn〉B|ψn〉 = bn|ψn〉
and we require that for any function |χ〉
[A, B]|χ〉 = (AB − BA)|χ〉
We can express the |χ〉 function as a linear combination of functionsbelonging to the complete configuration
|χ〉∑
l=1
Mcl|ψl〉
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Continuation ...
Then, introducing the combination into the commutation requirement
M∑
l=1
cl(AB − BA)|ψl〉 = 0 ⇔ ∀l∈(1;M)∩N(AB − BA)|ψl〉 = 0
Moreover, from previous theorem and from the assumption we have
AB|ψl〉 = BA|ψl〉 = albl|ψl〉Thus, generally
[A, B] = 0
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Properties of quantummechanical operators
Additional properties
• if two operators A and B representing two physical quantities commuteeach other then the both quantities can be measured.
• If A|ψn〉 = an|ψn〉 but the system is in state |φ〉 which is not aneigenfunction of the A operator then measuring of A quantity will givesome of eigenvalues of A. What can be calculated is only an averagevalue of the measurement
〈A〉 =〈φ|A|φ〉〈φ|φ〉
and if |φ〉 is normalized then
〈A〉 = 〈φ|A|φ〉
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Continuation ...
Function |φ〉 can be expressed as a linear combination of the completeconfiguration
|φ〉 =
M∑
l=1
cl|ψl〉
then the average value equals
〈φ|Aφ〉 =
M∑
k=1
M∑
l=1
c∗kcl〈ψk|A|ψl〉 =
=M∑
k=1
M∑
l=1
c∗kclal〈ψk|ψl〉 =M∑
k=1
M∑
l=1
c∗kclalδkl =M∑
k=1
|c|2kak
and for normalized function we have additionally
〈φ|φ〉 =
M∑
k=1
M∑
l=1
c∗kcl〈ψk|ψl〉 =
M∑
k=1
|c|2k = 1 |c|2k = probability
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5th axiom - Spin
Elementar paricles have additional degree of freedom - spin.
what can be measured is the square of the total spin
|S|2 = S2x + S2
y + S2z = s(s+ 1)~2
and one of its components, most likely the z - component
sz = ms~
s = spin quantum number. Can be an integer or half-integer.ms = magnetic spin number; ms = −s,−s+ 1, . . . , 0, . . . , s− 1, s.
It means that electrons (the spin number s = 12 ) have additional coordinate -
spin coordinate which may take form
σ(ms) = ms
For electrons we have one s = 12 and thus two different ms = ± 1
2 .
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The spin coordinates
Each i-th electron have three space coordinates ri = (xi, yi, zi) and onespin coordinate σi. It will be described as:
qi = (xi, yi, zi, σi)
and the wavefunction of N particles depends on 4N coordinates
ψ = ψ(q1, q2, . . . , qN )
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Spin functions - nonrelativistic mechanics
Orthonormal spin functions defined in the spin space
|α〉 =
(
1
0
)
|β〉 =
(
0
1
)
Orthogonality
〈α|β〉 ≡∑
σ
α(σ)∗β(σ) = α
(
−1
2
)∗
β
(
−1
2
)
+α
(1
2
)∗
β
(1
2
)
= 0·1+1·0 = 0
Normalization
〈α|α〉 ≡∑
σ
α(σ)∗α(σ) = α
(
−1
2
)∗
α
(
−1
2
)
+α
(1
2
)∗
α
(1
2
)
= 0·0+1·1 = 1
Normalization + Orthogonality = Orthonormality
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Spin operators
Sx =1
2~σx Sy = 1
2~σy Sz =1
2~σz
The Pauli’s matrices (2 × 2) are as follows
σx =
(
0 1
1 0
)
σy =
(
0 −ii 0
)
σz =
(
1 0
0 −1
)
Let’s act with the operator Sz on a basis vector, for example on |α〉
Sz|α〉 ≡ Sz
(
1
0
)
=1
2~
(
1 0
0 −1
)(
1
0
)
=1
2~
(
1
0
)
=1
2~|α〉
The α and β functions are eigenfunctions of the Sz operator witheigenvalues 1
2~ and − 12~ respectively.
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Spin operators - S2
S2|α〉 ≡ S2
(
1
0
)
=(
S2x + S2
y + S2z
)(
1
0
)
=1
4~
2
(
0 1
1 0
)2
+
(
0 −ii 0
)2
+
(
1 0
0 −1
)2
(
1
0
)
=1
4~
2
(
1 + 1 + 1 0 + 0 + 0
0 + 0 + 0 1 + 1 + 1
)(
1
0
)
=3
4~
2
(
1 0
0 1
)(
1
0
)
=3
4~
2
(
1
0
)
=
[1
2
(1
2+ 1
)
~2
]
|α〉
Identical result we will get for |β〉 function.
Thus, both functions are eigenfunctions of the S2 operator.
The |α〉 and |β〉 functions are not eigenfunctions of Sx and Sy operators.
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Spin of atoms and molecules
• More then one of elementar particles
S = s1 + s2 + · · · + sN
Sz = s1,z + s2,z + · · · + sN,z
We know: |S|2 = S(S + 1)~2 and Sz.
Systems with even number of fermions are bosons, systems with oddnumber of fermions are fermions.
