Roger Bello Romero

roger.bello@uam.es
Departamento de Química Física Aplicada
Módulo 14, 311
Table of Contents Topic 2

TOPIC 2: INTRODUCTION TO QUANTUM MECHANICS

2.1 The wave function and its interpretation

2.2 The postulates of Quantum Mechanics
2.3 The uncertainty principle
2.4 The Schrödinger equation
2.5 The free particle
2.6 The particle in a box and the particle in a ring
2.7 The harmonic oscillator
2.8 The rigid rotor
2.9 Separation of the center of mass
2.1 The wave function and its interpretation Topic 2
Postulate 1: The state of a physical system is completely specified by a complex function
(wave function or probability amplitude) of all the coordinates of the system (both
spatial and spin) and time.

Born (1926): If the wave function of a particle has the value ψ ( x , y , z ) at ( x , y , z )

then the probability of finding the particle in an infinitesimal volume dV =dxdydz in that
2
position is proportional to |ψ ( x , y , z )| dV

P ( x , y , z ,t ) =∭ Ψ 2 ( x , y , z ,t ) dxdydz Ψ 2= Ψ Ψ c
V

The normalized wave function is

Probability density
determined only within a constant
phase factor e iα

The probability of finding the particle at any place has necessarily to be 1 ( the particle must
be somewhere in space). This implies the NORMALIZATION CONDITION:
d
∫ Ψ 2 dV =1 ∫
dt V ∈ℝ
2
Ψ dV =0 The normalization integral is
V ∈ℝ
3 3
time-independent
2.1 The wave function and its interpretation Topic 2
Spherical coordinates

x=r sinθcosφ z= r cosθ

y =r sinθsinφ

dV =r 2 sinθ drdθdφ

∭ Ψ 2 ( r , θ , φ , t ) r 2 sinθ drdθdφ=1
V
0≤θ≤ π 0≤φ≤2 π 0≤r ≤∞

Cartesian coordinates
dV =dxdydz

∭ Ψ 2 ( x , y , z , t ) dxdydz =1
V

− ∞≤ x ≤ ∞ − ∞≤ y ≤ ∞ − ∞≤ z ≤ ∞
2.1 The wave function and its interpretation Topic 2
Born interpretation: (a) (b)
Impose severe restrictions on the
wave function properties
(a) Must not be infinite over a
finite region. L²-integrable
(square-integrable).

(b) Must be single-valued. (c) (d)

(c) Must be continuous.

(d) Must have a continuous

first derivative (slope). There
may be exceptions in some
special situations.
2.1 The wave function and its interpretation Topic 2
Example (1a): Carbon nanotubes are thin hollow cylinders of carbon with diameters
between 1 nm and 2 nm, and lengths of several micrometres. According to one simple
model, the lowest-­energy electrons of the nanotube are described by the wave function
sin(πx/L), where L is the length of the nanotube. Find the normalized wave function.

We write the wave function as: Ψ = Nsin

πx
L ( )
where N is the normalization factor. The limits of
integration are x=0 to x=L because the space
available to the electron spans the length of the tube.
It follows that:

√
L L

P=∫ Ψ ( x ) dx =∫ N sin
0
2

0
2 2
( )
πx
L
dx= 1 N
L
2
2
=1 N=
2
L

The normalized wave function is therefore: Ψ=

√ 2
L ( )
sin
πx
L
2.1 The wave function and its interpretation Topic 2
Example (1b): As seen in Example (1a), the lowest-energy electrons of a carbon nanotube
of length L can be described by the normalized wave function (2/L)1/2sin(πx/L). What is the
probability of finding the electron between x = L/4 and x = L/2?

