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NOTES

ON

QUANTUM MECHANICS

Prepared by

Dr. M. El-Banna

EE231

Prof. M. El-Banna EE231

2

Chapter:1

Introduction to Quantum Mechanics

1. Introduction

There are four main contradictions between theory and experiments that quantum mechanics have been brought up to resolve:

1. It was believed from electromagnetics that a moving charged particle emits continuous radiations, i.e. radiation can be at any value of frequencies, however experiments have only proved existence of discrete radiating frequencies.

2. It was comprehended that moving charged particles emit radiations, however experiments have showed that electrons on certain orbits around the nucleus, though moving charged particles, do not emit radiations.

3. It was thought that electrons usually come out from a metal surface upon irradiating by light, or heating up the metal, at any frequency value. However experiments have showed that there is a certain threshold frequency for each metal after which electrons start being emitted from the metal surface.

4. Classically, electrons can never be found beyond an energy barrier unless they posses an amount of energy equal to or greater than the barrier height, while experiments show that electrons of lesser amount of energy are found beyond the barrier.

Classical mechanics deal only with observables, quantities that can be measured. This is not the case in Quantum Mechanics where fictitious mathematical quantities are assumed. The central fictitious function is the state function φ, which is not measurable. The state function is a general complex function, but when being applied on by an operator, it results in observable, measurable and real quantity.

State Functionφ

• Velocity• Time• Distance• Energy• Momentum

Operator


3

2. Postulates of Quantum Mechanics Postulate 1: For any physical system there exists a state function φ(x, y, z, t) which contains all information about the system. The state function must posses the three following properties:

i- Finite ii- Continuous up to the 1st derivative. iii- Single-valued function

Example: which of the following function can be a state function?

φ3

x

φ2φ1

x

x

These properties have to be valid over the configuration space, the space through which the function is defined. For example if we have two particles P1 and P2 in the 3D Cartesian space and referred to by six coordinates (x1, y1, z1) and (x2, y2, z2) as shown:

. P1(x1, y1, z1)

. P2(x2, y2, z2)

x

x

y

z

In order to fully trace the two particles P1 and P2 we have to provide six coordinates (x1, x2, y1, y2, z1, z2). The elemental volume in the configuration space is:


4

d dxdydzτ = Generally we refer to a coordinate in a multi dimensional coordinate system by iq and hence:

i

ii

i 1 2 3 4i

i 1 2 3 4

d dq

a a *a *a *a ........

d dq dq *dq *dq *dq .......

τ = π

π =

τ = π =

Postulate 2: The probability of observing a system at an instant t inside an elemental volume dτ is the function *dϕϕ τ , therefore the integration over the whole configuration space is:

2*d d 1τ

ϕϕ τ = ϕ τ =∫ ∫

The is called the normalizing condition which imposes that the state function should be a square integrable function, i.e. belongs to the Hilbert space which composes all square integrable functions. Postulate 3: To every dynamic quantity L (energy, position, momentum) there corresponds a linear operator L which is Hermitian. Examples of operators of some basic quantities:

i- General position operator : iQ

i iQ q .

X x.Y y.

ϕ = ϕ

ϕ = ϕϕ = ϕ

ii- Time operator: T

T t.ϕ = ϕ

iii- Linear momentum operator: j− ∇


5

x y z

x y z

j P

j j ( a a a ) wherex y z

h2

and F(x, y, z) a a a F(x, y, z)x y z

− ∇ϕ = ϕ∂ ∂ ∂

− ∇ = − + +∂ ∂ ∂

=π

∂ ∂ ∂∇ = + + ∂ ∂ ∂

A point worth noting is that the algebraic formulae that relate the physical dynamic quantities also apply to operators, for example the velocity is calculated by dividing the distance travelled over the time taken to travel this distance, a velocity operator can be formed by just dividing the position operator over the time operator.

