ch40 young freedmancomplex.gmu.edu/ · 2020. 5. 7. · %odfnerg\ 5dgldwlrq &rqwlqxrxv 6shfwud,q frqwudvw wr olqh vshfwud hplwwhg iurp h[flwhg dwrpv lq d gloxwhg jdv udgldwlrqv iurp

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The Bohr ModelIt is a mixture of classical and new (quantum) ideas in trying to theoretically calculate the energy levels of a hydrogen or hydrogen-like atom.

Assumptions of the model:

1. e- moves in stable circular orbits around the nucleus under the influence of Coulomb Force.

2. Atom only radiates when e- jumps from a higher energy orbit to a lower one, i.e., .

3. Only certain circular orbits are allowed: angular momentum of e-

around nucleus are quantized, i.e., must be multiples of .

i fhf E E

L

2h

Electron Waves and the Bohr’s Model

In the Bohr’s model, angular momentum of the electron in a particular Bohr’s orbit is quantized.

Using the de Broglie wave hypothesis, one can imagine an electron as a standing wave in a given energy state n.

In order for the standing wave to fit around a given Bohr’s orbit, it must satisfy a periodic boundary condition (the wave must match back onto itself).

2 , 1,2,3,nr n n

Electron Waves and the Bohr’s Model

With 2 , 1,2,3,nr n n

Then, from de Broglie’s relation, we have /h mv Putting these together, we have

2 n

hr n

mv

Therefore, the stable energy levels in the Bohr’s Model are quantized with discrete angular momentum:

, 1,2,3,2n n n

hL mv r n n n

n is called the principal quantum number for the orbit.

Bohr’s Model (mathematical details)From a classical calculation, for a Hydrogen atom, the electron is bounded to the nucleus by the Coulomb’s Force:

2

20

1

4

eF

r

For an electron in circular orbit around a nucleus with radius r, we have

2

20

1

4

eF ma

r

2mv

r

Solving for r, we have,2

20

1 1

4

er

m v

From the angular momentum quantization, we have, , 1,2,nn

nv n

mr

Bohr’s Model (mathematical details)

Substituting this expression for v into the radius equation (boxed equation), we have,22

0

1

4n

n

mrer

m n

220

2n

hr n

me

and,2

0

1

2n

ev

h n

(orbital radii for a Bohr atom)

(orbital speeds for a Bohr atom)

2

20

1 1

4

er

m v

nr22

04nmre

2 2n


Writing this orbital speeds in the unit of c, we have

where

2 2

0 0

1 1

2 2n cc

e ev

h n h n

2 2

0 0

1

2 4 137

e eor

hc c

is called he Fine Structure Constant.

This gives and it justifies to treat the orbital speed being nonrelativistic.

1

1

137v c


The smallest radius for the Bohr atom corresponds to n = 1 and this minimum radius for the Bohr atom is called the Bohr’s radius,

2110

0 1 25.29 10

ha r m

me

With this fundamental length scale for an atom, the other radii can be written as,

20 , 1, 2,3,nr a n n


Now, consider the total energy for an e in an orbit at rn,

22

0

1 1

2 4

n n n

n nn

E K U

eE mv

r


Recall, from F = ma, we have2 2 2

22

0 0

1 1

4 4

e mv emv U

r r r

So, as in any circular motion under 1/r2 type of force, we have / 2K U

This gives the total energy for an e in an orbit at rn,

2

2 20

2 2

0 08 8

11

2n

nn

n n

me

r

U e eE

nU

hK

4

2 2 2 20

1 13.6

8n

me eVE

h n n

(energy levels of a Bohr’s atom)

Note: Energy levels are quantized as a consequence of the electron behaving as a wave in the atom (angular momentum quantization).

Bohr’s Model (notes)

• The ground state energy of the H-atom is given by E1 = -13.6 eV when n = 1.

4 47 1

2 2 2 30 0

1.097 108 8

me mehcR R m

h h c

Agree well with previous experimental value using wavelength measurements.

1n to n

1 13.6ionizationE E E eV

• Comparing the expression for H-atom’s energy with the empirical formula derived by Balmer, we can derive an explicit expression for the Rydberg Constant,

• The energy required to remove an electron completely is given by the transition from and it is called the ionization energy,

Hydrogen-like AtomsSingly ionized helium (He+), doubly ionized lithium (Li2+) are examples of hydrogen-like atoms with a single electron around the nucleus.

For hydrogen-like atoms, e2 in all equations from previous analysis is replaced by Ze2, where Z is the atomic number of the element.

