Review of Quantum Mechanics Part 1 for Students

Review of Quantum Mechanics p1 Tien-chang Lu

RREEVVIIEEWW OOFF QQUUAANNTTUUMM TTHHEEOORRYY

Content

Chapter 1 Schrdinger Equation -- Approach to quantum mechanics 1-1 Origin of Schrodinger Equation 1-2 Time independent and dependent Schrdinger Equation 1-3 Physical Interpretation and Implication of Schrdinger Equation

Chapter 2 Applications of Schrdinger Equation to Time Independent Potential 2-1 Free Particle in 1D space 2-2 A particle in a 1D infinite potential well 2-3 Particle in 3-D Box (with infinite potential walls)

2-4 Position wave function and momentum wave function -Fourier transform

2-5 Step Potential Wall 2-6 Single Potential Barrier: tunneling effect

Chapter 3 The Hydrogen atom 3-1 Schrodinger equation and solution of Hydrogen atom 3-2 Angular momentum 3-3 Zeeman effect 3-4 Spin of electron 3-5 Spin-orbit coupling Chapter 4 Total Angular Momentum Chapter 5 Harmonic Oscillator


Chapter 1 Schrdinger Equation -- Approach to quantum mechanics

1-1 Origin of Schrodinger Equation key observable properties of a particle mass m position x, y, z momentum px py pz energy E Based on the classical wave equation and De Broglies matter wave concept, Schrdinger postulated

[wave equation] must be valid and applicable in describing the matter wave like particle, regardless of the specific energy or momentum

We except the solution to the wave equation must have the familiar form (a) ( )( , ) j kx tx t Ae --------------- (1) (assume 1D case) (b) The wave equation contains appropriate derivatives of with respect to

coordinate and time 22

, ,d d ddx dtdx

(c) Wave equation must be valid independent of the specific energy or momentum total energy of particle

2

2PE Vm

(2)

kineticenergy Potentialenergy


(1) Lets take the first derivatives of with respect to time

( )j kx td jA edt

Energy ( E ) (2) Second derivatives with respect to coordinates

22 ( )

2j kx tAk e

x

2 2( )k P Therefore, we can set the wave equation as

22( )a bVt x

2

2PE Vm

(3) a, b constants to be determined substitute (1), (2) into (3) We obtain 2a j m

, jb The wave equation becomes

2 222j Vt m x

1D

22

2j Vt m

3D TimedependentSchrdinger Eq.


Solution of Schrdinger equation ( , ) ( ) (t)x t x X We obtain (1) ( , ) ( ) (t)x t x X

2

( ( ) ( )) ( ) ( )( ) ( )

( )hE jj t j th

j x X t E x X tt

X t Ej X tt

X t e e e

(2)2

2[ ] =E2 Vm

or 2

22

8( )( ) 0m E Vh

Time independent Schrdinger equation In classical mechanics, a particles total energy E can be expressed as

2 2 2 21 1( ) ( , ) P ( , ) ( , )2 2x y zE P P P V r t r t V r tm m Also the Hamiltonian of a system total energy i.e. 2 2 2x y z1 [P +P +P ]+V(r,t)2H E m Thus , (1) Schrdinger equation can be written in Hamiltonian form (operator form) as

( , ) ( , )r tj H r tt

where, 2

2( )2H Vm --- Hamiltonian operator

(2) Comparison of classical mechanics expression with the wave equation:


22

2

( )2

2

j Vt mpE Vm

Quantum Mechanics Classical

jt E 2 2 2P

V V x, y, z x, y, z m m Thus , Schrdinger proposed & defined Energy operator: E j

t

Momentum operator 2xp 2 2x

2yp 2 2y 2zp 2 2z

xp j x

yp j y

zp j z

Schrdinger equation in operator form becomes

E=H

observable!

Allobservablesareexpressedinoperationform

E:eigenvalueoftheoperatorH

eigenfunction


1-2 Time independent and dependent Schrdinger Equation Assume a solution of Schrdinger equation is

( , ) ( ) ( )x t x X t

Then , substitute it into Schrdinger equation We have (1)

( , ) ( , )x tj E x tt

1

( ) =Ej t j tX t e e

E (2)

2 22 ( )2 V x Em x

--- Time independent Schrdinger equation or If V is known, then all allowed and E can be calculated.

