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Chapter 1
Wave Function, Schrödinger equation and Postulates of quantum mechanics
In classical mechanics we have separate equations for wave motion and particle motion, whereas in
quantum mechanics, in which the distinction between particles and waves is not clear-cut, we have a
single equation—the Schr¨odinger equation. We have seen that the link between the Schrödinger
equation and the classical wave equation is the de Broglie relation.
If there is a wave associated with a particle, then there must be a function to represent it. This function
is called wave function.
Schrödinger in 1926 first proposed an equation for the broglies matter waves. This equation can not be
drived from other principles since it constitutes fundamental law of nature.
Before we start, we must agree on the meaning of words and on the methodology. Words can be traps
when discussing phenomena that are so new and unusual. We constantly use the following words:
physical system, state,wave function(eigenfunction, eigen state), eigenvalue, physical quantities,
observables.
In the previous chapter, we have described wave-particle duality and uncertainty principle.
Schrödinger and Born understood in 1926 that the complete description of a particle in space at time t
is performed with a complex wave function Ψ(𝑥, 𝑡).
The wave function can be obtained by solving Schrödinger equation:
𝑖ℏ
𝜕Ψ
ℏ2 𝜕 2 Ψ
=−
+ 𝑉(𝑥)Ψ
𝜕𝑡
2𝑚 𝜕𝑥 2
ℎ
where 𝑖 is complex number 𝑖 2 = −1, ℏ = 2𝜋, is reduced Planck's constant, V(x) is potential.
Physical interpretation of wave function
In classical physics momentum p and position x can be used to describe a particles physical quantities.
Position and momentum is function of time and at a certain time particles position or momentum can
exactly be determined. In quantum physics particles have wave properties and they are described by a
wave function. Therefore position of the particle is not function of time.
In the classical wave theory intensity of a wave is proportional to the square of its wave function. For
the photon the intesity is interepted as the probability of finding a photon in the given space.
According to Born's interpretation, probability of finding of an electron (particle) described by the
wave function Ψ(𝑥, 𝑡), in region [a,b] is given by
𝑏
𝑃 = ∫ |Ψ(𝑥, 𝑡)|2 𝑑𝑥
𝑎
where |Ψ(𝑥, 𝑡)|2 = Ψ(𝑥, 𝑡)∗ Ψ(𝑥, 𝑡). Ψ(𝑥, 𝑡)∗ is complex conjufate of the wave function. Probability
interpretation allow us to ynderstand electron interference.
Normalization
Total probability of finding the particle somewhere along x-axis is
𝑃 = ∫ |Ψ(𝑥, 𝑡)|2 𝑑𝑥
𝑥
If the particle exists , it must be somewhere on the x-axis . so the total probability of finding the
particle must be unity:
𝑃 = ∫ |Ψ(𝑥, 𝑡)|2 𝑑𝑥 = 1
𝑎𝑙𝑙
This is the normalization constant.
Example. The wave function for a particle confined to 0 ≤ 𝑥 ≤ 𝑎 in the ground state was
found to be:
𝜋𝑥
𝜓(𝑥) = 𝐴𝑠𝑖𝑛 ( )
𝑎
where A is the normalization constant. Find A.
Solution:
Using normalization condition
𝑎
𝑃 = ∫ |Ψ(𝑥, 𝑡)|2 𝑑𝑥 = 1
0
we can obtain
𝑎
𝜋𝑥
𝑃 = ∫ 𝐴2 sin2 ( ) 𝑑𝑥 = 1
𝑎
0
evaluate the integral, then you can obtain
𝐴=√
2
𝑎
Orthogonal state
The wave functions Ψ𝑛 (𝑥, 𝑡) 𝑎𝑛𝑑 Ψ𝑚 (𝑥, 𝑡) of a particle corresponding two different state satisfy the
orthogonality relation
∫ Ψ𝑛 (𝑥, 𝑡)∗ Ψ𝑚 (𝑥, 𝑡)𝑑𝑥 = 0, 𝑖𝑓 𝑚 ≠ 𝑛.