So far we have assume that the systems dealt with are all in what is known by eigen states. Being in eigen states each operator, time, velocity, energy, or momentum results directly in its corresponding dynamic quantity, time, velocity, energy or momentum, otherwise a quantity related to the namic quantity in question results. If the system is in an eigen state with respect to an observable L , its state function ϕ becomes an eigen function ψ of the operator L, it is then possible to determine the direct precise value of L, which is called an eigen value L = λ :

Lψ = λψ Postulate 4: The state function should satisfy the equation:

H jt

∂ϕϕ =

∂

Which is known an Schrödinger equation, Wave equation or more precisely time-dependent Schrödinger equation. H is known as the Hamiltonian Operator whose definition follows. Hamiltonian Operator:

If a system happens to be in an eigen state of H , then the state function ϕ is an eigen function ψ of H , hence

H jt

H E

∂ψψ =

∂ψ = ψ


6

Where E is the eigen value of H . The above equation is known as time-dependent Schrödinger equation an ca be rewritten as:

Ejt

∂ψ= − ψ

∂

Postulate 5: The expected or average value <A> of an observable A when a system in a state ϕ , which is not an eigen state ψ , is given by:

*A A dτ

< >= ϕ ϕ τ∫

3. Functions and Operators A function is a recipe to change a number into a new number , for example the function y = x2 takes a number x, squares it and assigns it to a new number y. In a much similar way, the operator changes a function into a new function. Here are examples of different operators:

1. Multiplication operator A This operator multiplies a function by its independent variable x, i.e. A F(x) = x F(x). Example:

F(x) = a + bx + cx2 .

then, A F(x) = ax + bx2 + cx3.

2. Differentiation operator D

ddxDf (x) f (x)=

3. Integration operator B

Bf (x) f (x)dx= ∫

4. Laplacian operator 2∇

2 2 22 f f ff (x, y, z)

x y z∂ ∂ ∂

∇ = + +∂ ∂ ∂

5. Identity operator E

E f (x) f (x)=


7

6. Inversion operator I I f (x) f ( x)= −

7. Translation operator aT

aT f (x) f (x a)= +

8. Rotation operator nC

n2C f ( ) f ( )nπ

ϑ = ϑ+

9. Partial derivative operator Dx

f (x, y)Dx f (x, y)x

∂=

∂

4. Linear operators

L is called a linear operator if it satisfies the following relation:

[ ]1 2 1 2L a f (x, y, z) b f (x, y, z) a L f (x, y, z) b L f (x, y, z)+ = +

Where a and b are constants. Linear operators are operators where superposition and homogeneity are satisfied during the operation. Example of a linear operator: The derivative operator is an example of a linear operator:

2

2

D(ax bx bc) a 2bxa D x b D x D bc a 2bx 0 a 2bx

+ + = +

+ + = + + = +

Example of a nonlinear operator: Assume an operator G which multiplies the function by its derivative:

d f (x)G f (x) f (x).dx

= 2

1 2Let f (x) ax, and f (x) b ( x c )= = + , therefore


8

[ ][ ]

21 2

2

2 3 2 2 2

G f (x) f (x) G ax bx bc

ax bx bc . a 2bx

2b x 3abx (a 2b c)x abc

+ = + + = + + +

= + + + +

Now applying G to each function separately,

[ ]

21 2

2 3 2 2

1 2

G f (x) G f (x) ax .a b (x c).2bx

2b x (a 2b c).xG f (x) f (x)

+ = + +

= + +

≠ +

therefore, G is not a linear operator. 5. Properties

We are going to discuss two main useful properties usually encountered in quantum mechanics: i- Distributive law The distributive law applies to linear operators. Example: Let A and B be two linear operators then:

( A B ) f (x) A f (x) Bf (x)+ = +

ii- Commutative law Commutation does not usually apply to linear operators. Example:

2

n 1 n

A B y A ( B y)(B y)is a new function,and wecan generallysay :

B B y B y

B B B y−

=

=

=

In general linear operators do not usually commute


9

[ ]

A B BAA B BA 0A,B A B BA, commutation of A and B

≠− ≠

= −

Example of non-commuting operators:

dA ,B xdx

d d dA B y ( B y) (x .y) x . ydx dx dx

d yB A y x. A B ydx

so, [A,B] y 0

= =

= = = +

= ≠

= ≠

i.e. A and B do not commute

6. Eigen functions and eigen values The state function ϕ is an eigen function ψ of an operator L if L ψ = λ .ψ , where λ is a constant. i.e. when applying L on ψ , it results in the function ψ itself multiplied by a constant λ . The function ψ is then called an eigen function of the operator L and the constant λ is the eigen value. If more than one eigen function have the same eigen values, then this eigen value is called a degenerate eigen value. You can never know to which eigen function this eigen value belongs. Example: Consider a function k

kxy eλ= defined over a configuration space x−π ≤ ≤π

π

y

x-π

This function ky is an eigen function of the derivative operator L = ddx

.

k k kkx

L y e .yλ

= λ = λ

Generally, the eigen value can be real or imaginary.


