ch 4. using quantum mechanics on simple systems ms310 quantum physical chemistry - discussion of...

Ch 4. Using Quantum Ch 4. Using Quantum Mechanics on Simple SystemsMechanics on Simple Systems

MS310 Quantum Physical Chemistry

- Discussion of constrained and not constrained - Discussion of constrained and not constrained particle motion particle motion ex) free particle, In 2-D or 3-D boxes, In vice versa ex) free particle, In 2-D or 3-D boxes, In vice versa

- Continuous energy spectrum of Q.M free particleContinuous energy spectrum of Q.M free particle

- Discrete energy spectrum and preferred position of - Discrete energy spectrum and preferred position of Q.M particles in the box (Quantized energy levels)Q.M particles in the box (Quantized energy levels)


4.1. The free particle

Free particle : no forcesClassical 1-dimension, no forces : 02

2

dt

xdmmaF

Solution : x = x0 + v0tx0 ,v0 : initial condition, constants of integrationExplicit value : must be known initial condition

How about the free particle in Q.M?

Time-independent Schrödinger Equation in 1-dimension is

)()()()(

2 2

22

xExxVdx

xd

m


constant V(x) : can choose the reference V(x)=0 (absolute potential reference doesn’t exist) → reduced to )(

2)(22

2

xEm

dx

xd

Solution is given by

ikxxmEi

ikxxmEi

eAeAx

eAeAx

)/2(

)/2(

2

2

)(

)(

Use these notations2/2

2,2

mE

pkmEp

Obtain Ψ(x,t) : multiply each e-i(E/ )tℏ or equivalently e-iωt (E = ℏω)

Eigenvalue : not quantized(all energy allowed : k is

continuous variable)m

kE

2

22

ikxikx eAeAxxx )()()(


Plane wave cannot be localized. → cannot speak about the position of particle.

Then, what about probability of finding a particle? → also cannot calculate (wave function cannot be normalized in interval -∞ < x < ∞)

However, if x is ‘restricted’ to the interval –L ≤ x ≤ L then

L

dx

dxeeAA

dxeeAA

dxxx

dxxxdxxP L

L

ikxikx

ikxikx

L

L

2)()(

)()()(

*

*

P(x) : independent of x → no information about position


)()()(ˆ

)()()(ˆ

xkekAeAx

ixx

ixp

xkekAeAx

ixx

ixp

ikxikxx

ikxikxx

What about the momentum of particle?

ψ+(x) : state of momentum + k(positive direction)ℏ ψ-(x) : state of momentum – k(negative direction)ℏ

4.2 The particle in a One-dimensional box


1-dimensional box : particle in the range 0<x<a only impenetrable : infinite potential

V(x) = 0 for 0 < x < a = ∞ for x ≥ a , x ≤ 0


Schrödinger Equation is changed by

)(])([2)(

22

2

xExVm

dx

xd

If ψ(x) ≠ 0 outside the box, then value of ψ’’(x) becomes infinite because value of V(x) is infinity outside the box.

However, 2nd derivative exists and well-behaved → ψ(x) must be 0 outside the box boundary condition : ψ(0) = ψ(a) =0

Inside the box : same as the free particle

We can write the solution by the sin and cos.

kxBkxA

kxAAikxAA

kxikxAkxikxA

eAeAx ikxikx

cossin

sin)(cos)(

)sin(cos)sin(cos

)(

): (noticesin)( nxa

nAxn

From the consideration of boundary conditions

a

nk

nnka

kaAa

kxAx

B

a

,3,2,1,0,

0sin)(

sin)(

00)0(

0)()0(

2

0

2

0

2

0

22

2

1

2

12cos1

2

1

sin)()(1

aAxAdxa

xnA

dxa

xnAdxxx

aa

a

nn

)sin(2

)( , 2

a

xn

ax

aA n

Normalization

2

2222

22

2

22

82

)()(2

)(2

)(ˆ

ma

hn

a

n

mE

xExa

n

mx

dx

d

mxH

n

nnnnn

Energy of the particle

‘Quantization’ arises by the boundary condition


Particle is ‘quantized’, n : quantum number

Ground state : n=1However, energy of n=1 is not zero : zero point energy(ZPE)

particle in a box : ‘stationary’ wave(not a traveling wave)

Also, n increase → # of node increase → wave vector k increase because

Finally, what about a classical limit? → same as result of C.M(same probability in everywhere)