𝑎𝑙𝑙
If the wavefunctions are normalized then above relation is said to be ortonormal state.
Properties of a wave function
A wavefunction Ψ(𝑥, 𝑡) must:
a) be single valued
b) be continuous
c) be differentiable
d) be square integrable
A wave function Ψ(𝑥, 𝑡) must be
a) Single valued.
A single-valued function is function that, for each point in the domain, has a unique value in the range.
It is therefore one-to-one or many-to-one. At every point in space have only one value associated with
it (single valued). If a wave function was not single valued, it would have multiple values for the same
position, thereby ruining the probabilistic interpretation of the wave function.
b) Continuous
The function has finite value at any point in the given space. Otherwise it implies that particle does not
exist at some points in the corresponding space.
c) Differentiable
Derivative of wave function is related to the flow of the particles. Therefore particle flow in the
corresponding space should be continuous.
d) Square integrable
The wave function contains information about where the particle is located, its square being
probability density. Therefore
∞
∫ |Ψ(𝑥, 𝑡)|2 𝑑𝑥 < ∞
−∞
Where |Ψ(𝑥, 𝑡)|2 = Ψ(𝑥, 𝑡)Ψ(𝑥, 𝑡)∗ . Ψ(𝑥, 𝑡)∗ is conjugate of the wave function.
𝑛𝜋
Example The wave function of a particle in the infinite well is obtained 𝜓(𝑥) = 𝐴𝑠𝑖𝑛( 𝑎 𝑥) and
defined 0 ≤ 𝑥 ≤ 𝑎. Is it valid wave function?
Solution:
The function is single valued, continuous, differentiable and square integrable if the constant A is
finite. Therefore the function is a valid wave function.
Example. Let two functions 𝜙 = 𝐴𝑥 and 𝜓 = 𝐴𝑒 −𝑎𝑥 be defined for 0 ≤ 𝑥 ≤ ∞. Explain why 𝜙 = 𝐴𝑥
cannot be a wavefunction but 𝜓 = 𝐴𝑒 −𝑎𝑥 could be a valid wavefunction.
Solution
Both functions are continuous and defined on the interval of interest. They are both
single valued and differentiable. However, consider the integral of x:
∞
∞
𝐴2 𝑥 3
∫ 𝐴 𝑥 𝑑𝑥 =
| → ∞.
2 0
0
2 2
is not square integrable over this range it cannot be a valid wavefunction.
Postulate 1. Quantum State postulate of quantum mechanics
Every physically realizable state of system is described in quantum mechanics by a state
function 𝚿 that contains all accessible physical information about the system in that state.
Conservation of probability and probability current
A continuity equation in physics is a differential equation that describes the transport of some kind of
conserved quantity. Since mass, energy, momentum, electric charge and other natural quantities are
conserved, a vast variety of physics may be described with continuity equations.
Continuity equations are the (stronger) local form of conservation laws. All the examples of continuity
equations below express the same idea, which is roughly that: the total amount (of the conserved
quantity) inside any region can only change by the amount that passes in or out of the region through
the boundary. A conserved quantity cannot increase or decrease, it can only move from place to place.
In quantum mechanics, the probability current (sometimes called probability flux) is a concept
describing the flow of probability density. In particular, if one pictures the probability density as an
inhomogeneous fluid, then the probability current is the rate of flow of this fluid (the density times the
velocity).