10

kk

j xy e λ=

For the 1st derivative operator D , the eigen value is an imaginary value.

k k kD y j y= λ

For the operator 2

2d

dxM = and the function k

kjy e± λ=

2

k kM y = −λ kje± λ i.e. 2

k−λ is a two-fold eigen value of the operator M and generally one can find n-fold eigen value.

7. Hermetian operators

If 1ϕ and 2ϕ are two functions belonging to the same class of functions and L is a linear operator acting on these two functions, L is said to be Hermetian if:

** *1 2 2 1( L ) d ( L ) d

τ τ

ϕ ϕ τ = ϕ ϕ τ∫ ∫

The integration is carried out over the whole configuration space τ . A different simpler way to express the Hermeticity of an operator is to write:

1 2 1 2( , L ) ( L , )ϕ ϕ = ϕ ϕ

The quantity to the left of the comma is always complex conjugate. Example:

The operator dL a jdx

= is Hermetian when applied to square integrable functions, i.e.

functions which belong to Hilbert space.


11

*1 2 1 2

* 21

*1 2

* *1 2 2 1

( , L ) ( L ) d

d(a j )dxdx

a j (d )

a j [ ( ) d ]

τ

∞

−∞

∞

−∞

∞∞

−∞−∞

ϕ ϕ = ϕ ϕ τ

ϕ= ϕ

= ϕ ϕ

= ϕ ϕ − ϕ ϕ

∫

∫

∫

∫

1ϕ and 2ϕ are assumed to be square integrable , i.e. their values go to zero ax x goes to ∞ , therefore:

*1

1 2

*1

2

* *2 1

1 2

d( , L ) a j dxdx

d( a j )dxdx

( L )dx

( L , )

∞

−∞

∞

−∞

∞

−∞

ϕϕ = − ϕ

ϕ= ϕ −

= ϕ ϕ

= ϕ ϕ

∫

∫

∫

Therefore L is Hermetian.

Theorem: The eigen values of Hermetian operators are real Proof: Let L be a Hermetian operator, ψ its eigen function and λ its eigen value. i.e.

L (1)ψ = λψ

Taking the complex conjugate of the last equation results in:

* * * *L (2)ψ = λ ψ Pre-multiply equation (1) by *ψ gives:

* *L (3)ψ ψ = ψ λψ

Integrating equation (3) over the whole configuration space


12

* *

*

L d d

(4)

d

τ τ

τ

ψ ψ τ = ψ λ τ

= λ ψ ψ τ

∫ ∫

∫

Premultiply Equation (2) by ψ

* * * *L(5)

ψ ψ = ψλ ψ

Integrate equation (5) over the whole configuration space

* * * * * *L d d dτ τ

ψ ψ τ = ψλ ψ τ = λ ψψ τ∫ ∫ ∫

(6) Comparing equation 4 and equation 6 Since L is a Hermetian operator * * *L d L d

τ τ

ψ ψ τ = ψ ψ τ∫ ∫

∴ * * *d dτ τ

λ ψ ψ τ = λ ψ ψ τ∫ ∫

* is realλ = λ ∴λ

Hermetian operators are very important in the study of quantum physics since in such study we are ultimately interested observable quantities which can be measured . i.e Real eigen values . It will be seen taker that physically observable quantities are always associated with Hermetian operators . 7. Hamiltonian (ENERGY) Operator We shall now develop the energy operator or the Hamiltonian operator H. for a conservative system corresponding to the energy of the single particle of mass M moving with a velocity V in potential fold V (x,y,z,t)

21E mv V (x, y, z, t)2

= +

Since 2

2 2 21 1 pmv m v2 2m 2m

= =

2pE V (x, y, z, t)2m

∴ = +

P j= − ∇


13

2pH V (x, y, z, t)2m

= +

2 2

H V(x, y, z, t)2m

−− ∇= +

The last equation is known as the Hamiltonian operator 8. Angular Momentum Operator: L linear momentum mv= Angular momentum I.W= I : 2rotational moment of inertia m r=