√
L L/2

Ψ=
2
L
sin
L( )
πx
P=∫ Ψ ( x ) dx =
0
2 2
L
∫
L/ 4
sin
2
( )
πx
L
dx

)|
L/2

P=
L 2
− (
2 x sin ( 2 π x / L )
4 π/L L/ 4
= 0.409

It follows that there is a chance of about 40.9% that the electron will be found in the region.
2.2 The postulates of QM Topic 2
Superposition principle: It states that a quantum-mechanical system which can take on the
discrete states ψ 1 ( r ) and ψ 2 ( r ) is also able to occupy the state ψ ( r ) = c 1 ψ 1 ( r ) +c 2 ψ 2 ( r )

ψ= c dead ψ dead +c alive ψ alive

2.2 The postulates of QM Topic 2
ψ= c dead ψ dead +c alive ψ alive Pdead = c 2dead c 2dead +c 2alive =1
Palive =c 2alive

1 c dead = c alive
c alive =c dead =
√2

1 1
State of the cat in the closed box: ψ= ψ dead + ψ alive
√2 √2

ψ alive Find the cat alive:

1
Measurement c alive = =∫ ψ calive ψ dV
c alive =1/ √ 2

collapse √2
collapse

Find the cat dead:

ψ
1
=∫ ψ dead ψ dV
c
c dead =
c dead =1/ √ 2 ψ dead √2
2.2 The postulates of QM Topic 2
The values which a given physical observable can take are called in QM its eigenvalues, and
the set of these is referred to as the spectrum of eigenvalues. The spectrum of eigenvalues can
be discrete (the magnitude can take only certain values) or continuous (the magnitude can
take any value).

The wave function of the system in the state where the observable f has the value fn by ψ n .
The wave functions ψ n are called eigenfunctions of the observable f.

∫ Ψ 2n dV =1 each of these functions

V ∈ℝ
3
are normalized
2.2 The postulates of QM Topic 2
Ψ =∑ a n ψ n ∑ n =1
a 2
Ψ c =∑ a cn ψ cn
n n n
superposition principle

∫ Ψ c
Ψ dV = ∑ n a n=1
a c
∫ Ψ Ψ c
dV = ∑ n ∫ n Ψ dV
a c
ψ c

n n

∑ n n ∑ n∫ n Ψ dV
a c
a = a c
ψ c

n n
Probability of finding the
system on the n-eigenstate
a n=∫ ψ cn Ψ dV is calculated by projecting
the total wave function
onto the n-eigenfunction

a n=∑ a m∫ ψ cn ψ m dV ∫ n ψ m dV =δ mn
ψ c

The set of eigenfunctions forms a

complete set of orthonormal functions
2.2 The postulates of QM Topic 2
Mean-value (or expected value) of physical observable f can be thought of as an average of all
the possible outcomes of a measurement weighted by their probability.
(operator)(function)=(new function)

f = ∑ f n a2n
⟨ f ⟩= ¯ f =∫ Ψ ( f^ Ψ ) dV
⟨ f ⟩= ¯ c

f̄ =∑ f n a n an =∫ Ψ ∑
( )
c c
an f n ψ n dV
n n

( f^ Ψ ) =∑ an f n ψ n f^ ψ n = f n ψ n
n

(operator)(function)=(constant factor)x(same function)

(operator)(eigenfunction)=(eigenvalue)x(eigenfunction)

Any physical observable has an associated operator

(part of postulate 2)
2.2 The postulates of QM Topic 2
observable mean value has to be real transposed operator
~
f̄ =∫ Ψ ( ^f Ψ ) dV =∫ Ψ ( f^ Ψ ) dV
c c c
∫ Ψ ( f Φ ) dV =∫ Φ ( ^f Ψ ) dV
^

~ ~
^
f = f^
c ^f = f^ c

∫ ψ ( f ψ m ) dV =(∫ ψ m ( f ψ n ) dV )
c
^c c ^
n Hermitian operator

Postulate 2: For any physical observable, there is a definite corresponding linear and
Hermitian operator.

ψ cm x f^ ψ n = f n ψ n ψ m f^ ψ n = f n ψ m ψ n
c c

ψ x f^ ψ = f ψ
n
c c
m m
c
m ψ f^c ψ c =f ψ ψ c
n m m n m

ψ m f^ ψ n −ψ n f^ ψ m=( f n − f m ) ψ m ψ n ( f n − f m ) ∫ ψ mc ψ n dV =0
c c c c

Hermitian operator has real eigenvalues and the

the eigenfunctions of different eigenvalues are ∫ ψ mc ψ n dV =δ mn
orthogonal
2.2 The postulates of QM Topic 2
Postulate 3: The result of the measurement of an observable A is always an eigenvalue, an ,
of the associated operator Â.