22[ ( )]2 V r Em

22

22

22

2

2 [ ] 08( )( ) 0

( ) 02

m E V

m E Vh

E Vm


Multiple solutions for & E

1 E1

2 E2

3 E3

Only very particular V(r) can be solved analytically e. g. Square Quantum well

Atom

Periodical atom ( ) ( )V x V x a

Parabolic well (Harmonic

oscillator)

12

2E

1E

a


1-3 Physical Interpretation and Implication of Schrdinger Equation (1) Schrdinger -- dont exactly know -- but can explain atomic spectrum well * -- cant explain photoelectric effect * -- cant fit plank formula (2) Max Born(1926) -- : probability amplitude of electron @ particular position in time

-- 2 physical probability density of the presence of associate particle

-- Atomic picture becomes

Bohr Characteristics of Schrdinger equation:

(1) Probability density suppose is a complex function of a particle 2* = = real quantity

2

r

x

0x

2


If 2 has a functional dependence (as previous pic.)

then 2* dx dx

then * .b

a

dx const In 3-D case (particle in confined volume-V)

3* 1dr --- Normalization 1 Dirac representation (bracket expression or Dirac notation)

( )x Kat one state in QM system

*( )x Bra conjugate of

(2) orthogonality In Cartesian coordinates 3-coordinates

1 2 3, ,e e e are orthogonal

jiji

ee ijji ,0,1

These unit vectors form a complete orthogonal set Similar iii EH

There are many 1 2,, eigen function

corresponding to 1 2,,E E eigen value

If s are normalized and form a complete orthogonal set then

* 3 1,0,i j ij

i jdr

i j


(3) Expectation value Define a position r for state (observable) < r > = expectation values

3 * 3 P(r,t)d ( , ) ( , )r r t r r t d r r 2 3rd r r similar

3*E j d r Et

3*( )P j d r From i i iH E ** i * * *i i i i i i i iH dV E dV E dV *j jE H dV H

H E Example If 122( ) sin x

L L 0 x L 1D infinite potential

*x x dx 2

0

2( ) sin ( ) 2L x Lx dx

L L

2( ) sin( ) cos( ) 0x xP j dxL L L L

1

2L


(4) Heisenberg's Uncertainty Principle (HUP)

Definition the root-mean-squared deviation from the mean

similar 2 2 2

x x xp p p where

*

2 * 2

22 2

2

j j

j j

jx j

jx j

x x dx

x x dx

p j dxx

p dxx

From previous example

122( ) ( ) sin( )xxL L

2 2 2 22

1 1 1( )3 2 4x x x L 0.17x L

2

2 2

2 2

2 2 2

2 2

( )2

22

x xx x x xx x x xx x xx x

2x


Similar

pL

So

2155.017.0

LLpx meet HUP!

(5) Probability density flow

For quantum system

--- (1) --- (2) We use )2()1( * to eliminate V

* 22 * 2 *

22 * *

2* *

2* *

[ ]2[ ]2[ ]2

[ ]2

jt m

jt m

m

m

The above equation is similar to the continuity equation of charge density

Jt 0Jt

2 * *[ ]2 Jt mj

22

* 22 * *

2

2

j Vt m

j Vt m

conjugate


* *[ ]=2J mj

Probability density current !!

We can also obtain probability density current from

* 3 31d r dr *

*

* *( )t t t t

From eqs. (1) and (2) and multiply the above eq. as * and with

t &

*

t

* *[ ( )]2t mj

J

(6) Non-Commuting Operators The momentum operator in the position representation is a differential operator p j

x

Thus [ , ]x p xp px

operator on

operator on x


Find [ , ]x p

or using bra-ket

or

[ , ]x p j

*( )xp px dx * ** *

* *

* * *

*

[ ( )]

[ ( )]

[ )]

[ , ]

xp px dx

j x x dxx x

j x x dxx x

j x x dxx x

j dx

jx p xp px j

j x xx x

j x xx x

j

[ , ]x p

( )( ) ( )