O

θ

α

V∧

θr∧

r v cosv = α

vsinvθ= α

2L mr= ω

mr .r= ω mr .vθ

=

mr .vsinvθ= α

L p.r.sin r p= α = ×

j (r )= − ×∇

L r p rx ( j )= × = − ∇


14

Components and properties of the angular momentum operator Components of L x x y y z zL L a L a L a= + +

j r= − ×∇

x y za a a

x y z

x y z

= ∂ ∂ ∂ ∂ ∂ ∂

xL j y zz y

∂ ∂∴ = − − ∂ ∂

yL j z xx z∂ ∂ = − − ∂ ∂

zL j x yy x

∂ ∂= − − ∂ ∂

x y x y y xL ,L L L L L = −

2

x yL L y z z xz y x z

∂ ∂ ∂ ∂= − − − ∂ ∂ ∂ ∂

2

2 22y z z xy xz

z x x y z y x ∂ ∂ ∂ ∂ ∂ ∂ ∂

= − − − + ∂ ∂ ∂ ∂ ∂ ∂ ∂

2 z x y zx z z y

∂ ∂ ∂ ∂= − − − ∂ ∂ ∂ ∂

2

2 22zy xy z x z

x z z x y z y ∂ ∂ ∂ ∂ ∂ ∂ ∂

= − − − + ∂ ∂ ∂ ∂ ∂ ∂ ∂

2

x y y xL L L L y z yz xz x zz x x z y z z y

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂− = − − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

2 z x y z y xz y x z x y

∂ ∂ ∂ ∂ ∂ ∂= − − + − ∂ ∂ ∂ ∂ ∂ ∂

z z2 jL z jLz

z z ∂ ∂

= − − ∂ ∂


15

z zz z

L Lj z z L j L

z z∂ ∂

= − − − = ∂ ∂

[ ] zLx ,Ly j L∴ =

i.e xL & yL don’t commute with each other and knowing tow components of the angular

momentum enables us from getting the third one .

L x L j L=

Inner product: 2 2 2x y zL.L L L L+ +=

2

x ,y,z, 0L L =

Even though the components of the angular momentum do not commute with one another they commute with

2L

aH jtφ

φ =∂

for any state function

for an eigenstate function ψ

H E jt

∂ψψ = ψ =

∂ and Ej

t∂ψ ψ

= −∂

therefore , ( )jE t

x.y.z e−

ψ = ψ

which means that we can solve the time independent equation and eventually we multiply

by jE t

e−

( )*A A dτ

< > = φ φ τ∫

Die: this resembles the case of throwing a die: suppose that a die is thrown very large number of time and each number ( 1 through 6 ) represents the temperature in degree Celsius so the average or expected temperature.

1 1 1 1 1 1 1T 1* 2* 3* 4* 5* 6* 3. C6 6 6 6 6 6 2

°< > = + + + + + =

9. Orthogonal Functions


16

A set of function is said to be orthogonal in the interval ( a , b ) if for any two

members of the set b

*j iji j

a

(x) (x) dx af f = δ∫

Kronecker: ij 1 for i jδ = = 0= elsewhere

Where ja is a non-zero positive constant and i jδ is the kρονεχκερ δ

• if ja = 1 the set is orthonormal • if a set is orthogonal but not orthonormal it is always possible to normalize the set

by dividing each function by the square root of j ja ; a

j*i j1/ 2 1/ 2 1/ 2 1/ 2 i j i j

j j j j

a1 f (x) f (x) dxa a a a

= =δ δ∫ = orthonormal function.

It is just the case of fourier series any periodic function can be expressed as a sum of sines and cosines which are orthogonal function. This is always done as long as the function to be expressed in terms of the orthogonal functions does not belong to the set of orthogonal function (the sine function is not expressible in terms of sines and cosines) likewise the state function φ can be expressed in terms of the orthogonal function.

n

ii ii 1(x) cons tan t (1)c cf=

φ = =∑

If i 'sc are all zeros except one this means that φ belongs to the set and can not be expressed in terms of the orthogonal function . Multiply both sides of equation (1) by *

jf and integrate from a to b

b bn*

j i jii 1a a

dx dxcf f f=φ = ∑∫ ∫

jjc A=

b

*jj

aj

1 dxc fa= φ∫

( )bn

*j i

i 1 j a

1 f . dx f (x)a=

φ = ϕ∑ ∫

THEOREM


17

Any two different eigen function with non-degenerate eigen values of a Hermetian operator are orthogonal.

