qmtp - notes

7/30/2019 QMTP - Notes

1/45

Aim: simplicity amidst complicatedness.

To understand is to project a known/understood phenomenon onto an unknown/unexplained phenomenon. Butunderstanding is not a rigourous thing. Whether something is true or not doesnt care whether you understand it

or not.

All of the following are (a) known phenomena, and (b) non-rigourous,

(1) dimensions; (2) scaling; (3) symmetries; (4) mathematical tricks (1.1)

Personal note: ion beam implantation experiment/theory interface. Maslov dug through Danish journal in order

to find an important result. Imagine incident particle with energy E0, and it impinges upon a lattice with atominc

binding energy Eb. How many atoms are displaced? After 10 pages, 0( / )bN c E E . Such a simple answer.Took answer to adviser, and he said Oh yeah, they just share energy in equal chunks! By the way, the actual

answer is0 / bN c E E , so the above result is wrong.

Short story about diffusion: how to describe in very simple, phenomenological way? Well, flux of particles:defined as,

2diffusion#

flux ; ; ;constantnL T

j D D (1.2)

This (1.2) should trigger symmetries of (1.1), because you built a vector out of a scalar (viz, del operator).

Time reversal: the LHS is odd under time reversal:j -j. But the RHS is not! Why? You implicity assumeddissipation; that is, an increase in entropy.

Continuity says n j ; putting this into (1.2) yields,2 ; ( , , );n n n n t rD D (1.3)

How fast does a point-source diffuse to a point where you have, say, 1 standard deviation of, say, a normal,

distribution1? Use dimensional analysis of (1.1).2

2 3

# 1[ ] [ ] ;

L L

L T L L T

D D (1.4)

Characteristic width of distribution: what is a formula for it? Actually, theres only one way to compute it: D

times some characteristic time. Therefore, 2 ~r tD , where ~ means you dont know the number t.

Now lets use scaling of (1.1). Make a guess about the form of a distribution. In three dimsneions, theconcentration-profile should be spherically-symmetric.

You have,

3 3/2

2 2 3 2 2 2 3/2 2

0 0

3/ 2/2 2

0 0

[unknown function] ( / ) ( , ) ( / ) ( ) ( / )

( , ) ( , )4 ( ) ( / ) 4

( )4 ( ) 4 ( )

x r t

g g r t n r t r g r t Dt f r t

r r n r t d r r n r t r dr r Dt f r t r dr

Dt tx f x dx t x f x dx

t

D

D D D

D

DD

D

(1.5)

The integral is just a number, despite it being from [0,infinity]. Also, it is of order 1.

Now execute some similar machinations with derivatives,

1Actually: Gaussian may be the concentration profile here, but Im not sure.


2/45

3/22 1/2 1 1/2 1/2 11 12 2 ( )

1/2 1/2

2 3/2

( ) ( ) ( ) ( ) ( ) ( )( ) ( ); ;

( ) ( )( )

( )( )

x x x x xx t tt t t tt

xx t

xx t

n n t f t t f t f t f f s s

n t f t

n f t

D

DD D D DD

D

D

D D D D

D D

D

(1

Actually, now we have an ODE; look,

2 2 2

throw away unbounded-solution;want to go to 0 at

1 1 1 1 1 102 2 2 2 2 2

/4 /(4 ) /(4 )1 1 12 2 2

0 ( ) 0

0 ( , ) ; ( , ) 1 ; ( , ) ;s x t x tc

t t

f f sf f sf f f sf f sf

f sf f ce n x t e n x t dx c n x t e

D DD D(1.7

Wednesday, January 09, 2013

- going to rearrange times

- book by migdal is available. Strongly recommended: Read first chapter of book.- gatman and Halpern = hopelessly out of print

- tim is making notes; those are reference too. The link to the notes is posted on the course-website.

- Maslov is trying to make a book.- homework #1 will be posted today; due in 2 weeks.

Consider bad/ugly integral,2

5/2 2

0

ln(ln )??

(1 )

xx edx

x

How do you estimate? Constraint: youre not allowed to sit down and do math. Do an order of magnitudeestimate,

2

5/2 2

0

ln(ln )

~1(1 )

xx e

dxx

Proof: what else could it be? QED. The function has no parameters. Integrand is positive-definite for wholerange, and is square-integrable thanks to the gausian.

Better estimate: turns out the integral is about 1.9

the point is that the order of magnitude estimate is ~ 1. Order of magnitude estimates are perfectly legitimatewhen you are working in theory. Not all physics questions require precision!

Diffusion: how width of diffusion changes. by units, the law is: 2 ~r tD in free space.

What happens when you have an interface? How does the law change?

The two concentration profiles would appear as,

mathematica: plot Gaussian vs. piecewise-Gaussian for various times


3/45

Implicity: we introduced the boundary conditioninterface

| 0x x

n

.

Reminder: do not use math or any sort of calculation.

Type III boundary condition: flux of particles proportional to number-density. The above extrema, vacuumand impermeable wall, were two extremes of an intermediary behavior were now going to try and capture. The

boundary condition is nowinterface

| /x xn n ; the is a characteristic-length that comes from the derivative-

operator.

Question: can you use dimensional analysis to claim that 1/2( , ) ( ) ( )xt

n x t t f D

D ? Can you still use this same

ansatz2?

Further guess: you will probably have a linear combination of the full Gaussian and the piecewise solution that

vanishes appropriately at the boundary.

N.B.: the method of images enters here, too. This is provocative.

Limiting cases: classical mechanics is limiting case of quantum theory. Somteims we know higher-ordertheory, sometimes we know lower order theory. For diffusion, we know lower-order theory: the Boltzmann

equation.

Problem with diffusion equation: too little information. Set up system in macroscopic. What koind of processin real life leads to diffusion? Happens by ballistics and relaxation-events w/probability.

Particle has definite position and velocity. diffusion equation does not ask about velocity; it only computes

concentration. You have averaged all velocity out. Higher order theory keeps information about velocity.

Collisions: elastic collisions on event of relaxation. Therefore, energy is conserved. Thus, velocity changes onlyin direction, not in magnitude.

So: compute function at constant energy,

2 20[const]

( , , ) | ( , , )v

f f t f x t

v

v r

Boltzmann equation: describes evolution of distribution function f,

0

0

mean freecos ; ;

timeF

f fdf f f p f f v

dt t x t x

2 Even so this is a Guess/ansatz of a solution, it is highly general; you are only asserting (1) a scale /x tD and (2) units 1 1L t

D

, in which there really isnt any sort of bad physics or reasoning.


4/45

Collisions randomize the angular distribution of velocities. If you sample velocity at two different times, youwill observe two different angles, but the same magnitude since you have energy conservation.

Relaxationtime approximation: you have0( ) /f f f .

Trick: expand function ( , , )f f x t in a complete basis. Since you have angular distribution, you need to

expand in basis complete overangels. Thats Legendre polynomials of the argument cos(theta),20 1( , , ) (cos ) ( , ) ( , ) 1 cos ( , )f f x t P f x t f x t f x t O

Derive diffusion equation from this! We actually only ned two Legendre polynomials: p0 and p1. Also, let a

large number of collisions take place: t .

Truncate the series. Then, you have another statement,1

1 1 10 1 0 0 02 2 2

1( , ) sin ( , , ) ( ) 2 ( , )n x t f x t d f pf dp f f n x t f

There: so the diffusion equation is the leading term in the series. The first order correction will tell us about

velocity. ut in flux,1 1

1 1 10 0 0 1 0 1 1 0 02 2 2

1 1cos sin ( ) 2 / ; ( , , ) ( , ) 2 ( , ) cos /j v f d v p f pf dp v f f j v f x t n x t j x t v

Second term disappears when we average over angles! Thats the behavior we want! We want leadingterm/diffusion-equation beaviour when we remove angle-dependence by averaging! (Cool).

Hm, dont know why, but,

0 0 0 0[deviation from equilibrium/Fick's law] cos 2 / cos ( 2cos / ) 2 cos / ( )

t t x xn j v v j v j v

Multiply whole of (1.8) by cos(theta). Find average value of function cos(theta) on both sides. Note the average

value of cos2(theta) is ,and that cos3(theta) has an average value of 0 due to its odd parity.1

0 02( / ) 0 [fick's law, which subsumes matter-conservation]t x t xn v v j n j

Extra cosine function kills all terms left in the angle. 2nd

term vanishes.

Now, see if you can obtain from (1.8), the,2 21 1 1

0 0 0 0 02 2 2/ / ( )

t x x t x t j v v n j v j v n j n j v D D

You computed diffusion-constant! Diffusion equation is limit when tau 0, or t tau.

Finally: note extra term t j . This imposes constraint on validity of diffusion equation. when this term is small,

ficks law is good approximation. Thus, current must change slowly in time. Neglect t j : go back to diffusion

equation, then you know ~xj n x t D D . Thus, x and t are related. Flux should evolve slowly in time.

Number density and flux will not have independent scales. There is no such thing as a scale where flux and

number density are not related. We need to look at times t . Derivative: ~ /t j j t , meaning that

~ /tj j t (and, of course, we mst have 21

02t xj v n ). That scale is the mean free path,

0~ ~ [mean free path]x t v D


5/45

Diffusion equation operates on long time scales compared to length-scales, as relatedby mean free path as

related to tD . When you do not have long time scales, the transience is captured by the termtj .

Note: in order to lsoe information about velocity, several collisions must take place. (after first collision, you

still should be able to track the trajectory of the particle; not yet randomized). This is why boltzman equation is

1st order in time: when you relatex

j n D , you have disparity under time-reversal (c.f., previous lecture),the paradox. The paradox is resolved in this information-about-velocity-direction point of view.

Example: strong explosion goes off3. How does blast front move in time, given the explosion is strong?

(1.9)

What are parameters at work? (1) Energy of the bomb. (2) since explosion is modeled as spherical front of hot

gas effusing (quickly) into cold gas, you have hot temperature and low temperature, forming a pressuregradient,

0 0parameters , , , ; ( ) [blast radius] ???R E t R R t (1.10)

Temperature inside (hot) should be function of E0, so possibly not independent. Temperature outside is

independent parameter (in which mass-density of air, specific heat, etc., will be subsumed).