Postulate 4: It is not possible, in general, to predict the result of an individual

measurement of an observable A when the system is in any state ψ. However, the
probability of obtaining the eigenvalue, an, when performing the same measurement in a
large number of identically prepared systems, P(a=an), is given by:

Pn =∫ ψ cn ψ dV

From this result, it is possible to prove that the expectation value or mean value of the
observable A in these measurements is:
⟨ A ⟩ = Ā =∫ Ψ c ( ^
A Ψ ) dV
2.2 The postulates of QM Topic 2
Physical observables and their operators:

In QM there are some observables/operators that don’t have a classical analogous, for instance
the spin.

Taken from Prof. Manuel Lara Garrido

2.2 The postulates of QM Topic 2
Example 2: What is the linear momentum of a free particle described by the wavefunctions
(a) ψ(x) = eikx and (b) ψ(x) = e−ikx?

This is an eigenvalue equation, with eigenvalue +kħ. It follows that a measurement of the
momentum will give the value px = +kħ.

Now the eigenvalue is −kħ, so px = −kħ. In case (a) the momentum is positive, meaning that the
particle is traveling in the positive x-direction, whereas in (b) the particle is moving in the
opposite direction.
2.2 The postulates of QM Topic 2
Example 3: Proving that the linear momentum operator is hermitian

The boxed term is zero because all wave functions are either zero at x=±∞ or the product
ψi*ψj converges to the same value at x=+∞ and x=−∞.
2.2 The postulates of QM Topic 2
Example 4: Two possible wave functions for a particle constrained to be somewhere on the
x axis between x=0 and x=L are ψ1=sin(πx/L) and ψ2=sin(2πx/L). Outside this region the
wave functions are zero. The wave functions correspond to different energies. Verify that
the two wave functions are mutually orthogonal.
L

∫ ψ 1 ψ 2 dx=0
c

The sine functions have been evaluated by using sin(nπ)=0 for n=0, ±1, ±2,.... The two
functions are therefore mutually orthogonal.
2.2 The postulates of QM Topic 2
Example 5: Calculate the mean value of the position of an electron in the lowest energy
state of a one-dimensional box of length L, with the (normalized) wave function ψ=(2/L)1/2
sin(πx/L) inside the box and zero outside it.

The expectation value of position is:

√
L
⟨ x ⟩ =∫ ψ c ( x ) x ψ ( x ) dx
0
ψ ( x )=
2
L
sin ( )
πx
L

( )) dx
L L
2
⟨ x ⟩ = ∫ x sin
L 0
2 πx
L
dx ( ) ⟨ x ⟩=
1
∫
L 0
x 1 − cos
2 πx
L (
L
⟨ x ⟩=
1 1 2 1
L 2
L − ∫ x cos
L 0
2 πx
L
dx ( ) ⟨ x ⟩=
L
2

This result means that if a large number of measurements of the position of the electron are
made, then the mean value will be at the center of the box.
2.2 The postulates of QM Topic 2
f^ ( Ψ 1 +Ψ 2 ) = ^f Ψ 1+ f^ Ψ 2
Operator is said to be linear if it
f^ ( aΨ 1 ) = a f^ Ψ has this properties.

