quantum mantras

8/6/2019 Quantum Mantras

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QUANTUM MANTRAS

Review of and supplement to P.C.W. Davies, Quantum Mechanics (1984)

Frank Munley, June 2011

CHAPTER 1: Preliminary concepts

1. Quantum mechanics is weird in the sense that it describes properties of matter that dont seem to

match and indeed contradict what we observe or are habituated to at the human level. One hallmark of this

weirdness is wave-particle duality, i.e., depending on the experimental circumstances, it is sometimes

convenient to treat a physical system as having wave properties, and at other times as having particle

properties. How something can simultaneously behave like a wave and a particle is counterintuitive to the

extreme. For example, just as a beam of light shows an interference pattern when it passes through two

slits in an opaque screen, so an electron beam exhibits an interference pattern even though electrons behave

like particles in many s ituations. The De Broglie relationship between the particle property of

momentum and wave property wavelength is: p = h /"= hk, where h = h / 2" , h = Plancks constant,

and k= 2" /#.

2. Another manifestation of wave-particle duality is the relationship between energy and frequency:

E= h" = h#. AlthoughE= h" holds for both a material particle which has a rest mass and a photon whose

rest mass is zero, an important difference arises in the relationship between the wave numberkand the

frequency " for these two particles. For a photon, "= kv , i.e., "= kc where c = speed of light, while for

a material particle, "= (h / 2m)k2 . This important difference leads to the Schrdinger equation for material

particles stated below.

3. Uncertainty in frequency: Frequency is most directly measured by counting the number of wave peaks

passing a fixed point. If the counting lasts a time " , andNpeaks are counted, then " = N /# . But the

minimal uncertainty in the number is 1, so "# =1/$ . Measuring frequency by counting: = 1/, where

is the time taken to count. " can also be a measure of the uncertainty in time " t.

4. Items 2 and 3 lead to the Heisenberg uncertainty relation for energy and time is: "E"t # h , so it is

impossible to carry out perfectly accurate measurements of both energy and time interval over which the

measurement is made.

5. " is the wave function which can be used to describe the energy, position, momentum, etc. of a

material particle. All measurements give real numbers, but " is generally complex. So by itself, " is not

observable; but "2

links us to physical reality:

In one dimension, "(x, t)2dx = probability that a particle can be found (when a position

measurement is made) between x and x + dx. In three dimensions, "(r, t)2d# =

probability that a particle can be found in the volume element d" about the position r.

(N.B.:

d" is a common notation for a 3-D volume element and has nothing to do with

counting time.) The probability that a particle can be found in a limited region is given

by the integral of "(r, t)2d# over that region. In particular, in one dimension, the

probability that a particle can be found between x1 and x2 is "(x, t)2dx

x1

x2

# . Since

probabilities should sum to 1,

"(x, t)2dx = 1

x1

x2

# (or, in 3-D, "(r, t) 2d# = 1x1

x2

$ ) withthe integration carried out over all space. In this case, we say the wave function is

normalized.

6. Consider a particle localized on a one-dimensional line in a region of size

" x. Since the particle also has

a wave nature, the question arises: What range of wavelengths have to be added together to form a wave

function, or in this context a localized wavepacket, of size " x? A rather straightforward application of

Fourier analysis gives us the answer. To suggest how this works, suppose the particle is known to be in a


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one-dimensional line box of size " x and we want to determine what range of wavelengths that are

required to cancel outside the box but add together in the box. This can be done precisely using Fourier

analysis, but we can guesstimate the range of wavelengths needed as follows.

The diagram shows the one-dimensional box containing the particle. The wave function has the

constant value "0

inside the box and zero outside. Also shown are three sinusoidal waves, each wave

representing a particular possible momentum of the particle if the momentum is measured. The figuresuggests that the waves add together inside the box but outside sometimes add but sometimes cancel.

These waves help out, because they are all non-zero fromL/2 to +L/2 and tend to cancel outside. In other

words, waves with "# L / 2 , if combined in the correct proportions as determined by Fourier analysis, canproduce a wave function which is equal to "

0betweenL/2 and +L/2. Contrast these waves with shorter-

wavelength waves:

As the diagram suggests, these waves tend to cancel out almost everywhere inside the box, and are largely

unnecessary to produce the desired ". But they cant be completely ignored, because combined in proper

proportion, as determined by Fourier analysis, they help the larger waves to cancel outside the box and to

produce the constant "0 in the box.

A precise Fourier analysis yields the amplitude as a function of wavelength in order to achieve

precisely the desired ". It is more convenient to do the analysis in terms of the amplitude as a function of

wave numberkrather than wavelength. The following diagram shows the result, whereA(k) is the

amplitude forL = 1 (and "0= 1):

"

x

"0

-L/2 L/2

2" /L

A(k)

k

"

x

"0

The particle is equally likely to be found

anywhere between L/2 and L/2. Theprobability of being anywhere in the boxis constant from point to point, so the

probability of being betweenx andx +dx is constant: Prob(x, x+dx) =

|0|2dx.

-L/2 L/2


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In general, for a 1-dimensional box between L/2 and +L/2, A(k) =2"0

2#

sin(kL /2)

k, and

"0 =1/ L . As the graph shows, the greatest amplitude is for waves between 2" /L for a spread of

4" /L . Butthe amplitude is small as one goes to the limits of this range, so a reasonable estimate of the

spread or uncertainty in kis about half of this, or 2" /L . Since p = hk, then "p = h"k# 2$h /L = h /L .

But the particle can be anywhere inside the box with equal probability, soL is a measure of the uncertainty

in the position of the box, i.e., "x = L . Therefore, "p"x # h .

In our box example, the particle is free and just bounces inside the box between its ends. In a

more complicated situation, the particle might be on a rubber band while bouncing back and forth, or

subject to some other type of potential energy function. In any case, the uncertainty relation is an order-of-

magnitude estimate, and the more general formulation of the uncertainty relation is framed in terms of h :

The Heisenberg uncertainty relation for position and momentum: "x"p # h .

The significance is that a more accurate measurement ofp creates a greater uncertainty inx, and vice versa.

Most significantly, perfectly accurate measurements of position and momentum are impossible to carry out:

the product of their uncertainties must always be h .

7. " must be single-valued, well-behaved (cant blow up), and it and its first derivative (or gradient in 3-D)must be continuous. One exception to the latter rule: where the potential is infinite, just set " =0.

8. Just as particles can flow from one place to another, so the probability of the particles being found at one

place or another can flow. The probability current density in one dimension is:

*)*(2

),( !!!! "#"#=m

it

hrj .

9. The weighted average of something is just a sum of the values it can have times the probability of

having the values. E.g., in one dimension,

xn= "

#x

n"dx

$%

%

& .

