•Particles are actually waves•Two de Broglie relations

•They have all the properties (and equations) that described waves before, together with some new ones

E hfp h

•Rewrite the de Broglie relations in terms of k and :

p h

2k

1

2f

T

2hk k p E hf 2h E

•We can also find a dispersion relation for these waves:•From this, we can find the group and phase velocity:

221

2 2

pE mv

m

2 2

2

k

m

2

2

k

m

g

dv

dk

k

m

p

m

pvk

1

22 g

kv

m

“velocity” is group velocity, not phase

velocity

p k

E

Quantum Uncertainty

Two uncertainty relations for any type of wave•Multiply by h-bar•Use quantum wave relations

12

12

x k

t

12

12

x p

t E

Uncertainty in Position/Momentum•Classical physics:

•Particle is defined by position x and velocity v = (dx/dt) •Or, if you prefer, exact x and exact p

•Quantum physics:•Particle is defined by wave function (x)•It cannot have a definite position and momentum – they tend to both be uncertain

12

12

x k

t

Uncertainty in Time/Energy•Things that last eternally can have definite energy•Things that last a brief time have uncertain energy

•Measurements of energy will have a spread in value

Time/Energy UncertaintyThe 0-meson has a rest mass of 134.977 MeV/c2 and lasts an average of 8.410-17 s. What is the spread in energies of a 0-

meson at rest due to its finite lifespan?

12t E 178.4 10 st

2E

t

15

17

6.582 10 eV s

2 8.4 10 s

39 eV

E - mc2 (eV)

EE

Consequences of Position Uncertainty

12x p

•Position and momentum of a particle cannotbe simultaneously specified or determined•We can often estimate onequantity if we know theother by treating this inequality as ifit were an equality

•Especially when the energy is being minimized LEnergy of Particle in a Box•By Carlson’s rule, theposition is uncertain by about x = ¼L. •By the uncertainty principle, momentum is uncertain by•We’d like the momentum to be zero, but we can’t•This causes the particle to have some energy, called Zero Point Energy

Harmonic Oscillator: 0.500

One-dimensional box: 0.568

Spherical 3D box: 0.557

Hydrogen Atom: 0.577

x p

x p

x p

x p

2

2p

x L

2

2

pE

m 2

2

p

m

2

2

2

mL

?

These methods give only estimates of the answer

2 2

22E

mL

Solving Uncertainty/Energy Problems

1. Write an expression for the energy: potential and kinetic2. Figure out what momentum (normally 0) and what position would have

the “ideal” lowest kinetic and potential energies3. Let x = a and p = /2a 4. Assume the momentum and position differ

from ideal by about x = a and p =/2a 5. Set derivative of energy function

equal to zero, solve for a6. Substitute a in to determine minimum energy

A particle of mass m lies above an impenetrable barrier in a gravitational field with acceleration g.

What is the minimum energy of the particle?

m2

2

pE mgy

m

0

0

p

y

and

2

y a

p a

2 28E ma mga

2 30 4ma mg 1/32

24a

m g

2/3 1/32 2 21/32 2

2 2

4 32

8 4 4

m gE mg mg

m m g

y

Uncertainty and the Hydrogen Atom12x p •There are two types of energy associated with the

Hydrogen atom•Potential Energy•Kinetic Energy

•Classically, these two energies are at a minimum when:•The electron is at rest, p = 0•The electron is at the origin, r = 0 (x = 0)

•In this case, the energy of the hydrogen atom E = –

2 2

2

p keE

m r

•Quantum mechanically, you can’t control both position and momentum•We will place the electron near the origin, r = 0, butthere is an uncertainty x associated with it•We will place the electron nearly at rest, p = 0, butthere is an uncertainty p associated with it•You have to compromise between these two choices to minimize E

•Let x = a, and let p be as small as possible by the uncertainty principle•Plug these into the formula for energy

2 2

r x a

px a

2 2

28

ke

ma a

Uncertainty and the Hydrogen AtomWe need to minimize the energy•Take derivative and set it to zero•Substitute result back in

2 2

28

keE

ma a

2 2

3 20

4

dE ke

da ma a

2

24a

mke

22 2 22

2 2

4 4

8

mke mkeE ke

m

2 4

22 4

mk e

2 4

2

2mk e

Correct answer is 4 times smaller than this. Why so far off?•It’s just an estimate•There is momentum in three dimensions

•This makes answer 3 times smaller•Then answer we get is 4/3 of correct answer

2 4

22

mk eE

2 4

2

2

3

mk eE

The Wave Function•We’ve been talking about waves, which require wave functions

•We’d better give it a name:•It would have more arguments in more dimensions

•It is a complex function; it has both a real and imaginary parts•What does it mean?•This turns out to be very hard

,x t

•Analogy: Electromagnetic Waves•Described by electric and magnetic fields

•What is an electric field?•It is defined in terms of a potential for something

,

,

x t

x t

E

B

American Heritage Science Dictionary“The distribution in space of the strength and

direction of forces that would be exerted on an electric charge at any point in that space”.