2

222

1 82 ma

h

amEn

2/22

mEp

k


graph of ψn(x) and ψn*(x)ψn(x)


Graph of ψn2(x) / [ψ1

2(x)]max

n increase : large energyLower resolution : cannot precise measure → near to C.MResult of Q.M ‘approach’ to the C.M when classical limit


4.3 Two- and Three- dimensional boxes

Boundary condition : similar to 1-dimensional box

V(x, y, z) = 0 for 0 < x < a, 0 < y < b, 0 < z < c = ∞ otherwise

Inside the box, Schrödinger Equation is given by

),,(),,()(2 2

2

2

2

2

22

zyxEzyxzyxm

Solving by separation of variable

And equation is changed by

)()()(

))()()()()()()()()((2 2

2

2

2

2

22

zZyYxEX

zZdz

dyYxXyY

dy

dzZxXxX

dx

dzZyY

m


Edz

zZd

zZdy

yYd

yYdx

xXd

xXm )

)(

)(

1)(

)(

1)(

)(

1(

2 2

2

2

2

2

22

Divide both side by X(x)Y(y)Z(z)

E : independent to coordinate → E = Ex + Ey + Ez and original equation(PDE) reduced to three ODEs.

)()(

2),(

)(

2),(

)(

2 2

22

2

22

2

22

zZEdz

zZd

myYE

dy

yYd

mxXE

dx

xXd

m zyx

Solution of each equation is already given.

c

zn

b

yn

a

xnNzyx zyx

nnn zyx

sinsinsin),,(

And energy is given by )(8 2

2

2

2

2

22

c

n

b

n

a

n

m

hEEEE zyxzyx


Normalization

2

000

2

0 00

2

0

2

0

2

0

22

8

1

2

1

2

1

2

1

2cos1

2

12cos1

2

12cos1

2

1

sinsinsin1

abcNzyxN

dzc

xndx

b

yndx

a

xnN

dzc

zndy

b

yndx

a

xnNd

cba

a cz

b yx

cz

b yax

nn

...)3,2,1:,,(

sinsinsin8

),,(,8

zyx

zyxnnn

nnnc

zn

b

yn

a

xn

abczyx

abcN

zyx


If total energy is sum of independent terms → wave function is product of corresponding functions

Solution has a three quantum numbers : nx, ny, nz

→ more than one state may have a same energy : energy level is degenerate and # of state is degeneracy

ex) if a=b=c, energy of (2,1,1), (1,2,1), and (1,1,2) is same.

in this case, state (2,1,1), (1,2,1) and (1,1,2) is degenerate and degeneracy of the level is 3.

2-dimensional box problem : similar to 3-dimensional problem(end-of-chapter problem)

2

2222

2

2

2,1,1 4

3)211(

8 ma

h

ma

hE

4. Using the postulate to understand the particle in the box and vice versa


Postulate 1 : The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.

We see the postulates of Q.M using the particle in a box.

Ex) 4.2 ψ(x) = c sin (πx/a) + d sin (2πx/a) a. Is ψ(x) an acceptable wave function of particle in a box? b. Is ψ(x) an eigenfunction of the total energy operator Ĥ? c. Is ψ(x) normalized?


Sol) a. Yes. ψ(x) = c sin (πx/a) + d sin (2πx/a) satisfies the boundary condition, ψ(0) = ψ(a) = 0 and well-behaved function. Therefore, ψ(x) is acceptable wave function.

b. No.

Result of Ĥψ(x) is not ψ(x) multiplied by constants. Therefore, ψ(x) is not a eigenfunction of the total energy operator.

c. No

)2

sin4sin(2

)2

sinsin(2

)(ˆ2

22

2

22

a

xd

a

xc

maa

xd

a

xc

dx

d

mxH

dxa

x

a

xdccddx

a

xddx

a

xc

dxa

xd

a

xc

aaa

a

2sinsin)(

2sin||sin||

|2

sinsin|

0

**

0

22

0

22

2

0


Third integral becomes zero because of orthogonality.

)|||(|2

]2

1[||]

2

1[||

2sin||sin||

220

20

2

0

22

0

22

dca

dc

dxa

xddx

a

xc

aa

aa

Therefore, ψ(x) is not normalized. However, the function

is normalized when |c|2+|d|2=1

Superposition state depends on time. Why?

Therefore, this state doesn’t describe the stationary state.

]2

sinsin[2

a

xd

a

xc

a

)()(]2

sinsin[2

),( // 21 tfxa

xde

a

xce

atx tiEtiE


Then, what about a probability of particle in the interval?