10. " can be determined from the Schrdinger Equation (V(r) is the potential affecting the particle of

mass m):

tiV

m !

!=+"

# $$$ h

h)(

2

22

r (A KEY EQUATION!)

CHAPTER 2: Wave mechanics I

1. The Schrdinger Equation is separable into time and space parts. The space part of(r,t) is u(r) or just

u:

"h2

2m#

2u +V(r)u = Eu (ANOTHER KEY EQUATION!)

This equation is a recipe for grinding out allowed energy values (oreigenvalue) for a particle bound in a

potential. The wave function associated with each energy eigenvalue is called an energy eigenfunction.

2. When a particle is free, the allowed energy values in the Schrdinger equation are continuous; if the

particle is bound (e.g., in an infinite or finite square well, by a harmonic force, etc.), the energy values are

discreteonly special values, E1, E2, etc., are allowed.

3. The time part of the Schrdinger equation tells us thath/)(),(

iEtexutx!

=" for a particle in a well-

defined energy state. Therefore, ||2

= |u|2

and to normalize u is to normalize .


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4. Most problems are very difficult to solve in quantum mechanics, but the infinite square well is an

exception. In one dimension (1-D), this corresponds to a line with rigid walls at both ends. The form of

the wave functions is completely similar to waves on a string, and that is the easiest way to generate them.

However, you should also know how to obtain them by a systematic solution of the 1-D Schrdinger

equation. The solutions are shown here (think of waves on a string with the endpoints fixed, so the wave is

zero there), and the general form is: for n odd, and for n odd.

un=

1

a

cosn"x

2a

#

$%&

'(forn odd, and u

n=

1

a

sinn"x

2a

#

$%&

'(forn even.

As the captions of the diagrams show, the de Broglie wavelengths are, in general, 4a/n.

5. The energies of the particle in an infinite square well are purely kinetic. Since k = 2/ and p = k, we

have the following equivalent expressions for En: En = pn2/2m = 2kn

2/2m = 2(42)/(2mn2) =

2

(42

)/[2m(4a/n)2

] = n2

2

2

/(8ma2

), n = 1,2,3,

6. Theorem: IfV(x) is symmetric, then the non-degenerate wave functions un have definite odd or even

parity. (Example: the infinite square well potential. The uns are either sines or cosines with definite

parity.)

7. The energy eigenvalues of the finite square well (from a to +a, i.e., V(x) is symmetric) are determined

by graphically solving the following equations:

tan(a) = / (even parity), cot(a) = -/ (odd parity), with

2/12 )/2( hmE=! , and .]/)(2[ 2/12hEVmo!="

CHAPTER 3: Wave mechanics II

1. The simple harmonic oscillator, for which V(x) = Kx2/2, has energy values !h)(21

+= nEn

, and

eigenfunctions )x(Hand)/m(with,e)x(H!n

un

//x

n

/

n/n!"=!!#$

%&'

(

)

!=

!* 212

21

21

22

2h is the

nth Hermite polynomial. Note that the lowest energy state has non-zero energy of/2. This is a reflection

of the uncertainty principle: zero energy would imply that momentum and position are both exactly equal

to zero, which is prohibited by the uncertainty principle.

V(x) oru(x)

-a +a

V(x) oru(x)

-a +a

V(x) oru(x)

-a +a

V(x) oru(x)

-a +a

u1; = 4a

(i.e., 4a/1)

u3; = 4a/3 u4; = a

(i.e., 4a/4)

u2; = 2a

(i.e., 4a/2)


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2. The hydrogen atom in its simplest treatment has the electron subjected to the Coulomb potential V(r) = -

e2/(4or). By neglecting angular motions of the electron, we can rather simply determine the energy levels

of the atom:

222

4

)4(2 n

meE

o

n

h!"#= .

Surprisingly, this is precisely what the simple Bohr theory of the H-atom predicts. The independence of the

energy eigenvalues on angular motions is analogous to the result from classical mechanics that the energy

of a planet or satellite under the influence of one other bodys gravitational force depends only on the major

axis of the elliptical orbit, not on its eccentricity. (It is the eccentricity which involves angular motions,

specifically angular momentum.)

3. The energy levels in the previous item can be determined by assuming that2/

e!"

!=! )F()u( (where

r!=" and2

mE/h82

!=" ). Then )F(! obeys the differential equation:

0Fd

dF

d

Fd2

2

=!!

"

#$$

%

&

'

()+

'!!

"

#$$

%

&(

'+

'

11

2,

where2

4

2

h!"#=$

o

2me

. Assume F is a polynomial of degree k and substitute it into the d.e. The

coefficient of the leading term in the differential equation is then1-k

! , and since this term is not canceled

by anything else on the left side of the equation, it must must be 0, and this means that 1k +=! , k = 0,

1, 2,, i.e., 3,..2,1,n =!" . Since ! (which is in ! ) has E in it, solving for E gives the energylevels.

4. Free particles. Free particles are bothersome, because strictly speaking, their wave functions, of the

form)/(),( hEtiet !"=# rr , are not normalizable. It is convenient, then, to imagine free particles to be

in fact confined to a very large cubical box with sides of length L. In one dimension, we take this to be a

very long line of length L. In these cases, the normalized wave functions for are:

"(x, t) =1

L1/ 2

ei(kx#Et/h )

(1# D), and for 3 - D, "(r, t) =1

L3 / 2

ei(k$r#Et/h )

(3 # D),

where the 1-D case is appropriate for a particle moving in the +x direction. For the 1-D case, the

probability current (see Chapter 1, item 8 above) is: j=dProb

dt=

v

L, where v is the velocity of the particle

andProb is the probability that the particle leaves the box. Therefore, Prob =vt

L

. Thus, when t = 0, Prob

= 0, since we assume the particle is initially in the box. After a time t = L/v, the particle must have left the

box. (If it starts at the left end of the box, it takes it this time to reach the right side and leave the box.)Therefore, when t = L/v,Prob = 1. More often, we are interested in the flux of particles in a beam, e.g., the

number of particles per second passing a given point or impacting a target. This can be represented in one

dimension by a wave function "(x, t) = Aei(kx#Et/h )

, where the flux is represented by the symbol A. It is

typical to take A = 1 for the incident beam wave function, since all we are interested in are ratios of fluxes.

(See next item.)

5. Scattering from a potential step:Considera beam of energy E = 2k2/2m coming towards a potential

step of height Vo


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classical physics cannot explain such a phenomenon. The reflection and transmission fractions, R and T,

are:

.)(

4,

)(

)(22

2

kk

kkT

kk

kkR

!+

!=

!+

!"=

6. Tunneling through a rectangular barrier. Consider a potential V(x) = Vo for 0 < x < a, and V(x) = 0everywhere else. Classically, a beam with energy E < Vo would be totally reflected, but in quantum

mechanics, part of the beam tunnels through:

.)(2

,)(4

sinh1

2/1

2

122

!"