Cobble parameters together to make a thing that you want: dimensions! Buckingham pi theorem.2 2 3

0 0 0 0[ ] [ ]( , , ) [ ][ ] [ ] ; [ ] [ ][ ] ;E E R t M L t M L (1.11)

Note: bomb may be chemical or nuclear. But both are the release of energy, and dimensions doesnt care which

it comes from. Using 212

E mv , and noting 0 0,E are material parameters: out of the three units that must be

on RHS of result, we cant make a dimensionless number. Therefore, each should enter as a power.

Now: the following is a true statement: order of magnitude estimate of R,

0 0~ ; [surprisingly-constraining statements!]a b cR E t (1.12)

The (1.12) is by no means rigorous, but it is very informative/helpful. Under what condition is it not helpful?When you have scaling function w/o dimensions,

0 0 0~ ( / ); [scaling function captured by dimensional analysis]a b cR E F t not (1.13)

So: we are linting ourselves to the scope (1.12). Step 2 = find the exponents. Substitute in the units,2 3 2 1 1 2

0 0 5 5 5[ ] [ ] [ ] [ ] [ ] [ ] [ ] 1 2 3 ; 0 ; 0 2 ( , , ) ( , , )a b c a b a b c aR E t M L t a b a b c a a b c (1.14

3Taken directly from research on the theoretical yield of an atomic bomb.


6/45

Hey, thats 3 equations and 3 unknowns! You get the radius as a function of time,

205

0

( ) ~ ;E

R R t t

(1.15)

Shortcomings: you dont have a characteristic distance coming out of (1.15). However, this does tell you thatthe explosion has a front,

This sacrifice of scales is manifest; recall that power law functions dont have scales. You are basically firstmaking a guess; no other units. Look at units, see power laws. See if youve found all qualitative behavior: (1) t

= 0 behaviour, (2) the expanding front, etc.; basically, units says you can only combine all parameters in oneway; mathematically speaking, you can find two solutions: (1) expanding front (@) bell-looking thing;

hwoever, neither solution has a scale.= due to power law.

Finally: fluid dnamics is nonlinear. This is a hard problem (nonlinear fluid dynamics equation, etc.). Before thismethod, nobody looked at strong explosions, since that was considered a difficult problem. This particular

solution was mostwelcome.

In the 1940s, propaganda. You saw news reels, not previews. One t ime, there was a news reel with a nuclearexplosion. John Archibald Wheeler happened to be in the audience. He observed that there were regularly-

spaced telephone poles in the explosion. He used the uniform-spacing, and made a computation. His result wasso scarily close to the actual result that he got in trouble with the FBI, because his result was scarily close to

classified information

More formal stuff symbols

Go to MM - 001 terminology.


2 1

2

2 2 3 2 2 1 2 2

[ ] ( )

( ) ;

( , ) ( ) ( ) ( , )4 ( ) ( )a

a a a

D L T R R t L L

R t x

x x xx x n x t d x D n x t d

D D D

Consider: on the basis of justdimensional analysis,

[dimensional analysis] ( ) ( ) ;a

tR t L Dt f


7/45

Now need to determine scaling function. What properties? Well, consider asymptotica

t vs.a

t :theres information in both these asymptotes. You have (0) ~ 1f and ( 1) ~ ?f s

Now: diffusion equation. you have absorption: 2a

n D n n . Assume a solution: /0( , ) ( , ) at

n x t n x t e as an

ansatz to handle the absorption term,2

/ /2

0

scaling function is1( , ) ( , ) exp( )

just an exponential42

a at t

a

xn D n n n x t n x t e e

DtDt

(1.16)

Then you have,/ / /2 2 2

0( , ) ( , )a a a

t t tx x n x t dx e x n x t dx e C tD C Dt e

(1.17)Particles mostly at surface. At long time, they most likely forget about the absorption term,

So: dimensional analysis only takes you as far as ( ) ( / )aR t Dt f t ; most of physics concerns itself with

finding this scaling function.

Asymptotic accuracy and parametric scaling

We only, only solve a model, which intrinsically neglects factors. All our results are only asymptotically exact.

Example: specific heat of Fermi gas: 2 312

/ ( )BF

k T

V B B F EC k n k T E O . This is true in the limit where thermal

energy is much smaller than the Fermi energy (as indicated). Our result always is where a parameter is turned to

one of its extremes (smallness or largeness). So: Boltzmann statistics: 12V B

C d k n , where dis thedimensionality

4. What does full behavour look like?

Width of asymptotic

4Note that Fermi energy is removed as scale.

1 2 3 4 5 6

0.05

0.10

0.15

0.20

0.25

0.30

0.35


8/45

Width of crossover region is about / [ ] [Kelvins]F B

E k K .

example: if a scaling function has a 0 near a point, we can ask if the function has linear, quadratic, or cubicbehavior near 0.

Example: a function blows up at the endpoint of an interval

Ratio of heat capacities: thatof Boltzmann and that of Fermi,2 21

, ,2

21, ,2

( / )( / ) 1 ;

V F V F B B FB F B F B F

V B B V B

C Ck n k T E d k T E k T E k T E

C nk d d C

(1.18)

Parameters you control, numbers you dont control (duh).

Functions

Dimensionless function f dimensionless argument. Could be two types of functions: (1) normal and (2)pathological.

Function-composition hides the pathology.

Friday, January 18, 2013


9/45

Brownian motion: Model as (1) random walk, (2) discrete steps that happen after a time (the mean free

time). (3) Once a time a elapses (over which time any number of discrete-steps may be taken), the particle

disappears, and gets absorbed,

(2.1)

Ficks second law then has an absorption term added to the time-derivative, / an n n . We make an ansatz:an exponentially-attenuated concentrated envelope which, if integrated, yields total particle-number at sometime (note: normalization is time-dependent for disappearing particles),

/ /2

0( / ); ( , ) ( , ) ; ( ) ( , )a at t

x an D n n n x t n x t e N t n x t dx e

(2.2)

Case 1: What if 1 ms and 1a s ( a )? Can this model still be used? Yes. The absorption time islong/absorption doesn't happen that often. diffusive trajectory is formed over many mean-free-times .

Diffusion is largely ballistic. Your concentration profile then appears as the usual solution,

2

2

3 2 2

1 122 2

2 2 2 1 3

102

( ) ( )( , ) ; ( ) ( ) ; ; ; ( ) ;

2 2

( )( )( ) ( ) 0 2 0;

( )

x xx x

xd x dxdt dt

g f d x d x d f n n x t f f s f s

r dt dt dx

f f ff ffd d fLn sf f f

dt dx

f sf f

D D

D D DD D

2 2/4 /(41

0

)12 2

ln ( , ) ( ) / /0 ;s x tsf d f s ds f ce n x t f s Ce D

(2

Case 2: Opposite limit:a

, in this limit, you have more traps than elastic scatterers. You can consider thisto be a lattice of a high density of traps, and a low density of elastic scatterers. For high density traps (frequent

absorption), diffusion equation becomes meaningless, and you have to go back to Boltzmann equation,

00 0

( , , ) 1{ } ; { } ; ( , ) ;

1

ii i i

C C

f t fF fI f f f v f I f f f T

t e

kk rk r

k

(2.4)

There are no forces ( 0iF ), we use the relaxation-time (as indicated in (2.4)), we retain the partial-derivativeof the nonequilibrium distribution function, and realize cosi i xv f v f r , and we have,

0 0cos ; cos ;a a a

f f f ff f f f f f fv v

t x t x

(2.5)

Then: effect an arrangement of ascewnding orders/perturbative-expansion ansatz, regarding isotropy as a

leading term, and having a correction 2 cos /j ,trial solution for f in "almost" isotropic: leading term2

( , , ) ( , ) cosboltzmann equation isotropic, plus -dependent correction

jf f x t n x t

(2.6)

Drawing the trajectory over two separate mean free paths, we have,

(2.7)

You could also have


10/45

0

/ /

0 0

1/ /1

0 021

( , ,0) ( ) (cos cos ) ( );

( , , ) ( cos ) (cos cos ) ( cos ) ;

( , ) (cos cos ) ( cos ) (cos ) ( cos )

a a

a a

t t

t t

f x x g x

f x t g x vt e x vt e

n x t x vt e d x vt e

Bunch of spikes.

Integrals

,

0,

( ) 0 ; ( ) ;

~ 1

I f x dx f x dx f x

Example,

0 00 0

2120 0

( ) ( ) | | 1; ~ 1;

0 ; 1; ; 1; ; ~ (1) 1;

vs. support from large region: | | 1;

vs. midd

x x x x

x

x x x A A

A AA

Pf x xe f x dx xe dx xe e x

Q

x A x A x A xe dx x dx

xe dx xe e Ae e

2

2 2

1 11

1 2le region: | | ; (x , ) ( , ) [interval where integral takes support]x

x xx x x

x xx

xe dx xe e x A

22

11

1 1

2 2 20 0

1 1

1 2

well-known ~ 1; 1; tan | tan ;integral 1 2 1 1 2

1 1tan | 1; tan | ~ 1;

2 1

AA

xx

A xx

dx dx dx x Ax x x

dxA x

A A x

Another example: function with an integrable singularity,

2 0

1 1( ) ( ) [finite] ?; [well: integrate tiny region]

( 1)f x f x dx

x x

0.5 1.0 1.5 2.0 2.5 3.0

0.2

0.4

0.6

0.8

1.0


11/45

220 0 0

1 1 1 1( ) 2 1

(0 1)( 1)f x dx dx

x x x

Therefore: the singularity we indicate isstillintegrable.

Convergence theorem from calculus: consider 0( ) | ~a

xf x x

. We must have 1a for convergence.

The upper bound a < 1 comes from avoiding logarithmic-divergence 0lim lnx x . This is basically thepower-law convergence that you are asked to remember.

2

00 0 0 0

divergent due toln ln ln 1; ln (ln ) (ln ) | ;

lower limit 0 2

x xx x xe dx e dx dx xd x xx x x

Question: how do you estimate lnx for ~ 1x ? Well, you have ln( ~ 1) ~ 1x . Note: for ln( 1) 0x .

