formalism of quantum mechanics

Formalism of Quantum Mechanics

• The theory of quantum mechanics is formulated by defining a set

of rules or postulates.

• These postulates cannot be derived from the laws of classical

physics.

The rules define the following:

1. How to describe a system (the wave function).

2. How to describe the time evolution of the wave function (theSchrodinger equation).

3. How to describe observable quantities (operators).

4. How to describe the outcomes of a measurement of observablequantities; how to describe the system after the measurement.


The wave function

• The state of a physical system is represented by a wave function

which contains all the information that can be known about the

system.

• The wave function is in general complex (it has real and

imaginary parts).

In configuration (coordinate) space the wave function of a particle is

a function of the position of the particle at a given time:

For N particles, this can be generalized to

! (x, y, z,t)

! (x1, x

2,...x

N, y1, y2....y

N, z1, z2...z

N, t)


Normalization

• The wave functions ! and c! represent the same state, where c is a

complex number. Hence we can always multiply the wave function

by an arbitrary complex number without changing it.

Consider the normalization integral

If I is a finite number, then the wave function is square integrable.

If we multiply the wave function by a constant

then the wave function is said to be normalized (to unity):

I = ! (x, t)"2

dx = ! *(x,t)! (x, t)" dx

c = 1 I

! (x, t)"2

dx = 1


The Born interpretation: Probabilities

• The absolute square of the normalized wave function ! :

is the probability of the particle being in the region dx around

position x at time t.

P(x,t) is called the probability density curve

P(x,t)dx = ! (x,t)2dx

Hence the normalization requirement is a statement that the

probability of finding the particle somewhere is 1.

! (x, t)"2

dx = P(x, t)dx =" 1


The Born interpretation: Probabilities

• The probability of finding a particle in a region a < x < b is then

! (x, t)2

a

b

" dx

This is the area under the probability density curve between x=a and

x=b:


Example: Find the value of the constant C of the normalized wave

function

where L = 1nm. Plot the wave function and the probability density

curves and find the probability of finding the particle in the region

! (x) = Ce" x /L, x # 0

! (x) = 0, x < 0

! (x, t)"2

dx = 1

# c2e$2x /L

dx = $c2L

2e$2x /L

0

%

"0

%

=c2L

2= 1

# c =2

L= 1.414

P(x) = ! (x)2= 2e

"2 x /1nm

x ! 1nm


Example: Find the value of the constant C of the normalized wave

function

where L = 1nm. Plot the wave function and the probability density

curves and find the probability of finding the particle in the region

! (x) = Ce" x /L, x # 0

! (x) = 0, x < 0

P(x ! 1nm) = P(x)dx1nm

"

#

$ c2e%2 x /L

dx = %c2L

2e%2x /L

1nm

"

#1nm

"

=c2L

2e%2

$ P(x) = 0.135 = 13.5%

P(x) = ! (x)2= 2e

"2 x /1nm

x ! 1nm


Superposition principle

• If the wave functions

represent two possible states of the system, then any linear

combination

also represents a possible state of the system. This is the superposition

principle.

This allows for superpositions of a particle wave function in two

different locations. (Recall double slit experiment).

Note:

!1and!

2

! = c1"1+ c

2"2

!2

= c1"1+ c

2"2

2

= c1

*"1

*+ c

2

*"2

*( ) c1" 1+ c

2"2( )

= c1

2

"1

2

+ c2

2

"2

2

+ c1

*c2"1

*"2+ c

1c2

*"1"2

*


Momentum space wave functions

Given a wave function in coordinate space representing the position

of a particle, the corresponding momentum wave function is the

Fourier integral of the the position wave function:

If the position wave function is normalized, the momentum wave

function is also normalized to unity:

!(p, t) =1

2"!( )1/2

e

#i!

xp( )$ (x, t)% dx

! (p, t)"2

dp = 1



We can interpret the square of the quantity

of the normalized wave function as the probability at time t of the

momentum of a particle being in the volume dp around p

P(p,t) = ! (p,t)2dp

Hence the normalization requirement is a statement that the

probability of finding the particle with some value of momentum is 1.

! (p, t)"2

dx = P(p, t)dp =" 1



• The probability of finding a particle with momentum a < p < b is

then

! (p, t)2

a

b

" dp

This is the area under the momentum probability density curve

between p=a and p=b:

p

!(p, t)2


Given a wave function representing the momentum of a particle, the

corresponding position space wave function is the inverse Fourier

integral of the momentum wave function:

If the momentum wave function is normalized, the position wave

function is also normalized to unity.

Note that the position and momentum wave functions represent the

same state of the system. All information about the state can be

obtained from either wave function.

! (x, t) =1

2"!( )1/2

e

i

!px( )

# $ (p, t)dp


Time-dependent Schrödinger equation

The wave function of a particle undergoing a force F(x) is the

solution to the Schrödinger equation:

U(x) is the potential energy associated with the force:

i!!

!t" (x,t) = #

!2

2m

!2

!x2" (x,t) +U(x)" (x,t)

F = !"U

"x


Time-dependent Schrödinger equation: Separation of variables

Since U(x) does not depend on time, solutions can be written in

separable form as a part that is only position dependent and a part

that is only time dependent:

i!!

!t" (x,t) = #

!2

2m

!2

!x2" (x,t) +U(x)" (x,t)

! (x, t) = "(x)#(t)

Inserting this into the above equation, we get

i!!(x)"

"t#(t) = $#(t)

!2

2m

"2

"x2!(x) +U(x)!(x)#(t)



! (x, t) = "(x)#(t)

Dividing by !(x,t),

i!!(x)"

"t#(t) = $#(t)

!2

2m

"2

"x2!(x) +U(x)!(x)#(t)

i!

!(t)

"

"t!(t) = #

1

$(x)

!2

2m

"2

"x2$(x) +U(x)

Left hand side (LHS) is a function of t alone

Right hand side (RHS) is a function of x alone

LHS=RHS only if LHS = E and RHS = E (E is a constant)



! (x, t) = "(x)#(t)

i!

!(t)

"

"t!(t) = #

1

$(x)

!2

2m

"2

"x2$(x) +U(x)

LHS=RHS only if LHS = E and RHS = E (E is a constant)

i!

!(t)

d

dt!(t) = E

"1

#(x)

!2

2m

d2

dx2#(x) +U(x) = E



! (x, t) = "(x)#(t)

Solutions for the time-dependent equation:

i!d

dt!(t) = E!(t)

!(t) = e" i#t , # =E

!

Check:

d

dt!(t) = "

iE

!e"i

!Et

# i!d

dt!(t) = Ee

"i

!Et

= E!(t)



! (x, t) = "(x)#(t)

Time-independent Schrödinger equation:

This equation is not always easy to solve analytically, but can be

solved numerically on a computer.

However we can analytically solve some special cases….

!!2

2m

d2

dx2"(x) +U(x)"(x) = E"(x)

formalism of quantum mechanics

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