All mathematical results follow through but with the following changes:

2

/n n

n n

r r Z

E Z E

Blackbody Radiation: Continuous SpectraIn contrast to line spectra emitted from excited atoms in a diluted gas, radiations from many thermally excited atoms in a solid will in general emit a continuous spectrum with characteristics depending on its temperature.

Blackbody Radiation: Continuous SpectraIn contrast to line spectra emitted from excited atoms in a diluted gas, radiations from many thermally excited atoms in a solid will in general emit a continuous spectrum with characteristics depending on its temperature.

Typically, a good light absorber will also be a good emitter so that a perfect emitter is typically called a “blackbody”.

Radiations emitted from a “blackbody” will be characteristic of that particular system since all other strayed radiations from the surrounding will be absorbed by the system and not reflected back out.

The continuous spectrum emitted by such a body is called blackbody radiation.

Blackbody RadiationThe total intensity (per unit surface area) of emitted light by a blackbody at absolute temperature T is given by the Stefan-Boltzmann Law:

4I T

where is call the Stefan-Boltzmann constant and it has the following value:

82 4

5.67 10W

m K

The Sun

Blackbody Radiation: Spectral Emittance

The intensity is not uniformly distributed over all wavelengths. The intensity distribution for a given range of wavelength is called the spectral emittance I().

(Experimental observed spectral emittance for different T’s.)

Note: as expected,0

( )I I d

There is also the experimentally obtained Wien displacement law:

3m 2.90 10T m K

As T , dominant (m) moves to shorter .

Blackbody Radiation: Cavity Radiation

Electric Forge

An example of a “blackbody” is a cavity with a small opening !

D_cavity

Blackbody RadiationA cavity with a small opening is a good candidate for a blackbody. Most importantly, one can derive the spectral emittance for a blackbody using such a cavity.

Model:

A metallic unit cube cavity filled with EM waves forming standing waves(normal modes) with nodes at the walls.

For a given range of wavelengths , the spectral emittance can be calculated as the combined energy from all the allowed standing waves (normal modes) within a given range of wavelengths.

, d

Blackbody Radiation

( ) ( )I d N E d

One can calculate the spectral emittance and it is given by:

where4

2( )

cN d d

is the # of EM modes (standing waves) allowed in the unit cube within a given wavelength range:

and is the average energy per normal mode (standing wave) within the cavity.

, d

E

As we will see, the different between the (classical) Rayleigh prediction (incorrect) and the (quantum) Planck prediction depends on how is calculated !E

Blackbody RadiationDistribution of energy states:

The energy of a given standing wave (normal modes) is distributed according to the Maxwell-Boltzmann distribution:

/

( )E kTe

P E dE dEkT

where k is the Boltzmann’s constant and T is the absolute temperature.(recall the special case for molecular speeds: the Maxwell’s distribution of molecular speeds)

Blackbody RadiationClassical Rayleigh Prediction (incorrect):

Assume E to be a continuous variable from 0 to , i.e., all energies are possible.

Then, 0

0

( )

( )

P E E dEE

P E dkT

E

Note: P(E) is already normalized. The denominator is 1. We are keeping it so that we will be consistent when we consider the discrete case later.

So, the spectral emittance is 4

2( , )T TI k

c

(Rayleigh-Jeans Law)

Problems with this classical prediction (“ultraviolet catastrophe”):

0

( , )0

( , )

I Tas

I T d

This is absolutely not physical! Intensity of the emitted light (radiated energy) must remain finite !

Blackbody RadiationMax Planck in 1900 proposed a solution to this problem.

A different way to calculate for the EM waves inside the cavity:E

E is not a continuous variable and E can only take on discrete values !

E hf

And, the discrete energy increment for the EM waves is given by:

0, ,2 ,3 , 0,1, 2,3,nE E E E n E n

Then, 0/

0

( )/

1( )

n nn

hc kT

nn

E P Ehc

EeP E

Blackbody Radiation

Then, the spectral emittance in this quantum calculation becomes:

2

5 /

2( , )

1hc kT

hcI T

e

(Planck’s Blackbody Radiation Law)

4 4 /

2 /

1

2( , )

hc kT

c c hcE

eI T

Notes: As (classical limit),0h / 1hc kT hc

ekT

so,/

/ /

1hc kT

hc hcE

e

1

hc

1

1kT

kT

Planck’s result reduces to the classical result if h = E = 0.

Planck’s Result in CM & QM Limits

0 : /hc kTe exponentially faster than 5 0 as 0

5 /

1, 0 0

1hc kTI T as

e

so

There is no ultraviolet catastrophe in QM limit !