The continuity equation is derived from the definition of probability current and the basic principles of
quantum mechanics. The probability is given by:
𝑝 = ∫|Ψ(𝑥, 𝑡)|2 𝑑𝑥 = ∫ 𝜌𝑑𝑥
And it is the probability that a measurement of the particle's position. The time derivative of this is
𝑑𝜌
𝜕Ψ ∗
𝜕Ψ∗
=(
Ψ +Ψ
)
𝑑𝑡
𝜕𝑡
𝜕𝑡
where the last equality follows from the product rule |Ψ|2 = ΨΨ∗ = 𝜌. In order to simplify this further
consider the time dependent Schrödinger equation
𝑖ℏ
𝜕Ψ
ℏ2 𝜕 2 Ψ
=−
+ 𝑉Ψ
𝜕𝑡
2𝑚 𝜕𝑥 2
Or
𝜕Ψ∗
ℏ2 𝜕 2 Ψ∗
−𝑖ℏ
=−
+ 𝑉Ψ∗
𝜕𝑡
2𝑚 𝜕𝑥 2
Eliminating
𝜕Ψ
𝜕𝑡
and
𝜕Ψ∗
𝜕𝑡
we obtain
𝑑𝜌
ℏ 𝜕2Ψ ∗
𝜕 2 Ψ∗
ℏ 𝜕 𝜕Ψ ∗
𝜕Ψ∗
=−
Ψ −Ψ
( 2 Ψ −Ψ 2 )=−
(
)
𝑑𝑡
2𝑖𝑚 𝜕 𝑥
𝜕 𝑥
2𝑖𝑚 𝜕𝑥 𝜕𝑥
𝜕𝑥
Comparing to the continuity equation we define:
𝑗(𝑥, 𝑡) =
ℏ 𝜕Ψ ∗
𝜕Ψ∗
Ψ −Ψ
(
)
2𝑖𝑚 𝜕𝑥
𝜕𝑥
is called probability current density. Then the continuity equation for probability:
𝜕𝜌
𝜕𝑗
=−
𝜕𝑡
𝜕𝑥
In three dimensions (3D) it can be generalised as:
𝜕𝜌
= −∇. 𝐽⃗
𝜕𝑡
Or
𝜕|Ψ|2
= −∇. 𝐽⃗
𝜕𝑡
This equation is called Cotinuity equation.
Schrödinger’s Equation for a Particle in a Potential
Let us consider first the one-dimensional situation of a particle going in the x-direction subject to a
“roller coaster” potential. What do we expect the wave function to look like? We would expect the
wavelength to be shortest where the potential is lowest, in the valleys, because that’s where the
particle is going fastest—maximum momentum. Perhaps slightly less obvious is that the amplitude of
the wave would be largest at the tops of the hills (provided the particle has enough energy to get there)
because that’s where the particle is moving slowest, and therefore is most likely to be found.
With a nonzero potential present, the energy-momentum relationship for the particle becomes the
energy equation
𝑝2
+ 𝑉(𝑥)
2𝑚
We need to construct a wave equation which leads naturally to this relationship. In contrast to the free
particle cases discussed above, the relevant wave function here will no longer be a plane wave, since
the wavelength varies with the potential. However, at a given x, the momentum is determined by the
“local wavelength”, that is,
𝐸=
𝑝 = −𝑖ℏ
𝜕𝜓
𝜕𝑥
It follows that the appropriate wave equation is:
𝜕Ψ
ℏ2 𝜕 2 Ψ
=−
+ 𝑉(𝑥)Ψ
𝜕𝑡
2𝑚 𝜕𝑥 2
This is the standard one-dimensional Schrödinger equation.
𝑖ℏ
In three dimensions, the argument is precisely analogous. The only difference is that the square of the
momentum is now a sum of three squared components, for the x, y and z directions, so
𝜕2
𝜕2
𝜕2
𝜕2
𝑏𝑒𝑐𝑜𝑚𝑒𝑠
+
+
= ∇2
𝜕𝑥 2
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
and the equation is:
𝑖ℏ
𝜕Ψ
ℏ2 2
=−
∇ Ψ + 𝑉(𝑥, 𝑦, 𝑧)Ψ
𝜕𝑡
2𝑚
This is the complete Schrödinger equation.
Postulate 2. Time evaluation postulate of quantum mechanics
The wavefunction or state function of a system evolves in time according to the timedependent Schrödinger equation
𝑖ℏ
𝜕Ψ
ℏ2 2
=−
∇ Ψ + 𝑉(𝑥, 𝑦, 𝑧)Ψ
𝜕𝑡
2𝑚
The central equation of quantum mechanics must be accepted as a postulate.
