Phys 341 Quantum Mechanics Day 3

1

2

Mon. 9/8

Tues 9/9

Wed. 9/10

Fri. 9/12

Mandl II 4-5, 1 (Q10) The Wavefunction (Q6) Computational: Setting Up Discretized Schrodinger

2.1-.2 Stationary States & Infinite Square Well (Q7.1-3); Computational: Using Discretized Schro.

2.3.0-.1 (to Ex 2.4) Harmonic Oscillator-algebraic part1 (Q7.4)

Daily 2.M

Weekly 2

Daily 2.W

Daily 2.F

3 Mon. 9/15

Tues 9/16

2.3.1 (rest of) Harmonic Oscillator – algebraic part2

Daily 3.M

Weekly 3

Equipment

Griffith’s text

Moore’s text

Printout of first computational reading.

Printout of roster with what pictures I have

Announcement:

Weekly # 5: Q6S.10 part b (taking the quoted result of part a as a given) note: Moore has a typo

in the wavefunction, should have x/a not a/x.

Weekly 4: part c, should include the condition that po = 0. Then you’ll find some constant plus

some functional dependence on x that you should recognize.

Check dailies

Daily 2.M Mon 9/8 Mandl II 4-5, 1 (Q10) The Wavefunction (Q6):

Not sure which problem you’re asking about: Could we go over the second problem prior to

tomorrow's daily set? When I calculated the standard deviation, I ended up getting a complex number as

a result. My way around this issue was to insert absolute values on the radicand." Jeremy,

Mandl II.5 The Schrodinger wave equation derivation

Mandl gives an argument for the Schrodinger equation based on the understanding that

there are waves for massive particles whose wavelengths and frequencies relate to

momentum and energy through Planks constant. I’ll take a slightly different approach

and reason by analogy to light.

From E&M, you know that Maxwell’s equations can be combined to produce light’s

wave equation

o txBx

ctxBt

,,2

22

2

2

Phys 341 Quantum Mechanics Day 3

2

(I’m phrasing it in terms of magnetic field just so we aren’t put in the

position of keeping track of when E means “electric field” and when E

means “energy.”)

o Now, consider the simplest of solutions

ft

xi

BetxB

2

,

(depending on known ‘boundary / initial’ conditions the exact solution

will be some linear combination of such general solutions.)

o You plug this in and take the derivatives on both sides and get a relation between

wavelength and frequency. I’m going to go step by step so we can come back to

this an borrow an intermediate result:

txBctxBf

txBix

ctxfBit

x

txB

xc

t

txB

t

txBx

ctxBt

,2

,2

,2

,2

,,

,,

2

22

2

2

2

22

2

2

2

22 1

cf

o So far, it’s quite classical. In general, the relationship between a wave’s

frequency and wavelength is called its “dispersion relation.” So that’s what

we’ve got.

o But now let’s use the new quantum mechanical relationships between frequency

and energy and between wavelength and momentum:

hp , hfE . We can

translate light’s ‘dispersion relation’ into an energy-momentum relation:

222 pcE

o This is the mass-less special case of Einstein’s energy-momentum relation:

2222 mcpcE . (recall, potential energy of a system is hidden inside the

mass term)

o So the chain of reasoning we have here is

Wave equation + wave function

Dispersion (f & ) relation +

hp , hfE

Phys 341 Quantum Mechanics Day 3

3

Energy-momentum relation

o If I recall, Schrodinger first tried to follow this path backwards, starting with

Einstein’s general energy-momentum relation (including mass), to generate a

general wave equation. He ran into some snags (Dirac got it right a few years

later.) So he switched to trying the non-relativistic approximation for the energy-

momentum relation.

Energy-momentum relation: Vm

pE

2

2

+

hp , hfE

Dispersion relation:

Vm

hhf

2

/2

Factor in solution

txVtxm

htxhf ,,

2

/,

2

Now, for light,

txfBit

txB,2

,

or

txfBt

txB

i,

,

2

1

and similarly

txBx

txB

i,

1,

2

1

.

What if this is true for all fundamental waves, for electrons

and such as well as light? Then we can substitute in the

derivatives in place of the frequency and wavelengths.

txVtxxim

htx

tih ,,

2

1

2,

2

12

222

Cleaning up, and using that 1/i = -i and i2 = -1

txVtxxm

htx

t

hi ,,

2

)2/(,

2 2

22

Finally, we define h-bar and have

txVtxxm

txt

i ,,2

,2

22

“Derived”? not exactly, but “reasoned by analogy.”

Phys 341 Quantum Mechanics Day 3

4

1. The Wave Equation

1.1 The Schrodinger Equation

txV

x

tx

mt

txi ,

,

2

,2

22

Analogous role in Q.M. to Newton’s in Classical – describe how the state of the system

evolves with time. Difference is, that rather than position’s being the ‘state’ that is directly

evolved through the equation, it’s the ‘wave function’ that is the state that the equation evolves.

Things get deduced from this.

1.2 The Statistical Interpretation

t.at time b, and abetween

particle thefinding ofy probabilit,

2b

a

dxtx .

