Wave-Particle Duality Leads to yUncertainty Principle

C t k iti & t Cannot know position & momentum or energy & lifetime precisely

Position (x) & Momentum (px) Uncertainty

2xx p

Energy (E) & Lifetime (t) Uncertainty

E 2

E t

Uncertainty in Electron PositionUncertainty in Electron PositionAn electron moving near an atomic nucleus has ga speed of 2 x 106 m/s. What is the uncertainty in its position?

Answer: Δx 3 x 10–10 m = 300 ÅThis speed is approximately that of a 1s l t i H d Δ i 300 t th electron in H, and Δx is 300 x greater than

the diameter of the atom (10–10 m), so we have no precise idea where in the atom the pelectron is located!

This result has profound implications for the atomic This result has profound implications for the atomic model - we cannot assign fixed paths for electrons.

Postulates for Quantum MechanicsPostulates for Quantum Mechanics

Consider a system with N degrees of freedom.y g1. There exists a function (q1 . . . qN ,t) that contains all information available for this system. It is called the state function or wave function Here q represents state function or wave function. Here q represents coordinate(s). Properties of :a. * is a probability density (* means complex conjugate)

“B h h i ”“Born hypothesis”

b. (normalized)* 1dq ( )

c. must be continuous.

q2

2, , and iq q

d. (q1 . . . qN,t) must be a single valued function of its

arguments

iiq q

NormalizationNormalizationThe probability of finding a particlehe probability of finding a particle

in volume element dx dy dz is proportional to 2(x y z) dx dy dz (x,y,z) dx dy dz

Total probability of finding the particle Total probability of finding the particle anywhere in space is unity

Requirement of unity probability leads to normalization: * d = 1

Jargon!* 1dq 2

2, , and

Jargon!q 2iiq q

Which of the following functions is an acceptable state function?Which of the following functions is an acceptable state function?

1. 2( ) a > 0 axf x e x

2

2( ) a < 0 axf x e x 2.

(edge bc ?)

5.3.

6.4.

7. ( ) a > 0 axf x e x

Operators H E ˆ ˆ ˆH L V p Operators H E

An operator is an instruction telling you what to do with a function,

H, , ,L V p

p ymuch like a function is an instruction telling you what to do with a number.Examples of operators include d/dx, *, ,, etc. We will denote an operator with a "hat" over the symbol, e.g. ÂS if  th  f( ) So if  = , then  f(x) =

Notation: ÂĈf(x) Â(Ĉf(x))Ĉ2 f(x) Ĉ [Ĉf(x)]

f x

Ĉ2 f(x) Ĉ [Ĉf(x)](Â + Ĉ) f(x) Â f(x) + Ĉ f(x)

The ORDER of operators can make a difference! A very important concept!The ORDER of operators can make a difference! A very important concept! Ĉ f(x) ≟ Ĉ  f(x)

Insuring your carRaise your left arm

Having an automobile accidentRaise your right army

Multiply by xMultiply by x

Ra se your r ght armMultiply by y sin3yDifferentiate with respect to x

Postulates for Quantum Mechanics2. For every physical observable A, there corresponds a linear Hermitian operator Â.

The operator  is obtained by writing A classically in terms of cartesian coordinates and momenta, and then replacingp g

, , , , , x y zx x y y z z p i p i p ix y z

e g KE ½ mv 2 ½ (mv )2/m p 2/2m e.g., KEx = ½ mvx2= ½ (mvx)2/m = px

2/2m 2 2 2

2 22 2 and KE

2x x x xp p pmx x

3. If  corresponds to physical observable “A”, then fora system with state function every measurement

2mx x

a system with state function , every measurement of the observable “A” must yield one of the eigenvalues of Â.

More Jargon!More Jargon!

OperatorpLinear OperatorLinear Hermitian operatorCommuting operatorsCommuting operatorsEigenvalueEigenfunctionO h l f iOrthogonal functions

Eigenvalue Equationsg qH E

The operator equation Â(f(x)) = c f(x), where c is a constant, is called an eigenvalue equation.

The function f(x) is called the eigenfunction of the operator Â, and the constant c is called the eigenvalue associated with f(x).

In quantum mechanics all physical properties are represented by operators and that the possible values of that property are related to the eigenvalues of that operator.

Thus much of our work will be related to finding eigenvalues and eigenfunctions of operators.

So look at when an operation returns a constant times the function:

#1ConcepTest #1

Is the function eax an eigenfunction of d/dx?Is the function eax2 an eigenfunction of d/dx?Is the function e an eigenfunction of d/dx?

A. no noA. no no

B. no yes

C yes noC. yes no

D. yes yes

ConcepTest #2ConcepTest #2

Is the function cos(ax) an eigenfunction of d/dx?Is the function cos(ax) an eigenfunction of d2/dx2?

A. no no

B. no yes

C. yes no

D. yes yes

Back to postulates

Postulates for Quantum MechanicsPostulates for Quantum Mechanics

3 If  corresponds to physical observable A then for3. If  corresponds to physical observable A, then fora system with state function , every measurement of the observable A must yield one of the eigenvalues of Â.