Time independent Schrödinger equation
If the potential time independent then time part of the Schrödinger equation can be seperated by
substituting
Ψ(𝑥, 𝑡) = 𝜓(𝑥)𝑓(𝑡)
𝑖𝐸𝑡
And defining 𝑓(𝑡) = 𝑒 − ℏ , the time dependent Schrödinger equation takes the form:
ℏ2 𝜕 2 𝜓(𝑥)
+ 𝑉(𝑥)𝜓(𝑥) = 𝐸𝜓(𝑥).
2𝑚 𝜕𝑥 2
Here 𝜓(𝑥) is named as wave function or eigen function and E is energy or eigenvalue. The equation
also called eigen value equation. The equation can also be written as
−
𝐻𝜓 = 𝐸𝜓
ℏ2 𝜕2
Where 𝐻 = −
+ 𝑉(𝑥) is known as Hamiltonian operator.(we will discuss later). Therefore two
2𝑚 𝜕𝑥 2
forms of the Schrödinger equation:
𝐻𝜓 = 𝐸𝜓
𝜕
Ψ
𝜕𝑡
Before going further let us mention some properties of the Schrödinger equation
𝐻Ψ = 𝑖ℏ
Schrödinger Equation
Is a wave equation
Based on Postulates of Deals with uncertainity Is
applied
to
QM
ptinciple
particle in a box
Probability calculation
Useful for calculating Wave particle duality
Energy eigenvalues
Expectation values
Free particle
Barrier penetration
Harmonic oscillator
Hydrogen atom
Nearly all QM problems consists of solving that equation.
Example: Particle in a box (Infinite well potential)
As an example consider infinite well potential describes a particle free to move in a small space
surrounded by impenetrable barriers. The model is mainly used as a hypothetical example to illustrate
the differences between classical and quantum systems. In classical systems, for example a ball
trapped inside a heavy box, the particle can move at any speed within the box and it is no more likely
to be found at one position than another. However, when the well becomes very narrow (on the scale
of a few nanometers), quantum effects become important. The particle may only occupy certain
positive energy levels. Likewise, it can never have zero energy, meaning that the particle can never
"sit still". Additionally, it is more likely to be found at certain positions than at others, depending on its
energy level. The particle may never be detected at certain positions, known as spatial nodes.
The particle in a box model provides one of the very few problems in quantum mechanics which can
be solved analytically, without approximations. This means that the observable properties of the
particle (such as its energy and position) are related to the mass of the particle and the width of the
well by simple mathematical expressions. Due to its simplicity, the model allows insight into quantum
effects without the need for complicated mathematics. It is one of the first quantum mechanics
problems taught in undergraduate physics courses, and it is commonly used as an approximation for
more complicated quantum systems.
Mathematically we can write:
∞
𝑉(𝑥) = {0
∞
𝑥≤0
0<𝑥<𝑎
𝑥≥𝑎
𝑉=∞
𝑉=∞
𝑉=0
0
𝑎
𝑥
We have three distinct regions, because the potential energy function (potential for short) changes
discontinuously. So, we have to solve the Schrödinger Equation three times.
Fortunately, for 𝑥 ≤ 0 and 𝑥 ≥ 0, the solutions are trivial: 𝜓(𝑥) = 0, since 𝑉(𝑥) = ∞.
Within the well, 𝑉(𝑥) = 0, so the particle is “free.”
−
ℏ2 𝜕 2 𝜓
= 𝐸𝜓
2𝑚 𝜕𝑥 2
The general solution is 𝜓 = 𝐴𝑠𝑖𝑛𝑘𝑥 + 𝑏𝑐𝑜𝑠𝑘𝑥. The parameters A, B, and 𝑘 =
by the boundary conditions on the 𝜓 and by the normalization requirement.
√2𝑚𝐸
ℏ
are determined
That is, we expect that 𝜓(𝑎) = 𝜓(0) = 0.