[ ]

xp px

x j j xx x

j x j xx x

j

[ , ]x p


Chapter 2 Applications of Schrdinger Equation to Time Independent Potential

2-1 Free Particle in 1D space If V = 0 throughout all space, then Schrdinger equation becomes

22 2

2 0d m Edx

Solution --- (1)

where

or --- (2) (1) All values of E0 are allowed (2) No quantization of E values (Continuous, not discrete energy) (3) All value of E


Parabolic curves

(6) De Broglie wave hp

2 2mEp k mE

mE2

2-2 A particle in a 1D infinite potential well

V V 0V

V

V=, xL : region III

Now, we list Schrdinger equations for all regions (I, II, III) In region I,

2 22 2

2 [ ( )] 0d m E V xdx --- (1)

E

k


Since in region II and III, V ,II III 0 to satisfy (1).

Inside the well 0V 2

2 22 0d m E

dx --- same as free particle

2jkx jkx

I Ae Be

mEk

Boundary conditions are (a) =0 x=0,L (b) * 1dx

From the above condition (1) We need to solve for coefficients A and B.

For non-zero solutions, we must have: 1 1 0jkL jkLe e 0

jkL jkLe e

sin( ) 0kL kL n , n is integer

nkL

So,

2 2 2 2 222 2n

k nEm mL

--- energy is quantized The wave function becomes

n nj x j xL L

n Ae Be

00 jkL jkL

A BAe Be

(2) (3)


From (2) , A+B=0 , A=B

n

[ ]=C [sin ]

n x n xj jL L

n A e en x

L

*

* 2 2

0 0

1

sin ( ) 1n n

L L

n n n

dx

ndx C x dxL

20

0

1 2sin [ sin ]2 4 2L

Ln x L n x nx LdxL n L L

122( )nC L

Thus

122( ) sin( )n n xL L Standing wave

121

122

123

2( ) sin( )2 2( ) sin( )2 3( ) sin( )

xL L

xL L

xL L

2 21 2

2 22 2

2 23 2

24292

EmL

EmL

EmL

Groundstate1stexcitedstate2ndexcitedstate


1

2

3

4

5

6

7

8

9

nE2 2

2( )2mL

21

22

23

1

2

3

0 x L

1E2E3E

4E

5E

6EnE

n


Physical interpretations gained from this example (1) For a confined particle, the solutions of Schrdinger equation

n n nH E

have a set of eigen functions n and eigenvalues nE where

(2) orthogonality of n exist for

*

0

L

n n mndx (3) Probability of finding the particle anywhere within the well n=1 : ground state

*1 1 1 1

0 2L Lx x x dx

n=21st excited state

2Lx

, but the highest probability density occurs at 3,4 4

L L n=32nd excited state

2Lx

, but the highest probability density occurs at 5, ,6 2 6

L L L (4) The allowed energies become discrete and separation becomes large as n

increases (5) As n increases to large number, 2n becomes near uniform distribution

12

2 2 22

2( ) sin( )

2

n

n

n xL LnE

mL


1 1 2 1

1 2 2 2

1 3 2 3

( ) cos sin( ) cos sin( ) C cos sin

x A k x A k xy B k y B k yz k z C k z

2-3 Particle in 3-D Box (with infinite potential walls) 0V inside the box V outside the box

Schrodinger equation inside the box

22 02 Em

( , , )x y z Separation of variables

set ( , , ) ( ) ( ) ( )x y z x y z Then

2 2 2 22 2 2

1 ( ) 1 ( ) 1 ( )[ ]2 ( ) ( ) ( )x y z E

m x x y y z z

Then 2 2

12

2 222

2 232

1 ( )2 ( )

1 ( )2 ( )

