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25/09/2014

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FI 3103 Quantum Physics

Alexander A. Iskandar

Physics of Magnetism and Photonics Research Group

Institut Teknologi Bandung

Basic Concepts in Quantum Physics

Probability and Expectation Value

Heisenberg Uncertainty Principle

Wave Function in Momentum Space

Alexander A. Iskandar Basic Concepts in Quantum Physics 2

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2

Review on Probability Recall the concept of probability density function.

Consider the following example • A manufacturer of insulation randomly selects 20 winter days and

records the daily high temperature

24, 35, 17, 21, 24, 37, 26, 46, 58, 30,

32, 13, 12, 38, 41, 43, 44, 27, 53, 27

• Put in a class table


Class Ti Freq. Rel. Freq. (fi)

10 < T 20 15 3 0.15

20 < T 30 25 6 0.30

30 < T 40 35 5 0.25

40 < T 50 45 4 0.20

50 < T 60 55 2 0.10

N = 20 1.00

0

5

10

5 15 25 35 45 55 MoreF

req

ue

nc

y

Histogram: Highest Temperature

Probability Density Function

Review on Probability Consider the following example

• A manufacturer of insulation randomly selects 20 winter days and records the daily high temperature

24, 35, 17, 21, 24, 37, 26, 46, 58, 30,

32, 13, 12, 38, 41, 43, 44, 27, 53, 27

• The average temperature can be calculated using the probability density function as

thus,


.)..)(.(

))(.(

functdistprobvaluemid

dataofnumber

frequencyvaluemidaverage

N

fPPT

N

fTT i

iii

ii

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Review on Probability • One other important statistical quantity is the standard deviation,

or,


22

22

22

2

22

2

2

2

TTTT

PTPTTPT

PTTTT

N

fTTTT

iiiii

iii

ii

222 TT

Probability Interpretation of Wave Funct. As stated by Born, the modulus of the wave function

of a particle is interpreted as the probability density function associated with the particle

Then the wave function has to satisfy the following

However, one can always normalize the wave function by multiplying it with a constant, .

Hence the condition needed to be satisfied by the wave function is that the initial state of the wave function must be a square integrable function


dxtxdxtxP2

,),(

),( tx

),( tx

1,),(2

dxtxdxtxP

),( txN

dxx2

0,

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Probability Interpretation of Wave Funct. Example : Normalize the wave function

To normalize the wave function, we multiply it with a number N, and impose the normalization condition

Thus, the normalized wave function is


elsewhere

LxxLxtx

0

0

)(),(

25

52

0

22342

222

303

1

4

2

5

12

,,1

NL

LNdxxLLxxN

dxtxNdxtxN

L

LxxLxL

txNtx 0,)(30

),(),(~5

Expectation Value from Probability In analogy with probability concept, the expectation value of

the particle’s position is

Or in general expectation value of any function f(x) should be calculated as

And uncertainty of the particle’s position measurement is


dxtxxtxdxtxxdxtxxPx ,,,),( *2

dxtxxftxxf ,)(,)( *

2

*2*

222

),(),(),(),(

dxtxxtxdxtxxtx

xxx

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Expectation Value from Probability Example : For the normalize the wave

function

Calculate , and hence x.


elsewhere

LxxLxtx L

0

0

)(),( 5

30

26

1

5

2

4

1302

30

)(30

,,

0

5432

5

0

23

5

*

LLdxxLxxL

L

dxxLxL

dxtxxtxx

L

L

x 2x

Expectation Value from Probability Example : For the normalize the wave

function

Calculate , and hence x.


elsewhere

LxxLxtx L

0

0

)(),( 5

30

7

2

7

1

6

2

5

1302

30

)(30

,,

22

0

6542

5

0

24

5

2*2

LLdxxLxxL

L

dxxLxL

dxtxxtxx

L

L

x 2x

724

1

7

222 LLxxx

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Conservation of Probability If we normalize the wave function at one time, will it stay

normalized? I.e. does

hold for all time? In short, is probability conserved?

Take a time derivative of the probability density function

Use the Schrodinger equation to replace the time derivative and assume that the potential function V(x) is real.

Recall the complex conjugate of the Schrodinger equation


1,2

dxtx

t

txtxtx

t

tx

t

tx

t

txP

),(

),(),(),(),(),(

2

),()(),(

2

),(2

22

txxVx

tx

mt

txi

Conservation of Probability Then

Define the probability flux or probability current as

Hence, we get


xximx

xxmi

t

txtxtx

t

tx

t

txP

2

2

1

),(),(),(

),(),(

2

2

2

22

xximtxj

2),(

0),(),(

x

txj

t

txP

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Conservation of Probability If we integrate over all space, we get

The last step follows from the fact that for square integrable function j(x,t) vanishes at .


0),(

),(

),(),(2

dxx

txj

dxt

txP

dxtxPt

dxtxt

Probability Current

The last relation is similar to the continuity equation found in classical mechanics or electromagnetism, which states that probability is conserved not only globally but also locally.

It means that if the probability of finding particle at a certain point decreases, this probability does not only turns up at another point, but instead it flows to this other region.

Hence the name probability current for


0),(),(

x

txj

t

txP

xximtxj

2),(

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Expectation Value of Momentum How do we calculate the expectation value of the momentum

of the particle associated with the wave function (x,t)?

Recall the classical expression for momentum

Take the expectation value of this expression, yields

Note that, the position x does not have a time dependence.