Question: of the zillions of special functions out there, how many have independentasymptotic behavior? The

point is: asymptotic behavior is indistinguishable from the function whence it came. If you captured asymptotic

behavior, you dont know what special function it came from. E.g., 0 0( ) | 1xJ x , but then again 0| 1x xe . Butthen again it could have come from

0| 0p

x

pe

.

In any event: a Bessel function is 20 4( ) sin( )

xJ x x

(oscillatory). Special functions have weird behavior,

and we are intererested in their properties at special points, such as 0s.

The function sin(x) goes to x at small x. So: handling asymptotic behavior of function means massaging it into

some form:0{ , ln , | }

pa c x

px x e

(??).

No class Mondaysleep well.


Pool ball example. But now the question is: find the thermal equilibrium,

(2.8)

Depends on whether pol table has walls or not. If walls, then after finite time allpoolballs will be involved inmotion. Without friction, the balls certainly will thermalize. Without walls, number of moving balls increases


12/45

intime, and at the same time the are of moving bals is also increasing. How does number and/or density ofmoving balls increase in time?

Assumption: spaerse-distribution of balls: d a . Total number densiy of balls for 2D table is 2~bn d .

Describe such a process in terms of classical cross section. Two balls collide. Each ball has radius of a. then, the

classical cross section is necessarily on order of: 2~ a . But total crss section really doesnt tell you anything

Note: in any process, incoming energy is shared/divided amidst the particles it collides with.2

2energy transfer

~ ; ( , , ) ( ) ; Ansatz: ~ ; 0 1; ;ratio

aaa d d f a d d x a x

(

The singularity at x = 0 is for forward-scattering. Peaked at small energies. Most probable collisions are suchthat the particle (1) keeps most of its energy and (2) keeps its original direction of motion. Particle goes through

array of stationary atoms. Keeps projectile motion. Motion, itself, is contained in narrow (conical?) regionencircling axis of original trajectory.

What if1 a ? Collision leads to rouighly dividing incoming fac

Question: whgich of the two scenarios dominates? Not all motion which spreads does spread ina Brownian

way. Brownian motion is a sub-case of this motion if the original ball (primary) has a mass very different fromother particles. Heavy/macroscopic particle is jostled/attacked by molecules which we do not see. Collisions

essentially elastic. So: in this scenario, is energy shared evenly, or is there a preference for small energytransfer?

Note: you must have a normalizable ( , , )f a . This necessity is realized as,

2 2 2 2

00 0

eye-catching yet-integrable-( / ) ( / ) ( ) ~ 1 ~

singularity, you have small number of thesec a d a x dx a a

Picture of equally-shared energy is simple and plausible. (recall Maslovs huge excursion into formalism, just to

get a laughably-simple result: particles equally-share energy). Estimate number of balls, 0~ /N . Look at

system as function of time. Consider system when average energy is . At the time where you measure the

average, you know number of balls that have been struck.

Thermalizing: at some point, the things must equilibrate/thermalize. Then, you have Boltzmann distributionfunction. Energies parameterized by just temperature. Does system reach thermal equilibrium?

Estimate mean free path of 1st

generation, vs. mean free path of 2nd

generaton,2 2 3 2 1 3 1 2 2 2

/ 1 1~ ; ~ ; ~ ~ ( ) ~ / / ( / ) | ;D D

b b d a xa n d n d a n d a a d a d a d x d d

(2.11)Area then increases in proportion to the number of circles,

(2.12)Then you have,


13/45

2 2 4 2

2 2

2 2 2 2 2 2

[#-density of moving-balls]1 /~ ~ ~ ~ ; ~ ( ) 1;

[number of balls? (per unit vol?)]1/

mmm

bb

nnN N d a d aR N n

nR R a n d d

Turns out: this does not thermally-equilibrate to some Boltzmann distribution, because it needs to finish its

transitory period illustrated in (2.12). When the walls reflect the transferred energy, then you get thermalizing.When mean free path is increasing function of energy, you have constant mean free path. The radii of the areas

of circles will go faster than energies. You then have that increase fater trhan the number of particles.

Central theorem of dimensional analysis

This is called the pi theorem. Already used this multiple times on homework. Worthless to repeat. Sludgethrough abstractions, and you see the stuf you used on your homework.

Friday, January 25, 2013 my birthday!

Example: what is resistive force as object moves through fluid?

3 1 1 2

1 1 3 1 1

0 0 1 1 1 ..;

0 1 0 1 2

a b c d a b c d c d b d

a

bF R v L M T L M Tc

d

See QM 11 - 002 - dimentional analysis for object in fluid and for fine structure const.

NOTE: since both force and velocity are vectors, then force must eb opposite of velocity. thus,w e make the

ansatz A F v , where scalar-function, ( )A A v . Symmetry: fixes direction between force and velocity, butdoesnt tell you what A(v) is. Assumption: analyticity: that means F depends on v to thefirstpower, allowingfor a Taylor expansion to be linear to lowest-order.

( ) ( ) (Re); (0) ~ 1;/

vR vRG G G G G

V

Expnanation for behavior of this: at boundary, fluid has same velocity as ball,

2small Reynold's

( / ) ~ / ; ~ ; ~ ~ ;number limits

u r R v R F vR F PR v R u

large Reynold's need drag-force

number limits stuff...

January 28, 2013

Estimate the air resistance on your car at 60 mph and assuming rolling friction of 0.01RFF N .


14/45

2 2

14

23 3

14

192

27

2 2 3 2 2(1 )

2(1 )

2 2 1

31 3 3(0.026 eV) (1.602 10 )

2 2 0.3 0.7 (0.3 16 0.7 14) 1.7 10

a b b c c d c a b c b d

B

B RMSO N

c ac a

b c b b aLK E L E K M K E L M

d c b a d aT

b c a

k T

mv k T v m m kg

2 10 2

710 ;

(2.5 10 )

m

s

m


Fractals: something that doesnt play well with dimensional analysis. It is an arrangement which thwartsdimensional analysis. Dimensional analysis (DA) uses SCA: spherical cow approximation. This is

inappropriate in the coastal line paradox. Solution to coastal line paradox is richardsonss law, which contains a

fractional-power of a dimensionless quantity, undetected by SCA/DA.

Koch snowflake:

1 2

1 1

ln(4 /3) ln

4 2 4 41 3 ( ) 4 ; 2 3 ( ) ; ... 3 ( ) 3 ( ) ; 3 ( ) ;

3 3 3 3

ln( / 3)3 3 ??

3 3

m m

m

n

nn

n P a a n P a n m P a a P a

a as n P ae ae

Topological dimension DT, and dimensionality D (e.g., surface in three dimensional space means DT = 2 and D= 3),

1 1

2 2 2 2 3 3 3

( ) ; 1; [line] ( ) ; 1 2;

( ) ; 2 3; ( ) 3 4;

T T TD D D D DD D D DT T

D D D D D

R RV R R s D D D L R R s Ds s

R RS R R s D V R R s D

s s

Friday, February 01, 2013

What is number of organisms vs. time, given that they breed? Start by thinking about parameters,

2

3

amount of O [ ] [ ][breathing rate] ; [weight] [ ]; [mass density] ;

Unit time [ ] [ ]

M MR W M

T L

(3.1)

Physicists introduced an additional parameter . You have,

amount of oxygen; [dimensionality of organism];

m

tL tL (3.2)

Putting (3.1) and (3.2) together, we have,

/3~ ( ) ; [mysids] 2.4; [crabs] 2.25; [humans] 2.4;W

R

(3.3)


15/45

Hence, there is some universality between the various species. It has to do with the ratio of lung-volume to therest of the body-mass.

[fractal]; 2 3; 2 3;DA L D (3.4)

Fractal dimensionality applied to slight modification of a game of pool

A pool ball hits a cluster, and causes the cluster to eject particles. Those secondary particles do likewise, and

cause a quadratically-larger number of tertiary particles. This sounds artificial, but this is like what happens innuclear reactions, and in lightning storms.

Should be done with dimensional analysis.

Next week: approximating integrals, differential equations, etc. you will be refreshed on any topics in math that

are relevant but you have forgotten.

Monday, February 04, 2013

You have integral: approximate it,

0

20

( )( ) cos ; ( 1) ?;

1

ax J xI a e xdx I ax

(3.5)

2

0

2 20

( )( ) ( ) cos ( )

( )

ax J xaI a e x d axa a ax

/00

2 20 0/

( )( ) 1( ) cos cos ( ) ~ ;

1 1 ( )

uaax u a u u

a auaa

JJ xI a e xdx e d

x a

All functions in the integral have a scale on the order of 1. You can strike out the integral in the region where

u/ainfinity, because of the decaying-exponential.

Similar example, simple, but in more detail. Consider an integral5 where we normally complete the square,2 2

212

0 0( ) (1 ( ) ...) ; ( 1) ?; ( 1) ?;ax x xI a e e dx ax ax e dx I a I a

(3.6)

The series generates a polynomial, whose coefficients are familiar,

2 21

1 ( 1)/2 2

0 0 0

the series( )1 1 1 1 1 1 1( ) 0

converges! ! ! 2 2 ! 2 2 ( 1)

n nx n x n y

n

x nC e dx x xe dx y e dy

n n n n n

(3.7)

Read up on asymptotic series; this is covered in homework

Example: the exponential vs. the Gaussian2

2 412

1 ...xe x x ,~1

0 0

1 1 1(0 1)

y axax yI e dx e dy

a a a

(3.8)The main logarithmic approximation

5Expanding the integral before integrating is actually bad: a mathematician would point out that the integral has a region where

x , so we have no right saying that the series-expansion converges. However: we have a Gaussian. We show the Gaussiancauses the series to converge in (3.7).


16/45

This is very

Example: consider,

02 2 2 2

0 0 0 0 0

(1 ...) ( ...)( ) ln |

1 1 2 1 2 1 2 2

axe dx ax dx ax dx x dx dxI I a

x x x x x

(3.9)

Thats trouble. But this series does converge. How to prove? Well, cut integrals first term to being,1 1

2

0 0

1 1( ) [ln(1 ) ln1]; ln1 0;2 1 2 2 1 2 2

a a

x dx dy aI a a ax y a

(3.10)

Now: consider 1 1

0 0

a ca

dx dx

, which yields 112( ) ( ln(1 ))I a a ca . Main logarithmicapproximation: requires that not only 1 1a but also 1ln 1a . Then, 1 1ln(1 ) ~ ln( )ca ca , and,

1 12 2

( ) ( [ln ln ]) ( ln ) ;I a a c a a a (3.11)

We sacrificed knowing the coefficient under the log.