1 ( )2 ( )

x Em x x

y Em y y

z Em z z

1 2 3E E E E

Solutions are (based on previous 1D example)

y

z

x

L

L

L

V 0V

11 2

22 2

33 2

2

2

2

mEk

mEk

mEk


Boundary conditions require that 0 at (x,y,z)=0 thus A1=B1=C1=0

& 0 at (x,y,z)=L thus

2 22

1 22 2

22 2

2 22

3 2

( )2( )2( )2

x

y

z

E nmL

E nmL

E nmL

1, 2,31,2,31,2,3

x

y

z

nnn

2 22 2 2

, , 2 [ ]2nx ny nz x y zE n n nmL

The eigen function

, , sin( )sin( )sin( )yx znx ny nz xyzn yn x n zA

L L L

using 2 1dxdydz 1238( )xyzA L

322( ) sin( )sin( )sin( )yx zn yn x n zL L L L

Ex 2, 2, 3x y zn n n

2 2 2 22 2 2

322 232 223 2 2[2 2 3 ] 172 2E E E mL mL

degenerate states !!


However, the wavefunctions are different!

23322

23232

23223

2 3 2 2sin( )sin( )sin( )

2 2 3 2sin( )sin( )sin( )

2 2 2 3sin( )sin( )sin( )

x y zL L L L

x y zL L L L

x y zL L L L

Thus, the state is three-fold-degenerate! 2-4 Position wave function and momentum wave function - Fourier transform Sometimes it is desired to represent wave function in spatial frequency domain

1( ) ( )2jkxx k e dk

where ( ) Fourier transform of ( )

spatial frequency related to momentum by

k xk

p k

Thus, ( ) momentum wave functionk Similarly ( )k can be transformed back to position wave function ( )x by

1( ) ( )2jkxk x e dx


Example

22( )4

1 14 21( ) : Gaussian wave function

(2 )x

x e

Find and x Sol:

(1) 2

2( )212

1 0(2 )

x

x e xdx

(2)

2 2 2 2x x x x x 2 2x

x

Find p Sol:

2 214

1( ) ( )22( )

jkx

k

k x e dx

e

k 2 2

2 2

1 22

2122

( ) ( )

2( )

2( ) [ ]40

k

x

k k k dk

ke dk

e

0 xx

x


p k

2p

HUP

2 2x p

2p 2 22 2 2

22

22

[ ]1[ 0]4

4

kk k

2 2

2

1 2 22

1212

3

2

( ) ( )

2( )

( )2 2( ) 41

4

k

k k k dk

k e dk

2k


2-5 Step Potential Wall Apply to: Schottky barriers Thermionic emission Field emission

0

0

( ) ( )( ) 1, 0( ) : Heavyside step function ( ) 0, 0

0

V x V x

x xx

x xV

Assume 0 : incident energy


A B Cjk A B rC

We can express two coefficients &B C in term of A : rA B C

jk

2

-

C jkA jk rB jk rA jk r

From2 2

12 2

2

k rBA k r

where 12 tan rk

incident & reflected probability are equal

1 : phase different no transmitted probability

BA

@ 0x

2 2( )( ) 2 1 cos 2

standing wave

j kxjkxx A e e

x A kx

2k

2k

2

222

kA

x


2 2

2

2 2

1 0

02

( ) 42 ( ) 02

0 ( ) 2 1 cos 2 tan 1

0 when , (0) 0

x x Ak

x xk

x x A

VE

x V

:

:

:

:

Infinite barrier case!

ExamplePenetration depth

Assume the incident electron is traveling at a velocity 5 010 / sec into the barrier 2 at 0m V E x

2 21 21 4.56 10 2.85 102E m J eV

2

02

( )2

rxII x A e

m V Er

Define Penetration depth as 1e point at x d then 1rd

2 2 34

1/ 231 210

1.054 10 11.62 2 2 2 9.11 10 4.56 10d

m V E m E E

~2 lattice constantd If 0 10V E

2

2.92 10d m E E

0 2V E

I II


2-6 Single Potential Barrier: tunneling effect

0( ) ,00( ) 0,

V x V x ax

V xx a

I II III

:a Potential barrier width assume it is small! Case(1): 0E V

1 1

2 2

1

1 1 1 1 2

00 2 2 2 2 2

3 3

2Region I : 0, , 2Region II : , ,

Region III: 0, ,

jk x jk x

jk x jk x

jk x

mEV A e B e k

m E VV V A e B e k

V A e

1 22

only out-going wave

mEk

Boundary conditions:

2 2 1

2 2 1

1 1 2 2

1 1 1 2 2 2

2 2 3

2 2 2 1 3

0 jk a jk a jk a

jk a jk a jk a

A B A Bx

k A B k A B

A e B e A ex a

k A e B e k A e

1 1 2 2 3, , , , 5 unknowns, 4 equationsA B A B A By eliminating 2 2, ,A B

E

0 ax

( )V x

0V11

jk xA e

11jk xB e

13jk xA e

2

2

2 1

2

22 21 21

2 2 21 1 2 1 2

3 1 22 2 21 1 2 1 2

1 : Reflectance

4 : Transmittance

jk a

jk a

j k k a

jk a

k k eBA k k k k e

A k k eA k k k k e


Definition: Probability of transmissionTransmission coefficient Transmission coefficient

122 2 2 21 2 23

2 21 1 2

12 20 2

0

02 2

0 2 0

sin1 :Tunneling probability4

sin 1 44 - sin 4

Similar to FP interfere

k k k aATA k k

V k aE E V

E E VV k a E E V

nce!!

Reflection coefficient

1201

2 21 0 2

2 20 2

2 20 2 0

0

41 sinsin :Reflection probabilitysin 4

reflection exist even

E E VBRA V k a

V k aV k a E E V

E V

Of course 1R T

220

20

02 2

2 20 22

11 sin4

2

2

TV k a

E E V

m E Vk a a

E V k ama

max 2

min 220

0

22 202

2

1 @ , 1, 2,3...1 1 3 5 @ ' , ' , , ...2 2 21 4

1 11 2

T k a n n

T k a n nV

E E V

ma VE k a


Example: V0 = 3 eV, a = 10 A;

0 5 10 15 20

0.2

0.4

0.6

0.8

1.0

x axis is normalized to 02 2

22

E V

ma

Example: V0 = 1 eV, a = 5 A;

0 5 10 15 20

0.2

0.4

0.6

0.8

1.0

x axis is normalized to 02 2

22

E V

ma

1

/ 2 3 / 2 2 5 / 2 30

Resonant scatting

ak2


2

0

2 2

As

2

like a free electron!!

k aE V

mEk

Case(2): 0E V

From the above result,

2

202

0

11 sin4

TV k a

E E V

Since 0E V , 02 22 very smallm E Vk a a

So, we obtain 2

22

0

2k mE V

, and 2 2sin ,k a k a and 0E V

2 2

20 02 2

0

1 11 ( ) 14 2

TV mV ak a

E E V

20

2 depends on 2mV aT

Case(3): 0E V

022

2' m V Ek jk

Thus

12 20

0

02 2

0 0

sinh ' 1 44 sinh ' 4

V k aTE V E

E V EV k a E V E

Note:

2 2

Hyperbolic sin:sin sinh

1sinh 2x x

jk j k

x e e

102 2

0

41 sinh 'E V E

RV k a


(1) When potential barrier is weak i.e. ' 1k a 22sinh ' 'k a k a

2 20

0

1 1 '4

TV k a

E V E

02 22' m V Ek

2 2 00

20

22 20 00

22

1 21 41 1

11 22

Tm V EV a

E V E

ma V Vma VEE

2

00 2plot vs / for 2ma VT E V

(2) Strong potential barrier ' 1k a

2 2

1sinh , 121 2

1sinh 4

x x

x

x

x e e x

e

x e

22 '0

0

1 11 4 4k a

TV e

E V E

0.2 0.4 0.6 0.8 1EVo

0.2

0.4

0.6

0.8

1T


0 2 '2

clearly 0 exponential 1 decay!!

16 k aE V ET eV

Note that, for the original transmission equation: 0.1 1 5

Please refer to the following link: http://phet.colorado.edu/en/simulations/translated/zh_TW

12 20

0

02 2

0 0

sinh ' 1 44 sinh ' 4

2 5,1, 0.1,

V k aTE V E

E V EV k a E V E

mIf a then

0.2 0.4 0.6 0.8 1 EVo

0.2

0.4

0.6

0.8

1T

Documents

Review of Quantum Mechanics Part 1 for Students