In other words, dx

txtx

,Pr,

2 is the “probability density”.

Could we go over the problem solving process for the first problem? I didn't fully understand how to figure

out how to find the probability of of x>0. None of the equation I saw worked nor could I figure out how to

find it using the graphical methods." Anton

Daily Conceptual: Suppose you have a particle with

wavefunction , where(t=0) is graphed at right. What is

the probability of measuring x>0 at t=0?

Weekly: Q6S.10 part b (taking the quoted result of part a as a given) : for

21

12

axa

x

, if a = 4.0 nm, what is the probability that the object is found

between x = 0 an 8.0nm?

Brief word on probability in physics

o We cannot use the Schrodinger equation to say with

absolute certainty where the object is at time t. There are

two possibilities:

Realist: the object’s position is well-defined but our

knowledge is incomplete, or

Orthodox: the object’s position isn’t well-defined

and our knowledge is complete.

Phys 341 Quantum Mechanics Day 3

5

o Bell’s Theorem. It can’t be well-defined (unless you’re

willing to allow information to travel within a system

instantaneously.) Return to in the end.

o “collapsing”: Orthodox means that when you do measure

and get a well-defined value, you must be changing the

state; ‘collapsing” it to having a well-defined position.

Decoherence (http://arxiv.org/ftp/quant-

ph/papers/0306/0306072.pdf) is the name given to

the study of the environment’s interactions with a

quantum mechanical system – how the

wavefunction can change from one state to another

because of oft-ignored interactions. Measurement

itself inherently involves such an interaction.

1.3 Probability

Q: In quantum mechanics, “expectation value” is synonymous with which

property of a probability distribution?

A: Average. In spite of the name suggesting ‘most probable’.

1.3.1 Discrete Variables

Given the statistical interpretation, it’s good to brush up on the basics of

statistics or probabilities. Griffith’s uses the example of ages of people in a

room.

1. Probability. The probability that you randomly pick a person of a given

age is

totalN

jNj Pr

If there were 10 people who were 20 years old in a room of 30 people,

then the odds of picking someone who was 20 would be 10/30 = 1/3.

The probability that you pick on someone of one or another age, say 20 or

24, would be the sum of the probabilities of picking either. For that matter, the probability of

picking someone of any age at all, would be 1: when you pick someone, he/she has some age.

1Pr1

j

j ; to be general, can let the maximum go to infinity (we may

simply not have any one, and so 0 probability, for many of the ages.)

2. Most probable: that with the greatest probability.

3. Median: the value for which just as many folks have less as have more.

4. Average: jjjj

1

Pr

a. Note: This is not necessarily either the most probable or the

median. Say you had 4 people, one was 5years old one was 18

Phys 341 Quantum Mechanics Day 3

6

years old and two were 20 years old. 20 years old would be the

most popular (with two people sharing that age), 19 would be the

Median (with two younger and two older) and 16 would be the

average .

Measures of how broad the distribution is: Varieance = 2j (we settle

for this as a characterization of the breadth because the sum of each

point’s deviation from the averge is 0.)

Standard deviation = 222jjj . This relationship is

handy since it may be easier to do one or the other sum.

Daily Math: Let s be the number of spots shown by a die thrown at random. Calculate

<s> and σs.

Say I have a 20 sided dice and rather than having a 20, I have two 5’s.

Someone walk us through finding the average (9.75)

1.3.2 Continuous Variables

For a continuous function, it doesn’t make sense to ask the probability of having

one value out of an uncountably infinite number of values. Instead we say the

probability of having a value in an infinitesimal range.

dx) (x andbetween x lies chosen)

(randomly individualan y that probabilit)( dxx

b

a

dxxbxa Pr

dxx1

dxxxx

dxxxfxf

222xxxx

Phys 341 Quantum Mechanics Day 3

7

"#3 from the daily. I understand what each thing we need to find is, but putting it into practice confuses me

with all of the summation notation involved." Bradley W

Daily Math: (1.3) Consider the Gaussian distribution2( )( ) x ax Ae ,where A, a, and

are positive real constants.

Find the normalization constant A.

Find <x>, <x2>, and σ.

Sketch the graph of ρ(x).

Could we go over some of the grunt mathematics of problem 4 on the weekly? I understand what in general what we must do (ex. taking the square integral of the wave function and setting equal to one to find the normalization constant)." Gigja

Perhaps this is a question for outside of class? Gigja

I heard the trick is to convert the coordinates from rectangular into polar, and by inserting a dummy variable. Jeremy,

Weekly: Consider the wavefunction

20

2

( )

4( , )o

o

x x ip x

i tax t Ae e e

What is the normailization constant A?