At 𝑥 = 0; 𝐴𝑠𝑖𝑛0 + 𝐵𝑐𝑜𝑠0 = 0 => 𝐵 = 0
Then
At 𝑥 = 𝑎; 𝐴𝑠𝑖𝑛𝑘𝑎 = 0 => 𝑘𝑎 = 𝑛𝜋; 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1,2,3, …
The choices of A=0 also solution but in that case 𝜓 = 0 is not a valid solution.
From this we obtain the discrete allowed energy levels for the particle confined in the well.
𝑛 2 𝜋 2 ℏ2
2𝑚𝑎2
The n is known as the principle quantum number. It labels the energy levels, or energy states of the
particle. In other from classical physics we have obtained discrete energy values instead of continuous
energy.
𝐸=
Superposition of wave functions
Quantum superposition refers to the quantum mechanical property of a particle to occupy all of its
possible quantum states simultaneously. Due to this property, to completely describe a particle one
must include a description of every possible state and the probability of the particle being in that state.]
Since the Schrödinger equation is linear, a solution that takes into account all possible states will be a
Linear combination of the solutions for each individual state. This mathematical property of linear
equations is known as the superposition principle.
Consider the superposition of stationary states Ψ1 (𝑥, 𝑡 ), Ψ2 (𝑥, 𝑡), . . . , Ψ𝑛 (𝑥, 𝑡) which are solutions of
the Schrödinger equation for a given potential V . Since these are stationary states, they can be written
as:
Ψ𝑛 (𝑥, 𝑡) = 𝜓𝑛 (𝑥) 𝑒𝑥𝑝 (−
𝑖𝐸𝑛 𝑡
)
ℏ
At time t = 0, any wavefunction Ψ(𝑥, 0) can be written as a linear combination of these
states:
𝜓(𝑥, 0) = 𝑐1 𝜓1 + 𝑐2 𝜓2 + … + 𝑐𝑛 𝜓𝑛
Time evaluation of this state is
𝑖
Ψ(𝑥, 𝑡) = ∑ 𝑐𝑛 𝜓𝑛 𝑒 −ℏ𝐸𝑛 𝑡 = ∑ 𝑐𝑛 𝜓𝑛 𝑒 −𝑖𝜔𝑛 𝑡
𝑛
𝑛
Where 𝐸𝑛 = ℏ𝜔𝑛 .
Inner product and probability
The inner product of two wave functions 𝜓1 (𝑥) and 𝜓2 (𝑥) is defined by:
〈𝜓1 |𝜓2 〉 = ∫ 𝜓1∗ 𝜓2 𝑑𝑥
The particle can not be found in two state at the same time, then the product of basis states:
〈𝜓1 |𝜓2 〉 = ∫ 𝜓1∗ 𝜓2 𝑑𝑥 = 0
This property is known as the orthogonality of the wave functions. If we have normalized basis
functions then they are orthonormal, which means that:
1 𝑖𝑓 𝑛 = 𝑚
〈𝜓𝑛 |𝜓𝑚 〉 = ∫ 𝜓𝑛∗ 𝜓𝑚 𝑑𝑥 = {
0 𝑖𝑓 𝑛 ≠ 𝑚
Since product of wave function gives probability of finding of particle in the given interval, the
probability can be written as:
𝑃 = 𝑝1 + 𝑝2 + ⋯ + 𝑝𝑛 = 1
Where 𝑝𝑖 = 𝑐𝑖 𝑐𝑖∗.
The fact that basis states are orthogonal allows us to calculate the expansion coefficients
using inner products.
Calculating a Coefficient of Expansion
The coeeficients of superposition of the wave functions can be calculated by using orthonormality and
orthogonality properties of them. It is obvious that
< 𝜓|𝜓𝑖 >= ∫ 𝜓 ∗ 𝜓𝑖 𝑑𝑥 = 𝑐𝑖
Example. A particle of mass 𝑚 is in a one-dimensional quantum well of width 𝑎. The wavefunction is
known to be:
𝜓=
𝑖 2
𝜋
2
2𝜋
1
3𝜋
√ sin ( 𝑥) + 𝑐2 √ sin ( 𝑥) + √ sin ( 𝑥)
2 𝑎
𝑎
𝑎
𝑎
𝑎
𝑎
Calculate 𝑐1 . If the energy is measured, what are the possible results and what is the probability of
obtaining each result? What is the most probable energy for this state?