#$%

& '=

!!"

#

$$%

&

'+=

'

h

EVm

EVE

aVT

o

o

o(

(

7. Consider a set of N non-interacting distinguishable particles, described by wave functions

N!!! ,...,

21. The joint probability that #1 is in volume element

1!d , #2 is in

2!d ,etc., is the product

of the separate particle probabilities: Prob =NN

d...dd !"!"!"2

2

2

21

2

1. We will get this result if

we take the combined wave function to be the product of the individual ones:N

...!!!="21

. But if

the particles are indistinguishable, we must have a combined wave function which reflects thisindistinguishablity. Two cases arise. For some particles, called bosons (e.g., photons), the wave function

must be symmetric under an exchange of particle coordinates. For other particles, called fermions (e.g.,

electrons, protons, and neutrons), the wave function must be antisymmetric under an exchange of particle

coordinates. See Davies for simple examples of each for two particles.

CHAPTER 4: The formal rules of quantum mechanics

1. Ordinary functions as vectors in infinite-dimensional space. In ordinary geometrical space, we have

three dimensions, which we can label 1, 2, and 3 (usually corresponding to x,y, and z), so a vector in 3-

space can be written as A = (A1,A2,A3). The scalar or dot product between this A and another vectorB =

(B1,B2,B3) is, as everyone knows, just A "B= A1B1 , A2B2, A3B3.

A function ofxcan be thought of as a vector in an infinite-dimensional Hilbert space withthe variablex playing the indexing role of 1,2,3 except that it can take on an infinite number of values.

To understand how this works, consider first a five-dimensional space defined atx = 1.5,x = 2.2,x = 2.4,

x = 3.5, andx = 3.8, and a functionf(x) =x2, defined over these values. Then everything we want to knowabout this function is represented in the vectorf= (2.25, 4.84, 5.76, 12.25, 14.44). Similarly, a function

g(x) =x1.5 isg= (1.837, 3.263, 3.718, 6.548, 7.408). And the dot product offandgis just (2.25*1.837 +

+ 14.44*7.408) = 228.5. Now imagine we define functions over not five values ofx between 1 and 4 but

over a billion values, more or less equally spaced. Thenfandgwould be billion-component vectors, and

the scalar product between them would be very close to the integral of (x2)(x1.5) from 1 to 4. Going to the

limit of continuousx over the entire real line leads to a vector representation of the functions, which we can

just represent as (fx) and (gx), or more conventionally as f(x) =x2 andg(x) =x1.5, with the scalar product

betweenx = a andx = b being:

f(x)g(x)dx =a

b

" x3.5

dxa

b

" .

2. Two functions, f(x) and g(x), are considered mutually orthogonal (or just orthogonal) on the interval

[a,b] if!b

a

f(x)g(x)dx = 0, with an analogous equation for three dimensions.. Usually, a = - and b = +.

3. Solutions to the Schrdinger equation corresponding to different energies are mutually orthogonal:


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! =b

a

mnnmdx ,* "## (A KEY EQUATION!)

with an analogous equation for three dimensions.

4. For a given physical system, the solutions (i.e., eigenfunctions) of the Schrdinger equation form a basis

(analogous to z,y,x

), so that just as any three-dimensional vectorr can be written as a sum of z,y,x

, socan any arbitrary function f(x) be written as a sum of the solutions to the Schrdinger equation. Thus, (x)

= cnn, and the sum is in general, over an infinite number of eigenfunctions. Fourier analysis is a good

example of this for a free particle. The cn numbers are called expansion coefficients.

5. To find the expansion coefficients in (x) = cnn, we can use the orthogonal property of the n

eigenfunctions. Thus, if we want cm, multiply the equation by m*, integrate, and we get:

cm= ! m*dx, (A KEY EQUATION!)with an analogous equation for three dimensions.

6. The probability that a measurement on a system in state will yield the energy Em is:

Pmc = m2. (A KEY EQUATION!)

7. Collapse of the wave function: If a measurement shows that the system has energy En, then the system

is in the staten

! right after the measurementi.e., the wave function collapses from (x) = cnn ton

! .

7. The time-independent Schrdinger Equation can be written in operator form H un = Enun, where

H = -(h2/2m)

2! + V(r), and

2! =

2

2

2

2

2

2

zyx !

!+

!

!+

!

!.

8. An eigenvalue equation has the form A = , where A is any operator and is a number which in

general can be complex but in quantum mechanics must be real if it is to correspond to a measured

quantity. The function is an eigenfunction of the operator A . So the eigenvalue equation shows a

special relationship between A and : thinking of as a vector, when A operates on it, it doesnt change

its direction; it only changes its length by the factor.

9. Any legitimate operator in quantum mechanics, i.e., any operator which corresponds to an observable

(like position or momentum) must be Hermetian, i.e., it must satisfy the following relationship for any two

arbitrary functions and :

! !"" #$=$# %% dAdA )()( .

In other words, if we always assume the operator operates to the right in the above expression, A can

operate on , but it can also operate on *if we change A to A

*.

10. Another way to state the Hermetian relationship in the previous item is to think ofA

being able tooperate right or left, but if it goes to the left and operates on * , we must complex-conjugate A first.

This symmetry of going left or right, with everything on the left being understood to be complex-

conjugated, leads to Diracs elegant bracket notation, whereby the integrals in the previous item can both

be represented as:

!"="!=!"# #$$

AdAdA )()( %% .

In this notation, ! is the bra part of the backet and ! is the ket part.


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11. In the bra and ket notation, the expected value of an operator in a state ! is:

!!= AA . (KEY EQUATION)

12. The commutator, [ BA , ] , of two operators is defined as ABBA ! . Usually, we must operate onsome actual function to obtain the correct properties of the commutator.

13. If two operators commute, i.e., if their commutator is zero, then the operators can have simultaneous

eigenfunctions. If they dont commute, then they cannot have simultaneous eigenfunctions.

14. Two physical observables can be simultaneously measured without any error if and only if their

operators commute. In other words, a measurement of one will not affect the measurement of the other if

their operators commute.

15. The operators for position and momentum do not commute: [ ] hipx =, , i.e., hh ix

i,x =!"

#$%

&

'

'( .

Because they dont commute, they cant be simultaneously measured without disturbing each other.

Therefore, they obey the Heisenberg uncertainty principle. (See Chapter 1 Quantum Mantras.)

16. In general, for any two operators A and B , which have eigenvalues and respectively, we define

the uncertainty of the measured values as =2

2 AA ! , and likewise for.

17. In general, for any two operators A and B , which have eigenvalues and respectively,

>1

2A,B[ ] . This again implies that they are simultaneously measurable, i.e., = 0, only

if the operators commute. If the operators dont commute, they obey the uncertainty principle.