1.4 Normalization

Could we go over the proof on page 13?"Spencer

He uses Schrodinger’s equation and its complex conjugate applied to the

wavefunction’s complex conjugate to demonstrate that once-normalized, always-

normalized.

dx

t

txtx

t

txtxdxtx

tdxtx

dt

d ,,

,,,,

**22

V

i

xmi

t

2

2

2 and *

2

*2*

2

V

i

xmi

t

So

V

i

xmi

t

*

2

2**

2

and *

2

*2*

2

V

i

xmi

t

Plugging in,

dxV

i

xmiV

i

xmidxtx

dt

d *

2

*2*

2

2*2

22,

So the potential terms cancel, and you’re just left with

Phys 341 Quantum Mechanics Day 3

8

dx

xxmidx

xmi

xmidxtx

dt

d2

*2

2

2*

2

*2

2

2*2

222,

two paths from here: integrate by parts and assume that the wavefunction goes to

0 at infinity (which it must to be normalizable.) so you’re just left with

02

,**

2

dxxxxxm

idxtxdt

d .

Alternatively, you can observe that

xxxxx

**

2

*2

2

2*

and invoke the fundamental theorem of calculus to cancel the integration and

differentiation, leaving

xxm

idxtxdt

d **2

2,

=0 since the wavefunction must be 0

at infinity.

1.5 Momentum

The statistical interpretation means ),(,2

txtx

So

dxtxxdxtxxx

txtx

),(,

),(,

2

2

Can we discuss what the expectation value of x (1.28) mean in terms of the wave function? I am a little

confused about griffith's explanation." Jessica

Then the rate at which the average position changes is

dxtxt

xdxtxxdt

dx

dt

d

txtx

22

2

,,

),(,

Note: becomes partial and skips the x because, when on the outside and

done after the integral is taken, x is just a dumby integration variable.

Now, while showing that the normalization doesn’t change over time we’d dealt

with a similar integral and we can adopt that

xxxmitx

t

**2

2,

. So,

dx

xxxx

mix

dt

d **

2

.

Integration by parts cancels the x and the derivative (once again, invoking that the

wavefunction is 0 at infinity so the uv term vanishes, leaving just –vdu.

Phys 341 Quantum Mechanics Day 3

9

dx

xxmix

dt

d **

2

Performing integration by parts on the second term essentially flips which is

getting the derivative taken (and flips sign)

dx

xmidx

xxmix

dt

d ***

2

Multiply across, we have

dx

xix

dt

dmp * .

He asserts that

dx

xmT

2

2*

2

2

. I’m not sure how that average T would

be defined in terms of x and p to derive this integral. It may simply be straight

from the Schrodinger Equation and defining VET , both of which are

explicitly present in the Schrodinger equation, so if you just multiply by * and

integrate, you pretty clearly get the right hand side, leaving the left hand side.

From the next chapter, we find that

txVx

tx

mtxE

txVx

tx

mt

txi

,,

2,

,,

2

,

2

22

2

22

So,

dxxm

TVE

dxxm

dxVE

xmVE

xmVE

2

2*

2

2

2*

22

2

2*

2*

2

22

2

2

2

2

Weekly (1.7) Derive an expression for d<p>/dt in terms of V(x).

Weekly: Consider the wavefunction

20

2

( )

4( , )o

o

x x ip x

i tax t Ae e e

Calculate the expectation of x, x2, p and p

2.

a. For what potential energy function V(x) would satisfy the Schrodinger equation

if po = 0?

Phys 341 Quantum Mechanics Day 3

10

1.6 The Uncertainty Principle

He argues qualitatively that there must be a trade-off between spread of positons

and wavelengths. He doesn’t define the equation though (defers to Ch 3)

2

px

Weekly: Consider the wavefunction

20

2

( )

4( , )o

o

x x ip x

i tax t Ae e e

b. Calculate x and p. Are they consistent with the uncertainty principle?

Setting up the discrete Schrodinger Equation.

ExVxm

2

22

2

1. gets used to evaluate wave function only at specific locations x apart.

2. Thus the derivative has to be replaced with a finite difference.

jjj

jjjxExxV

x

xxx

m

2

112 2

2

3. In anticipation of using a computer, convenient to ‘non dimensionalize’

jx

jjx

jjj xmExxVmxxx 22

11 222

Define jx

j xVmxv2

2~ and Em x 2

2

can we work on number 4 from to daily, I wasn't quite understanding it. Maybe I'm overthinking it?" Sean M, I wasn't too sure on how or where to start with those two exercises. Jeremy,

x

Phys 341 Quantum Mechanics Day 3

11

Daily Computational: show that e can be phrased in terms of de Broglie wavelength for no V or

constant V.

jjjjj xxxxvx 11~2

4. Lay out all N equations in form suggestive of matrix

11~2 xxv 2x = 1x

1x 22~2 xxv 3x 2x

2x 33~2 xxv 4x 3x

3x 44~2 xxv 5x 4x

Etc.

5. Written in matrix form

N

j

N

j

N

j

x

x

x

x

x

x

x

x

xv

xv

xv

xv

...

...

...

...

~21

...

1~21

...

1~21

1~2

2

1

2

1

2

1

6. Have computer solve the matrix equation. It will find that there are N possible energies,

and corresponding with each is a wave function, x . Of course, it won’t come up with

an analytical expression for each wave function, rather, it will come up with, for each

energy, a list of the wave function’s values at each discrete location, x1, x2, x3,…

Wednesday

Don’t forget your Discussion Prep – 8a.m.

More reading on line

Phys 341 Quantum Mechanics Day 3

12