Solution
We begin by recalling that the 𝑛th excited state of a particle in a one-dimensional
box is described by the wavefunction:
2
𝑛𝜋
𝜓𝑛 = √ sin ( 𝑥)
𝑎
𝑎
With the energy
𝑛 2 𝜋 2 ℏ2
𝐸=
2𝑚𝑎2
Then we can write the superposition of the wave functions:
𝜓=
𝑖
1
𝜓1 + 𝑐2 𝜓2 + √ 𝜓3
2
2
Using normalization
< 𝜓|𝜓 >= 𝑐1 𝑐1∗ + 𝑐2 𝑐2∗ + 𝑐3 𝑐3∗ = 1 =>
1
1
1
+ 𝑐2 𝑐2∗ + = 1 => 𝑐2 = ∓
4
2
2
Therefore the wave function takes the form
𝜓=
𝑖
1
1
𝜓1 + 𝜓2 + √ 𝜓3
2
2
2
The probablities and energies can be tabulated as:
State number probability energy
1
1
4
𝜋 2 ℏ2
2𝑚𝑎2
2
1
4
4𝜋 2 ℏ2
2𝑚𝑎2
3
1
2
9𝜋 2 ℏ2
2𝑚𝑎2
Physical quantities: Observables and their operators
In this lecture we study the relation between dynamical variables in classical and quantum mechanics
and show that in the new quantum formalism these must be represented by operators acting on the
wave function. We introduce the concept of commutator and derive the fundamental commutation
relations holding for the position and momentum operators.
Let’s start from the beginning. Physical quantities are things corresponding to aspects of the reality of
a system that can be measured, that is, they can be characterized by numbers. In classical physics,
considering a particle, the result of the measurement of a quantity A is a number a and this single
number at time t.
In quantum mechanics, what we said concerning position and momentum holds for any physical
quantity. There is some probability law for the quantity. Consequently, the complete result of a
measurement of a quantity A on the state ψ is the whole set of issues 𝑎𝑖 and associated probabilities 𝑝𝑖 .
Expectation values and uncertainty relations
The classical view
Examples of dynamical variables (or observables 𝑂) in classical physics are the position x, the
momentum p, the energy 𝐸 and the angular momentum 𝑳 = 𝒓 × 𝒑. According to the classical view a
particle has always univocally defined values for all possible observables at any time 𝑡. They define
indeed the state of the particle and evolve according to a deterministic equation (Newton equation).
Moreover any classical observable can vary in a continuous way.
The quantum view
We know that in the microscopic world the position of a particle is known only in probabilistic terms.
The outcome of a single measurement is random and we can only foresee an average value resulting
from many identical experiments. The probability density for a position measurement can be
calculated as follows: After a large number of measurements, will performed then we will have a
collection of results whose mean value is can be related to probability 𝑃 = |Ψ|2 :
∞
∞
〈𝑥〉 = ∫ 𝑥𝑃𝑑𝑥 = ∫ 𝜓 ∗ 𝑥𝜓𝑑𝑥
−∞
−∞
This is called as well expectation value of the position. Note that this is not the most probable value of
x. That occurs where 
2
is a maximum. The variance of the measurements of x is
𝜎𝑥 = ∆𝑥 = √〈𝑥 2 〉 − 〈𝑥〉2
Using this relation we can calculate uncertainty between the operators.