• Nucleus
12C and 16O have S = 0, while 13C, 15N, 19F have S = 12
• Atoms and molcules
is atom or molecule a fermion or a boson?
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Spin of atoms and molecules
• Hydrogen atom: 2 fermions (proton and electron), both have spin 12 . Thus,
hydrogen atom is a boson.
• Sodium atom: 23 nucleons (spin 12 ) and 11 electrons (spin 1
2 ). Thussodium atom is also a boson.
• Hydrogen molecule: two electrons (M′
s = ms,1 +ms,2 = 12 + 1
2 = 1, or
M′
s = ms,1 +ms,2 = 12 − 1
2 = 0) and we have two possibilities:
1. (S = 1) Ms = −1, 0,+1 - three orientations of total spin - triplate state(parallel spins).
2. (S = 0) Ms = 0 - one orientation of total spin - singlet state (antiparallelspins).
Additionaly - nucleus: two protons.
1. antiparallel spins - S = 0 - (parahydrogen)
2. parallel spins - S = 1 (orthohydrogen)
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Dirac’s equation - relativistic quantum mechanics
(E − qφ− βm0c2 − cα · π)Ψ = 0
q = charge of a particleφ = skalar potentialE = total energy
β ≡ α0 =
(
1 0
0 −1
)
α · π ≡∑
µ αµπµ, where αµ and πµ are determined by Pauli’s matrices.Ψ = realtivistic quantum state
Using the spinors and bispinors, in matrix form the equation can be rewritten
(
V c(σ · π)
c(σ · π) V − 2m0c2
)(
Ψ
φ
)
=
(
E −m0c2 0
0 E −m0c2
)(
Ψ
φ
)
, 8 sierpnia 2010 - p. 49/??
Relativistic equation - when it is important?
Average value of kinetic energy for 1s electron in hydrogen-like atom
T =1
2Z2
where Z is the charge of nuclei.
For hydrogen atom ZH = 1 and v/c ≃ 0.7%. It means that for hydrogen atomthe relativistic effects are neglectable.
For gold (for example), however
ZAu = 79, and v/c ≃ 51%.
In gold atom the mass of 1s electron is about 15% larger then the invariantmass.
, 8 sierpnia 2010 - p. 50/??
6th axiom - Symmetry of the wave function
Wave function of N bosons of the same type is symmetrical when only wewill shift the coordinates of two of them, for example l and k
Ψ(q1, . . . , qk, . . . , ql . . . qN , t) = Ψ(q1, . . . , ql, . . . , qk . . . qN , t)
while the wavefunction of N fermions of the same type is antisymmetrical
Ψ(q1, . . . , qk, . . . , ql . . . qN , t) = −Ψ(q1, . . . , ql, . . . , qk . . . qN , t)
Particles with integer spin are bosons, while particles with half-integer spinare fermions.
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Bosons
Satyendra Nath Bose (1 January 1894 - 4
February 1974). Bose did not have a docto-
rate, but obtained an MSc from the Univer-
sity of Calcutta in 1915 and therefore did
not have a doctoral advisor.
Albert Einstein (14 March 1879 - 18 April
1955). Nobel Prize in 1933 „for his services
to Theoretical Physics, and especially for
his discovery of the law of the photoelec-
tric effect”.
Bosons are particles which obey Bose-Einstein statistics
Bosons may be either elementary, like the photon, or composite, like mesons. Most bosons are compositeparticles.
All observed bosons have integer spin. Several bosons can occupy the same quantum state.
, 8 sierpnia 2010 - p. 52/??
Fermions
Enrico Fermi (29 September 1901 - 28 No-
vember 1954). Nobel Prize in 1938 „for his de-
monstrations of the existence of new radio-
active elements produced by neutron irradia-
tion, and for his related discovery of nuclear
reactions brought about by slow neutrons”
Paul Adrien Maurice Dirac (8 August
1902 - 20 October 1984). Nobel Prize in
1933 „for the discovery of new produc-
tive forms of atomic theory” (together
with Erwin Schrödinger).
Fermions are particles which obey Fermi-Dirac statistics.
Fermions can be elementary, like the electron, or composite, like the proton.
All observed fermions have half-integer spin, as opposed to bosons.
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Symmetry - consequences
• Probability that two fermions are in the same place and with same spincoordinate.
Ψf (1, 1, 3, 4, . . . , N) = −Ψf (1, 1, 3, 4, . . . , N)
Ψf (1, 1, 3, 4, . . . , N) = 0, and |Ψf (1, 1, 3, 4, . . . , N)|2 = 0
Two electrons possesing the same spatial AND spin coordinates don’t likeeach other.
• Probability that two bosons are in the same place and with same spincoordinate
Ψb(1, 1, 3, 4, . . . , N) = Ψb(1, 1, 3, 4, . . . , N)
Ψb(1, 1, 3, 4, . . . , N) > 0, and |Ψb(1, 1, 3, 4, . . . , N)|2 > 0
Experimental proove - 2001, Eric A. Cornell, Wolfgang Ketterle, Carl E. Wieman. Nobel Prize „for the
achievement of Bose-Einstein condensation in dilute gases of alkali atoms, and for early fundamental
studies of the properties of the condensates”.
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