18. The commutator of an operator with the Hamiltonian operator tells us what the average time rate of

change of that operator and its observable are:

d

dt"A" =

i

h"[H,A]" . (A KEY EQUATION!)

In particular, if the commutator of an operator with H is zero, the observable corresponding to the

operator is conserved, i.e., it is constant in time.

CHAPTER 5: Angular Momentum

1. Classically, the angular momentum is L = r!p, and Lx = ypz zpy, etc. We obtain the quantum

mechanical operator by substituting the operators for y, pz, etc., soy

ziz

iyLx

!

!+

!

!"= hh )( , etc.

2. The operators for Lx, Ly, and Lz do not commute: zyx LiLL, h= , etc. Therefore, these components

cannot be measured with perfect accuracy simultaneously. In a sense, a measurement of one interferes ordisturbs a measurement of the other.

3.2L does commute with all three components ofL. Since

zL is very simple in spherical coordinates

(!"

"#= hiL

z ), we usually choose to work with simultaneous eigenfunctions of

2L andz

L .


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9

un=

1

a

cosn"x

2a

#

$%&

'(

4. The eigenfunctions of2L and

zL are designated by the notation m,l , where l and m are the

quantum numbers of2L and

zL respectively. For well-behaved eigenfunctions (i.e., single-valued &

finite), we must have l a positive integer: l =0,1,2,3,. For each l , m = - l , - l +1, - l +2,-2, -1, 0, 1,

2, l -2, l -1, l , for a total of 2 l +1 values ofm . The eigenvalues of2L and

zL are, respectively,

2)1( hll + and hm .

5. In spherical coordinates, m,l =m

imm

mYePN

lll=

!")(cos , where

mN

lis the normalization

constant:

2/1

)!(

)!(

4

12)1( !

"

#$%

&

+

'+'=

m

mN

m

m

l

lll

(

, and )(cos!m

Pl

is an associated Legendre polynomial.

The entire wave function is the well-known spherical harmonic,m

Yl

.

6. The mYl functions are orthonormal: ''',', mmmm !!llll = .

7. Them

Yl

functions can be used to construct matrices for2L and

zL . For each value ofl , the 2 l +1

values ofm define a 2 l +1 by 2 l +1 matrix representation for2L and

zL . Therefore, for l =0, we have a

1x1 matrix (the sole element is 0); for l =1, a 3x3 matrix; for l =2, a 5x5 matrix; etc. (Where are the

even matrices?? See below!) The matrices for2L and

zL are:

!!!!!!

"

#

$$$$$$

%

&

+=

1......00

0...::......:

0......10

0......01

)1( 22 hllL

, and zL

=

!!!!!!

"

#

$$$$$$

%

&

'

+'

'

l

l

l

l

h

....00

01......0

0...0:

0..010

0....0

.

Note that the matrices for2L and

zL are both diagonal, reflecting the fact that they commute and hence

have common eigenfunctions when expressed in terms of the eigenvectors ofz

L . Also, note that the

diagonal elements are their eigenvalues. A similar process allows matrices forx

L and yL to be obtained.

For l =1, the complete set of matrices are as follows.

!!!

"

#

$$$

%

&

'

'

=

!!!

"

#

$$$

%

&

=

!!!

"

#

$$$

%

&

'

=

!!!

"

#

$$$

%

&

=

00

0

00

2,

010

101

010

2,

100

000

001

,

100

010

001

222

i

ii

i

LLLL yxzhh

hh

Note that the matrices matrices forx

L and yL are not diagonal; the off-diagonal elements reflect the

fact that we have chosen to express everything in simultaneous eigenstates of2L and

zL , but

xL and


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10

yL do not commute with

zL . Note also that all four matrices are Hermetian, which for a matrix

means that the complex conjugate of the transpose of the matrix leaves the matrix unchanged.

8. For a givern l , the most general eigenfunction is a 12 +l -component vector:

!!!!!!

"

#

$$$$$$

%

&

=

+12

2

1

.

.,

l

l

a

a

a

Cm ,*

12

*

2

*

1 ....*, += ll aaaCm

where C is a normalization factor. (N.B.: the bra form is complex-conjugated!) The expectation of any

matrix operator A is just mAm ,, ll .

9. The eigenvectors of2L and

zL are just like m,l shown in #19, but with all zeros except for the

particular value ofz

L which is measured. For example, if l =5, there are 10 eigenvalues ofm , from +5

to -5. The eigenvector which will yield an zL

value of +3 will have a3 = 1 while all othera-values are

zero; the eigenvector for 5 will have a10 = 1 while all others are 0.

10. As we saw above, the r!p type of momentum yields only integral values of l and square matrices

with an odd number of elements in a row or column. But not all angular momentum is of the r!p type.

Intrinsic angular momentum, more commonly called spin, fills in the gaps with half-integral values of l .

For example, if l =1/2, we get a two-by-two matrix, meaning that only two values ofz

L are possible.

Thats experimentally true for an electron. Any particle with a half-integral value ofl is called a

fermion, while one with an integral value is called a boson. The crucial difference is that they obey

different types of statistics, but this will not be explored here.

11. For an electron, we designate its 2 by 2 matrices by the letter S to remind ourselves that were

dealing with spin. Because of the great importance of these matrices, they are called the Pauli spin

matrices after their formulator Wolfgang Pauli (1900-1958).

,10

012

4

32

!!"

#$$%

&= hS !!

"

#$$%

&

'=

10

01

2

1

hz

S , !!"

#$$%

&=

01

10

2

1

hx

S , !!"

#$$%

& '=

0

0

2

1

i

iSy h .

12. MEASUREMENT RULE: If a spin is in a state !!"

#$$%

&

b

a, and a measurement of

zL is made, the state

immediately changes to either !!"

#$$%

&

0

1if the measurement yields +/2 or !!

"

#$$%

&

1

0if the measurement yields

/2. This is true in general. For example, if a system is in a state in which the probability of obtaining a

momentum between xp

and xp

+d xp

is some continuous function of xp

, then if a measurement ofmomentum yields the magnitude px, the system immediately jumps into the wavefunction or eigenstate

corresponding to this value, i.e., the wave function immediately changes toh/xipxe .

13. If the axis z! makes an angle with the z axis, then the state corresponding to +/2 along z, i.e., the

state ! , is : !"#$"=$ )2/sin()2/cos( %% . Therefore, if a measurement puts the system in the

state ! , and then a measurement is made along z! , the probability of finding the spin up along z! is


11/24

11

)2/(cos2 ! . The probability of finding the spin down along z! is )2/(sin 2 ! . Similar results apply if

we start with a spin along z or + z! : !"+#"=! )2/cos()2/sin( $$ ;

!+"="# )2/sin()2/cos( $$ ; !+"#=!$ )2/cos()2/sin( %% .

