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7.1 Application of the SchrdingerEquation to the Hydrogen Atom

7.2 Solution of the SchrdingerEquation for Hydrogen

7.3 Quantum Numbers7.4 Magnetic Effects on Atomic

SpectraThe Zeeman Effect

7.5 Intrinsic Spin

7.6 Energy Levels and Electron

Probabilities

CHAPTER 7

The Hydrogen Atom

The atom of modern physics can be symbolized only through a partial differential

equation in an abstract space of many dimensions. All its qualities are inferential; no

material properties can be directly attributed to it. An understanding of the atomic world

in that primary sensuous fashionis impossible.

- Werner Heisenberg

Werner Heisenberg

(1901-1976)

Prof. Rick Trebino, Georgia Tech, www.physics.gatech.edu/frog/lectures


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7.1: Application of the SchrdingerEquation to the Hydrogen Atom

The potential energy of the electron-proton system is electrostatic:

Use the three-dimensional time-independent Schrdinger Equation.

For Hydrogen-like atoms (He+ or Li++), replace e2 withZe2 (Zis theatomic number).

In all cases, for better accuracy, replace m with the reduced mass, m.


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Spherical Coordinates

The potential (central force)

V(r) depends on the distancerbetween the proton andelectron.

Transform to spherical polarcoordinates because of the

radial symmetry.


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TheSchrdinger

Equation inSphericalCoordinates

Transformed intospherical coordinates,the Schrdingerequation becomes:


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Separable Solution

The wave function yis a function of r, q, f. This is a potentiallycomplicated function.

Assume optimistically that yis separable, that is, a product of threefunctions, each of one variable only:

This would make life much simplerand it turns out to work!


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7.2: Solution of the SchrdingerEquation for Hydrogen

Start with Schrodingers Equation:

Multiply both sides by -r2 sin2q/R f g:

Substitute:


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Solution of the Schrdinger Equation for H

rand qappear only on the left side and fappears only on the right side.

The left side of the equation cannot change as fchanges.

The right side cannot change with either ror q.

Each side needs to be equal to a constant for the equation to be true.

Set the constant to be m2

Sines and cosines satisfy this equation, but it is convenient to choose the

solution to be .

azimuthal equation


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satisfies the azimuthal equation for any value of m.

The solution must be single valued to be a valid solution for any f:

m must be an integer (positive or negative) for this to be true.

Specifically:

So:


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Now set the left side equal to m2:

m2

Now, the left side depends only on r, and the right side depends onlyon q. We can use the same trick again!

Rearrange it and divide by sin2(q):


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Set each side equal to the constant ( +1).

Radial equation

Angular equation

Weve separated the Schrdinger equation into three ordinary second-order differential equations, each containing only one variable.


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Solution of the Radial Equation for H

The radial equation is called the associated Laguerre equationand thesolutionsR are called associated Laguerre functions. There areinfinitely many of them, for values of n = 1, 2, 3,

Assume that the ground state has n = 1 and = 0. Lets find this solution.

The radial equation becomes:

The derivative of yields two terms:


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Solution of the RadialEquation for HTry a solution

A is a normalization constant.

a0 is a constant with the dimension of length.Take derivatives ofR and insert them into the radial equation.

To satisfy this equation for any r, both expressions in parentheses mustbe zero.

Set the second expression equal to zeroand solve for a0:

Set the first expression equal to zero andsolve forE:

Both are equal to the Bohr results!


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PrincipalQuantum

Numbern

There are many solutions to the radial wave equation, one foreach positive integer, n.

The result for the quantized energy is:

A negative energy means that the electron and proton are bound

together.


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7.3: Quantum Numbers

The three quantum numbers:

n: Principal quantum number

: Orbital angular momentum quantum number

m: Magnetic (azimuthal) quantum number

The restrictions for the quantum numbers:n = 1, 2, 3, 4, . . .

= 0, 1, 2, 3, . . . , n 1

m= , + 1, . . . , 0, 1, . . . , 1,

Equivalently:

n > 0

< n

|m|

The energy levels are:


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Hydrogen Atom Radial Wave Functions

First fewradialwavefunctionsRn

Sub-scriptsonR

specifythevalues ofn and .


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Solution of the Angular and AzimuthalEquations

The solutions to the azimuthal equation are:

Solutions to the angular and azimuthal equations are linked

because both have m.