This is used in theory of superconductivity, where powers of a logarithm are,

2 3 2 2 3 2

1 2(ln ) (ln ) ... ~ ln ln ... ;1 ln

T T T T

T

gg g c g c g g g

g

(3.12)

There is a lot of pysics contained in the condition for a series to converge.

Theories which have logs are knwn as renormalization group theories. Bare interaction is then dressed.Watching for logs has become the favourite pastime of theorists; a log means the potential has a chance to be

nontrivial.

Added notes

When expanding the integrand doesnt work

On series-expansion of the integrand, you could wind up with a divergent term in the series that issupposedto

be small, e.g., 2 221 10 0

axe dx x dx

x xa

, in which the term leading in a is divergent even so 1a .

Method 1: you could use a cutoff procedure,

Taylor2 20 0series...

2 20

2 20 0

leading-order[1 ( )] 1(0)

term...1 1 2( ) ; ( 1)

1 divergence from[1 ]

infinite-order terms1 2 1

ax

ax dx dxI

x xe dxI I a I a

x ax dx x dxa

x x

O

O

(3.13)

remaining integral is dominated by large x. non-analytic series in a: the closed-form integral is not a power

series as nnc a , but rather ( )n nc f a , where ( )n nf f a does not have a series-expansion for small a.

solution: re-express the upper bound of the integral, 10lima a

, but let a be close to zero, but finite, as a

cutoff,,1 2 11/

220 0 0 020 0

1 1| | ln(1 ) | ln | ln ln

2 1 1 2 2

a a

a a a a a a a a

dyx dxI a a I a a

x y a


17/45

Ambiguity: what if we multiplied the upper limit by ca instead ofa ? Turns out this ambiguity is immaterial;

consider 0 00 10 ~ 10c c ; then 2 2( 1) ln (ln ln )I a a ca a a c in which ln lna c for 0a .

Recapitulating our treatment of the integral,

Wednesday, February 6, 2013

BCS theory of type-I superconductivity

Convergence vs. divergence of integrals: integral in problem 1 that was2

ln0

0( )x

xJ x dx

. Joke problem: once yousee that its convergent, you get the integral of order-1. Some said it was divergent. Thats wrong. How do weknow its not divergent? Rule of thumb: logs neverchange convergence or divergence of integral.

Consider diagram with ( , ) p interacting with ( , )p , and emerging as ( , ) p and ( , ) p (respectively6).

You have the Hamiltonian with the energy gap g a a p pp

2 2 *1 12 2 ; ;BCSm mH a a g a a a a H a a a a a a p p p pp p p p p p p pp p p p pp p (3.14)

Approximate . .a a a a h c p p p pp , which is sort of a mean field approximation, and thats where ourenergy-gap * comes from. Replace a bp p , bosons. Then,

BCSH b b p p pp , and the energies are

modified by the split2 2

2 2

2 2[ ]

p p

F Fm m p . You get energy gap by self-consistency relation. The self-

consistency relation appears7

as 0ln( / ) 2 ( / )I T , in which T , the zero-temperature gap is1

0 (0)ge

and2 2

2 2

1

0( ) ( 1)

x y dx

x yI y e

. Obviously, 0ln( / ) 2 ( / )I T is a transcendental

equation.

Limit of large y: take out a factor of y, and expand the exponential,

212

2

/

2( / ) 10 0

( ) 21( ) ;

2 2( / ) 11

x y yx yy y

y x y

d ye eI y e dx e

y y yx ye

(3.15)

You can clearly see in (3.15) that as y 0, you have divergence; specifically, a logarithmic divergence as you

get to the lower-end of the integral, x 0 (e.g., 2 2 10lim ( )x x y y

, which integrates to a log).

Meanwhile, use 0 0 0 00 0 0 0

0

21

1ln ln ln(1 ) ln(1 )

O , and we can then get home free and

realize 0 00

/

0 02(1 2 / )

Ty

ye T e

.

6 It should be explicit by this respectiveness that theres no spin-flip, for simplicity.7

The coupling constant g of the BCS interaction was absorbed into the band-gap equation


18/45

Now: the aim is to find the critical temperature. Let the integrand of (3.15) be a function, f(x,y), so you have2 2

2 2

1 1( , ) ( 1)x y

x yf x y e

You have (1) ?CT T . You have / ( ) / 1Cy T T T , and,

1 1 1( , ) ~ ; ( 1, ) ~ ; ( 1, ) ~ ;

1

x

x

ef x y f y x y f x y

y x e x

(3.16)

These three regions appear as,

2 111 12

ln( )I y y C I I I In which

2 21 1

2 20 0

tanh tanh1 1 1 1 1( ) ; ( )

2 2 1 2 2

x x

zI dx I I dx

x z e z xx y

Massaging,

21

2 20 0

tanh1 1lim( )

2 2

A A x

A

dxdxI

x y


Warm up: is the folowiung integral convergent or divergent?

3/2 2

0sin (ln )x x x dx

Well, at the upper limit infinity, the functions x-3/2 kills the 2(ln ) sinx x .

Also: the lnx function goes to infinity faster at x 0 faster than 1/x goes to infinity.

1 2 3 4 5

0.05

0.10

0.15

0.20

0.25


19/45

Main logarithmic approximation,

0 0 0 00

( ) ln ln ln ; ln ~ 1;Q

Q P P P Q Q QQ

Where is log-approximation relevant? Solid state physics; extremely relevant there. Example: Drude model of

conductivity has 2 /ne m , in which is computed by the Boltzmann equation. at lowest temperatures,whats main scattering mechanism of atoms? Impurities and defects. (Temperature-dependence of resistivity isVERY important),

0 0 0ln ln ln ln ;D D DT

TT

Example: frictional forces; they scale as the log of the velocity,0 0ln( / )frf f v v ; this is the friction a series of

OOOOOO atoms exerts on a V point, pointing downward.

Example: fluid dyamics, if you have a flow across a surface with friction, you have an x component and y

component of velocities, xu and yu respectively. You have 0 0ln( / )xu y y y .

In theory, all these dependencies are obtained using MLA, or by slugging through some pretty heavy formalism

Cutting stuff off: an integeral-functio n of a that has an integrand that goes to 1/x at the lower limit of 0requieres treatment by cutoff,

0 2 0

0

trouble at1( ) ( ) ( ) ( ( )) ~ ln

lower limit

a

a

aI I a f ax g x dx x dx

x a

O

How to effect a cutoff treatment: consider,2

2 21 .

/2 2. /1

2 2 2

0 0

1 ...( ) ~1 1 2 1 2 2

ax vsa aax ax vs x aa

a a

e ax x dxI a dx dx a a dx ax x x

This integralsaturates to the value 1 as x infinity. Thats where the divergence comes from.

dI/da: When you have exponentials under the integral ( )I I a , its a good idea to play with the integral bycomputing dI

da. You can have,

2 22 21

2 2

0 0

??1 1

ax axadI x e x edx dx

da x x

2 4 6 8 10

0.2

0.2

0.4

0.6


20/45

Well, consider x infinity. That means the function2

21~ 1x

x. Meanwhile, the Gaussian falls off as 1/ a . The

Gaussian is a slow function. Nothing happens to the Gaussian (it is ~1) until 1/x a . So basically, youare breaking up the integral into 1 1[0, ] [0, ] [ , ]

a ax ; e.g., breaking up the interval in a manner entirely

around the properties of the Gaussian.

BUT WAIT: thats not all we could possibly do. Use 2 2y ax , and the above intervals are mapped to,2

2 2

2

2

0

0

0 1 1

1 1 1 1~ ( )

1

ayy y

yy ya

dIdI y edx e dy I I e dy da

da a daa a a

Closing question: what happens when we consider a 1 in the above integral? The Gaussian is much more like

a spike/Dirac delta which samples the function2

21

x

xclose to the origin, at x = 0. Then, 12

2( ) (1 )dI

daa a

,

and we see this goes to 0 as a infinity. (this is an interpolative function).


Integrals with oscillatory term inside, times some envelope function. You get these integrals when you have

Fourier transforms,

1 20 0

( ) ( )cos ; ( ) ( ) sin ; ;I f x x dx I f x x dx

Suppose ( ) xf x e . Then you could have,

1 (1 )

2 0 0 0

2

2 2 1

( ) cos Re Re( )

( ) sin Im Im

Re Re Re 10 1 1 1 1

~ ;Im Im Im(1 ) 1 1 1

x x x xI x

I e dx e e dx e dxI x

i i

i

i i

Consider plot of functions cosxe x and sinxe x . The number of oscillations the function undergoes is ill-

defined for infinite bounds of integration.

Integration by parts: consider cosine integral, and integrate by parts,

2 2

1 1 1 1

0 0 0 0

1 1

0 0

cos sin sin ( ) ( cos ) ... ...

sin cos ?? ...

x x x x

x x

e x dx d e x x e dx d e x

e x dx d e x

(3.17)

And continuing, let 21

1( ) xf x ,2

2 2 2 2

0 0 0

doesn't have power-lawcos 1 sin 1 1 1( ) sin ( ) ; 1 ...

series...1 1 1 1

x x dI dx d x dx x

x x dx x x

Instead, consider the following contour integral with two poles x = + i, as.

2 2

1 cos 1 1 2 1Re Re exp[ ]

2 1 2 1 2 2 2 2 1/

xx e dx edx e

x x

ii

i(3.18)

You put in a non-analytic exponential function,


21/45

1cos

2

xI e x dx

The 21

1 xis analytic over the entire real axis. That means real axis is free of singularities.that means we can do

integral by contour integral. Cauchy theorem just then says that value of integral determined by singularities

(specifically, the abveoemntioned x i ), as,

As long as you substitute complex umber into oscillatory function, they become exponential functions.