CHAPTER 7: Approximation methods

1. If Eno is the unperturbed energy of the non-degenerate nth energy state u

nofor the Hamiltonian H

o,

and En is the energy of this state when the Hamiltonian isH

o+ !H ( !H is the perturbation), then for

!H


12/24

12

For example, assume we have a one-dimensional system and ut has just two adjustable parameters:

ut(,;x). Then the variational recipe is as follows:

(1) Calculatett

uHuE ),(0 =!" .

(2) Minimize this with respect to and :

.0),(

,0),( 00

==

!"

#"!

!"

#"! Eand

E

(3) The two equations in step (2) can be used to solve for the minimizing values of and , call

them min and min.

(4) Now the minimizing values from step (3) are plugged into the average for0

E in step 1:

),( minmin0,0 !"EE estimate = .

This estimate is always greater than or equal to the actual ground state energy. It cannot be less.

CHAPTER 8: Transitions

The following review includes a discussion of electromagnetic (EM) transitions, analogous to the simple

Bohr model in which EM radiation is produced by an electron undergoing a transition from one state to

another, or in which EM radiation induces a transition from one state to another. In the latter case, there are

two possibilities. Either the radiation pumps an electron from a lower to a higher state, i.e., the radiation is

absorbed, or the radiation induces a transition from a higher to a lower state, i.e., we have stimulated

emission which results in additional EM radiation. In all of these cases, the EM radiation will be treated as

a classical continuous electromagnetic field (which you are familiar with from Maxwell's equations),

because a full quantum mechanical treatment of the radiation field is beyond the scope of this course. In

other words, the treatment here is "semi-classical": classical for the EM radiation and quantum mechanical

for the atomic transition itself. (A few brief remarks on the full quantum mechanical treatment can be

found at the end of Section 11.) The discussion starts off with a general treatment of transitions due to a

time-dependent perturbation. Electromagnetic radiation will not be specifically discussed until Section 8.

So keep in mind that everything before that is wondrously general, particularly Fermi's Golden Rule, which

has a multitude of applications throughout modern physics.

1. TIME-DEPENDENT PERTURBATION THEORY. Consider a system with Hamiltoniano

H and

eigenvectors htiE

nono

no

eut!

= )(),( rr" . The system is in eigenstate kand at t= 0 it is suddenly

subject to asmallperturbation )t,(H r! . This changes the wave functions and energy levels, and the new

wave functions can be expanded in terms of the original wave functions:

!="#

h

tiE

nonn

no

e)(u)t(at,( rr .


13/24

13

In what follows, the subscript o will be dropped, since it will be understood that everything now refers to

the original unperturbed wavefunctions and energies. To a first-order approximation, the time-dependent

expansion coefficient, am(t), is:

hh

km

mk

tti

mkm

EEdteH

ita mk

!="= # $

$

where,1

)(0

.

Therefore, the probability of a transition from state kto state m is am(t)

2.

2. HARMONIC PERTURBATION. The most common type of a time-dependent perturbation is a

harmonic one which is turned on at t= 0: tcos)(H)t,(H !=" rr . This leads to:

am= "

Hmk

2h

ei(#

mk"#)t"1

#mk

"#+

ei(#

mk+#)t"1

#mk

+#

$

%&

'

().

We expect transitions to be significant when " #"mk

, in which case the first term in the brackets is much

bigger than the second. The so-called rotating wave approximation consists in neglecting the second term

for" #"mk

. In this case, the probability )t,(Pm ! of making a transition in time t to a state m starting

from a state k withEm > Ek is, to first order, for a small perturbation:

./)EE(with,t)(

t)(sin

tHa)t,(

mP

kmmk

mk

mk

mk

mh

h

!="

##$

%

&&'

( "!"

##$

%

&&'

( "!"

=="2

2

2

2

2

24

2

2

Since the system goes from a lower energy,Ek, to a higher one,Em, this represents absorption.

Please note th at the perturbation frequency is by definition positive, but mkcan be positive or

negative. Specifically, mkis negative ifEk > Em, in which case !"!mk in the preceding

equat ion is replaced by !+!mk

. This representsstimulated emission (predicted by Einstein in 1917,

before quantum mechanics was developed in the mid-1920s!) in whic h t he incident radiation induces

a tr ansition from the sta te wit h energy Em down to the lower energy Ek. In this case,

Pm

(", t) = am

2=

Hmk

2t

2

4h2

sin2 ("mk+")t

2

#

$%&

'(

("mk

+")t

2

#

$%&

'(

2, with "

mk= (E

m) E

k) /h < 0.

3. PERTURBATION FREQUENCY !EQUAL TOmk

! . Suppose th at t he frequency of a

perturbation precisely matches the energy difference between two levels: " ="mk

. Then the

probabi li ty of making a transition to the sta te kgrows wit h t ime. This probabil it y can be

obtained from the equat ion for )t,(Pm ! by taking the limit asmk, which yields:

(ABSORPTION)

(STIMULATED

EMISSION)


14/24

14

2

22

4h

tH)t,(P

mk

mkm =! .

Please remember that t his ho lds only for a smal l perturbat ion, i.e.,since ),( tPmkm

! increases

with time, it holds only for t imes sufficiently small so that ),( tPmkm

!


15/24

15

If the perturbation has a frequency "o, then the probability of a transition at time tis given by the

magnitude of the )t,(Pm ! curve at that point. As shown in this diagram, o! is between the first and second

zeros to the left of mk! , so the probability is rather small compared to the maximum 2

22

4h

tHmk

.

Note from the equation for )t,(Pm ! that at t= 0, the sine function is identically zero, so the

Pm(",0) = 0everywhere for any arbitrary ! . This is consistent with the graph of )t,(Pm ! as a function of

tshown in Figure 8.5.1. This is quite reasonable, since we can't expect a perturbation to cause something to

happen immediately after it is turned on. Note also the first zeros to the left and right of the central

resonance peak, which are separated from that peak by t/2! . Therefore, as tincreases, these zeros get

closer and closer tomk

! , and for a fixedo

! , these zeros sweep through this value of! , again consistent

with the graph of )t,(Pm ! vs. t. Simultaneously, as tincreases, all of the peaks must grow in height,

because as the )t,(Pm ! vs. tgraph shows, by the time the very low peaks initially far from it sweep through

a given ! , they all have the same height! This is exactly consistent with Figure 8.4.1.