PROOF

Let

1 2andψ ψ be two eigen functions of a Hermetian operator L with eigen values

1 2 1 2and ( is not equal )λ λ λ λ

11 1L =ψ ψλ

22 2L =ψ ψλ

*

12 1 1L d d

τ τ

τ = τψ ψ ψλ∫ ∫

( )* *

21 2 1 2L d d

τ τ

τ = τψ ψ ψ ψλ∫ ∫ L is a Hermetian operator

( )**

2 1 1 2L d L d

τ τ

τ = τψ ψ ψ ψ∫ ∫

*

1 2 1 2( ) d 0

τ

∴ − τ =ψ ψλ λ ∫

*

1 1d 1

τ

∴ τ =ψ ψ∫

s∴ ′ψ are orthonormal functions.

So any well behaved function can be expressed as a sum of non-degenerate eigen functions of a Hermetian operator. Exercise Prove that the state function

1 21 2c cφ = +ψ ψ is not an eigen function of an

operator L whose non-degenerate eigen functions are1 2

andψ ψ .

1 21 2c cφ = +ψ ψ

( )1 21 2L L c cφ = +ψ ψ

1 21 2L Lc c= +ψ ψ

1 1 2 21 2c c= +ψ ψλ λ


18

Since 1λ not equal

2λ L not equalφ λφ φ is not an eigen function of L this is also true for a general combination of n-eigen function since φ is a state function

*d 1φ τ =φ∫

( ) ( )*

1 2 1 21 2 1 2d 1c c c c

τ

+ + τ =ψ ψ ψ ψ∫

( ) ( )* ** *

1 2 1 21 2 1 21c c c c

τ

+ + =ψ ψ ψ ψ∫

* * * ** * * *

1 1 1 2 2 1 2 21 1 1 2 2 1 2 2d d d d 1c c c c c c c c

τ τ τ τ

τ + τ + τ + τ =ψ ψ ψ ψ ψ ψ ψ ψ∫ ∫ ∫ ∫

* *

1 1 2 2 1c c c c+ = 2 2

1 2 1c c+ = n

ii 1 ic=φ = ψ∑

Where *

i idc

τ

= φ τψ∫

( )* n n*

i ii 1 i 1A d d A i dc c= =

τ τ

φ τ = τ ψ τφ ψ∑ ∑∫ ∫

( ) ( )*n n*

i ii 1 i 11 iA dc c= =

τ

= τψ ψ∑ ∑∫

= ( )*n n*

i i ii 1 i 11 idc c= =

τ

τψ ψ∑ ∑ λ∫

* n *

i i ii 1A d c c=

τ

∴ φ τ =φ ∑ λ∫

Knowing that

*

i id 1

τ

τ =ψ ψ∫

**

i j j i jd 0c c

τ

τ =ψ ψλ∫

<A>* 2n

i ii 1A d c=

τ

= φ τ =φ ∑ λ∫

2n

ii 11c=

=∑


19

Applications

Assume a particle of energy E inside the shown one dimensional well and we would like to determine

1. The possible energy the particle can possess 2. The possible locations of the particle.

x

IIIII I 0V0V

V

.E0V 0=

0 a

H Eψ = ψ

22 v E

2m v −

+ ψ = ψ

22

2xav =∂

2 2

2 v E2m x− ∂ ψ

+ ψ = ψ∂

2 2

2 v E2m x

∂ ψ− φ = − ψ

∂

( )2 2

2 E v 02m x

∂ ψ+ − ψ =

∂

( )2

2 2

2m E v 0x

∂ ψ+ − ψ =

∂


20

Region I: 0 x a, and V 0< < =

( )2

112 2

221

12 2

1 1 2

2m E 0x

2mE0 ,x

A cos x A sin x

∂ ψ+ ψ =

∂

∂ ψ+ α ψ = α =

∂

∴ψ = α + α

Region II: o ox a, V V , and E V≥ = <

( )

( )

22

o 22 2

22

o 22 2

221

1 02 2

x x2 1 2

2m E V 0x

2m V E 0x

2mE0 , (V E)x

B e B eβ −β

∂ ψ+ − ψ =

∂

∂ ψ− − ψ =

∂

∂ ψ− β ψ = β= −

∂

∴ψ = +

Region III: o ox 0, V V , and E V≤ = <

x x3 1 2C e C eβ −β∴ ψ = +

1 1 2A cos x A sin xψ = α + α x x

2 1 2B e B eβ −βψ = + x x

3 1 2C e C eβ −βψ = + Since ψ has to be finite:

1 2B C 0= =

1 1 2A cos x A sin xψ = α + α x

2 2B e−βψ =


21

x3 1C eβψ =

ψ has also to be continuous up to the first derivative

1 3 1 1(0) (0) A Cψ = ψ ⇒ =

31 d (0)d (0)dx dx

ψψ=

x

1 2 1x 0 x 0

2 1 1

1

2

A sin x A cos x C e

A C AAthereforeA

β= =

∴ − α α + α α =β

α = β = βα

=β

Boundary conditions at x = a:

1 2a

1 2 2a

1 2 2

(a) (a)

A cos a A sin a B e .....................................(1)

A sin a A cos a B e ..........................(2)

−β

−β

ψ =ψ

α + α α =

− α α + α α = −β

Dividing (1) over (2) results in:

1 2

1 2

A cos a A sin a 1A sin a A cos a

α + α α −=

− α α + α α β

Dividing numerator and denominator by 2A cos aα yields:

2

2

22

2 2

2 2

o

o

/ tan a 1/ tan a

tan a tan a

2 ( ) tan a ( ) tan a

2 ( ) tan a2tan a

( )

2 E (V E)2E V

α β+ α −=

−α β α +α β

αα −α=α+β α

β

αα= − β α = α − β α

β

αβ = α − β ααβ

α =α − β

−=

−


22

Therefore,

o2

o

2 E (V E)2mtan E .a2E V

−=

−

1 2f (E) f (E)=

2E 3E1E

1 2f , f

E

It is clear now that the particle possesses quantized values of energy, not continuous. A second interesting point is that where the particle can most likely be located? This is obtained by integrating the probability density function 2ψ over a region within which the particle is searched for. It is clearly that the integration over the whole configuration space should be unity, i.e.

0 a* * * *

3 3 1 1 2 20 a

dx dx dx dx 1∞ ∞

−∞ −∞

ψψ = ψ ψ + ψ ψ + ψ ψ =∫ ∫ ∫ ∫

Below is a sketch of the wave function ψ for a given energy value E and the associated

probability density function 2ψ .


23

ψ

x0 a

x0 a

2ψ

Remarks:

1. 2ψ gives the probability of finding the particle in a distance dx within the

configuration space. Integrating 2ψ dx from over a distance 1 2(x , x ) gives the probability that the particle can be located within this distance.

2. Even though E is less than oV , there is a nonzero probability of finding the particle in regions III and II.

Example: Let us assume that oV =∞ . Recall the solution we obtained before and get its limit when

oV goes to ∞ .

o

o2 V

0

2 E (V E)2m Etan a 02E V →∞

−= =

−


24

therefore,

2

2

2

2m Etan a 0 which means

2m Esin a 0

2m Ehence, a n , n 1,2,3,4,..................

=

=

= π =

Which again confirms that the energy E takes on discrete quantized values.

2 2 22

2 22

n 2

2mEa n

therefore, E n ,n 1,2,3,..2m a

∴ = π

π= =

Now, how about 2 3andψ ψ

0

02 V

2 3

2m (V E)

0

→∞

−β = = ∞

∴ψ = ψ =

i.e. the particle can never be found behind the well neither in the positive nor in the negative directions of the x-axis. Boundary conditions:

1 3 1

1 2 1

at x 0, 0 A 0at x a, (a) (a) 0 A 0

= ψ = ψ = ⇒ == ψ = ψ = ⇒ =

1 2nA sin .xaπ

∴ψ =

A2 can be obtained by integrating the probability density function over the whole configuration space:


25

( )

( )

*1 1

2 22

a2 22

0a 2

2

022

d 1

A sin x dx 1

A sin x dx 1

A 1 cos 2 x dx 12

A a 12

τ

∞

−∞

ψ ψ τ =

α =

α =

− α =

=

∫

∫

∫

∫

Which gives: 22

2

1

2Aa

2Aa

2 nsin xa a

=

=

π∴ψ =

Below is a sketch of the wave function for n=1, and n=2.

n=1n=2

oV =∞ ∞

x

3 0ψ = 2 0ψ =

0 aa / 2

1ψ

Squaring the wave function gets the probability density function


26

n=1n=2

oV =∞ ∞

x0 aa / 2

21ψ

22ψ

23ψ

The probability distribution function shows positions where the particle can most likely be found, positions of maximum probability. In region I, the particle can possess any value of the infinite discrete values of available energies. Example:

For n=1, the energy 2 2

1 2

n aE , and the likely hood is to find the particle at2m a 2

= .


