Anomaly
Zephir
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Quantum Mechanics can be formulated in terms of a few postulates (i.e., theoretical principles
based on experimental observations). The goal of this section is to introduce such principles, to-
gether with some mathematical concepts that are necessary for that purpose. To keep the notation
as simple as possible, expressions are written for a 1-dimensional system. The generalization to
many dimensions is usually straightforward.
P ostulate 1 : Any system in a pure state can be described by a wave-function , ψ(t, x), where t is
a parameter representing the time and x represents the coordinates of the system. Such a function
ψ(t, x) must be continuous, single valued and square integrable.
Note 1: As a consequence of Postulate 4, we will see that P (t, x) = ψ∗(t, x)ψ(t, x)dx represents
the probability of finding the system between x and x + dx at time t, first realized by Max Born. 2
P ostulate 2 : Any observable (i.e., any measurable property of the system) can be described by
an operator. The operator must be linear and hermitian.
What is an operator ? What is a linear operator ? What is a hermitian operator?
Definition 1: An operator ˆO is a mathematical entity that transforms a function f (x) into another
function g(x) as follows, R4(96) ˆOf (x ) = g(x ),
where f and g are functions of x.
Definition 2: An operator ˆO that represents an observable O is obtained by first writing the clas-
sical expression of such observable in Cartesian coordinates (e.g., O = O(x, p)) and then substi-
tuting the coordinate x in such expression by the coordinate operator ˆx as well as the momentum p
by the momentum operator ˆp = −i~∂/∂x.
Definition 3: An operator ˆO is linear if and only if (iff),
ˆO(af (x) + bg(x)) = a ˆOf (x) + b ˆOg(x),
where a and b are constants.
Definition 4: An operator ˆO is hermitian iff,
∫
dxφ∗
n(x) ˆOψm(x) =
[∫
dxψ∗
m(x) ˆOφn(x)
]∗
,
2Note that this probabilistic interpretation of ψ has profound implications to our understanding of reality. It es-
sentially reduces the objective reality to P (t, x). All other properties are no longer independent of the process of
measurement by the observer. A. Pais’ anecdote of his conversation with A. Einstein while walking at Princeton em-
phasizes the apparent absurdity of the implications: [Rev. Mod. Phys. 51, 863914 (1979), p. 907]: ’We often discussed
his notions on objective reality. I recall that during one walk Einstein suddenly stopped, turned to me and asked whether
I really believed that the moon exists only when I look at it.’
8
where the asterisk represents the complex conjugate.
Definition 5: A function φn(x) is an eigenfunction of ˆO iff,
ˆOφn(x) = Onφn(x),
where On is a number called eigenvalue.
Property 1: The eigenvalues of a hermitian operator are real.
Proof: Using Definition 4, we obtain
∫
dxφ∗
n(x) ˆOφn(x) −
[∫
dxφ∗
n(x) ˆOφn(x)
]∗
= 0,
therefore,
[On − O∗
n]
∫
dxφn(x)∗φn(x) = 0.
Since φn(x) are square integrable functions, then,
On = O∗
n.
Property 2: Different eigenfunctions of a hermitian operator (i.e., eigenfunctions with different
eigenvalues) are orthogonal (i.e., the scalar product of two different eigenfunctions is equal to
zero). Mathematically, if ˆOφn = Onφn, and ˆOφm = Omφm, with On 6 = Om, then ∫ dxφ∗
nφm = 0.
Proof: ∫
dxφ∗
m ˆOφn −
[∫
dxφ∗
n ˆOφm
]∗
= 0,
and
[On − Om]
∫
dxφ∗
mφn = 0.
Since On 6 = Om, then ∫ dxφ∗
mφn = 0.
P ostulate 3 : The only possible experimental results of a measurement of an observable are the
eigenvalues of the operator that corresponds to such observable.
P ostulate 4 : The average value of many measurements of an observable O, when the system is
described by ψ(x) as equal to the expectation value ̄O, which is defined as follows,
̄O =
∫ dxψ(x)∗ ˆOψ(x)
∫ dxψ(x)∗ψ(x) .
P ostulate 5 :The evolution of ψ(x, t) in time is described by the time-dependent Schr ̈odinger
equation :
9
i~ ∂ψ(x, t)
∂t = ˆHψ(x, t),
where ˆH = − ~2
2m
∂2
∂x2 + ˆV (x), is the operator associated with the total energy of the system, E =
p2
2m + V (x).
Expansion P ostulate : R5(15), R4(97)
The eigenfunctions of a linear and hermitian operator form a complete basis set. Therefore,
any function ψ(x) that is continuous, single valued, and square integrable can be expanded as a
linear combination of eigenfunctions φn(x) of a linear and hermitian operator ˆA as follows,
ψ(x) = ∑
j
Cj φj (x),
where Cj are numbers (e.g., complex numbers ) called expansion coefficients.
Note that ̄A = ∑
j Cj C∗
j aj , when ψ(x) = ∑
j Cj φj (x),
ˆAφj (x) = aj φj (x), and
∫
dxφj (x)∗φk(x) = δjk.
This is because the eigenvalues aj are the only possible experimental results of measurements of ˆA
(according to Postulate 3), and the expectation value ̄A is the average value of many measurements
of ˆA when the system is described by the expansion ψ(x) = ∑
j Cj φj (x) (Postulate 4). Therefore,
the product Cj C∗
j can be interpreted as the probability weight associated with eigenvalue aj (i.e.,
the probability that the outcome of an observation of ˆA will be aj ).