(short wavelengths, high photon energy) QM regime

:

/ 1hc kT hce

kT

4 4

2 2( , )

c cI T E kT

Then, again, and

(long wavelengths, low photon energy) CM regimeClassical R-J Law

Again, we have

/

/

1hc kT

hcE kT

e

Blackbody Radiation (Review)

4

2( , )

cI T kT

(Rayleigh-Jeans Law)

2

5 /

2( , )

1hc kT

hcI T

e

(Planck’s Blackbody Radiation Law)

NOT COMPLETE Classical Prediction:

RIGHT QM Prediction:

Light must come in discrete quanta !

PHYS 262George Mason University

Professor Paul So

Chapter 40/41: Quantum Mechanics Wave Functions & 1D

Schrodinger Eq

Particle in a Box

Wave function

Energy levels

Potential Wells/Barriers & Tunneling

The Harmonic Oscillator

The H-atom

Electron Waves

In the Bohr’s model, angular momentum of the electron in a particular Bohr’s orbit is quantized.

In the de Broglie wave hypothesis, one can imagine an electron as a standing wave in a given energy state n.

So, what is the equation which describes these matter wave?

Wave Equation for a Mechanical String

For a wave on a string (1D) moving with speed , a wave function must satisfy the wave equation (Ch. 15):

,y x t

2 2

2 2 2

, ,1y x t y x t

x v t

,y x t

v

It has the following sinusoidal form as its fundamental solution:

, cos siny x t A kx t B kx t

where is the wave number and is the angular frequency of the wave. [A and B determines the amplitude and phase of the wave.]

2k 2 f

By substituting the fundamental wave function into the PDE, we can arrive at the algebraic relation (dispersion relation) that and must satisfy:

Wave Equation for a String

22

2k

v

,cos sin sin cosk

y x tA kx t B kx t A kx t B x

xk k t

x

or vk

k

check...

1. Each spatial derivative of will pull out one k: ,y x t

So, the 2nd order spatial derivative gives,

22

22,

cos siny x t

A kx t x tk B kx

k

(Obviously, don’t forget the signs.)

Wave Equation for a String

,cos sin sin cos

y x tA kx t B kx t A kx t B kx t

t t

check...

2. Each time derivative of will pull out one : ,y x t

So, the 2nd order time derivative gives,

22

22,

cos siny x t

A kx t B kx tt

(Again, don’t forget the signs.)

Putting these back into the wave equation, we then have,

2 2

2 2 2

, ,1y x t y x t

x v t

2

2 2

2 2

cos sin

1cos sin

k kA kx t B kx t

A kx t B kx tv

22

2k

v

is intimately linked to the form of the wave equation !

Thus, the fundamental property of a mechanical wave

Wave Equation for a StringPutting the definitions for and back into the dispersion relation, we have the familiar relation for wavelength, frequency, and wave speed.

2 2

2 2 2

, ,1y x t y x t

x v t

v f2 2f v

k

22

2or v f k

v

Now, we will try to use the same argument to find a wave equation for a quantum wave function.

vk

Now, from the de Broglie relations, the energy and momentum of this quantum free particle can be related to its wave number and angular frequency through:

Since the reference point for is arbitrary, we can simply take .

Then, the total energy E of a free particle will simply be its kinetic energy,

Wave Equation for a Quantum Free Particle

A free particle has no force acting on it. Equivalently, the potential energy

( )U x

k

0xF dU x dx ( )U x

( ) 0U x

2 2 221

2 2 2

m v pE mv

m m

must be a constant for all x, i.e., or U(x) is a constant.

22

hE hf f

2

2

h hp k

2 2

2

k

m

2

2

pE

m

(non-relativistic)

We now assume the same fundamental sinusoidal form for the wave function of a quantum free particle with mass m, momentum and energy :


Thus, a correct quantum wave function for a quantum free particle must satisfy this quantum dispersion relation for and :k

E

2 2

*2

k

m

, cos sinx t A kx t B kx t

p k

(non-relativistic)

Recall from our discussion on the mechanical wave, we have the following:

t

x

take out an overall kfactor from ,x t

take out an overall factor from ,x t

2 2

2

k

m

We can deduce that the PDE for the quantum wave function for this free particle must involves:


So, from the quantum dispersion relation,

Putting in other constants so that units are consistent and one additional dimensionless “fitting” constant C, we have this trial wave equation,

t

2

2x

22

2

, ,

2

x t x t

m x tC


Now, we substitute our trial quantum wave function

into the proposed wave equation to solve for the “fitting” constant C:

, cos sinx t A kx t B kx t

22 2

2 22

,cos sin

2 2

x tAk kx t Bk kx t

m x m

2 2

cos sin2

kA kx t B kx t

m

,sin cos

x tC C A kx t B kx t

t

cos sinCB kx t CA kx t


Equating the two terms and using the equality , we have,

2 2

2

k

m

cos sinA kx t B kx t cos sinCB kx t CA kx t

A CB

B CA

In order for this equality to be true for all , all coeff’s for cos and sin must equal to each other,

,x t

Substituting the first eq into the second, we have,

2 1B C CB C

Thus, the “fitting” constant is where .