Figure 8.5.1 illustrates the energy-time uncertainty principle. Imagine that the energy

differenceEm Ek, or equivalently the frequency "mk , is unknown. To measure it, we subject the system

to a perturbation of frequency ! . If! is sufficiently close tomk

! , then the absorption will be sizable. If

it is far away, then the absorption will be small. But most of the probability for the transition is in the main

central peak, which covers a frequency interval ]/2,/2[ ttmk

!"!" +# . In other words, a perturbation

which is within this frequency interval has a good chance of inducing a transition from state kto state m.

We see that if our experiment lasts a time t, then a spread of input frequencies in that interval, i.e., an

energy range of h("mk

# 2$ / t) to h("mk

+ 2$ / t) , will be strongly absorbed. Therefore, the spread of

energies strongly absorbed is tE /4 h!=" , which is also a measure of the uncertainty in the measured

value of the energy difference between the two levels. Thinking of the measuring time tas the

uncertainty in time, t! , we can cross-multiply by tt !" and we get hh !"=## 4tE , QED.

Why is the sky blue? An interesting consequence of the uncertainty principle involves scattering

of radiation from a molecule. If the radiation is close to a resonant frequency, i.e., if the energy of the

radiation is close to the energy difference between two molecular levels, then E! is small, and

"t= 4#h /"E is large. On the other hand, if E! is large, then "t is small. But what is"t? Assume

that there is absorption of the radiation energy by the molecule. Then there are two interpretations of"t,

depending on if one is looking at the wave or particle aspect of the interaction between radiation and

molecule. From the wave aspect, i.e., considering the radiation to be wave-like, "t is the time the

molecule is subject to the perturbation. From the particle aspect in which the radiation consists of photons

and we are considering absorption of a photojn, the absorption is for all practical means instantaneous, but


16/24

16

now "t represents the time the molecule holds on to the radiation. For small E! , the molecule holds

onto the radiation for a long time, long enough for collision with another molecule to occur. The collision

will typically thermalize the energy, i.e., the energy absorbed by the molecule is transformed into

molecular kinetic energy, so the colliding molecules rebound with greater kinetic energy than they started

with. On the other hand, if E! is large, then the molecule holds onto the radiation for a shorter time, sothe radiation is can be quickly emitted. If emission happens before a collision occurs, then we have

scattering of the radiation. The small- E! situation helps tell the story of absorption of solar uv radiation

ozone in the stratosphere. Ozone, and also O2 and N2, have absorption peaks in the uv so the uv energy is

resonantly absorbed, a collision occurs, and the energy is safely thermalized. Blue light, on the other hand,

is farther removed from the uv resonant peak, so the large E! case holds, and it is scattered. Of course,

green and red light is also scattered, but they are farther removed from the resonant peak so they arent

absorbed and scattered as much as blue. In fact, the absorption is proportional to the inversefourth power

of the wavelength, so blue is scattered about 16 times as much as red. Thats why we have a b lue sky!

6. A FOURIER TRANSFORM INTERPRETATION OF )t,(Pm ! .The graph of )t,(Pm ! for fixed time

is precisely a graph of the Fourier transform of a cosine segment as shown in Figure 8.6.1. This is not a

coincidence. Assuming that a harmonic cosine perturbation in the form of an electromagnetic wave is

turned on at t= 0, then at time tit looks like the cosine segment shown in the figure. (The question mark

just indicates we dont know what the future holds past time tthe perturbation might continue or it might

abruptly end.) If

we assume this segment is of the form tcoso

! for a particular value ofo

! , then the Fourier transform of

the segment will be centered ato

! .

Figure 8.6.2 shows the Fourier transform of this segment. A comparison of this figure with Figure

8.5.1 shows that the probability of a transition, now given by the s trength of the Fourier frequency atmk

! ,

is exactly what Figure 8.5.1 shows it to be ato

! . This is because the equation forPm(", t) is symmetric

in mko !"! , so it depends only the distance from the center, i.e., on mko !"! . And as before, for

longer times, the longer the cosine segment will be and the more tightly its Fourier transform will be

located near the central frequency (in this case,o

! ).

t

FIGURE 8.6.1

),( tH r!

?

time


17/24

17

7. FERMIS GOLDEN RULE. Consider the absorption of light by atoms. Say they are all hydrogen

atoms. We are prone to think that the energy levels of one hydrogen atom are the same as for another

hydrogen atom, but this is true only at absolute zero, i.e., for motionless atoms, with lots of space betweenthe atoms. If the atoms are warm, i.e., at any temperature above absolute zero, the levels and the

frequencies corresponding to differences between energy levels are Doppler-shifted because of the thermal

motion. Broadening of energy levels can be caused by a number of other factors, such as crowding the

atoms together (pressure broadening) or just the natural lifetime of the state which, by the uncertainty

principle, causes a smearing out of the energy level. Fermis "Golden Rule" leads to a simple result for the

absorption in cases where the broadening is sufficiently large.

To derive the Golden Rule, assume to begin with that we have only a few energy levels, say 4,

labeled m1, m2, m3, and m4, separated but not too much so, as shown in Figure 8.7.1 (a). We can think of

these as really distinct levels or just as an approximation to the result of broadening a single level. Also

shown is the initial state,Ek, and the energy of incident radiation, mk!h . This harmonic perturbation can

induce transitions fromEk to one of the four higher states.

Figure 8.7.1(b) shows the probability curves, )t,(Pim! , for the four levels as a function of! ,

with their bases placed for comparison at the energy levels shown in 8.7.1(a). (Note that the curves have

different heights, corresponding to different matrix elements for )(H kmi r ). The peak of )t,(P im ! is at

"m

i

=

Emi# E

k

h. The incoming radiation of frequency

mk! has probabilities ,a,a,a

2

3

2

2

2

1and

2

4a , of

inducing a transition to each of the four excited states. Just looking at the curves, it is easy to see that theprobability that a transition to state 3 has the greatest probability. This is to be expected, since 8.7.1(a)

shows Em3 to be closest to mk!h . The probability that a transition will occur to any of the four excited

states is just the sum of the separate probabilities, i.e.,2

4

2

3

2

2

2

1aaaa +++ , with a1, a2, and a4 very

small compared to a3.

2

22

4h

tHmk

),( tPm

!

!= resonant

absorption

frequencymk

!

Area under entire

curve =2

2

2h

tHmk!

Probability of absorption as a function of resonant absorption

frequency for fixed time and fixed perturbing frequency

o!

"o# 2$ / t

"o+ 2# / t

FIGURE 8.6.2


18/24

18

Now suppose all of the levels have the same values of km iH between states kand mi (i = 1,2,3,4).

Then all of the four probability curves are identical. As mentioned in the previous item, the probability of

radiation of frequencymk! being absorbed depends only on the difference

mimk!"! , so we can replace

the four probability curves by one centered onkm! (and for convenience, placed right on the ! axis). This

is shown in Figure 8.7.1(c).