Hilbert-Space
According to the Expansion Postulate (together with Postulate 1), the state of a system described
by the function Ψ(x) can be expanded as a linear combination of eigenfunctions φj (x) of a linear
and hermitian operator (e.g., Ψ(x) = C1φ1(x) + C2φ2(x) + . . .). Usually, the space defined by
these eigenfunctions (i.e., functions that are continuous, single valued and square integrable) has
an infinite number of dimensions. Such space is called Hilbert-Space in honor to the mathematician
Hilbert who did pioneer work in spaces of infinite dimensionality.R4(94)
A representation of Ψ(x) in such space of functions corresponds to a vector-function,where C1 and C2 are the projections of Ψ(x) along φ1(x) and φ2(x), respectively. All other
components are omitted from the representation because they are orthogonal to the “plane” defined
by φ1(x) and φ2(x).
3 Continuous Representations
Certain operators have a continuous spectrum of eigenvalues. For example, the coordinate operator
is one such operator since it satisfies the equation ˆx δ(x0 − x) = x0 δ(x0 − x), where the eigenvalues
x0 define a continuum. Delta functions δ(x0 − x) thus define a continuous representation (the so-
called ’coordinate representation’) for which
ψ(x) =
∫
dx0Cx0 δ(x0 − x),
where Cx0 = ψ(x0), since
∫
dxδ(x − β)ψ(x) =
∫
dx
∫
dαCαδ(x − β)δ(α − x) = ψ(β).
When combined with postulates 3 and 4, the definition of the expansion coefficients Cx0 =
ψ(x0) implies that the probability of observing the system with coordinate eigenvalues between x0
and x0 + dx0 is P (x0) = Cx0 C∗
x0 dx0 = ψ(x0)ψ(x0)∗dx0 (see Note 1).
In general, eigenstates φ(α, x) with a continuum spectrum of eigenvalues α define continuous
representations,
ψ(x) =
∫
dαCαφ(α, x),
with Cα = ∫ dxφ(α, x)∗ψ(x). Delta functions and the plane waves are simply two particular
examples of basis sets with continuum spectra.
Note 2: According to the Expansion Postulate, a function ψ(x) is uniquely and completely defined
by the coefficients Cj , associated with its expansion in a complete set of eigenfunctions φj (x).
However, the coefficients of such expansion would be different if the same basis functions φj
depended on different coordinates (e.g., φj (x′) with x′ 6 = x). In order to eliminate such ambiguity
in the description it is necessary to introduce the concept of vector-ket space.R4(108)
4 Vector Space
Vector-Ket Space ε: The vector-ket space is introduced to represent states in a convenient space
of vectors |φj >, instead of working in the space of functions φj (x). The main difference is that
the coordinate dependence does not need to be specified when working in the vector-ket space.
According to such representation, function ψ(x) is the component of vector |ψ > associated withvide infra). Therefore, for any function ψ(x) = ∑
j Cj φj (x), we can define a ket-vector
|ψ > such that,
|ψ >= ∑
j
Cj |φj >.
The representation of | ψ > in space ε is,
6
-
........................................
....................
Ket-Space ε
C2
|φ2 >
C1 |φ1 >
|ψ >
Note that the expansion coefficients Cj depend only on the kets | ψj > and not on any specific
vector component. Therefore, the ambiguity mentioned above is removed.
In order to learn how to operate with kets we need to introduce the bra space and the concept of
linear functional. After doing so, this section will be concluded with the description of Postulate
5, and the Continuity Equation.
Linear functionals
A functional χ is a mathematical operation that transforms a function ψ(x) into a number. This
concept is extended to the vector-ket space ε, as an operation that transforms a vector-ket into a
number as follows,
χ(ψ(x)) = n, or χ(|ψ >) = n,
where n is a number. A linear functional satisfies the following equation,
χ(aψ(x) + bf (x)) = aχ(ψ(x)) + bχ(f (x)),
where a and b are constants.
Example: The scalar product,R4(110)
n =
∫
dxψ∗(x)φ(x),
is an example of a linear functional, since such an operation transforms a function φ(x) into a
number n. In order to introduce the scalar product of kets, we need to introduce the bra-space.
Bra Space ε∗: For every ket |ψ > we define a linear functional < ψ|, called bra-vector, as follows:
< ψ|(|φ >) = ∫ dxψ∗(x)φ(x).
12Note that functional < ψ| is linear because the scalar product is a linear functional. Therefore,
< ψ|(a|φ > +b|f >) = a < ψ|(|φ >) + b < ψ|(|f >).
Note: For convenience, we will omit parenthesis so that the notation < ψ|(|φ >) will be equivalent
to < ψ||φ >. Furthermore, whenever we find two bars next to each other we can merge them into
a single one without changing the meaning of the expression. Therefore,
< ψ||φ >=< ψ|φ > .
The space of bra-vectors is called dual space ε∗ simply because given a ket |ψ >= ∑
j Cj |φj >,
the corresponding bra-vector is < ψ| = ∑
j C∗
j < φj |. In analogy to the ket-space, a bra-vector
< ψ| is represented in space ε∗ according to the following diagram:
6
-
........................................
....................
Dual-Space ε∗
C∗
2
< φ2|
C∗
1 < φ1|
< ψ|
where C∗
j is the projection of < ψ | along < φj |.
Projection Operator and Closure Relation
Given a ket | ψ > in a certain basis set |φj >,
|ψ >= ∑
j
Cj |φj >, (1)
where < φk|φj >= δkj ,
Cj =< φj |ψ > . (2)
Substituting Eq. (2) into Eq.(1), we obtain
|ψ >= ∑
j
|φj >< φj |ψ > . (3)
From Eq.(3), it is obvious that
∑
j
|φj >< φj | = ˆ1, Closure Relation
13where ˆ1 is the identity operator that transforms any ket, or function, into itself.
Note that ˆPj = |φj >< φj | is an operator that transforms any vector |ψ > into a vector pointing
in the direction of |φj > with magnitude < φj |ψ >. The operator ˆPj is called the Projection
Operator. It projects |φj > according to,
ˆPj |ψ >=< φj |ψ > |φj > .