Ex) 4.3probability of ground-state particle in the central third?

Sol) ground state : a

x

ax

sin2

)(1

609.0)]3

2sin

3

4(sin

46[

2sin

2)()(

3/2

3/

23/2

3/

1*1

aa

adx

a

x

adxxxP

a

a

a

a

Probability of finding a particle in central third is 60.9%.

However, we cannot obtain this result by one individual measurement. We can only predict the result of large number of experiment(60.9%).


Postulate 3: In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.

Postulate 4 : If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by

dxtxtx

dxtxAtx

a

),(),(

),(ˆ),(

*

*

We can understand the two postulates together.


2) wave function is not an eigenfunction of operator. → each measurement gives different value

Ex) normalized superposition state

2

222

1

111

22

12

22

1

82

)()(2

)(2

)(ˆ

ma

h

amE

xExam

xdx

d

mxH

1) wave function is an eigenfunction of operator. → all measurement gives same value and it is average value

Ex) ground state of particle in a box

1||||],2

sinsin[2

)( 22 dca

xd

a

xc

ax


]2

sinsin2

sinsin[2

]2

sin||sin|[|2

)2

sinsin](2

[)2

sinsin(2

)()](2

)[()(ˆ)(

00

1*

2*

0

22

2

0

21

2

2

22

0

**

2

22**

aa

aa

a

dxa

x

a

xEcddx

a

x

a

xdEc

a

dxa

xEddx

a

xEc

a

a

xd

a

xc

dx

d

ma

xd

a

xc

a

dxxxVdx

d

mxdxxHxE

Last two integrals are zero by orthogonality and final result is

22

12

0

22

2

0

21

2 ||||]2

sin||sin|[|2

EdEcdxa

xEddx

a

xEc

aE

aa

where2

22

8ma

hnEn


However, result of individual experiment is only E1 or E2 by the postulate 3. How can represent the result?

→ by the postulate 4, result of the large number of individual experiment, probability of E1 is |c|2 and probability of E2 is |d|2, and the ‘average value’ of energy <E> = |c|2E1| + |d|2E2.

More generally, we can think about this case

Ψ(x) = cΨ1(x) + dΨ2(x) + 0(Ψ3(x) + Ψ4(x) + …)

All coefficient except Ψ1(x) and Ψ2(x) is zero. Therefore, no other energy is measured except the E1 and E2.


0)0sin(sin2

cossin2

)](sin[sin2

)(ˆ)(

sin2

)(

222

02

00

*

na

nidx

a

xn

a

xn

a

ni

dxa

xn

dx

di

a

xn

adxxpxp

a

xn

ax

a

aa

Now, consider the momentum.

We know and can calculate the average value of Momentum of nth state.

dx

dipx ˆ

In the Q.M, momentum of particle : cannot be zero(energy E = p2 / 2m cannot be zero in Q.M)

→ average of two superposition state is zero!

We can rewrite the wave function by complex form.

(use the )i

eex

ixix

2sin


axinaxinaxin

axinaxinaxin

axinaxin

ea

ne

dx

diep

ea

ne

dx

diep

i

ee

aa

xn

ax

///

///

//

ˆ

ˆ

)2

(2

sin2

)(

In the case of momentum, two probability of positive momentum and negative momentum is same. Therefore, the average value seems to zero.


Ex) 4.4Particle in the ground state. a. Is wave function the eigenfunction of position operator? b. calculate the average value of the position <x>.

Sol) a. position operator :

a

x

ax

sin2

)(

)(sin2

sin2

)(,ˆ xca

x

ac

a

xxa

xxxx

Therefore, wave function is not an eigenfunction of position operator.

b. expectation value is calculated by postulate 4.

2]

8)0

84[(

2]

4

2sin

)(8

2cos

4[

sin2

sinsin2

)(ˆ)(

2

2

2

22

02

2

0 0

2

0

*

aaaa

aa

a

xx

a

a

xx

dxa

xx

adx

a

xx

a

x

adxxxxx

a

a aa

Average value of particle is half, the expected position.


- The motion of particle which is not constrained shows continuous energy spectrum however, the particle in a box has a discrete energy spectrum.

- The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.

- In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.

Summary Summary


If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by

dxtxtx

dxtxAtx

a

),(),(

),(ˆ),(

*

*

ch 4. using quantum mechanics on simple systems ms310 quantum physical chemistry - discussion of...

Documents