Physical example: consider a charge q in 3D. laplaces equation is 2 34 4 ( ( ) )e iq r . Chargedensity of ions vs. electrons. Born approximation: ions are rigidly held in place. Then, charge density of

electrons is function of position, ( )e e r , resulting in a complicated and nonlinear PDE.

But what if charge was small enough to only slightly perturb the background charge? Well, chemical potentialgoverns how much energy youintroduce per unit charge. Chemical potential is,

2 2 0 2

0

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

series expand ( ) ( ) ; ( ) ; ( ); ( ) ;

F F e e e F e

e e e e i

e e e

n n

e e N n N d N e N

r r r r r r r r

So then we have Poisson equation that is a sum of a Green function attackable part, with other one,2 3 24 4 ( ( ) ( ) )q e N r

Apply Fourier transform, letting image function be 3( ) ( )e d r i k rk r . We have2 2 2 2( 4 ( )) 4 4 / ( )k e N q q k , in which we have 2 24 ( )e N , the Thomas-Fermi

screening wavevector. Inverting the Fourier transform,13 2 cos

3 3 2 20 1

1

cos

1

4( ) ( ) 2 (cos );

(2 ) (2 )

1 sinIn which: (cos ) ( )( ) 2 ;

kr

kr kr kr

d k k dk q ee d

k

kre d e e

kr kr

ii k r

i i i

r r

i

So we finally have,

3 2 20

4 2 1 sin 1( ) ; screening radius ; 1

(2 )

q krk dk r k

r k

r

Now: you already know the answer: you get a Yukawa potential. Note: sinx x . You are able to analytically

continue this ito the complex plane,

2 2

1 1 sin( ) ; ( ) ;

2 2

r rk kr dk qe er k r

r r

You see that poles are complex.

1) look at integrand on real axis. (2) look for points of non-analyticity on real axis. If there are, cant go tocomplex plane. (3)

Poles in complex plane substituted into


22/45

Use the Cauchy theorem exponential. No Cauchy theorem/non-analyticity power law.

Example: consider plane of charge in 3D, and consider z-axis going through normal. you have log behavior. Go

back to the Poisson equation, and say 2 3 24 4 ( ( ) ( ) ( ) )q e N z r , in which3 2

( )D D

n n z . Thedelta function sends the z-argment to 0, and you have homogeneous equation everywhere else. You basically

have ( ) ( ) ( ) ( , ) ( ) (0, )z z z r z r r . We must have,2

02 2

0 2 24 4 ( ) ( )(0, ) ( ) (0, ) ( ) ( ); ( , ) ;z

k z

z

z

q e N k dz r e e z d r r e z d r k k k

ik r ik ri

kk

Fourier transform,

2

0

0

2

0 2

0 02 2

2 ( )

0 0 1 1

2 2

( , ) ( , ) ( ,0) ( , );2 2

4 4 ( ) ( )( ) 4 4 ( ) ( )

2 2

2 2 2( ) 1 ( )

z

zk z z z

z z

z z

z

e N

k

D D

dk dk z k e k

q e N dk dk q e N

k k

q q q

k k k

ik k k k

kk k

k k

You then have the potential,

2

3

2

/ ; 1;( )

/ ; 1;

D

D

q r r

r r

r

p

Yukawa potential in 2D is a dipole potential. (interesting). Very important qualitative result. Once a Fouriertransform has an absolute value of something, look for power laws. Notice in 3D, you had denominator of

2 2

2Dk , wheras in 2D, you have1 1

2Dk .


Last time, we thought about screening radius of electronic impurity in charge jellium,1

2

32 2

12( ) ; ( ) ;

1

D

D D

r rqk r

k r r

Try inverse-transofrming, you hit a snag,

cos2 22 cos1

02

2 2 2 20 0 0 0 0

00

00

2[ ( )] 2 ( )

(2 ) 2 2 2

( )( )

k r k r

D D D D

k e dk d k dk k dk qe d k q q qF k e d J k r

k k k k

J x x dxq qI x

r x x r

ii k ri

So now, we have to do an asymptotic expansion on the function 0( )I x as,

0 1 10 2

0 0

( ) ( ) ( )( ) | ...???

( )

J p p dp J p J p dpI a

p a p a p a


23/45

Feynman trick,

2 3/2

( )

0 0 2 3/2

0 0 0 0 0

(1 ) , from table

2 3/2 2 2

0

Let: ( ) ( ) ( ) ;(1 )

1 11 ;

(1 )

aSp a S aS pS

S S

aS

eI a pJ p dp e dS e dS J p e p dp dS

s

SSe dS

S a a

Hard cutoff of function, becomes non-analytic, e.g.,2

1

( ) kxf k e dk

i . Example where this comes up is friedel

oscillations. You have 1D Fermi gas. Finite Fermi energy. Consider solution of single particle SE for one

electron. Free motion, and then hard wall. Wavefunction should vanish at wall, and so the wavefunction shouldbe a sine-function. Look at probability corresponding to this sine function,

Mathematica: plot sine vs. sin2

Far away from boundary, you have no trace of the boundarys effect (free planewaves). Near boundary, you hve

sine function. What happens inbetween? Well, look at electron density,

( ) ; [equilibrium or non-equilibrium distribution function]2

dkn f x f

k k k

Difference between Fermi vs. Boltzmann statistics is felt here. Which wavelength makes what contribution to

the density? How does wave-interference influence the number density? Etc.

The function fk

is a box for zero temperatures, so integrate over the box only,

2 2 2 2 2

0 0 0

22

steady- oscillatory1 cos 2( ) 2 sin 2 2

state part part2 2

Normalization of wavefunction vs.sin 2 sin 221 ;

2 2 no

F F F F

F

k k k k

k

F FF F

F

dk kxn x A kx dk A dk A dk

k x k xAk A k

x k x

k

2rmalization of e- density FA k n

So then you have oscillatory electron density ( ) (1 sinc2 )Fn x n k x . You can then plot this vs. kFx and see that

both boundary conditions are satisfied. In 2D, this is 1( ) sinc2 Frn x k r and in 3D we have 21( ) sinc 2 Frn x k r .

Important effect, Altshuler-Aronov effect.

Luttinger liquid: 1D wires, quantum hall (integer and fractional), etc. work with system of single-particle-

description of electrons.single particle picture. In 2D, more or less ewre home free and we have simple pictureof single particle motion. Fermi liquid theory. FLT says electrons move, and they have free particle motion with

renormalized parameters. But in 1D system, you dont have this luxury. You cant just renormalize parametersin 1D. either your way is obstructed, or it isnt. collective solids take on much diferent parameters. Going from

2D to 1D is going from a traffic jam with alternate routes, and then shutting off alternate routes all of a sudden.

Potential: study scattering problem in 1D system upont his potential,

0( ) ( ) ( ) ( ) scattering amplitude ln ;F FV x V x x n x dx T

Example: liquid helium 3, let it cool down to temperature where it flows without dissipation. BCS theory formotion.


24/45

Liquid he3 is a p-wave superconductor. Most other things are d-wave superconductor. Unconventionalsuperconductivity; we do not suspect phonons as main culprit of ??

Simple story of Friedel oscillations has far-reaching implications. Can put piece of copper, place atom on top,

and look at distribution of number density about the system.


Physics of the mott problem; we will introduce method of steepst descent. Consider mott problem: a bunch of

square wells randomly arranged in space. Eeach well has 1 bound state. The bound states are at random energy.Question: how do particles move from well to well? Partial anwer: they hop, as quantum particles do.

Mechanism: tunneling. The tunneling happens over known distance. Typcal electron will tunnel to potential

well closestto it in space. Forsimplicity, let the electron concentration be low: / 1e wellsn n , so we can useBoltzmann statistics, not Fermi/Dirac statistics.

Once you tunnel: you are in enw energy level. New energy levels dont match. Energy level mismatchbetgween old and new positions has to be absorbed.

So: lets say wells couple to infinite heat bath that it can borrow energy from. But large energy mismatch means

transition-frequency is limited by both (1) space/distance and (2) energy-disparity. So, there must becompromise between these two factors in constituting a hopping frequency.

Conductivity of this medium is product of two exponentials. Let typical distance between wells be R , and let

the extent of localization of the wavefunction8

be . Meanwhile, in energy-space, let there be energy mis-match

of , and let the probability be Boltzmann of this energy scale. then

exp expB

R

k T

Process of tunneling needs to borrow the energy from the thermal bath. Energy difference large means lowprobability of hopping.

Let there be independence between R and . Then the only way to optimize conductivity is to make R = 0. Inreal life, you have ( )R . There is a relation between the energy gap and thecharacteristic spacing R,

8 Colloquially: the size of the wavefunction. For a point-like wavefunction, you have 0 , and no hopping.


25/45

( )

3 3

# of energy levels 1 1

exp( ) exp[ ]Volume energy available

f R

B

R

N f eNR NR k T

Want to minimize this function,

40

44

0 03 4

0

/30 04440 3

0

1 1 3 3 3( ) 0 ;

1 3 1 1( ) ( ) ( ) ;

3

B B B B

T TB

B B B

Rf f R f R R

NR k T NR k T N k T Nk T

R TNk Tf R T e

NR k T Nk T N k T T

This is mechanism of conduction in amorphous solids! Traps and electrons. Only way to go between traps is to

hop.

Gamma function: is tautology of factorial function. Now: lets find out where stirlings approximation comes

from using method of steepest descent1

ln ( , )

0 0 0( 1) ???

pp x p x x f x p

p x e dx e dx e dx

Plot the function ( ) lnf x p x x

Steepest descent: approximate function by Taylor series. Replace f in exponential, noticing that you are at anextreme value,

0 0 2 2

0 0

2 310 0 02

1( ) ln ( ) 1 0; ( ) 0

( ) ( ) ( )( ) ;

p p pf x p x x f x f x

x x p p

f x f x f x x x

O

(3.19)

So then,

2 201 102 2

( )ln ln ln ln

0 0( 1) 2 2 2 2 2

2

p p

y x xx x yp p p p p p p p p p p p p p

p e e dx e e dy e p e p p e p

(3

You can clearly see stirings approximation if you take the log of both sides of (3.20).