( f^ + g^ ) Ψ = f^ Ψ + g^ Ψ
Sum and multiplication.
( f^ g^ ) Ψ =f^ ( g^ Ψ )

[ ^f , g^ ] Ψ =( f^ ^g − g^ ^f ) Ψ Commutator

The product of two operators in general does not commute: f^ g ^ f^ ≠ 0

^−g

f^ g
^ ψ n = ^f g n ψ n= f n gn ψ n ^g f^ ψ n = ^g f n ψ n= g n f n ψ n

[ ^f , g^ ] ψ n =( f^ ^g − g^ ^f ) ψ n= ( gn f n − gn f n ) ψ n= 0

[ ^f , g^ ]=( f^ ^g − g^ f^ ) =0
If two observables f and g can take simultaneously take definite values,
then their operators commute
2.2 The postulates of QM Topic 2
Example 6: The Commutator of Position and Momentum Operators

[^
pi , x^j ] = ( ^
pi ^
xj−^ pi ) =−i ℏ δ ij
xj ^
In general the coordinates and the
momentum satisfy the commutation [^
pi , ^p j ] =( ^
pi ^
pj− ^ p i )= 0
pj ^
rules:
( i , j )=( x , y , z ) [^
x i , x^j ] = ( ^
x i x^j − ^ x i ) =0
xj ^
2.2 The postulates of QM Topic 2
Example 7: Justifying the expression for the expectation value of an operator.

Now suppose the (normalized) wave function is the linear combination of two eigenfunctions of
the operator Ω, each of which is individually normalized to 1. Then:
2.2 The postulates of QM Topic 2
Example 8: What are the operators associated to the components of the orbital angular
momentum.

l^x = ^y ^
p z − z^ ^
py l^x =−i ℏ ( ^y ∂ z − ^z ∂ y )
^ r^ × ^
l= p l^y = ^z ^
p x − ^x ^
pz l^y = −i ℏ ( ^z ∂x − x^ ∂ z )
l^z = x^ ^
p y − ^y ^
px l^z =−i ℏ ( x^ ∂ y − ^y ∂x ) l^z =− i ℏ ∂ φ

in spherical
coordinates

In general the angular momentum components satisfy the commutation rules:

[ l^x , l^y ] =i ℏ l^z [ l^y , l^z ]=i ℏ l^x [ l^z , l^x ] =i ℏ l^y
Thus the components of angular momentum are not measurable at the same time. In contrast, the
square of the angular-momentum operator commutes with all components of the angular-
momentum operator.
2.3 The Uncertainty Principle Topic 2
Let f and g be two physical observable whose operators fulfill the following condition:

[ ^f , g^ ]=i q^
Then we can prove that for any vector (state) ψ of the system the following relation holds:

2 2 2
Δf Δg ≥ ⟨ q
^ ⟩ /4

Δf = f^ − ⟨ f^ ⟩ g −⟨ g
Δg = ^ ^⟩

For the particular case of the position and momentum operators we get:

[^
pi , x^j ] = ( ^
pi ^
xj−^ pi ) =−i ℏ δ ij
xj ^

ΔpΔx ≥ ℏ/ 2

Heisenberg's Uncertainty Relations

It is impossible to find (and thus to prepare) a state of the system where Δx and Δp are
both arbitrarily small
2.3 The Uncertainty Principle Topic 2

Heisenberg's Uncertainty Principle

2.4 The Schrödinger equation Topic 2
Postulate 5: The evolution in time of the state of the system is governed by the time
dependent Schrödinger equation:
∂Ψ
iℏ =^
HΨ
∂t
If the Hamiltonian and the initial conditions ψ(t=0) are known, this linear equation allows one
to obtain the state of the system at any time ψ(t).

Reminder:
2
^ ℏ 2
H=− ∇ +U ( r , t ) Hamiltonian (total energy operator)
2m

kinetic energy operator p²/2m: potential energy operator: interaction of

translational motion the system with any field
2 ∂2 ∂2 ∂2
∇ = 2
+ 2
+ 2
∂x ∂ y ∂z
2.4 The Schrödinger equation Topic 2
If the system is CONSERVATIVE, i.e. the Hamiltonian of the system does not explicitly
depend on time:
^
H (r , t)=^ H (r)

∂
iℏ Ψ n (r , t)=^
H Ψ n ( r , t )= E n Ψ n ( r ,t )
∂t

t t
d Ψ n (r , t) Ψ n (r , t )
∫ Ψ n(r , t )
=− ( i / ℏ ) En ∫ dt ln
Ψ n (r , 0)
=− iE n t / ℏ
0 0

Ψ ( r , t ) =ψ ( r ) e −iEt /ℏ Ψ n ( r , t )=Ψ n ( r , 0 ) e
− iE n t /ℏ

|Ψ ( r , t )|
2
=|ψ ( r )|
2
Stationary states: The probability distribution
is independent of time. The same is valid for
the mean values.