Note that ˆP 2
j = ˆPj , where ˆP 2
j = ˆPj ˆPj . This is true simply because < φj |φj >= 1.
4.1 Exercise 1
Prove that
i~ ∂ ˆPj
∂t = [ ˆH, ˆPj ],
where [ ˆH, ˆPj ] = ˆH ˆPj − ˆPj ˆH.
Continuity Equation
4.2 Exercise 2
Prove that ∂(ψ∗(x, t)ψ(x, t))
∂t + ∂
∂x j(x, t) = 0,
where
j(x, t) = ~
2mi
(
ψ∗(x, t)∂ψ(x, t)
∂x − ψ(x, t)∂ψ∗(x, t)
∂x
)
.
In general, for higher dimensional problems, the change in time of probability density, ρ(x, t) =
ψ∗(x, t)ψ(x, t), is equal to minus the divergence of the probability flux j,
∂ρ(x, t)
∂t = −∇ · j.
This is the so-called Continuity Equation .
Note: Remember that given a vector field j, e.g., j(x, y, z) = j1(x, y, z)ˆi+j2(x, y, z)ˆj+j3(x, y, z)ˆk,
the divergence of j is defined as the dot product of the “del” operator ∇ = ( ∂
∂x , ∂
∂y , ∂
∂z ) and vector j
as follows:
∇ · j = ∂j1
∂x + ∂j2
∂y + ∂j3
∂z .
145 Stationary States
Stationary states are states for which the probability density ρ(x, t) = ψ∗(x, t)ψ(x, t) is constant
at all times (i.e., states for which ∂ρ(x,t)
∂t = 0, and therefore ∇ · j = 0). In this section we will show
that if ψ(x, t) is factorizable according to ψ(x, t) = φ(x)f (t), then ψ(x, t) is a stationary state.
Substituting ψ(x, t) in the time dependent Schr ̈odinger equation we obtain:
φ(x)i~ ∂f (t)
∂t = −f (t) ~2
2m
∂2φ(x)
∂x2 + f (t)V (x)φ(x),
and dividing both sides by f (t)φ(x) we obtain:
i~
f (t)
∂f (t)
∂t = − ~2
2mφ(x)
∂2φ(x)
∂x2 + V (x). (4)
Since the right hand side (r.h.s) of Eq. (4) can only be a function of x and the l.h.s. can only be a
function of t for any x and t, and both functions have to be equal to each other, then such function
must be equal to a constant E. Mathematically,
i~
f (t)
∂f (t)
∂t = E ⇒ f (t) = f (0)exp(− i
~ Et),
− ~2
2mφ(x)
∂2φ(x)
∂x2 + V (x) = E ⇒ ˆHφ(x) = Eφ(x) .
The boxed equation is called the time independent Schr ̈odinger equation.
Furthermore, since f (0) is a constant, function ̃φ(x) = f (0)φ(x) also satisfies the time independent
Schr ̈odinger equation as follows,
ˆH ̃φ(x) = E ̃φ(x) , (5)
and
ψ(x, t) = ̃φ(x)exp(− i
~ Et).
Eq. (5) indicates that E is the eigenvalue of ˆH associated with the eigenfunction ̃φ(x).
5.1 Exercise 3Prove that ˆH is a Hermitian operator.
5.2 Exercise 4
Prove that -i~∂/∂x is a Hermitian operator.
155.3 Exercise 5
Prove that if two hermitian operators ˆQ and ˆP satisfy the equation ˆQ ˆP = ˆP ˆQ, i.e., if P and Q
commute (vide infra), the product operator ˆQ ˆP is also hermitian.
Since ˆH is hermitian, E is a real number ⇒ E = E∗ (see Property 1 of Hermitian operators), then,
ψ∗(x, t)ψ(x, t) = ̃φ∗(x) ̃φ(x).
Since ̃φ(x) depends only on x, ∂
∂t ( ̃φ∗(x) ̃φ(x)) = 0, then, ∂
∂t ψ∗(x, t)ψ(x, t) = 0. This demonstration
proves that if ψ(x, t) = φ(x)f (t), then ψ(x, t) is a stationary function.
6 Particle in the Box
The particle in the box can be represented by the following diagram:R1(22)
6
-
6
V (x) ∞ Box
V = ∞ V = 0 V = ∞
0 a
x
Particle
r
The goal of this section is to show that a particle with energy E and mass m in the box-potential
V(x) defined as
V (x) =
{
0, when 0 ≤ x ≤ a,
∞, otherwise,
has stationary states and a discrete absorption spectrum (i.e., the particle absorbs only certain
discrete values of energy called quanta). To that end, we first solve the equation ˆH ̃φ(x) = E ̃φ(x),
and then we obtain the stationary states ψ(x, t) = ̃φ(x)exp(− i
~ Et).
Since ̃φ(x) has to be continuous, single valued and square integrable (see Postulate 1), ̃φ(0) and
̃φ(a) must satisfy the appropriate boundary conditions both inside and outside the box. The bound-
ary conditions inside the box lead to:
− ~2
2m
∂
∂x2 Φ(x) = EΦ(x), ⇒ Φ(x) = A Sin(K x). (6)
Functions Φ(x) determine the stationary states inside the box. The boundary conditions outside the
box are,
− ~2
2m
∂
∂x2 Φ(x) + ∞Φ(x) = EΦ(x), ⇒ Φ(x) = 0,
16and determine the energy associated with Φ(x) inside the box as follows. From Eq. (6), we obtain:
~2
2m AK2 = EA, and, Φ(a) = ASin(K a) = 0,
⇒ Ka = nπ, with n = 1, 2, ... ⇒
Note that the number of nodes of Φ (i.e., the number of coordinates where Φ(x) = 0), is equal to
n − 1 for a given energy, and the energy levels are,
E = ~2
2m
n2π2
a2 , with n = 1, 2, ...
e.g.,
E(n = 1) = ~2
2m
π2
a2 ,
E(n = 2) = ~2
2m
4π2
a2 , ...