Postulate 3. Expectation value of an operator in quantum mechanics
The average value of many measurements of an observable 𝑂̂, when the system is described
by function Ψ(𝑥, 𝑡), is equal to the expectation value 𝑂̂, which is defined as follows,
∞
〈𝑂〉 = ∫ 𝜓 ∗ 𝑂𝜓𝑑𝑥
−∞
But what about now if we want to know the expectation value of the momentum 𝑝 ? This is a
dynamical variable on its own and doesn’t depend upon x so that an integral such as the one above
cannot hold in this case. To learn how to do this we remember that the new quantum theory we are
studying must not contradict the old classical theory in the situations where the latter can be
successfully applied. In other words quantum theory must include Newton theory as a special case
holding within the appropriate limits. In Newton theory position and momentum are related by
𝑑𝑥
𝑑𝑡
We thus expect the new quantum theory to be such that
𝑝=𝑚
〈𝑝〉 = 𝑚
𝑑
〈𝑥〉
𝑑𝑡
The fact that Newton equation holds, at least in an average sense, constraints the mathematical form of
the momentum in the new quantum formalism. To obtain expectation value of momentum let us
calculate:
𝑚
〈𝑝〉 = 𝑚
𝑑
𝑑 ∞
〈𝑥〉 = 𝑚 ∫ 𝜓 ∗ 𝑥𝜓𝑑𝑥
𝑑𝑡
𝑑𝑡 −∞
∞
𝑑
𝜕𝜓 ∗
𝜕𝜓
〈𝑥〉 = 𝑚 ∫ 𝑥 (
𝜓 + 𝜓 ∗ ) 𝑑𝑥
𝑑𝑡
𝜕𝑡
𝜕𝑡
−∞
We have shown before(see probability current density)
ℏ 𝑑 𝜕Ψ ∗
𝜕Ψ∗
𝜕𝜓 ∗
𝜕𝜓 ∗
−
Ψ −Ψ
𝜓 + 𝜓∗
)
(
)=(
2𝑖𝑚 𝑑𝑥 𝜕𝑥
𝜕𝑥
𝜕𝑡
𝜕𝑡
After some tedious calculations we can show that
〈𝑝〉 = −
∞
ℏ
𝑑 𝜕Ψ ∗
𝜕Ψ∗
𝑚∫ 𝑥 (
Ψ −Ψ
) 𝑑𝑥
2𝑖𝑚
𝜕𝑥
−∞ 𝑑𝑥 𝜕𝑥
〈𝑝〉 =
ℏ ∞ ∗ 𝜕Ψ
∫ 𝜓
𝑑𝑥
𝑖 −∞
𝜕𝑥
Then momentum operator can be expressed as
ℏ 𝑑
.
𝑖 𝑑𝑥
Therefore we can obtain an operator for a physicsl observables of the quantum physics. As an example
the kinetic energy operator can be obtained as follows:
𝑝=
𝐾=
𝑝2
1 ℏ 𝑑 ℏ 𝑑
ℏ2 𝑑2
=
=−
2𝑚 2𝑚 𝑖 𝑑𝑥 𝑖 𝑑𝑥
2𝑚 𝑑𝑥 2
As expected.
Postulate 4. Operator representation of observables of quantum mechanics
To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum
mechanics.
In general in wave mechanics all dynamical variables are represented by operators that act on the wave
function.
The discussion above was in 1D but the generalization to 3D is straightforward. We note that in
quantum mechanics all observables are represented by a suitable operator.
Operator Algebra and Eigenvalue equations
We must be careful when replacing the classical 𝑥 and 𝑝 with operator x and p since the order in
which they appear is important. This is because two given operators A and B do not in general
commute. Classical variables always commute, that is
𝐴𝐵 = 𝐵𝐴
Operators, similar to the matrices and they may not commute:
𝐴𝐵 ≠ 𝐵𝐴
We can define the commutator (or commutation relation)
[𝐴, 𝐵] = 𝐴𝐵 − 𝐵𝐴
If [𝐴, 𝐵] = 0 we say that A and B commute this physical means that the observables A and B can be
measured simultaneously, there is no uncertanity relation between them.
Example:
Find the commutators of the momentum and position operator [𝑝, 𝑥].
Solution
The commutator can be written as
[
ℏ 𝑑
ℏ 𝑑
ℏ 𝑑
, 𝑥] =
𝑥−𝑥
𝑖 𝑑𝑥
𝑖 𝑑𝑥
𝑖 𝑑𝑥
Important!!!!
