chapter 7 7.1: application of the schrödinger equation...
TRANSCRIPT
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7.1 Application of the Schrödinger
Equation to the Hydrogen Atom
7.2 Solution of the Schrödinger Equation for Hydrogen
7.3 Quantum Numbers
7.4 Magnetic Effects on Atomic
Spectra–The Zeeman Effect
7.5 Intrinsic Spin
7.6 Energy Levels and Electron Probabilities
CHAPTER 7
The Hydrogen AtomThe 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 fashion…is impossible.
- Werner Heisenberg
Werner Heisenberg
(1901-1976)
7.1: Application of the Schrödinger Equation to the Hydrogen Atom
The potential energy of the electron-proton system is electrostatic:
Use the three-dimensional time-independent Schrödinger Equation.
For Hydrogen-like atoms (He+ or Li++)
Replace e2 with Ze2 (Z is the atomic number).
Replace m with the reduced mass, µ.
Spherical Coordinates
The potential (central force)
V(r) depends on the distance rbetween the proton and
electron.
Transform to spherical polar coordinates because of the
radial symmetry.
The
Schrödinger
Equation in
Spherical
Coordinates
Transformed into spherical coordinates,
the Schrödinger
equation becomes:
Separable Solution
The wave function ψ is a function of r, θ, φ. This is a potentially complicated function.
Assume instead that ψ is separable, that is, a product of three functions, each of one variable only:
This would make life much simpler, and it turns out to work.
7.2: Solution of the Schrödinger
Equation for Hydrogen
Separate the resulting equation into three equations: R(r), f(θ), and g(φ).
The derivatives:
Substitute:
Multiply both sides by r2 sin2 θ / R f g:
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Solution of the Schrödinger Equation for H
r and θ appear only on the left side and φ appears only on the right side
The left side of the equation cannot change as φ changes.
The right side cannot change with either r or θ.
Each side needs to be equal to a constant for the equation to be true.
Set the constant to be −mℓ2
It is convenient to choose the solution to be .
azimuthal equation
Solution of the Schrödinger Equation for H
satisfies the azimuthal equation for any value of mℓ.
The solution must be single valued to be a valid solution for any φ:
mℓmust be an integer (positive or negative) for this to be true.
Now set the left side equal to −mℓ2 and rearrange it [divide by sin2(θ)].
Now, the left side depends only on r, and the right side depends only on θ. We can use the same trick again!
Solution of the Schrödinger Equation for H
Set each side equal to the constant ℓ(ℓ + 1).
Radial equation
Angular equation
We’ve separated the Schrödinger equation into three ordinary second-
order differential equations, each containing only one variable.
Solution of the Radial Equation for H
The radial equation is called the associated Laguerre equation and the
solutions R are called associated Laguerre functions. There are infinitely many of them, for values of n = 1, 2, 3, …
Assume that the ground state has n = 1 and ℓ = 0. Let’s find this solution.
The radial equation becomes:
The derivative of yields two terms.
This yields:
Solution of the Radial
Equation for HTry a solution
A is a normalization constant.a0 is a constant with the dimension of length.
Take derivatives of R and insert them into the radial equation.
To satisfy this equation for any r, both expressions in parentheses must
be zero.
Set the second expression
equal to zero and solve for a0:
Set the first expression equal
to zero and solve for E:
Both are equal to the Bohr results!
⇒
Principal
Quantum
Number n
There are many solutions to the radial wave equation, one for
each 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
ℓ: Orbital angular momentum quantum number
mℓ: Magnetic (azimuthal) quantum number
The restrictions for the quantum numbers:
n = 1, 2, 3, 4, . . .
ℓ = 0, 1, 2, 3, . . . , n − 1
mℓ= −ℓ, −ℓ + 1, . . . , 0, 1, . . . , ℓ − 1, ℓ
Equivalently:
n > 0
ℓ < n
|mℓ| ≤ ℓ
The energy levels are:
Hydrogen Atom Radial Wave Functions
First few
radial
wave functions Rnℓ
Sub-
scripts on R
specify
the values of
n and ℓ.
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 mℓ.
Physicists usually group these solutions together into
functions called Spherical Harmonics:
spherical harmonics
Normalized
Spherical Harmonics
Solution of the Angular and Azimuthal
Equations
The radial wave function R and the spherical harmonics Y determine
the probability density for the various quantum states. The total wave function depends on n, ℓ, and m
ℓ. The wave function
becomes
Orbital Angular Momentum Quantum
Number ℓ
Energy levels are degenerate with respect to ℓ (the energy is
independent of ℓ).
Physicists use letter names for the various ℓ values:
ℓ = 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 ℓ.
A state with n = 2 and ℓ = 1 is called a 2p state.
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Orbital Angular Momentum Quantum Number ℓ
It’s associated with the R(r) and f(θ) parts of the wave function.
Classically, the orbital angular momentum with L = mvorbitalr.
L is related to ℓ by
In an ℓ = 0 state,
This disagrees with Bohr’s semi-classical “planetary”
model of electrons orbiting
a nucleus L = nħ.
Classical orbits—which do not
exist in quantum mechanics
Example: ℓ = 2:
Only certain orientations of are
possible. This is called space
quantization.
And (except when ℓ = 0) we just
don’t know Lx and Ly!
Magnetic Quantum
Number ml
The angle φ is the angle from the z axis.