Conclusion: The energy of the particle in the box is quantized! (i.e., the absorption spectrum of
the particle in the box is not continuous but discrete).
6.1 Exercise 6
(i) Using the particle in the box model for an electron in a quantum dot (e.g., a nanometer size
silicon material) explain why larger dots emit in the red end of the spectrum, and smaller dots emit
blue or ultraviolet.
(ii) Consider the molecule hexatriene CH2 = CH − CH = CH − CH = CH2 and assume that
the 6 π electrons move freely along the molecule. Approximate the energy levels using the particle
in the box model. The length of the box is the sum of bond lengths with C-C = 1.54 ̊A, C=C = 1.35
̊A, and an extra 1.54 ̊A, due to the ends of the molecule. Assume that only 2 electrons can occupy
each electronic state and compute:
(A) The energy of the highest occupied energy level.
(B) The energy of the lowest unoccupied energy level.
17(C) The energy difference between the highest and the lowest energy levels, and compare such
energy difference with the energy of the peak in the absorption spectrum at λM AX =268nm.
(D) Predict whether the peak of the absorption spectrum for CH2 = CH − (CH = CH)n − CH =
CH2 would be red- or blue-shifted relative to the absorption spectrum of hexatriene.
7 Commutator
The commutator [ ˆA, ˆB] is defined as follows:R4(97)
[ ˆA, ˆB] = ˆA ˆB − ˆB ˆA.
Two operators ˆA and ˆB are said to commute when [ ˆA, ˆB] = 0.
7.1 Exercise 7
Prove that [ˆx, −i~ ∂
∂x ] = i~. Hint: Prove that [ˆx, −i~ ∂
∂x ]ψ(x) = i~ψ(x), where ψ(x) is a function
of x.
Note: Mathematically, we see that the momentum and position operators do not commute simply
because ˆp = −i~∂/∂x, so ˆpx = −i~(1+x∂/∂x) is not the same as xˆp = −i~x∂/∂x. Conceptually,
it means that one cannot measure the position without affecting the state of motion since measuring
the position would perturb the momentum. To measure the position of a particle it is necessary
to make it leave a mark on a sensor/detector (e.g., a piece chalk needs to leave a mark on the
blackboard to report its position). That process unavoidably slows it down, affecting its momentum.
8 Uncertainty Relations
The goal of this section is to show that the uncertainties ∆A =
√
< ( ˆA− < ˆA >)2 > and ∆B =
√
< ( ˆB− < ˆB >)2 >, of any pair of hermitian operators ˆA and ˆB, satisfy the uncertainty rela-
tion:R3(437)
(∆A)2(∆B)2 ≥ 1
4 < i[A, B] >2 . (7)
In particular, when ˆA = ˆx and ˆB = ˆp, we obtain the Heisenberg uncertainty relation :
∆x · ∆p ≥ ~
2 . (8)
Proof:
ˆU ≡ ˆA− < A >, φ(λ, x) ≡ ( ˆU + iλ ˆV )Φ(x),
ˆV ≡ ˆB− < B >, I(λ) ≡ ∫ dxφ∗(λ, x)φ(λ, x) ≥ 0,
18I(λ) =
∫
dx[( ˆA− < A >)Φ(x) + iλ( ˆB− < B >)Φ(x)]∗[( ˆA− < A >)Φ(x) + iλ( ˆB− < B >)Φ(x)],
I(λ) =< Φ|U 2|Φ > +λ2 < Φ|V 2|Φ > −iλ < Φ|U V − V U |Φ >≥ 0, (9)
The minimum value of I(λ), as a function of λ, is reached when ∂I/∂λ = ∂I/∂λ∗ = 0.
This condition implies that
2λ(∆B)2 = i < [A, B] >, => λ = i < [A, B] >
2(∆B)2 .
Substituting this expression for λ into Eq. (9), we obtain:
(∆A)2 + i2 < A, B >2
4(∆B)2 − i2 < A, B >2
2(∆B)2 ≥ 0,
(∆A)2(∆B)2 ≥ i2 < A, B >2
4 .
8.1 Exercise 8
Compute < X >, < P >, ∆X and ∆P for the particle in the box in its minimum energy state and
verify that ∆X and ∆P satisfy the uncertainty relation given by Eq. (7)?
8.2 EPR Paradox
Gedankenexperiments (i.e., thought experiments) have been proposed to determine “hidden” vari-
ables. The most famous of these proposals has been the Einstein-Podolski-Rosen (EPR) gedanken-
experiment [Phys. Rev. (1935) 47:777-780] , where a system of 2 particles is initially prepared
with total momentum pt. At a later time, when the two particles are far apart from each other, the
position x1 is measured on particle 1 and the momentum p2 is measured on particle 2. The paradox
is that the momentum of particle 1 could be obtained from the difference p1 = pt − p2. Therefore,
the coordinate x1 and momentum p1 of particle 1 could be determined with more precision than
established as possible by the uncertainty principle, so long as the separation between the two par-
ticles could prevent any kind of interaction or disturbance of one particule due to a measurement
on the other.
The origin of the paradox is the erroneous assumption that particles that are far apart from
each other cannot maintain instantaneous correlations. However, quantum correlations between
the properties of distant noninteracting systems can be maintained, as described by Bohm and
Aharonov [Phys. Rev. (1957) 108:1070-1076] for the state of polarization of pairs of correlated
photons. Within the Bohmian picture of quantum mechanics, these quantum correlations are estab-
lished by the quantum potential VQ(q), even when the particles are noninteracting (i.e., V (q) = 0).