27

Assignment #1

1. We have postulated that the quantum mechanical wave functions satisfy certain mathematical conditions of good behavior. Examine the following functions for this good behavior, pointing out why some of them do not qualify.

a- ψ(x) = x for x ≥ 0

= 0 elsewhere b- ψ(x) = x2 c- ψ(x) = e-x

2

d- ψ(x) = e-x

2. Consider two operators O1 and O2 defined by the following operations: O1 ψ(x) = ψ(x) + x O2 ψ(x) = dψ(x)/dx + 2ψ(x) Check for the linearity of O1 and O2 .

3. Test the following operators for linearity: a- O1 ψ(x) = x2 ψ(x) c- O3 ψ(x) =ψ*(x)

b- O2 ψ(x) = eψ(x) d- O4 ψ(x) = x2 dψ(x) /dx

4. Evaluate the following commutator , by operating on a wave functions: a- [ x , d/dx ]

b- [ x2 , d/dx]

5. Prove that the position operator X and the linear momentum operator dP jdx

=−

do not commute, and find their commutator. 6. A certain system is described by the Hamiltonian operator

H = - d2/dx2 + x2 Show that ψ(x) = Ax e-x

2/2is an eigen function of H and determine the eigen value.

7. A particle in a one dimensional box of length a is in the ground state . Calculate the probability of finding the particle in the interval ∆ x = 0.01a at the point x = a/2.

8. A one-dimensional impenetrable box of length a contains an electron that suffer a small perturbation and emits a photon of frequency γ = 3E1/h , where E1 is the energy of the ground state . Would it be correct to conclude that the initial state of the electron is the n = 2 box . Why or why not? .

9. Consider an electron in a one dimensional box with impenetrable walls of width

10-8cm. a- Calculate the three lowest allowed values of energy.


28

b- The frequency of light that would cause the electron to jump from the ground to the third excited energy level .

c- When the electron de-excites, what are the frequencies of the emitted photons .

d- Plot the probability distribution for all three states and comment on where the electron is most likely to be found.

10. Find the probable current-density (probability current for the plane wave

ej(kx-ωt).


29

Assignment # 2 1. Derive an expression for the flux , current density , of free particles moving along the

x-direction. First, start by writing down the continuity equation which is a conservation law, expressing the fact that a change in the particle density in a region of space is compensated for by a net change in flux into that region. Then write down the time-dependent Schrödinger wave equation for free particles. Eventually, by way of analogy you get the expression of the current density or flux.

2. The wave function for a particle is given by :

ψ(x) = A ejkx + B e-jkx What flux does this represent . 3. What is the flux associated with a particle described by the eigen function : ψ(x) = U(x) ejkx Where U(x) is a real function . 4. Show that the flux , the particle current density J may be written as : J = (1/2m) [ψ*P ψ + (ψ*Pψ)* ] Where P is the linear momentum operator. 5. For the step potential barrier shown in figure, solve Schrödinger wave equation and find :

i- The eigen-function of the system. ii- Both the reflection and transmission coefficient

Hint: Assume the free-particle Schrödinger wave equation to possess a form of propagating waves , incident , reflected , and incident . V(x) Ji Vo Jt Jr x 0


30

Assignment # 3 1. Derive an expression for the density of available energy states of a particle trapped

inside a three dimensional potential well and Fermi-Dirac statistics. 2. Find the total number of electrons per unit volume obeying Fermi-Dirac statistics at

0oK, from which deduce an expression for the Fermi energy Ef . 3. Find the average energy of electrons obeying Fermi-Dirac statistics inside solids at

0oK. 4. Prove that the average energy of an electron inside a metal at 0ok is equal to three

fifths of the Fermi level. Find the average energy of an electron at 0oK inside silver if it is a mono-valent and its gram molecular volume is 10.28 cm3. Given that:

Avogadro’s number Nav:6.023*1023 mole-1 Planck’s constant h :6.626*10-34 J.s 5. If the E-K diagram of the conduction band is approximated by: E = AK2 + B Find the effective mass.

notes on quantum mechanics - alexandria …eng.staff.alexu.edu.eg/~mbanna/solis_state...

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