Assumptions: we assumed integral came from narrow region about maximum. We see that20 1/2

0

0 0

1 1~ ~ ~ ; 1 ; ~1 ; ~ ;

2 2

zx x p y

x x y p p p p z e dz y px p p p p

Steepest decent for complex integrals: suppose you have integral of a smooth function ( )g z in complex

plane. Let it be a contour between points z1 and z2. Let f(z) and g(z) be analytic functions. (note: this excludes

the innocent-looking 21 z ). Then,


26/45

01 2

1

( )

1 2[ , ]

2 2 1

1

( , ) ( ) ; ( ) | 0; ( , ) ( , ) ( , );

( ) ( )0 ; lim ( );

f z

z zz z

z z

I I z z g z e dz f z f x y u x y v x y

f z f zu v f z

z z

i

Recalling a bit of complex analysis, thining about derivatives,

2 2 2 2

; ;

( , ) ( , ) ( , ) ( , );

; 0

x x

y y

x y x x y y x y y x

z x f u v z y

u x y dy u x y v x y dy v x yf v u

dy

u v u v u v u v u v

i i

i ii

i

Electrostatics: Earnshaws theorem! Cant form a stable system.

0

1( )

2u z u d

R

That is the ultimate freedom of complex analysis. Idea: deform contour such that (1) it passes through saddle

point and (2) in direction of steepest decent. Then integral becomes integral over real line. Do same thing aswith gama function: expand thunctionnear saddle point. And do Gaussian integral for small deviations.

Example: Bessel, or any cylindrical function. All are given by contour-integral representation. Integral is from

0 to pi band in complex plane. Trick: textbook has particular choice of axis. Hed ask you to draw contour ofintegration. But deformed contour is equally valid. Bessel function: saddle point on this interval jut happens to

be Pi/2.

All large-value formulas from mathematical physics, from Bessel functions to whatever, Legendre polynomials,hermite polynomials, they all come from the same interval.


Gotta use tims notes


Consider sums. Diamagnetism of 2D electron gas. You have B Fk T and C F . You haveeB

C c

and

the spectrum 12

( )n C n . Since you have A B , the Hamiltonian decouples in this uniform vector

potential, ec

P A . You have A must be lienarin the potential. On squaring the 2( )ec

P A , you have a term

that is proportional to2

A , an this one is simple. The cross term is an anticommutator which is a shift (?).recall: ( / ) p i , and you have /e i p x being the generator of translations (?) by x .

Free energy: compute intensive quantity free energy per unit area; this appears as,

2 20

ln(1 ); ; ;2 2

nBB n n

nB B

k TF F ce

A eB

(3.21)

Sum over positions of circles, in which you have localized states. You have to estimate the sum.


27/45

Discrete energy levels at fixed chemical potential. Change field, stretch the levels. When you stretch levels,

some pass through the chemical potential. If you measure BM F , you will see oscillations in theoscillations. Anytime the chem potential crosses a level, you have some undulation, but the next level is

equivalent (note: 1n n C ), and you get oscillation. Lab people: put graphene in magnetic field. Measurecurrent. Current starts to oscillate, because the resistance is oscillating as a function of the field.

Effect when a quantity oscillates as a function of the field: de Haas/von-Alphen oscilltaions. (See Ashcroft and

Mermin). Tells you the shape of the Fermi surface. This is the reason why we built the high field lab intalhassee, FL: we put materials in a magnetic field and we watch for oscilations. Thats the purpose. To seegood oscillations, we need strong magnetic fields.

Two things (1) estimate sum and (2) expect sum to be oscillatory function of B. (note: B-dependence is buried

inside of the ( )n n C and ( )C C B .

Look at,

2

1coth 3.15 ~ 1

1n n

Rules for sum-convergence is same as rules for integral-convergence, because integral is a sum.

2 4 23

2 2 22 2

[ ] 1[ ] ; ;

1

3 45n

Coth a a dn

n ana

a a aa

O

Note that in the integral, dependence on coth[a] is lost.

In general,

1

2 2 2 4 3

2 2 2 2 2 2 2

( )1; 1;

( )

( )2 1 1example: ~ 1 ~ ~ ; ~ ; ~ ;

( ) ( )

1 1 2 2 1 1 11 ... 2 ... ~

1 4 2

n n

n

n

f nf f

f f n

f nn n ay y n a f n a

a n a a a f n a

an a a a a a a

But replacing sumby integral may kill some important physics: e.g., quantum effects. In that regard, you cant

do the integral-approximation.

Poisson formula: wonderful invention. consider identity where you sum infinite number of delta functions.

This formula appears like our formulae for densities of states,

2( ) ; ;

kx

n n

x n e n

i

Exercise: compute,

2[ ( )] ?; [ ] ?;m m

kx

n nm m

x n dx e dx

i

Anyway, consider 2 2 1kme e i i


28/45

Poisson formula says that the sum of the function is equal to the sum of Fourier transforms. K plays the role of

the wavenumber.

2 2( ) ( ) ( )[ ] ; ( ) ( )kx kx

k n k

x m f x x n dx f x e dx f n f x e dx

i i Thing you gained: it is often easier to find the asmptoi value of the sm of transforms, rather than the original

sum itself. In the original example,

2 2 2 21 1( ) ; ( )f n f x

n a x a

Fourier transform of this function,2 2

2 2 2 2 2 2 2 21 1

cos22 2Re

kx kx

k k k

e dx dx kx dx e dx

x a x a x a a x a

i i

Now do contour integration: you have contour along real axis. K is positive. Therefore, make integrand vanishwhen circle of pole goes to infinity. Carefully choose the orientation of each contour so that the exponential

vanishes at infinity. residues are2 2

22 ( / )

ka kaea

I a e

i

i .

Therefore,2 2

2 2

2 2 2 2 21 1

2 1Re ... 1 2 coth

1

kx kxka a

ak k k

e dx e dxe e a

x a a x a a a a e a

i i

In the opposite limit, when we go to 1/a, you get behavior that goes as 1/a2. We got both of these limits in our

estimation! The icing on the cake is when we plug in a = 1, and get what we got before

Returning to the sum over landau levels:

( )

0

ln(1 )n

n

S e

Problem: sum is asymmetric. So: consider a limit: C F . Large population between occupied levels.Occupy large number of levels semiclassical approximation regime. This allows us to use a trick on this

semi-infinite sum. Go back to the Poisson-formula, 2( ) kxn k

x n e

i . Multiply both sides by f(x)

and integrate over x from [ , ]x a .

Imagine sum of delta functions vs. x (e.g, the Dirac comb you treated in the very first solid state physics

problem last semester). Ignore the interval below x = a,

2

0( ) ( )

kx

n k af n e f x dx

i

Homework questions: integrate by parts on problem 2a.21/

2 2 2 2 2 21 10 0/ 0

1 1 1 1( )

11 ( 1)( )

rescale the variable;

a

a

a a a

a dx dI a dx

x a ax a



29/45

Electrons in quantized magnetic fields. You have 12

( )n C n . Add kinetic energy of z-axis motion is justconstant shift tohamiltonian, so not interesting physics here. Regime where populated by landau-levels: ratio

ofchemical potentials tospacingbetween landau levels is large, 1C

. Also consider Bk T and

B

C

k T

being

arbitrary. Let free energy be, and we have the su

( )/

0 0 20

ln(1 ); ;2

n Bk T B

n B

k TF F e F

(3.22)

There is a cut to confine the sum to be finite. The exponential of 0n very rapidly becomes small. Thatinvites, nota step-function treatment, but rather a Poisson formula treatment,

0 1

( ) ( ) 2 ( ) cos(2 ) ; 1 0;c oscl n ka a

f n f x dx f x kx dx F F F a

(3.23)

Problem difficult to formula mathematically: not interested in leading/largest term, but ratheroscillatory term.

Each time you cross chemical potential, you have oscilation. Experiment: manifestation of magnetization. Wedont know how mathematicians split into oscillatory vs. non-oscillatory. So how do we split? First term:

although large, is non-oscillatory, on physical grounds. This contribution to free energy just is a constant shift toF: it is likefree energy of gas in absence of magnetic field, and were not interested in that.

Argument: integral term is actually classical contribution to freenergy which neglects quantization due to landau

levels. Sum keeps discrete nature of spectrum. Sum consists of two parts as indicated in (3.23). first term is asmearing out term (?). if you were asked to find free energy of classical gas in B field, you should see first term.

Bohr-von Lewween theorem: one of the few useful theorems in theoretical physics. There is no diamagnetism

in classical systems. (Actually: there is no magnetism in a classical system, period). Magnetism is a purelyquantum/relativistic effect. Any induced magnetism is due to an external field. Spin is hopelessly excluded by

classical mechanics; it is purely relativistic. The effect of the spin is due to an expansion in 1/c, where c is thespeed of light. The leading term is the Zeeman term. If you didnt have the Dirac equation, we would have no

reason to even suspect spin existed if it werent for experiment (e.g., the structure of the periodic table).

Hamiltonian for such a system,2 2

( , ) /( ) ( , )/( )3 3 3 3( )( , ) ; ;

2 2B B

ep r k T p r k Tcp r Z e d r d p e d r d

m m

p A H H

H (3.24)

Now: in quantum mechanics, position and momentum are non-commuting operators. However, in classical

mechanics, position and momentum do comute, and ec

p p A

and r r r is just a canonicaltransform. One then integrates over canonical variables, and you cannot tell about magnetism,period.

Magnetic effects just enter in as a shift in momentum, and thats just a Galilean transform which measurement

cannot detect. Therefore: magnetism is apurely quantum and/or relativistic effect.