19
@Over @subhuman incel
based on experimental observations). The goal of this section is to introduce such principles, to-
gether with some mathematical concepts that are necessary for that purpose. To keep the notation
as simple as possible, expressions are written for a 1-dimensional system. The generalization to
many dimensions is usually straightforward.
P ostulate 1 : Any system in a pure state can be described by a wave-function , ψ(t, x), where t is
a parameter representing the time and x represents the coordinates of the system. Such a function
ψ(t, x) must be continuous, single valued and square integrable.
Note 1: As a consequence of Postulate 4, we will see that P (t, x) = ψ∗(t, x)ψ(t, x)dx represents
the probability of finding the system between x and x + dx at time t, first realized by Max Born. 2
P ostulate 2 : Any observable (i.e., any measurable property of the system) can be described by
an operator. The operator must be linear and hermitian.
What is an operator ? What is a linear operator ? What is a hermitian operator?
Definition 1: An operator ˆO is a mathematical entity that transforms a function f (x) into another
function g(x) as follows, R4(96) ˆOf (x ) = g(x ),
where f and g are functions of x.
Definition 2: An operator ˆO that represents an observable O is obtained by first writing the clas-
sical expression of such observable in Cartesian coordinates (e.g., O = O(x, p)) and then substi-
tuting the coordinate x in such expression by the coordinate operator ˆx as well as the momentum p
by the momentum operator ˆp = −i~∂/∂x.
Definition 3: An operator ˆO is linear if and only if (iff),
ˆO(af (x) + bg(x)) = a ˆOf (x) + b ˆOg(x),
where a and b are constants.
Definition 4: An operator ˆO is hermitian iff,
∫
dxφ∗
n(x) ˆOψm(x) =
[∫
dxψ∗
m(x) ˆOφn(x)
]∗
,
2Note that this probabilistic interpretation of ψ has profound implications to our understanding of reality. It es-
sentially reduces the objective reality to P (t, x). All other properties are no longer independent of the process of
measurement by the observer. A. Pais’ anecdote of his conversation with A. Einstein while walking at Princeton em-
phasizes the apparent absurdity of the implications: [Rev. Mod. Phys. 51, 863914 (1979), p. 907]: ’We often discussed
his notions on objective reality. I recall that during one walk Einstein suddenly stopped, turned to me and asked whether
I really believed that the moon exists only when I look at it.’
8
where the asterisk represents the complex conjugate.
Definition 5: A function φn(x) is an eigenfunction of ˆO iff,
ˆOφn(x) = Onφn(x),
where On is a number called eigenvalue.
Property 1: The eigenvalues of a hermitian operator are real.
Proof: Using Definition 4, we obtain
∫
dxφ∗
n(x) ˆOφn(x) −
[∫
dxφ∗
n(x) ˆOφn(x)
]∗
= 0,
therefore,
[On − O∗
n]
∫
dxφn(x)∗φn(x) = 0.
Since φn(x) are square integrable functions, then,
On = O∗
n.
Property 2: Different eigenfunctions of a hermitian operator (i.e., eigenfunctions with different
eigenvalues) are orthogonal (i.e., the scalar product of two different eigenfunctions is equal to
zero). Mathematically, if ˆOφn = Onφn, and ˆOφm = Omφm, with On 6 = Om, then ∫ dxφ∗
nφm = 0.
Proof: ∫
dxφ∗
m ˆOφn −
[∫
dxφ∗
n ˆOφm
]∗
= 0,
and
[On − Om]
∫
dxφ∗
mφn = 0.
Since On 6 = Om, then ∫ dxφ∗
mφn = 0.
P ostulate 3 : The only possible experimental results of a measurement of an observable are the
eigenvalues of the operator that corresponds to such observable.
P ostulate 4 : The average value of many measurements of an observable O, when the system is
described by ψ(x) as equal to the expectation value ̄O, which is defined as follows,
̄O =
∫ dxψ(x)∗ ˆOψ(x)
∫ dxψ(x)∗ψ(x) .
P ostulate 5 :The evolution of ψ(x, t) in time is described by the time-dependent Schr ̈odinger
equation :
9
i~ ∂ψ(x, t)
∂t = ˆHψ(x, t),
where ˆH = − ~2
2m
∂2
∂x2 + ˆV (x), is the operator associated with the total energy of the system, E =
p2
2m + V (x).
Expansion P ostulate : R5(15), R4(97)
The eigenfunctions of a linear and hermitian operator form a complete basis set. Therefore,
any function ψ(x) that is continuous, single valued, and square integrable can be expanded as a
linear combination of eigenfunctions φn(x) of a linear and hermitian operator ˆA as follows,
ψ(x) = ∑
j
Cj φj (x),
where Cj are numbers (e.g., complex numbers ) called expansion coefficients.
Note that ̄A = ∑
j Cj C∗
j aj , when ψ(x) = ∑
j Cj φj (x),
ˆAφj (x) = aj φj (x), and
∫
dxφj (x)∗φk(x) = δjk.
This is because the eigenvalues aj are the only possible experimental results of measurements of ˆA
(according to Postulate 3), and the expectation value ̄A is the average value of many measurements
of ˆA when the system is described by the expansion ψ(x) = ∑
j Cj φj (x) (Postulate 4). Therefore,
the product Cj C∗
j can be interpreted as the probability weight associated with eigenvalue aj (i.e.,
the probability that the outcome of an observation of ˆA will be aj ).