The time dependence of comes from the time dependence of (x,t).


p

dt

dxmmvp

dxt

txxtxtxx

t

txm

dxtxxtxdt

dmx

dt

dmp

),(),(),(

),(

),(),(

x

Expectation Value of Momentum Use the Schrodinger equation, to obtain

Note that


dxx

txxtxtxx

x

tx

i

dxt

txxtxtxx

t

txmp

2

2

2

2 ),(),(),(

),(

2

),(),(),(

),(

2

2

2

2

xx

xxx

x

xxx

xx

xx

xxx

xxx

x

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Expectation Value of Momentum Thus,

Since the wave function is square integrable function, it means that it vanishes at , hence the first term does not contribute to the evaluation.

Thus, we have

which suggest to associate the momentum to the operator


dxxix

xxxi

dxxi

dxx

xxxxi

p

2

222

dx

xip

xip

Expectation Value of Momentum Example : For the normalize the wave

function

Calculate , and hence p.


elsewhere

LxxLxtx L

0

0

)(),( 5

30

p 2p

03230

)2)((30

),(),(

0

223

5

0

5

L

L

dxxLLxxiL

dxxLxLxiL

dxtxxi

txp

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10


function



elsewhere

LxxLxtx L

0

0

)(),( 5

30

p 2p

2

2

0

2

5

2

0

5

2

2

222

1022

30

)2)((30

),(),(

LdxLxx

L

dxxLxL

dxtxx

txp

L

L


function



elsewhere

LxxLxtx L

0

0

)(),( 5

30

p 2p

Lppp

10

22

72

Lx Recall the previous results of , then

Which is consistent with .

6.072

10

L

Lxp

2 xp

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Expectation Value of Momentum The expectation value of position x is a real value, as can be

easily seen from the definition,

However, because of the form of momentum operator that involves the imaginary number as well as a differentiation,

will the expectation value of the momentum be a real value?

In fact, it can be proved that the momentum expectation value is always a real number.


dxtxxdxtxxPx2

,),(

dxtx

xitxp ),(),(

Expectation Value of Momentum

In the last step, the square integrability property of the wave function, (x,t), has been used, where it states the the wave function is a localized function so that (x,t) 0 as x .


0),(),(),(),(

),(),(

),(),(*

),(),(

),(),(*

),(),(*),(),(*

**

*

*

*

txtxi

dxtxtxxi

dxx

txtx

x

txtx

i

dxx

tx

itx

x

tx

itx

dxtxxi

txdxtxxi

txpp

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Heisenberg Uncertainty Principle The result of the last example on product of the expectation

values of x and px, it was found that

This relation is called the Heisenberg Uncertainty principle.

It states one of the fundamental concept of quantum world, that is measurement of position and linear momentun cannot be done simultaneously with the highest accuracy.

If measurement of position is done very accurately, x = 0, then the value of the linear momentum is not known, since according to Heisenberg uncertainty principle yields px , and vice versa.

Position and momentum are said to complementary variables.


2

xpx

Heisenberg Uncertainty Principle One observation can be used to

see this principle.

From wave optics, the spread of the diffraction pattern as

Considering the light as photons, at the slit, the position of the


2

2

2xp

2xpa

photon is within the measurement’s uncertainty of x a, but the momentum’s uncertainty has spread.

Thus

where .

hxppapap

pxxx

x

x

2 kpx

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Heisenberg Uncertainty Principle Note that the Heisenberg uncertainty principle between the

position and momentum holds for all direction (x, y and z).

One other Heisenberg uncertainty relation is the uncertainty between energy and time


2

tE

E

t

21

E

ground state

excited state

E

ground state

excited state

Wave Function in Momentum Space Recall that the spectral distribution function of the wave

packet is none other than the Inverse Fourier transform of the wave function at t = 0,

Calculate the following


dxexppx

i

)0,(

2

1)(

1)()(

)(2

1)(

)(2

1)()()(

dxxx

dxdpepx

dpdxexpdppp

pxi

pxi

25/09/2014

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Fourier transform of a normalized wave function is normalized.

Next, consider the following

Which is a statement that the momentum expectation value can also be calculated from (p) using the momentum operator p itself.


1)()()()(

dxxxdppp

dppppdpdxexpp

dxdpepdx

d

ix

dxxxi

xp

pxi

pxi

)()()(2

1)(

)(2

1)(

)()(


This last statement is similar to

Thus if (x,t) is the wave function in spatial domain, (p) should be interpreted as the wave function in momentum space, with the probability density function of finding a particle with momentum p is given by |(p)|2.

In this momentum space, the position operator is given by

Hence, the expectation value of position is


dppppp )()(

dxtxxtxx ,,*

dpp

pipx )()(

pi

25/09/2014

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Wave Function in Momentum Space Example 2.4

Consider a particle whose normalized wave function is


elsewhere

Lxxetx

x 0

0

2),(

x 2x

p 2p

a. For what value of x does P(x) = |(x)|2 peak?

b. Calculate and .

c. What is the probability that the particle is found between x = 0 and x = 1/?

d. Calculate (p) and use this to calculate and .

x 1/

(x)

Summary Physical quantity (an observable) is represented by an

operator.

Measurement of observable is evaluated as calculating expectation value

There are two ways to calculate the expectation value, in spatial space or in momentum space.


dxtxtxquantityphys ),(ˆ

),(. O

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Summary


Spatial space Momentum

space

Wave function

Position

Momentum

),( tx )( p

xi

pi

x

p

dpeptxEtpx

i)(

)(2

1),(

dxexppx

i

)0,(

2

1)(

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