Dr. Carlson, PHY 114“Electric Field is

potential for force at a distance on a charge”.

The Wave Function: what does it mean?•We are talking about one particle – but it is not at one location in space

•If we measured its position, where would we be likely to find it?

The Wave Function is also called the probability amplitude•Clearly, where the wave function is small (or zero), you wouldn’t expect to find the particle•Where it’s negative or imaginary, wouldn’t expect to have negative or imaginary probability

•We’d better make darn sure that the probability is always positive•For electric fields, the energy density is proportional to the field squared

•If working with complex waves, take amplitude first•How about we make probability density proportional to wave function magnitude squared:

2u E

2,x t

,x t

Sample ProblemA wave in the region 0 < x < a has the wave function above.

What is the probability density at all locations x at all times t?

1 21 2

, sin sini t i tx xx t e e

a aa

1 22 1 2

, sin sini t i tx xx t e e

a aa

1 2 2 12 21 2 2sin sin sin sin i t i t i t i tx x x x

e ea a a a a

Helpful Identity:

2cosi ie e 1 2

1 2sin sini t i tx x

e ea aa

Sample ProblemA wave in the region 0 < x < a has the wave function above.

What is the probability density at all locations x at all times t?

1 21 2

, sin sini t i tx xx t e e

a aa

2 2

2

1 2

2sin sin

1,

22sin sin cos

x x

a ax t

a x xt

a a

What Does Probability Density Mean?•The probability density (in 1D) has units of m-1

•In a small region of size dx, the probability of finding the particle is there is given by ||2dx.

•To find probability over a largerregion, you have to integrate it

2,

b

aP a x b x t dx

Normalization: The probability that the particle is somewhere must be 1•If we integrate over all x, we must get 1•In some cases, the problem implies that we restrict to some region

21 ,x t dx

… in the region 0 < x < a …

2 2

0, ,

ax t dx x t dx

2 2

0

1 2 2sin sin 2sin sin cos

a x x x xt

a a a a a

10cos

2 2

a at

a

1

Sample Problem

At t = 0, the wave function is given by the expression below.(a) What is the normalization constant N?(b) What is the probability that the particle is at x > ½a?

for 0

,0 for 0

0 elsewhere

N a x x a

x N a x a x

21 ,0x dx

0 2 22

0

a

aN a x dx a x dx

02 2 2 2 2

02 2

a

aN a ax x dx a ax x dx

02 2 2 3 2 2 31 1

3 3 0

a

aN a x ax x a x ax x

2 3 3 3 3 3 31 13 3N a a a a a a

2 323 N a

3

3

2N

a

Sample Problem

At t = 0, the wave function is given by the expression below.(a) What is the normalization constant N?(b) What is the probability that the particle is at x > ½a?

for 0

,0 for 0

0 elsewhere

N a x x a

x N a x a x

212

/2

,0a

P x a x dx

3

3

2N

a

22

/2

a

a

N a x dx 2 2 2 313 /2

a

aN a x ax x

2 3 3 3 3 3 31 1 1 13 2 4 24N a a a a a a

33

3 1 1 1 11 1

2 3 2 4 24a

a

3 1

2 24

1

16

Sample ProblemAt t = 0, the wave function is given by the expression above.(a) What is the most likely / least likely places to find the particle?(b) What is the normalization constant N?(c) What is the probability that the particle is at 0 < x < a?

2 2

Nxx

x a

•Least likely when function vanishes, at x = 0•Most likely when function is largest positive or negative

0d

dx

2 2

22 2

2x a N Nx x

x a

2 2

22 2

N a x

x a

Normalization:•Let x = atan

2 2

22 21

N x dx

x a

2 2 2 2 2

22 2 22

tan sec

tan

N a a d

a a

2a

N

x a

2

2 2

2

sinN a d

2 2N a 22 1 12 4 2

sin 2N a

Sample ProblemAt t = 0, the wave function is given by the expression above.(a) What is the most likely / least likely places to find the particle?(b) What is the normalization constant N?(c) What is the probability that the particle is at 0 < x < a?

2 2

Nxx

x a

2 2

22 20

0a N x dx

P x ax a

2

41 12 4 0

sin 2N

a

2 1

8 4

2aN

Quantum Wave Equations You Need:

, i kx tx t Ae 2b

aP a x b x dx

12

12

x p

t E

p k h

E hf

End of material for Test 2

21 x dx