Hilbert-Space
According to the Expansion Postulate (together with Postulate 1), the state of a system described
by the function Ψ(x) can be expanded as a linear combination of eigenfunctions φj (x) of a linear
and hermitian operator (e.g., Ψ(x) = C1φ1(x) + C2φ2(x) + . . .). Usually, the space defined by
these eigenfunctions (i.e., functions that are continuous, single valued and square integrable) has
an infinite number of dimensions. Such space is called Hilbert-Space in honor to the mathematician
Hilbert who did pioneer work in spaces of infinite dimensionality.R4(94)
A representation of Ψ(x) in such space of functions corresponds to a vector-function,where C1 and C2 are the projections of Ψ(x) along φ1(x) and φ2(x), respectively. All other
components are omitted from the representation because they are orthogonal to the “plane” defined
by φ1(x) and φ2(x).
3 Continuous Representations
Certain operators have a continuous spectrum of eigenvalues. For example, the coordinate operator
is one such operator since it satisfies the equation ˆx δ(x0 − x) = x0 δ(x0 − x), where the eigenvalues
x0 define a continuum. Delta functions δ(x0 − x) thus define a continuous representation (the so-
called ’coordinate representation’) for which
ψ(x) =
∫
dx0Cx0 δ(x0 − x),
where Cx0 = ψ(x0), since
∫
dxδ(x − β)ψ(x) =
∫
dx
∫
dαCαδ(x − β)δ(α − x) = ψ(β).
When combined with postulates 3 and 4, the definition of the expansion coefficients Cx0 =
ψ(x0) implies that the probability of observing the system with coordinate eigenvalues between x0
and x0 + dx0 is P (x0) = Cx0 C∗
x0 dx0 = ψ(x0)ψ(x0)∗dx0 (see Note 1).
In general, eigenstates φ(α, x) with a continuum spectrum of eigenvalues α define continuous
representations,
ψ(x) =
∫
dαCαφ(α, x),
with Cα = ∫ dxφ(α, x)∗ψ(x). Delta functions and the plane waves are simply two particular
examples of basis sets with continuum spectra.
Note 2: According to the Expansion Postulate, a function ψ(x) is uniquely and completely defined
by the coefficients Cj , associated with its expansion in a complete set of eigenfunctions φj (x).
However, the coefficients of such expansion would be different if the same basis functions φj
depended on different coordinates (e.g., φj (x′) with x′ 6 = x). In order to eliminate such ambiguity
in the description it is necessary to introduce the concept of vector-ket space.R4(108)
4 Vector Space
Vector-Ket Space ε: The vector-ket space is introduced to represent states in a convenient space
of vectors |φj >, instead of working in the space of functions φj (x). The main difference is that
the coordinate dependence does not need to be specified when working in the vector-ket space.
According to such representation, function ψ(x) is the component of vector |ψ > associated withvide infra). Therefore, for any function ψ(x) = ∑
j Cj φj (x), we can define a ket-vector
|ψ > such that,
|ψ >= ∑
j
Cj |φj >.
The representation of | ψ > in space ε is,
6
-
........................................
....................
Ket-Space ε
C2
|φ2 >
C1 |φ1 >
|ψ >
Note that the expansion coefficients Cj depend only on the kets | ψj > and not on any specific
vector component. Therefore, the ambiguity mentioned above is removed.
In order to learn how to operate with kets we need to introduce the bra space and the concept of
linear functional. After doing so, this section will be concluded with the description of Postulate
5, and the Continuity Equation.
Linear functionals
A functional χ is a mathematical operation that transforms a function ψ(x) into a number. This
concept is extended to the vector-ket space ε, as an operation that transforms a vector-ket into a
number as follows,
χ(ψ(x)) = n, or χ(|ψ >) = n,
where n is a number. A linear functional satisfies the following equation,
χ(aψ(x) + bf (x)) = aχ(ψ(x)) + bχ(f (x)),
where a and b are constants.
Example: The scalar product,R4(110)
n =
∫
dxψ∗(x)φ(x),
is an example of a linear functional, since such an operation transforms a function φ(x) into a
number n. In order to introduce the scalar product of kets, we need to introduce the bra-space.
Bra Space ε∗: For every ket |ψ > we define a linear functional < ψ|, called bra-vector, as follows:
< ψ|(|φ >) = ∫ dxψ∗(x)φ(x).
12Note that functional < ψ| is linear because the scalar product is a linear functional. Therefore,
< ψ|(a|φ > +b|f >) = a < ψ|(|φ >) + b < ψ|(|f >).
Note: For convenience, we will omit parenthesis so that the notation < ψ|(|φ >) will be equivalent
to < ψ||φ >. Furthermore, whenever we find two bars next to each other we can merge them into
a single one without changing the meaning of the expression. Therefore,
< ψ||φ >=< ψ|φ > .
The space of bra-vectors is called dual space ε∗ simply because given a ket |ψ >= ∑
j Cj |φj >,
the corresponding bra-vector is < ψ| = ∑
j C∗
j < φj |. In analogy to the ket-space, a bra-vector
< ψ| is represented in space ε∗ according to the following diagram:
6
-
........................................
....................
Dual-Space ε∗
C∗
2
< φ2|
C∗
1 < φ1|
< ψ|
where C∗
j is the projection of < ψ | along < φj |.
Projection Operator and Closure Relation
Given a ket | ψ > in a certain basis set |φj >,
|ψ >= ∑
j
Cj |φj >, (1)
where < φk|φj >= δkj ,
Cj =< φj |ψ > . (2)
Substituting Eq. (2) into Eq.(1), we obtain
|ψ >= ∑
j
|φj >< φj |ψ > . (3)
From Eq.(3), it is obvious that
∑
j
|φj >< φj | = ˆ1, Closure Relation
13where ˆ1 is the identity operator that transforms any ket, or function, into itself.
Note that ˆPj = |φj >< φj | is an operator that transforms any vector |ψ > into a vector pointing
in the direction of |φj > with magnitude < φj |ψ >. The operator ˆPj is called the Projection
Operator. It projects |φj > according to,
ˆPj |ψ >=< φj |ψ > |φj > .