In practice, physical effects like Doppler broadening produce not just four or ten or twenty levels,

but a virtual continuum of energy levels, which we initially assume for convenience to be very large in

number but not infinite. This is depicted in Figure 8.7.2 (a), with the density, ! , of the smear of states

represented by the darkness of the shading. If there are dNstates in the frequency interval [ ]!+!! d, ,

then )(!" is defined by the relationship!

=!"d

dN)( , i.e., the number of states in the interval is just

!!"= d)(dN . If we reasonably assume that the probability curves of all the states making up the

continuum are equal, then as before we can determine the probability of a transition by looking at the

overlap of the continuum with one single absorption curve centered onmk! . This is shown in Figure

8.7.2(b) where, in a small interval aroundmk!h , )(!" is approximately constant. Both )t,(Pm ! and the

density of the states, )(!" , are plotted on the vertical axis.

The approximation which makes up Fermis Golden Rule consists in noting, first, that the

separation between adjacent energy levels, shown much expanded in Figure 8.7.1 above, is very small

compared to the width of the absorption curve, as suggested in Figure 8.7.2. And second, the density of

states doesnt change very much across the range !" . So approximately, the number of states absorbing

significant amounts of energy is approximately N)( mk !"#!#$ . The probability of a transition to any of

the N! states is simply the sum of the separate probabilities:

Em

1

Em2

Em

3

Em

4

Ek

E

mk!h

!

E, Pm

(", t)

1m!

2m!

3m!

4m! mk

!

FIGURE 8.7.1

!

1m!

2m!

3m!

4m!

E, Pm(",t)

(a)

(c)(b)

mk!


19/24

19

Prob = ai

2"

i=1

#N

$ ( ai2(1)

i=1

#N

$ )

The reason for writing the number (1) in each term of the sum is as follows. Let the separation between

adjacent states i! and 1+!i be i!" , which for a virtual continuum of states will be very small. Then the

density of states ati! is

i

ii )(!"

=!#1

since there is just one state in this frequency interval, i.e.,

1=!"!# ii )( . Now we can substitute ii )( !"!# for the number 1, and going to the limit of a continuum

of states, and taking the different )( i!" all to be approximately )( mk!" , we get:

! """#$"% ! ""#&"'"#% ==(

()

'

=

(

()

'

=

d)t,(P)(d)(a)()(a)(a)t(obPr mkN

imkiii

N

ii

1

22

1

2

1 .

But the last integral is just2

2

2h

tHkm

!

, where2

km

H is the matrix element between state k and a typical state

with an energy in the neighborhood ofkm!h . Therefore, we get:

Prob(t) ="H

km

2

t

2h2

#($mk).

More usually, were interested in the transition rate," which is the transition probability per unit time:

t

)t(obPrw = . SinceProb(t) is proportional to t, w is constant:

)(H

wmk

km!"

#=

2

2

2h

. (FERMIS GOLDEN RULE)

8. THE DIPOLE APPROXIMATION AND BROADBAND RADIATION. Consider an atom bathed

in monochromatic electromagnetic radiation (i.e., radiation of single frequency, ! ) for which the

wavelength is much greater than the linear size of the atom. (As mentioned at the beginning of this review,

no attempt will be made to treat the radiation field quantum mechanically. A "semi-classical" treatment

will be sufficient.) Then at any instant of time the electric field, which we assume to be varying

Ek

km!h

E

Em

(a)

FIGURE 8.7.2

k!

!" ),t,(mP

!

!" )(!"

)( k!"

(b)

P(", t)


20/24

20

harmonically, is constant over the atom. Therefore, "H = #er $Eocos(%t) , so rEr !"= oe)(H where r is

the coordinate of the electron (taking the origin conveniently to be at the nucleus of the atom, which is so

massive we can neglect its motion). For most electromagnetic radiations of interest, e.g., anything longer

than hard ultraviolet, the dipole approximation is excellent. It fails only when we get into the X-ray region.

An electromagnetic wave, by its very name, includes a magnetic component,B(t), too. The

perturbation )tcos(eH o !"#=$ Er ignores the magnetic part of the wave because in the dipole

approximation, it is mainly the electric field that affects the motion of the electron around the nucleus. This

can be appreciated by noting that, in this approximation, the field changes only slowly compared to the

time it takes the electron to react to the field, but by Faraday's law, the magnetic effect, which depends on

the (slow) rate of change ofB, is small. In essence, then, at any instant of time, the atom is subject to a

quasi-static electric field which produces the change in potential energy given by H! .

The equation for )t,(Pm ! involves the matrix element mkH . In the present case, this is equal to

kme rEo!" . Thenthe equation for )t,(P

m! is:

2

2

oz

22

2

2

2

24

EEE

!!"

#

$$%

& '('

!!"

#

$$%

& '('

++

='t)(

t)(sin

kzyxmte)t,(P

mk

mk

oyox

m

h

.

For plane-polarized light, it is convenient to take the direction of polarization along the z axis (Eo = Eo z )

(and the dirction of propagation perpendicular toz), so we have only one integral to do in the matrix

element for )t,(Pm ! .

Suppose again that the atom is bathed in monochromatic radiation but with random (isotropic)

polarizations and directions of propagation. Then Eox

2= E

oy

2= E

oz

2= E

o

2/3 . Therefore, the matrix

element is justE

o

2

3m r k

2

, so:

2

22

o

22

2

2

2

212

E

!

!

"

#

$

$

%

& '('

!!"

#

$$%

& '('

='t)(

t)(sin

kmte)t,(P

mk

mk

m

h

r

.

Our result for )t,(Pm ! is not realized in practice because radiation, even from a laser, is never

strictly monochromatic. Instead, we must integrate )t,(Pm ! over" to obtain a frequency-independent

transition probability Pm

(t) . As a first step towards obtaining Pm

(t) , Eo2

can be expressed in terms of the

energy density of the wave, )(u ! . For a single-frequency plane wave, u is the sum of the electric and


21/24

21

magnetic energy densities:o

oBE

u

+!

=22

22

. But for a plane wave,c

EB = , where c = speed of light.

Also, the speed of light is given by

oo

c

!

=1

. Therefore, )t(cosEEu ooo !"="=222

. Averaged over a

cycle of time, we get the average energy density2

2

ooEu

!

= . Solving for Eo2

and substituting into the

previous equation for )t,(Pm ! gives:

2

2

2

222

2

2

6

!!"

#

$$%

& '('

!!"

#

$$%

& '('

)='

t)(

t)(sin

kmtue)t,(P

mk

mk

o

mh

r

.