Hencethe first term in (3.23) does not affect the energy levels! The second term is the oscillatory part of trhe

free energy! Integrate by parts using sin22

cos(2 ) ( )kxk

kx dx d

, and you get,

( )/

( )/

0 0 ( )/1 1

sin2ln(1 )cos(2 )

2 1

x B

x B

x B

k Tk T C

osc k Tk k Ba a

kx eF F e kx dx F d dx

k e k T

(3.25)

Introduce dimensionless energy ( ) /x By k T ; this is the measurement of the width of the energydistribution where we have occupied states. We have


30/45

12

we lose information about "a", and( )~

consequently about few lowest landau-levelsCx C

B B B B

ay

k T k T k T k T

(3

You are losing information about 312 2

0, , ,...C C . These levels play no role; they are too deep. So: in the

integral (3.25), we have y,

22 /(2 ) 2 / (2 )2

0 0

1 1

Im[ ] Im[ ] ;

1 1

C C

y kyk kkyCB

osc y y

k kC B

k T e eF F e e dy F e dy

k T e e

i

i ii

(3.27)

We have2

1C

. Quantum mechanics is lost in two ways: either (1) you heat the system up (you still have

kBT mu, but you spread out the thermal spread such that you encompass more landau levels; window

becomes large, and you lose oscillations, and you do so in an exponential way; see the pole ), or you (2)decrease the field. At zero temperature, a weak field going to 0 still makes you lose oscillations in the free

energy. For reference, 1 Tesla produces 1 kBT of energy/frequency (think hbar = 1).

Caution: midterm on Wednesday. Review session on Monday at 6 PM.

2 2

2kRym

2

02

3 240 03

86

keak ke

Ea a

Cosine is analytic over the real.

Problem number 2: Taylor series expansion!

3

1series expansion...

1x

Problem number 3:

2

1 0 2 20

2( 1) ( )

y dyI J y

y

Must put 0 . Two ways to do this (1) strike out delta2 in denominator. BUTsubsequently integrate byparts and evaluate J0(y)

2at a small argument, approximating it as 1. Upper limit of integration can be kept as

infinity! ORless formal way (2) you could strike out the upper limit of integration and replace it with 1. Ok todestroy informationabout delta. Going to put delta in lower limit anyway.

Two ways to proceed.

Monday, March 11, 2013 A Robinson Curusoes approach to special functions

Final stretch to end of semester. APS meeting will interfere. Class will be canceled on Wednesday, march 20,2013. First make up class shall be Wednesday, march 13, 2013, 6 PM

Current topic: differential equations. Estimate regions of relevant behavior, matching boundary conditions.

Consider problem from QM: you have hard-disk of radius a in plane. Potential well. Electrons move in the

plane, so x yk x k y k . You have potential 0( )U r a U and ( ) 0U r a . Get fundamental solutions of

Schrdinger equation Bessel functions. Butpretend you dont know anythingabout Bessel functions.


31/45

Simplifying assumptions: s-wave scattering. That means 1 ~k

a (e.g., long-wavelength scattering). We alsoassume isotropic scattering.

Height of potential well is U0; let it be that2 2

0 2km

U E . Small but very-high potential-barrier. Another note:

consider thin, high potential; knee-jerk reaction is to model it as a delta function! Legitimate replacement, butyoull never be able to investigate physics of finite radius a, or barrier-height U0.

Isotropic scattering: wavefunction only depends on radial distance.

Two Schrdinger equations: one inside disk, one outside disk,

2 2

2

1 2 2 1 2 2 2 2 2

0

( ); ( ); ;

; 0 ; 2 / ; 2 / ;

mr a r a U E

r k r k mU k mE

(3.28)

Now lets get scattering cross section of the disk, 0( , , )a U . Simplification: k inside the disk. Eachof these is a 2

ndorder differential equation. pretend we dont know what Bessel functions are. Get properties of

solutions. Decide which fundamental solutions to keep.

1 2 {1,2} 1 2 {1,2}; ;r k r (3.29)

Inconvenience: having both 1st and 2nd derivatives. So introduce ansatz ( )ar , and let x r ,1 2 1 1

2 1 2 1

1/23/2 1/2 3/2 1/2 2 21 1 1 1

2 2 2 4

; ( 1) 2 ; ;

( 1) 2

( )

a a a a a

a a a a a a

a

x a a x ax x ax x

a a x ax x ax x x

x x x x x x

(3.30)

Thus, the ansatz helped us make this into something easy to integrate,1

2; ; ; / / 1;a

x x x x x x

(3.31)

We arrived at 1 for the solution if we neglected the 2x on the RHS of (3.30).

Now, rewrite the derivative and the factors as an operator,

0 1

2 2 2 2 2 2 2 2 5/2 2 5/21 1 14 4 4

0x

x x xx x x x x x x x x x

(3

Perturbation theory: must discard terms of higher order in a consistentmanner. Use power-law ansatz Ax . That gives you,

14

5/2;

5/2 5/2 21 1 14 4 4

1( 1) ( 1) 1 ... ;A

A x Ax x x x x xx

(3.33)

This gives you solution behaiour near the origin.

Other solution: should not behave well. Well-behavedness at the origin is orthogonal to divergence at orgin.

You need a log-function for that,

2

1 1 1 (2) (2) (1) (2)11 2~ ~ ln (???); ( ) ~ ln ; ;xx x x x x x A A

(3.34)

Let y kr , and write,


32/45

2

1 1 12 4

( ) ; ; 0;y

y x

(3.35)

Our solution is then,

sin cos sin( ) (sin cos cos sin )

(sin cot cos ) (sin cot cos )Cy

A y B y C y C y y

C y y y y

(3.36)

Now, play the matching game,

( ); ( ) cot ( ) | ; | | ; | | ;y kr r a r a r a r aS x G y F y (3.37)

Turn this into a matching game of their log-derivatives. This is convenient, because normalization as an

unknown drops out, and we can focus on finding the unknown phase shift,

case-1 1; 1; case-3 1;( ) ( )cot ( );

case-2 1; case-4 ~1;( ) ( )cot ( )

ka a aS a G ka F kak

a aS a G ka F ka

(3.38)

The radius of the disk enters everywhere. Thus, we definitely sacrifice pysics by using a delta function.

Wednesday, March 13, 2013

Summarily, we had, for dimensionless 2 202 /mU and

2 22 /k mE , and variables ( ) ( )x r S x C x and y kr , the eigenfunctions,

21

1 2 1 241 ... 1

; cot sin cos ;ln

xr r k y y

x y

(3.39)

Wavefunction should be localized in wavefunction of the disk, but there is zero-point motion happening inside

the disk,2 2

0 0 02 / ~ /x a mU a ma U (3.40)

Mapping: dont want to solve the problem again. Compare two solutions inside and outside the disk. Infer howsolution of outside behaves in same region. Note the two equations in (3.39): they differ only in sign. Map one

equation onto another by switching two dimesionless distances. Consider replacement: r r i . Both theoperators

2 2 2

2 2 2

2d d d

dr dr dr

i and 1 1d ddr dr

r r change sign under this switch. What happens when we effect

this switch in (3.39)? map the 2nd

expansion 214

( ) ~ 1x x i , and ( ) ~ ln ln ln ~ lnx x x x i i i .

Now, note also that the transform x x leaves the differential equation 1f x f f unchanged. Thisdifferential equation is also homogeneous, which means you can multiply through by a constant and leave thesolution unchanged. Multiply through by i ,

1 2 142 2 11

0 4

2 2 1 ;

1 ; 2 ; 2 ;

x a a ax a

ax x a x a

(3.41)

Go back to (3.39) and concern ourselves with the phase-shifts, and get,

1 12

1

1

cot 0( ) cot ( ) ( )| | ~ ~

( ) cot ( ) ( ) cot 1 ln cot ln

( ) 1 1cot ln ~ ln 1; cot big 0 cot( ) ; 1 ;

( ) ln

kaka ka

a a

ka

S S a G ka F kak k k

S S a G ka G ka ka ka

S aka ka dx

aS a dx

O

(3.42)


33/45

This is actually a very weird solution. Recall this is 2D scattering off a disk, not a sphere. In 3D, you have a

phase shift which is also small, but it is small as a power-law, 1ka . The log-divergence is an

oddity of 2D quantum mechanics.

Whats 3D scattering cross section? Must estimate! Consider the wavenumber and phase-shift: then,

( , )k . Cross section and phase shift tell you everything. The k is the initial incident energy, and the phaseshift is the change due to the interaction.

In two dimensions: note that the units of are 1 2 1 1DL L L . So, 1 2 1 21~ ~ / (ln )ka

k k . The exact

expression is 1 24 sink . Know that k E . So, small particle becomes large. 2D quantum mechanics isvery weird.

Example: graphene. If you measure the drudian collision-frequency 1/ vs. electron-number density n, and you

imagine a renormalized-Drude-conductivity 2 */ne m , you get 1 1 2( ) ~ / lnn n n . Awesome. 2Dquantum mechanics is bizzare, but real and observable.

Opposite limit: consider 1a ; we need ( 1)S x . This is like making our disk-radius really huge compared

to the wavelength. Use mapping again oryou could start from scratch. Recourse to the original differentialequation1 21

1 14~ 0 | ; 1 ... | ;x xS x S S S S e x S x

(3.43)

Lets imagine that S is a power-series, 220

n

nnS a x

. Put this back into the first equation 1S x S S of

(3.43), and we get,

1

2 2 2 2

2 2 2 2 2 2 20 0(2 2)(2 1) (2 2)(2 1) 2 ( !) ;n n nn n n n nn na n n x a x a n n a a n

(3.44)

Opposite limit: you could have tiny disk, 1a , so 214

( ) 1S x x , and ( ) 1S a . Youthen give yourself the

derivative1

2( ) ( )S a a , and you get,

22

21 2

2 ( )1 2

( )

22 1

1 2 1 1cot ln 1; ; ~ ln ;

( ) ln

ln( )two regimes: ( ~ 1) ~ ; ( 1) ~

ka aka a

ka

Sa a k

aka ka

k k

(3.45)

Thomas Fermi model: should be part of standard graduate QM. You say how big is the atom?,

( / )Bf r a (3.46)

Getting the Bohr radius without heavy formalism: should scale as the Bohr radius,2

2B mea , ratio of quantum-

force to Coulomb-force. You have 2[ / ] /Be a J m m J . Zero point moton kinetic energy is2

0 ~ ( / ) /BT a m

; then, 20 ~ / BU e a and

2 2

0 0 ~ / ( ) 0.522BT U a me A .

Now consider Z-proton atom with one electron. It must be a much-smaller atom. You have ~ 0.525 /Ba A Z .