Note that ˆP 2
j = ˆPj , where ˆP 2
j = ˆPj ˆPj . This is true simply because < φj |φj >= 1.
4.1 Exercise 1
Prove that
i~ ∂ ˆPj
∂t = [ ˆH, ˆPj ],
where [ ˆH, ˆPj ] = ˆH ˆPj − ˆPj ˆH.
Continuity Equation
4.2 Exercise 2
Prove that ∂(ψ∗(x, t)ψ(x, t))
∂t + ∂
∂x j(x, t) = 0,
where
j(x, t) = ~
2mi
(
ψ∗(x, t)∂ψ(x, t)
∂x − ψ(x, t)∂ψ∗(x, t)
∂x
)
.
In general, for higher dimensional problems, the change in time of probability density, ρ(x, t) =
ψ∗(x, t)ψ(x, t), is equal to minus the divergence of the probability flux j,
∂ρ(x, t)
∂t = −∇ · j.
This is the so-called Continuity Equation .
Note: Remember that given a vector field j, e.g., j(x, y, z) = j1(x, y, z)ˆi+j2(x, y, z)ˆj+j3(x, y, z)ˆk,
the divergence of j is defined as the dot product of the “del” operator ∇ = ( ∂
∂x , ∂
∂y , ∂
∂z ) and vector j
as follows:
∇ · j = ∂j1
∂x + ∂j2
∂y + ∂j3
∂z .
145 Stationary States
Stationary states are states for which the probability density ρ(x, t) = ψ∗(x, t)ψ(x, t) is constant
at all times (i.e., states for which ∂ρ(x,t)
∂t = 0, and therefore ∇ · j = 0). In this section we will show
that if ψ(x, t) is factorizable according to ψ(x, t) = φ(x)f (t), then ψ(x, t) is a stationary state.
Substituting ψ(x, t) in the time dependent Schr ̈odinger equation we obtain:
φ(x)i~ ∂f (t)
∂t = −f (t) ~2
2m
∂2φ(x)
∂x2 + f (t)V (x)φ(x),
and dividing both sides by f (t)φ(x) we obtain:
i~
f (t)
∂f (t)
∂t = − ~2
2mφ(x)
∂2φ(x)
∂x2 + V (x). (4)
Since the right hand side (r.h.s) of Eq. (4) can only be a function of x and the l.h.s. can only be a
function of t for any x and t, and both functions have to be equal to each other, then such function
must be equal to a constant E. Mathematically,
i~
f (t)
∂f (t)
∂t = E ⇒ f (t) = f (0)exp(− i
~ Et),
− ~2
2mφ(x)
∂2φ(x)
∂x2 + V (x) = E ⇒ ˆHφ(x) = Eφ(x) .
The boxed equation is called the time independent Schr ̈odinger equation.
Furthermore, since f (0) is a constant, function ̃φ(x) = f (0)φ(x) also satisfies the time independent
Schr ̈odinger equation as follows,
ˆH ̃φ(x) = E ̃φ(x) , (5)
and
ψ(x, t) = ̃φ(x)exp(− i
~ Et).
Eq. (5) indicates that E is the eigenvalue of ˆH associated with the eigenfunction ̃φ(x).
5.1 Exercise 3Prove that ˆH is a Hermitian operator.
5.2 Exercise 4
Prove that -i~∂/∂x is a Hermitian operator.
155.3 Exercise 5
Prove that if two hermitian operators ˆQ and ˆP satisfy the equation ˆQ ˆP = ˆP ˆQ, i.e., if P and Q
commute (vide infra), the product operator ˆQ ˆP is also hermitian.
Since ˆH is hermitian, E is a real number ⇒ E = E∗ (see Property 1 of Hermitian operators), then,
ψ∗(x, t)ψ(x, t) = ̃φ∗(x) ̃φ(x).
Since ̃φ(x) depends only on x, ∂
∂t ( ̃φ∗(x) ̃φ(x)) = 0, then, ∂
∂t ψ∗(x, t)ψ(x, t) = 0. This demonstration
proves that if ψ(x, t) = φ(x)f (t), then ψ(x, t) is a stationary function.
6 Particle in the Box
The particle in the box can be represented by the following diagram:R1(22)
6
-
6
V (x) ∞ Box
V = ∞ V = 0 V = ∞
0 a
x
Particle
r
The goal of this section is to show that a particle with energy E and mass m in the box-potential
V(x) defined as
V (x) =
{
0, when 0 ≤ x ≤ a,
∞, otherwise,
has stationary states and a discrete absorption spectrum (i.e., the particle absorbs only certain
discrete values of energy called quanta). To that end, we first solve the equation ˆH ̃φ(x) = E ̃φ(x),
and then we obtain the stationary states ψ(x, t) = ̃φ(x)exp(− i
~ Et).
Since ̃φ(x) has to be continuous, single valued and square integrable (see Postulate 1), ̃φ(0) and
̃φ(a) must satisfy the appropriate boundary conditions both inside and outside the box. The bound-
ary conditions inside the box lead to:
− ~2
2m
∂
∂x2 Φ(x) = EΦ(x), ⇒ Φ(x) = A Sin(K x). (6)
Functions Φ(x) determine the stationary states inside the box. The boundary conditions outside the
box are,
− ~2
2m
∂
∂x2 Φ(x) + ∞Φ(x) = EΦ(x), ⇒ Φ(x) = 0,
16and determine the energy associated with Φ(x) inside the box as follows. From Eq. (6), we obtain:
~2
2m AK2 = EA, and, Φ(a) = ASin(K a) = 0,
⇒ Ka = nπ, with n = 1, 2, ... ⇒
Note that the number of nodes of Φ (i.e., the number of coordinates where Φ(x) = 0), is equal to
n − 1 for a given energy, and the energy levels are,
E = ~2
2m
n2π2
a2 , with n = 1, 2, ...
e.g.,
E(n = 1) = ~2
2m
π2
a2 ,
E(n = 2) = ~2
2m
4π2
a2 , ...