^
H ψ = Eψ Time-independent (stationary)
Schrödinger equation
2.4 The Schrödinger equation Topic 2

⟨ ψ 1|ψ 2 ⟩ =∭ ψ c1 ψ 2 dV bra-ket or Dirac

V notation

‖ψ‖= √ ⟨ ψ|ψ ⟩ norm

bra ket

⟨ ψ m|ψ n ⟩ =∭ ψ cm ψ n dV =δ mn orthonormality relation

V

⟨ ψ m|f^|ψ n ⟩ =∭ ψ cm f^ ψ n dV = f mn operator matrix

V element

c
⟨ ψ 1|ψ 2 ⟩ =( ⟨ ψ 2|ψ 1 ⟩ ) ⟨ ψ|ψ ⟩ ≥ 0 ⟨ ψ|ψ ⟩ =0 if and only if ψ= 0

⟨ ψ 1|α ϕ 1 + β ϕ 2 ⟩ = α ⟨ ψ 1|ϕ 1 ⟩ + β ⟨ ψ 1|ϕ 2 ⟩

⟨ α ϕ 1 + β ϕ 2|ψ 1 ⟩ = α c ⟨ ϕ 1|ψ 1 ⟩ + β c ⟨ ϕ 2|ψ 1 ⟩

2.4 The Schrödinger equation Topic 2

associative axiom of multiplication

completeness relation (closure)

projection operator

Both (bra and ket) are vectors (state vectors) in a linear vector space. By using this notation,
many relations in quantum mechanics can be expressed more succinctly than by using integral
representation.
2.5 The free particle Topic 2
A free particle is a particle that moves freely, in the absence of forces (V=0). Remember that,
classically speaking, such a particle would move at constant speed (let’s assume along the X
axis), thus having a constant momentum and constant kinetic (thus total, H=T+V=T) energy.

^ ℏ2 d2
H ψ = Eψ − ψ ( x ) = Eψ ( x )
2 m dx 2

d2 2 mE
2
ψ ( x ) =− 2
ψ (x)
Energy levels: double degeneracy dx ℏ
Spectrum:
E=[0,∞)
ψ ( x ) = Aeikx + Be− ikx

k=
√ 2 mE
ℏ2

Energies in a continuum
(E is not quantized)
2.6 The particle in a box & ring Topic 2
0 0< x< L
V =0 ∞ x⩾ L
∞ x⩽0

ψ ( x< 0 ) =ψ ( x> L ) =0

The particle cannot penetrate if the

potential is infinite

ψ ( x=0 )=ψ ( x= L )=0

Inside the box: V=0
The wavefunction has to be continuous
^
H ψ = Eψ boundary conditions

−
ℏ2 d2
2 m dx 2
ψ ( x ) = Eψ ( x ) ψ ( x ) = Aeikx + Be− ikx k =
√ 2 mE
ℏ2
2.6 The particle in a box & ring Topic 2
ψ ( x ) = Aeikx + Be− ikx k=
√ 2 mE
ℏ2

ψ ( x ) =Ccos ( kx ) + Dsin ( kx )

ψ ( x=0 )=ψ ( x= L )=0

C =0 Dsin ( kL )=0

√ 2 mE n2 π 2 ℏ2
kL =n π 2
L=nπ En = 2
ℏ 2 mL

Only some particular values of the energy are allowed: the energy is quantized. We
number the solutions using a quantum number, n = 1,2, ... , ∞.

√
L

ψ n ( x ) = Dsin ( nπx
L ) ∫ m ψ n dx= δ mn
ψ
0
c
D=
2
L
2.6 The particle in a box & ring Topic 2