Watch out: you dont have any dimensional control over the pure-number Z.( ) ( ); 1;zB Ba a f z Z (3.47)


34/45

Many nodes: for large-Z, you have many nodes in the wavefunction. That means youre far away from the zeropoint energy. That means you can actually use semiclassical approximation. So: consider orbiting electrons as a

Fermi gas. The gas species are nearly-free particles,

34deg 3

3

1/3 2/32 2 2

2 ( )0 ( ); ( ) ;

(2 )

1 1( ) 3 ( ) ; 3 ( ) ;

2 2

F

F

F F F

k rNT E E r n r

V

k r n r k n r

(3.48)

Chemical potential, 0( ) ( ) const 0 ( ) ( );F Fr r r r (3.49)

Now, Poisson equation is 2 (3)4 ( )n Z r . Zeroth order solution ( 0) ( ) /r r Z r . You havethe density related to potential in a nonlinear way,

2 3/2 (3)8 2( );

3Z r

(3.50)

Um,2 3/2

0; | ~ / ; | 0;r rc Z r r (3.51)

Weak localization: If you shine a laser at a gas-cloud, the probability of backscattering isfour times the classical

value. This was discovered in the 60s (when the laser was invented), but it was not paid attention to. Shine lighton a medium, and the medium, itself, seems to trap the light.

In a seemingly-separate vein: they observed the Bloch resistivity vs. temperature, T5

law. At low temperatures,

resistivity deviated from this T5

law in thin metallic films, and the deviation was extremely-strong in 1Dsystems. (Iron whiskers: pretty much one-dimensional).

Essentially, you have,2 2 2 2 2*

1 2 1 2 1 2 1 2 1 2 1 22 Re 2 cos( )ABP C C C C C C C C C C (3.52)

And,3/2 1 2 3 3/2~ ( ) ; ~ ; ~ ; ~ ( ) ;P t R t A t V t D D D D (3.53)

Probability is divergent:2 2

3 2 13/2 3/2

0

divergent in certain; ln ; ;

dimensionalities( ) ( )

T T

D D D

v dt v dt v dt T v dt P P P T

t t t t

D D D D

(3.54)

VERY successful theory in solid state physics, even more so than BCS theory.

Monday, march 18, 2013

You missed first 20 minutes. Consider continuous symmetries in classical mechanics: symmetry under (1)translation in space (2) translation in time (3) rotation (4) gauge invariance.

Lets look at gauge invariance. You have,

; ;f A A A (3.55)


35/45

Four vectors,

( , ) ( , ); ( , ); ;xx ct x ct j c A A A

r r j (3.56)

Now, think about Lagrangian densities,3 1 1

0 int int; ( );

c cd x j A

j AL L L L L (3.57)

Subject the interaction-part of the Lagrangian to a gauge transform,

1 1int int int int( ) ( ) ;x x xc cj j j L L L L (3.58)

Electrodynamics of superconductors: natural consequence of gauge invariance. Wavefunctions are not invariantunder gauge-transform: you get a relative phase,

; exp( );e

c

A A A

i(3.59)

To prove this, consider what happens to the canonical momentm,

( )e e ec c c

i i

p A A A (3.60)

Aharanov bohm effect; persistent current.

Wavefunction of cooper pairs will similarly suffer this phase,

2 ( ) ( ) exp( ); ( );

eF F e

c

ir r r

i(3.61)

Supercurrent of cooper pairs related to gradient ofF in the same way that number current is related to the

gradient of the wavefunction * *2

( )m

i

j , and this appears as,

* *2

( ) ( )2 2

)2

(S S S S S S

e eF F F F n e n e n e n e

m m

en

m

i i i ii

ji

(3.62)

Make a gauge-invariant combination,

2

2 22 2 /

e eS S S S S S c c

e e e e en n n n

m m m c mc j A j A B B

(3.63)

Meanwhile: maxwells equations appear as,2 2

2 24

2 2 2

plasma1; 0; ; ; ;

frequency4S Pc

S P

mc c

n e

B j B B B (3.64)

Monday, April 08, 2013

Assignment 4 posted. Due in 2 weeks. Last homework of semester. Nearing end.

Lets talk about qualitative methods in high energy physics/quantum field theory. Consider decay of K-mesons.

Tells us symmetry: e.g., parity, charge-conjugation, time-reversal (PCT). Four K-mesons: 0, , , oK K K K . You

have 10~ 10 s . You have quantum number strangeness: 0( , ) 1K K S , and 0( , ) 1K K S .

You could have following processes,


36/45

0 0 0 0 0 0

0 0 0 0

; ; ;

; ; ; ;

K K K

K K K K

(3.65)

Hamiltonian to capture decay-physics (3.65) then appears as,

1 0 2 0 3 0 4 0 . .H g K g K g K g K h c (3.66)

0

K and0

K are energy-degenerate eigenstates. There are two superpositions: the short-living superposition S

and the long-living superposition L. you have,

1 2 1 0 2 0 1 2 2 0 1 0 1 2 2 21 2

1( , ) ; ( , ) ; ( , ) ;S LK N g g g K g K K N g g g K g K N N g g

g g

(3.67)

Omitting the 3-roder terms, we can write the kets 0SK and 0LK , and thereby get,

1 2 0 0 1 2 0 0

1 2 1 2 1 2

0 0 0 ( )

(1) 0 (1) 0 1 1 0 0

L LH HK g g K K g g K K

g g g g g g

O(3.68)

In contrast to (3.68), we have

0 0SHK ; we have (3 ) (2 )L S . In experiment, we have LK . Wewould have

(2 ) 3

(3 )~ 10L

L

.

Parity to inversion: you obviously have the following properties,2; ( ) ( ); ( ) ( ); 1; [ ] ;P P P e e i q r i q rr r r r r r q q (3.69)

Book: Lipkin, quantum mechanics

Wednesday, April 10, 2013

Could you have the reaction 0 ? I do not think so. It is not time-reversal invariant.

Recall kaon decays 0K and 0K . The Hamiltonian for these decays was (3.66), e.g.,

1 0 2 0 3 0 4 0 . .H g K g K g K g K h c (3.70)

Parities: the parity of the interaction is -1. Then, 0 0PK P K andP P . Time-reversal,

* * * *( ) ( ) ( ) ( ) ( ) ( );

t t tt H t t H t t H t i i i (3.71)

Suppose you have *H H ; this is ordinary circumstance of quantum mechanics. Then you have,* ( ) 1 *( ) ( ); ; ;T t t Te e e TAT A i q r i q r i q r (3.72)

How does time-reversal affect the kaon? Consider the action of time-reversal as described by (3.72) upn theFourier-decomposition of the state of the kaon,

1

0 0( ) ( ) ( ) ( );K q K e TK T K i q rr q q (3.73)

Similarly: you can time-reverse a 180-degree rotation,1

180 180 0 180 0( ) ( ) ( );T T K T K

q q q qD D D (3.74)


37/45

What if we had*

H H ? Would be a more general interactionmight not conserve probability density, whichcould represent the transmutation of one particle into another (?).

Charge conjugation: if this is symmetry of Hamiltonian, consider transform,1

1 0 1 0 2 0 2 0( ); ( ) ; ( ) ;CAC C A g C K g K g C K g K (3.75)

Bring back long-living and short-living combinations, (3.67), but rewrite as,

1 2

1 2

1 0 2 0 1 0 0

1 22 2

1 2

2 0 1 0 1 0 0

; 1( , ) ;

;

g g

S

g g

L

K N g K g K Ng K KN N g g

g gK N g K g K Ng K K

(3.76)

Action of charge-conjugation (3.75) upon (3.76) appears as,

( ) ; ( ) ;S S L LC K K C K K (3.77)

How does PC apply to the decays? Parity and charge conjugation,

( ) ; ( ) ; ( ) | ; ( ) | ;S S L L S S PC K K PC K K PC K PC K (3.78)

This is why you have short living and long living combinations. That was terminology before 1964. Then,

decays were discovered. After 1964, we learned that LK . That means that CP is not conserved.

CPT: Luden-Pauli theorem is that T is broken. No CPT. CP at T interactions CPT.

Friday, April 12, 2013

Last time, looked at time-reversal. Transformation appears as *( ) ( )T t t . But suppose Hamiltonian is not-

Hermitian,

H H . This happens in the Zeeman effect, where 212

Bm

H I p B

, in which,

0 1 0 1 0;

1 0 0 0 1

z x y

x y z

x y z

B B BB B B

B B B

iiB

ii

(3.79)

Spin orbit interaction: couples spin to orbital angular momentum. Can only rotate the two together. Yyou do nothave axis-freedom. Must consider 3 components of spin.

Contributions,2

2

1 1; ( )

2SO SO B SO

eH

c c m mc

p iB E v E B E i H

(3.80)

Here, we have a Hermitian H H , and we have,

* * * * 1 *

* 1 * * 1 * *

; ( ) ( );

; ; ;

SO x x y y z z H H S H ES S H S S

SH S ES H SH S HS ES

i H H(3.81)

That means the thing solves the same Schrdinger equation dn thus gives the same physics. Explicit form of

Hamiltonian, using 1 1x xS S and 1y yS S

and 1z zS S , is as follows,

x x y y z zH H i H H (3.82)


38/45

This is a linear system. A lot of algebra could be done to show the following result: yS .

Note, also, that,1 1 1; ; ;y x y y x y y z x x y y y y y z y z

i i i (3.83)

Spin-orbit interaction,*

y yT K (3.84)

Complex conjugation operator,* 1[complex-conjugaton operator]; ; ; ;yK Ka a T K THT H

(3.85)

This means: spin in Hamiltonian will cause intricacies under time-reversal symmetry.

Exercise: go back to rashba Hamiltonian, and check time-reversal symmetry. It fits into the above paradigm ofthe Hamiltonian, but it doesnt contain certain terms.

Discrete symmetries

Bibliography: QM 06, ch 12; Tinkhams bookGroup Theory and Quantum Mechanics, Dresselhauss Group

Theory: Applications to the Physics of Condensed Matter, and Ramonds bookGroup Theory: A PhysicistsSurvey.

Group properties: closure, associativity, identity, inverse.

Monday, April 15, 2013

Symmetry operations. Irreducible representations of D3. You aso have D4.

Example: CO2 molecule. Symmetries of it are important. Has D2 symmetry.

Group theory doesntsolve the moton. It specifies transformation criteria. If one was to tra

qmtp - notes

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