Physicists usually group these solutions together intofunctions called Spherical Harmonics:

spherical harmonics


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NormalizedSpherical

Harmonics


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Solution of the Angular and AzimuthalEquations

The radial wave functionR and the spherical harmonics Ydetermine the probability density for the various quantum states.The total wave function depends on n, , and m.

The wave function becomes:


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Probability Distribution Functions

We use the wave functions to calculate the probability distributions

of the electrons.

The position of the electron is spread over space and is not well

defined.

We may use the radial wave functionR(r) to calculate radial

probability distributions of the electron.

The probability of finding the electron in a differential volumeelement dt is:

where the differential volume element in spherical polarcoordinates is


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At the moment, well consider only the radial dependence.So we should integrate over all values of qand f:

The qand fintegrals are just constants.

So the radial probability density is P(r) = r2|R(r)|2 and it depends only

on n and .


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R(r) and P(r) forthe lowest-lyingstates of thehydrogen atom.

Note thatRn0 ismaximal at r= 0!But the r2 factorreduces the

probability there to0. Nevertheless,theres a nonzero

probability that theelectron is insidethe nucleus.

ProbabilityDistribution

Functions


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The probability density for the hydrogen atom for three different

electron states.


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Orbital Angular Momentum QuantumNumber

Energy levels are degenerate with respect to (the energy isindependent of ).

Physicists use letter names for the various values:

= 0 1 2 3 4 5 . . .Letter = s p d f g h. . .

Atomic states are usualy referred to by their values of n and .

A state with n = 2 and = 1 is called a 2pstate.


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Its associated with theR(r) andf(q) parts of the wave function.

Classically, the orbital angular momentum is withL = m vorbital r.

But quantum-mechanically,L isrelated to by:

In an = 0 state,

This disagrees with Bohrs semi-

classical planetary model of

electrons orbiting a nucleusL = n,where n= 1, 2,

Orbital Angular Momentum QuantumNumber

Classical orbitswhich do notexist in quantum mechanics

L r p


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Example: = 2:

Only certain orientations of arepossible.

And (except when = 0) we just

dont knowLx andLy!

Magnetic QuantumNumberm

The solution for g(f) specifies that

m is an integer and is related tothez component ofL:


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Rough derivation of L2 = (+1)2

We expect the average of the angular momentum componentssquared to be the same due to spherical symmetry:

But

Averaging over all mvalues (assuming each is equally likely):

2 ( 1)(2 1) / 3

n

m

-

because:


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In 1896, the Dutch physicist Pieter

Zeeman showed that spectral linesemitted by atoms in a magneticfield split into multiple energylevels.

Think of an electron as an orbiting

circular current loop ofI= dq /dtaround the nucleus. If the period isT= 2pr/ v, then:

I = -e/T = -e/(2pr/ v) = -e v /(2pr)

7.4: Magnetic EffectsThe Zeeman Effect

The current loop has a magnetic moment m=IA:

Nucleus

2

eL

mm -

whereL = mvristhe magnitude ofthe orbital angularmomentum.

= [-e v /(2pr)]pr2= [-e/2m] mrv


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If the magnetic field is in thez-direction, all that matters is thez-component of m:

where mB = e / 2m is called the Bohr magneton.

The potential energy due to themagnetic field is:

The Zeeman Effect

( )2 2

z z B

e eL m m

m mm m - - -

2

eL

mm -


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B z BV B m Bm m - -

The Zeeman Effect

We find that the magnetic field

splits the m levels. Thepotential energy is quantizedand now also depends on themagnetic quantum number m.

m Energy

1 E0 + mBB

0 E0

1 E0mBB

Technically, we need to go backto the Schrdinger Equation andre-solve it with this new term in

the total energy.

When a magnetic field is applied, the 2plevel of atomic hydrogen issplit into three different energy states with energy difference of DE=mBB Dm.

z Bmm m -


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TheZeeman

Effect

The transition

from 2pto 1s,split by amagnetic field.


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An atomic beam of particles in the = 1 state pass through a

magnetic field along thez direction.

The Zeeman Effect

The m = +1 state will be deflected down, the m= 1 state up, andthe m = 0 state will be undeflected.

( / ) ( / ) ( / )z B z B

F dV dz dB dz m dB dzm m - -

B zV Bm -


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7.6: EnergyLevels and

ElectronProbabilities

For hydrogen, the energy

level depends on the prin-cipal quantum number n.

An electron can make atransition from a state of any

n value to any other.

But what about and m

quantum numbers?


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Selection Rules

The probability is proportional tothe mag square of thedipole moment:

Allowed transitions:

Electrons absorbing or emitting photons can change stateswhen D= 1 and Dm= 0, 1.