2 2

2

k

m

1i C i

With , the free particle quantum wave function can also be written in a compact exponential form using the Euler’s formula,


Then, finally, putting everything together, we have the desired wave equation for a quantum free particle,

B CA iA

22

2

, ,

2

x t x ti

m x t

, cos sinx t A kx t i kx t

, i kx tx t Ae

This is the 1D Schrodinger’s Equation for a free particle.

(quantum wave function for a free particle)

And the corresponding wave function for a quantum free particle is given by,


For a quantum free particle, the wave equation is,

22

2

, ,

2

x t x ti

m x t

, cos sinx t A kx t i kx t

, i kx tx t Ae

This is the 1D Schrodinger’s Equation for a free particle.

(quantum wave function for a free particle)

Free Particle Wave Function &Uncertainty Principle

The wave function for a free particle is a complex function with sinusoidal real and imaginary parts

A quantum free particle exists in all space , ,

& (wave function extends into all space & time)

but 0 & 0 (energy and momentum is fixed)

x t

p E

Note: 2x p can still be satisfied.2t E &

More Realistic Particle (Wave Packets)

Under more practical circumstance, a particle will have a relatively well defined position and momentum so that both x and p will be finite with limited spatial extents.

A more localized quantum particle can not be a pure sine wave and it must be described by a wave packet with a combination of many sine waves.

( , ) ( ) i kx tx t A k e dk

The coefficient A(k) gives the relative proportion of the various sine waves with diff. k (wave number).

(a linear combination of many sine waves.)

Wave Packets

Recall: Combination of two sine waves more localized than a pure sine wave.

Wave Packets (characteristic)

p smaller x bigger

Wave Packets (characteristic)

p bigger x smaller

The is consistent with: !x p

Quantum Wave FunctionIn QM, the matter wave postulated by de Broglie is described by a complex-valued wavefunction (x,t) which is the fundamental descriptor for a quantum particle.

x,t

Re/Im (x,t)

(x,t) is a complex-valued function of space and time.

1. Its absolute value squared gives the probability of finding the particle in an infinitesimal volume dxat time t.

2( , )x t dx

( , )O x t

( , )O x t( , )x t

( , )x t2. For any Q problem: The goal is to find for the

particle for all time. Physical interactions involves

“operations” (O) on this wave function:

Experimental measurements will involve the “products”,

The General 1D Schrodinger Equation for a Quantum Particle (not free)

As we have seen,

- the first term (2nd order spatial derivative term) in the Schrodinger equation is associated with the Kinetic Energy of the particle

- the last term (the 1st order time derivative term) is associated with the total energy of the particle

- together with the Potential Energy term U(x) (x)

the Schrodinger equation is basically a statement on the conservation of energy.

KE PE Total E+ =

2 2

2

( , ) ( , )( ) ( , )

2

x t x tU x x t i

m x t

The Schrodinger EquationIn Classical Mechanics, we have the Newton’s equation which describes the trajectory x(t) of a particle:

mF x

In EM, we have the wave equation for the propagation of the E, B fields:

2 2

2 2 2

, 1 ,E B E B

x c t

(derived from Maxwell’s eqns)

In QM, Schrodinger equation prescribes the evolution of the wavefunctionfor a particle in time t and space x under the influence of a potential energy U(x),

2 2

2

( , ) ( , )( ) ( , )

2

x t x tU x x t i

m x t

(general 1D Schrödinger equation)

U(x)

Quantum Wave Function

The wave function for a quantum particle is a complex function with sinusoidal real and imaginary parts

( , )x t

Since is a probability, it has to be normalized !

Wave Function and Probability2 *( , ) ( , ) ( , )x t x t x t

is the probability distribution function for the quantum particle.

In other words,

is the probability in finding the particle in the interval at time t. [ , ]x x dx

(shaded area)

2( ) ( , )p x dx x t dx

2( ) ( , ) 1p x dx x t dx

(At any instance of time t, the particle must be somewhere in space !)

2( , )x t dx

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ch40 young freedmancomplex.gmu.edu/ · 2020. 5. 7. · %odfnerg\ 5dgldwlrq &rqwlqxrxv 6shfwud,q frqwudvw wr olqh vshfwud hplwwhg iurp h[flwhg dwrpv lq d gloxwhg jdv udgldwlrqv iurp