The first term can be evaluated as follows:
ℏ 𝑑
ℏ 𝑑
ℏ 𝑑𝑓
(𝑥𝑓) = (𝑥
𝑥, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑤𝑖𝑡ℎ 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓,
+ 𝑓)
𝑖 𝑑𝑥
𝑖 𝑑𝑥
𝑖
𝑑𝑥
We have used derivative of product of two function. Then
ℏ 𝑑
ℏ 𝑑𝑓
ℏ 𝑑
ℏ
𝑑
(𝑥𝑓) = (𝑥
+ 𝑓) , 𝑟𝑒𝑚𝑜𝑣𝑒 𝑓,
𝑥 = (𝑥
+ 1)
𝑖 𝑑𝑥
𝑖
𝑑𝑥
𝑖 𝑑𝑥
𝑖
𝑑𝑥
Therefore the commutator can be written as
ℏ 𝑑
ℏ
𝑑
ℏ 𝑑
ℏ
, 𝑥] = (𝑥
+ 1) − 𝑥
=
𝑖 𝑑𝑥
𝑖
𝑑𝑥
𝑖 𝑑𝑥 𝑖
!Remember uncertainity relation
[
Eigenvalue Equation
The operators act on a function. The eigenvalue equation can be written as:
𝑂Ψ = 𝑎Ψ
Where O is operator, Ψ is eigenvector and a is eigenvalue.
Example: Determine the eigenfunction of the momentum operator
Solution
We can write the eigenvalue equation
ℏ 𝑑
Ψ = 𝑝Ψ
𝑖 𝑑𝑥
Where p is eigenvalues of the momentum operator. We solve the equation and we obtain
𝑖𝑝
Ψ = 𝑐𝑒 ℏ
𝑥
Operators whose expectations are real < 𝑂 >=< 𝑂 >∗ are called Hermitian operators.
Example: Check weather the momentum operator is hermitian or not?
Solution
Lets check if < 𝑝 > −< 𝑝 >∗ = 0 𝑜𝑟 𝑛𝑜𝑡?
ℏ
𝑑𝜓
ℏ
𝑑𝜓 ∗
∫ 𝜓∗
𝑑𝑥 − (− ) ∫ 𝜓
𝑑𝑥
𝑖
𝑑𝑥
𝑖
𝑑𝑡
ℏ
𝑑𝜓
𝑑𝜓 ∗
ℏ 𝑑
< 𝑝 > −< 𝑝 >∗ = ∫ (𝜓 ∗
+𝜓
) 𝑑𝑥 = ∫ (𝜓𝜓 ∗ )𝑑𝑥
𝑖
𝑑𝑥
𝑑𝑡
𝑖 𝑑𝑥
< 𝑝 > −< 𝑝 >∗ =
The integral is evaluated at -∞ and ∞, where (𝜓𝜓 ∗ )vanishes therefore momentum operator is
hermitian.
Hermitian operators also gives real eigenvalues.
Postulate 5. Eigenvalue equation and measurement
In any measurement of the observable associated with operator 𝑂̂, the only values that will
ever be observed are the eigenvalues 𝑎, which satisfy the eigenvalue equation
𝑂̂Ψ(𝑥, 𝑡) = 𝑎Ψ(𝑥, 𝑡)
This postulate captures the central point of quantum mechanics--the values of dynamical
variables can be quantized (although it is still possible to have a continuum of eigenvalues in
the case of unbound states). If the system is in an eigenstate of 𝑂̂ with eigenvalue 𝑎, then any
measurement of the quantity 𝑂̂ will yield 𝑎.
*This lecture notes collected from the references given below
http://galileo.phys.virginia.edu/
http://en.wikipedia.org/wiki/Particle_in_a_box
An introduction to Quantum Mechanics COURSE NOTES by: Paolo BONIFACIO
Quantum Mechanics Demystified (Mcgraw-HILL) DAVID McMahon