Next, turning to the problem at hand of radiation covering a continuous range of frequencies, we

must suppose we have a differential of probability, )t,(dPm ! , proportional to the differential of energy in

the frequency interval !+!! d, . In other words, )t,(dPm ! is the probability that radiation in that

frequency interval induces a transition. The differential of energy can be written as du(") = u1(")d",

which defines u1(") as the radiation density (per unit volume per unit frequency interval). We can get the

transition probability from the previous equation by replacing u by u1(")d" and integrating over! . (In

probability terms, we want the probability of a transition due to frequencies in the range [ ]!+!! d, , plus

the probability due to frequencies in the range [ ]!+!!+! d,d 2 , etc.) We assume the radiation density

covers a wide range of frequencies compared to the region over which )t,(Pm ! is appreciable, as shown in

Figure 8.8.1. When the integration is carried, out, most of the contribution comes from the narrow region

aroundmk! , as suggested

by the figure. In this narrow region, u1(") is practically constant, of magnitude u

1("

mk) , which can be

taken out of the integral. Taking the limits of the integral with respect to ! from 0 to +! is equivalent to

taking the limits of the relevant variable in the integrand, tmk !"

#$%

& '('

2, from !"+! to , since "

mkis

!

mk!

!"

u1(") FIGURE 8.8.1


22/24

22

essentially infinite in the narrow region over which )t,(Pm ! is appreciable. The result of integrating

sin2("

mk

#")t

2

$%&&

'())

("mk

#")t

2

$%&&

'())2

ist

!2. Therefore, we end up with:

Pm(t) =

"e2u1(#mk) m r k

2

t

3$oh

2= B

mku1(#mk)t,

which defines the constant Bmk

. This constant will be met again in section 11. As with Fermi's Golden

Rule, this leads to a transition rate, w, which is constant;

w =Pm(t)

t= B

mku1("mk) .

Note that Davies chooses to express w in terms of the radiant intensity I(") # u1(")c , where c is the

speed of light.

The foregoing result is really another form of Fermi's Golden Rule, which we should now

understand to apply if narrow-band radiation causes a transition to a continuum of states or if broad-band

radiation, essentially a continuum of radiation states, causes a transition to a narrow s tate.

9. SELECTION RULES FOR THE HYDROGEN ATOM. As the equations for )t,(Pm ! and

)t(Pm show, they depend on the matrix element22

k)(HmHmk r! . For the hydrogen atom, m and k

each consist of four quantum numbers, n, l, m, ands, so we could write

Hmk

2" nml mmmsm H(r) nkl kmksk

2, but with this understood, the simpler m H(r) k

2notation can

be used. Forgetting spin, this matrix element, which depends only on space variables, is very important

because it is zero for the dominant dipole interaction rEr !"= oe)(H except for certain values ofm and k.

For example, it can be shown that transitions for any system with a spherically symmetric potential (e.g.,

the H atom or hydrogen-like atoms) must obey the following selection rules:

"l = 1

"m = 0 for z polarization

"m = 1 for polarization in the xy plane

As an application, consider the H-atom states 012013 ,,and,, . Can a transition occur between these

states? No, because the selection rule on l! is violated (i.e., it didnt change at all), even though the

selection rule on m is satisfied. What about a transition from 003 ,, to 112 !,, ? This can occur for


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radiation polarized in thexy plane because the conditions on both l! and m are satisfied (m = -1 and

l! = +1).

10. WHERE IS THE LORENTZIAN? The absorption curves so far are similar to the famous,

ubiquitous Lorentzian that appears in mechanics, optics, and nuclear radiation (in the Mssbauer effect),

but they have the oscillating wings which the Lorentzian doesn't have. This is because all of our resultshave been for a truncated sinusoidal radiation, i.e., radiation that has a pure sinusoidal form for a finite

period of time. A pure Lorentzian does appear for a perturbation of the form "H = #erEocos($t)e

#%t,

i.e., an exponentially dampedsinusoidal electric field. (You might remember, we got a Lorentzian shape in

Davies, Problem 5, Chapter 1.) We expect such an exponentially damped classical field to result from an

atomic transition. The Lorentzian is also appropriate for the inverse case of radiation being absorbed by a

gas.

11. SPONTANEOUS EMISSIONEINSTEIN'S A AND B COEFFICIENTS.

In 1917, Einstein developed an argument based on statistical mechanics and thermal equilibrium

which related the spontaneous transition rate to the absorption rate Bmk

, which was calculated above in

section 8. His great achievement was to do this before the development of quantum mechanics almost 10

years later. Consider a collection of identical atoms in an excited state n inside a cavity in which there is

black body radiation at temperature T. For simplicity, it is assumed that none of the levels are degenerate.

As Planck showed in 1900, the energy density of the radiation as a function of frequency is:

u1(") =

h"3

#2c

3

1

eh" / kT

$1,

where kas usual is Boltzmann's constat. As in Section 8, u1(")d" = radiation energy per volume in the

frequency range ",

"+ d"[ ] . Let Amn be the probability per unit time that the atom decaysspontaneously from state n to a lower-energy state m, let B

mku1(") be (as in Section 8) the absorption

probability per unit time for a transition from state m to state n, and let Bkmu1(") be the stimulated

emission probability per unit time for a transition from state n to state m. In thermal equilibrium, let Nm

=

the number of atoms in the lower state and Nn

= the number atoms in the excited state. Then thermal

equilibrium requires that the number of transitions from m to n equals the number of transitions from n to

m:

NmB

mnu1(") = Nn Anm + Bnmu1(")[ ] .

Now we can solve for u1("):

u1(") =A

nm

(Nm/N

n)B

mn# B

nm

.

But the Boltzmann equation gives:

Nn

Nm

=

e"En / kT

e"Em / kT

= e"h#

mn/ kT

.


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Therefore,

u1(") =A

nm

Bmneh" / kT

# Bnm

.

Comparison with Planck's law shows that:

Bnm = Bmn ,

Anm=

h"3

#2c3Bmn

Using the previous value of Bmn

for a dipole transition, Anm

is:

Anm=

e2"

3

3#c3$3h

mr n2

.

Einstein provided a clever and accessible way to determine the spontaneous transition rate. A

strictly quantum mechanical treatment explains spontaneous transition in terms of the quantized

electromagnetic field. In this picture, the EM field is viewed as an infinite number of harmonic oscillators,

each oscillator corresponding to a radiation frequency. The amplitude of a harmonic oscillator mode with a

particular frequency is represented by the number of photons of that frequency. The elementary quantum

mechanical theory of the harmonic oscillator shows that the ground state does not have zero energy but

rather has an energy of h" /2 . This "zero point energy" is always there for every mode of the quantized

EM field, and this energy interacts with an electron in an excited state to "rattle" it out of that state, with the

consequence that it falls into a lower-energy state and in the process (a la Bohrs or iginal idea)

spontaneously emits radiation.

quantum mantras

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