Conclusion: The energy of the particle in the box is quantized! (i.e., the absorption spectrum of
the particle in the box is not continuous but discrete).
6.1 Exercise 6
(i) Using the particle in the box model for an electron in a quantum dot (e.g., a nanometer size
silicon material) explain why larger dots emit in the red end of the spectrum, and smaller dots emit
blue or ultraviolet.
(ii) Consider the molecule hexatriene CH2 = CH − CH = CH − CH = CH2 and assume that
the 6 π electrons move freely along the molecule. Approximate the energy levels using the particle
in the box model. The length of the box is the sum of bond lengths with C-C = 1.54 ̊A, C=C = 1.35
̊A, and an extra 1.54 ̊A, due to the ends of the molecule. Assume that only 2 electrons can occupy
each electronic state and compute:
(A) The energy of the highest occupied energy level.
(B) The energy of the lowest unoccupied energy level.
17(C) The energy difference between the highest and the lowest energy levels, and compare such
energy difference with the energy of the peak in the absorption spectrum at λM AX =268nm.
(D) Predict whether the peak of the absorption spectrum for CH2 = CH − (CH = CH)n − CH =
CH2 would be red- or blue-shifted relative to the absorption spectrum of hexatriene.
7 Commutator
The commutator [ ˆA, ˆB] is defined as follows:R4(97)
[ ˆA, ˆB] = ˆA ˆB − ˆB ˆA.
Two operators ˆA and ˆB are said to commute when [ ˆA, ˆB] = 0.
7.1 Exercise 7
Prove that [ˆx, −i~ ∂
∂x ] = i~. Hint: Prove that [ˆx, −i~ ∂
∂x ]ψ(x) = i~ψ(x), where ψ(x) is a function
of x.
Note: Mathematically, we see that the momentum and position operators do not commute simply
because ˆp = −i~∂/∂x, so ˆpx = −i~(1+x∂/∂x) is not the same as xˆp = −i~x∂/∂x. Conceptually,
it means that one cannot measure the position without affecting the state of motion since measuring
the position would perturb the momentum. To measure the position of a particle it is necessary
to make it leave a mark on a sensor/detector (e.g., a piece chalk needs to leave a mark on the
blackboard to report its position). That process unavoidably slows it down, affecting its momentum.
8 Uncertainty Relations
The goal of this section is to show that the uncertainties ∆A =
√
< ( ˆA− < ˆA >)2 > and ∆B =
√
< ( ˆB− < ˆB >)2 >, of any pair of hermitian operators ˆA and ˆB, satisfy the uncertainty rela-
tion:R3(437)
(∆A)2(∆B)2 ≥ 1
4 < i[A, B] >2 . (7)
In particular, when ˆA = ˆx and ˆB = ˆp, we obtain the Heisenberg uncertainty relation :
∆x · ∆p ≥ ~
2 . (8)
Proof:
ˆU ≡ ˆA− < A >, φ(λ, x) ≡ ( ˆU + iλ ˆV )Φ(x),
ˆV ≡ ˆB− < B >, I(λ) ≡ ∫ dxφ∗(λ, x)φ(λ, x) ≥ 0,
18I(λ) =
∫
dx[( ˆA− < A >)Φ(x) + iλ( ˆB− < B >)Φ(x)]∗[( ˆA− < A >)Φ(x) + iλ( ˆB− < B >)Φ(x)],
I(λ) =< Φ|U 2|Φ > +λ2 < Φ|V 2|Φ > −iλ < Φ|U V − V U |Φ >≥ 0, (9)
The minimum value of I(λ), as a function of λ, is reached when ∂I/∂λ = ∂I/∂λ∗ = 0.
This condition implies that
2λ(∆B)2 = i < [A, B] >, => λ = i < [A, B] >
2(∆B)2 .
Substituting this expression for λ into Eq. (9), we obtain:
(∆A)2 + i2 < A, B >2
4(∆B)2 − i2 < A, B >2
2(∆B)2 ≥ 0,
(∆A)2(∆B)2 ≥ i2 < A, B >2
4 .
8.1 Exercise 8
Compute < X >, < P >, ∆X and ∆P for the particle in the box in its minimum energy state and
verify that ∆X and ∆P satisfy the uncertainty relation given by Eq. (7)?
8.2 EPR Paradox
Gedankenexperiments (i.e., thought experiments) have been proposed to determine “hidden” vari-
ables. The most famous of these proposals has been the Einstein-Podolski-Rosen (EPR) gedanken-
experiment [Phys. Rev. (1935) 47:777-780] , where a system of 2 particles is initially prepared
with total momentum pt. At a later time, when the two particles are far apart from each other, the
position x1 is measured on particle 1 and the momentum p2 is measured on particle 2. The paradox
is that the momentum of particle 1 could be obtained from the difference p1 = pt − p2. Therefore,
the coordinate x1 and momentum p1 of particle 1 could be determined with more precision than
established as possible by the uncertainty principle, so long as the separation between the two par-
ticles could prevent any kind of interaction or disturbance of one particule due to a measurement
on the other.
The origin of the paradox is the erroneous assumption that particles that are far apart from
each other cannot maintain instantaneous correlations. However, quantum correlations between
the properties of distant noninteracting systems can be maintained, as described by Bohm and
Aharonov [Phys. Rev. (1957) 108:1070-1076] for the state of polarization of pairs of correlated
photons. Within the Bohmian picture of quantum mechanics, these quantum correlations are estab-
lished by the quantum potential VQ(q), even when the particles are noninteracting (i.e., V (q) = 0).
19
@Over @subhuman incel