3a) If many repeated measurements are made of A, then fora system with state function ,the average value(or expectation value) of A, A, is given by the integral

* A AA dq qThe expectation value need not be any of the eigenvalues of Â, just as the average weight of our class need not bej g gthe weight of any one of us.

Examples to come . . .

Postulates for Quantum MechanicsQ

4. The time evolution of (q, t) is given by

where is the Hamiltonian operator.

h l l l d h l

(q,t)ˆ (q,t) = it

The classical Hamiltonian corresponds to the total energy(kinetic and potential) of an isolated system. This equation is the time-dependent Schrödinger equation.q m p g q

(Weird symbols and operations should become clear in future lectures)We will see shortly that when we wish to calculate the stationary statesof a system the time-dependent Schrödinger equation for a singleof a system, the time dependent Schrödinger equation for a singleparticle of mass m moving in one dimension reduces to the simplertime-independent Schrödinger equation,

2 2 ( ) ˆ2

(x)2m

2

2(x)d +V(x) (x) = E

dx ˆ E or

Total Energy is Sum of Kinetic gyEnergy & Potential Energy

E = Ek + Ep

H = E = px2/2m + Ep(x)px p( )

2 2 2 2

p2 2ˆH E (x) V(x)

2 2

p2 2( ) ( )2m 2mx x

is known as the Hamiltonian operatorH is known as the Hamiltonian operatorH2 2 2

2 In 3 dimensions:

2 2 2x y z In 3 dimensions:

Schrödinger EquationSchrödinger Equation

Time-independent:2

2 V(x y z) (x y z) E (x y z)

V(x,y,z) (x,y,z) E (x,y,z)

2m

H E

For many problems, such as those concerned with h f d l l the structure of atoms and molecules, we are not

concerned with time-dependent wave functions and energies.g

QM of "Particle in a Box"QM of Particle in a Box

Potential energy Ep= 0 within the box (0 x a) Potential energy Ep 0 within the box (0 x a) and Ep= outside the box (x<0 and x>a)

Particle in a Box

I II III

V(x) = V(x) =

V(x) = 0V(x) = 0

The QM postulates insist that (x) be continuous and single valued.We will solve the Sch. Eqn separately in regions I, II and III q p y gand then put the pieces together to obey the above constraints.

Particle in a Box (2)( )REGIONS I and IIIIn these regions, the Schrödinger equation is given by

22 (x)d

which implies (x) = 0 for x < 0 or x > aNow consider Region II, where V(x) = 0

2(x)d + (x) = E (x)

2m dx

Now consider Region II, where V(x) 0

so a solution in this region is

22

2(x)d = E (x)

2m dx

g

for 0 x a .

h h l l d l h b d d

sin cos2 22mE 2mE(x) = A x + B x

We now have the total solution, and must apply the boundary conditions. The state function (x) is certainly well behaved except possibly at x = 0 or x = a. To be continuous, we must require that the Region II solution vanish at these points Thus for x 0 vanish at these points. Thus, for x = 0,

2 22 20 sin 0 cos 0mE mEA B

Particle in a Box (3)From the first equation, we have

( )

2 22 20 sin 0 cos 0mE mEA B

or 0 = A (0) + B(1) so B = 0

2 2 20 i imE mE mEA B A and at x = a 2 2 20 sin cos sinmE mE mEA a B a A a

and at x = a,

There can only be TWO possibilities:There can only be TWO possibilities:

1) A = 0 , which says (x)= 0 for all x (no good) Why??

or2mE

2) 22sin 0mE a

Particle in a Box (4)Can we make this happen?? What parameters are available to us?

ONLY the Energy, E !!!!HOW? Wh th f th i f ti ?

( )

HOW? Where are the zeros of the sin function?We know that sin z = 0 for z = n , where n = 1, 2, 3, . . So we must have

22 where n = 1,2,3 mE a n

Solving for E,

2 wh r n , , a n

22 2 2 2

n 2 2n π n hE = = where n = 1,2,3, …2ma 8ma

2 22ma 8ma

and the state function is

sin for 0 , and 0 otherwisen nn xx A x a

a

Particle in a Box (5)The imposition of boundary conditions and periodic motion has forced quantized energy states upon us!!

( )

q gy pThe Born Postulate says that * is a probability density and thus its integral over all space must equal 1.

a * 2 2 2

0

( ) ( ) 1 sin2

a

n nn x ax x dx A dx A

a

So 2nA

a

What do these functions look like???Particle in a Box n and n

2

We next plot n and n2 for the first few states

1,2,3 for Particle in a Box

Postulates for Quantum MechanicsPostulates for Quantum Mechanics

3 If  corresponds to physical observable A then for3. If  corresponds to physical observable A, then fora system with state function , every measurement of the observable A must yield one of the eigenvalues of Â.

3a) If many repeated measurements are made of A, then fora system with state function ,the average value(or expectation value) of A, A, is given by the integral

* A AA dq qThe expectation value need not be any of the eigenvalues of  just as the average weight of our class need not beof Â, just as the average weight of our class need not bethe weight of any one of us.

Examples . . .