Forbidden transitions:

Other transitions are possiblebut occur with much smallerprobabilities.

*

f id er

We can use the wave functionsto calculate transition probabilities

for the electron to change from

one state to another.

where i and fare the initial andfinal states of the

transition.


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7.5: Intrinsic Spin

In 1925, grad students, SamuelGoudsmit and George Uhlenbeck,in Holland proposed that theelectron must have anintrinsic angular momentum

and therefore a magnetic moment.

This seems reasonable, but Paul Ehrenfest showed that, if so,the surface of the spinning electron would be moving faster thanthe speed of light!

In order to explain experimental data, Goudsmit and Uhlenbeckproposed that the electron must have an intrinsic spinquantum number s = .

S


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Intrinsic Spin

The spinning electron reacts similarly to the

orbiting electron in a magnetic field.

The magnetic spin quantum numbermshas only two values, ms = .

And Sz = ms .

The electrons spin is either up (ms = +)

or down (ms = -) and can never be

spinning with its magnetic momentSexactly along thez axis.

S

( 1) 3 / 4S s s


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What about Sx and Sy?

Quantum mechanics says that, no

matter how hard we try, we cant alsomeasure them!

If we did, wed measure , just aswed find forSz.

But then this measurement wouldperturb Sz, which would then becomeunknown!

S

The total spin is ,

so itd be tempting to conclude that every component of theelectrons spin is either up (+ ) or down (ms = - ). Butthis is not the case! Instead, theyre undetermined. Well see

next that the uncertainty in each unmeasured component isequal to their maximum possible magnitude ( )!

2 2 21 1 12 2 2

( 1) 3 / 4 ( ) ( ) ( )S s s


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Generalized Uncertainty Principle

Define theCommutatorof two operators,A andB:

,A B AB BA -

Then the uncertainty relation between the two correspondingobservables will be:

*12 ,A B A BD D So ifA andB commute, the two observables can be measured

simultaneously. If not, they cant.

Example: ,p x px xp i x x ix x

xi i x x i i

x x x

- - - -

- - - - -

,p x i -So: / 2p xD D and


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Uncertainty in angular momentum andspin

Weve seen that the total and z-components of angular momentumand spin are knowable precisely. And the x and y-componentsarent. Heres why. It turns out that:

,x y zL L i L

*12 ,A B A BD D Using:We find:

2

* *1 12 2 ( )

2L

x y z L mL L i L i mD D So theres an uncertainty relation between the x and y components

of orbital angular momentum (unless m = 0). And the same for spin.

,x y zS S i S and


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Two Types of Uncertainty in QuantumMechanics

1. Some quantities (e.g., energy levels) can, at least in principle,be computed precisely, but some cannot (e.g.,Lx, Ly, Sx, Sy).

Even if a quantity can be computed precisely, the accuracy oftheir measured value is limited by the Uncertainty Principle.

For example, energies can only be measured to an accuracy of/Dt, where Dtis how long we spent doing the measurement.

2. And there is another type of uncertainty: we often simply dont

know which state an atom is in.

For example, suppose we have a batch of, say, 100 atoms,which we excite with just one photon. Only one atom isexcited, but which one? We might say that each atom has a1% chance of being in an excited state and a 99% chance ofbeing in the ground state. This is called a superposition state.


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Superpositions of states

Stationary states are stationary. But an

atom can be in a superposition of twostationary states, and this state moves.

1 1 1 2 2 2( , ) ( ) exp( / ) ( ) exp( / )r t a r iE t a r iE t y y - -

2 2 2

1 1 2 2

* *

1 1 2 2 2 1

( , ) ( ) ( )

2 Re ( ) ( ) exp[ ( ) / ]

r t a r a r

a r a r i E E t

y yy y

-

where |ai|2 is the probability that the atom is in state i.

Interestingly, this lack of knowledge means that theatom is vibrating:

E

nergy

Groundlevel,E1

Excitedlevel,E2

DE = hn

Vibrations occur at the frequency difference between the two levels.


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Writing H atom states in the bra-ketnotation

The bra-ket notation provides a convenient short-hand notation forH states. Since n, , m, and ms determine the state, we can write astate as a ket:

sn m m sn m my

Theres no need to write the value ofs, since its always for

electrons.

The specific mathematical functions involved are well known, soeveryone knows what this means.

And when relevant, we can write the bra form for the complexconjugate, as well:

*

sn m m sn m my

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