The solution for g(φ) specifies that mℓis an integer and is related to
the z component of L:
Rough derivation of ‹L2› = ℓ(ℓ+1)ħ2
We expect the average of the angular momentum components
squared to be the same due to spherical symmetry:
But
Averaging over all ml values (assuming each is equally likely):
2 ( 1)(2 1) / 3
n
m
= −
= + +∑l
l
l l lbecause:
In 1896, the Dutch physicist Pieter Zeeman showed that spectral lines emitted by atoms
in a magnetic field split into multiple energy
levels. It is called the Zeeman effect.
Consider the atom to behave like a small magnet.
Think of an electron as an orbiting circular current loop of I = dq / dt
around the nucleus. If the period is T = 2π r / v,
then I = -e/T = -e/(2π r / v) = -e v /(2π r).
The current loop has a magnetic moment µ = IA = [-e v /(2π r)] π r2 =
[-e/2m] mrv:
where L = mvr is the magnitude of the orbital angular momentum.
7.4: Magnetic Effects on Atomic
Spectra—The Zeeman Effect
2
eL
mµ = −
rr
If the magnetic field is in the z-direction, we only care about the z-
component of µ:
where µB = eħ / 2m is called the Bohr magneton.
The potential energy due to the
magnetic field is:
The Zeeman Effect
( )2 2
z z B
e eL m m
m mµ µ= − = − = −
l lh
2
eL
mµ = −
rr The Zeeman Effect
A magnetic field splits the mℓlevels. The potential energy is quantized
and now also depends on the magnetic quantum number mℓ.
When a magnetic field is applied, the 2p level of atomic hydrogen is
split into three different energy states with energy difference of ∆E =
µBB ∆mℓ.
E0 − µBB−1
E00
E0 + µBB1
Energymℓ
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The
Zeeman Effect
The transition
from 2p to 1s,
split by a magnetic field.
An atomic beam of particles in the ℓ = 1 state pass through a
magnetic field along the z direction.
The Zeeman Effect
The mℓ= +1 state will be deflected down, the m
ℓ= −1 state up, and
the mℓ= 0 state will be undeflected.
( / )Bm dB dzµ= −l
7.6: Energy
Levels and Electron
Probabilities
For hydrogen, the energy
level depends on the prin-cipal quantum number n.
In the ground state, an atom cannot emit radiation. It can
absorb electromagnetic
radiation, or gain energy through inelastic
bombardment by particles.
Selection Rules
The probability is proportional to
the mag square of the
dipole moment:
Allowed transitions:
Electrons absorbing or emitting photons can change states when ∆ℓ = ±1 and ∆ml = 0, ±1.
Forbidden transitions:
Other transitions are possible
but occur with much smaller probabilities.
*d er= Ψ Ψ∫r r
We can use the wave functions
to calculate transition probabilities for the electron to change from
one state to another.
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 function R(r) to calculate radial
probability distributions of the electron.
The probability of finding the electron in a differential volume
element dτ is .
Probability Distribution Functions
The differential volume element in spherical polar coordinates is
Therefore,
At the moment, we’re only interested in the radial dependence.
The radial probability density is P(r) = r2|R(r)|2 and it depends only on n
and l.
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R(r) and P(r)
for the lowest-lying states of
the hydrogen
atom.
Probability Distribution
Functions
Probability Distribution Functions
The probability density for the hydrogen atom for three different
electron states.
7.5: Intrinsic Spin
In 1925, grad students, Samuel Goudsmit and George Uhlenbeck,
in Holland proposed that the
electron must have an intrinsic angular momentum
and therefore a magnetic moment.
George Uhlenbeck, Hendrik Kramers,
and Samuel Goudsmit around 1928 in
Ann Arbor. Uhlenbeck and Goudsmit
proposed the idea of electron spins three
years earlier when they were studying in
Leiden with Paul Ehrenfest.
Ehrenfest, Paul (1880-1933)
Paul Ehrenfest showed that, if so, the surface of the spinning electron should
be moving faster than the speed of light!
In order to explain experimental data, Goudsmit and Uhlenbeck proposed
that the electron must have an intrinsic spin quantum number s = ½.
Goudsmit, Samuel Abraham
(1902-1978)George Eugene Uhlenbeck
1900 - 1988
Intrinsic Spin
The spinning electron reacts similarly to
the orbiting electron in a magnetic field.
The magnetic spin quantum number ms
has only two values, ms = ±½.
The electron’s spin will be either “up” or
“down” and can never be spinning with its
magnetic moment µs exactly along the zaxis.
Intrinsic Spin
The magnetic moment is .
The coefficient of is −2µB and is a consequence of relativistic quantum mechanics.
Writing in terms of the gyromagnetic ratio, g: gℓ= 1 and gs = 2:
The z component of .
In an ℓ = 0 state:
Apply ms and the potential energy becomes:
no splitting due to .
there is space quantization due to the intrinsic spin.
and
2L
eL
mµ = −
rrRecall:
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Multi-electron atoms
When more than one electron is involved, the potential and the wave function are functions of more than one position:
1 2( , ,..., )N
V V r r r=r r r
1 2( , ,..., , )N
r r r tΨ = Ψr r r
Solving the Schrodinger Equation in this case can be very hard.
But we can approximate the solution as the product of single-
particle wave functions:
1 2 1 1 2 2( , ,..., , ) ( , ) ( , ) ( , )N N N
r r r t r t r t r tΨ = Ψ Ψ Ψr r r r r r
L
And it turns out that, for electrons (and other spin ½ particles), all
the Ψi’s must be different. This is the Pauli Exclusion Principle.
Molecules have many energy levels.
A typical molecule’s energy levels:
Ground
electronic state
1st excited
electronic state
2nd excited
electronic state
Energ
y
Transition
Lowest vibrational and
rotational level of this electronic “manifold”
Excited vibrational and
rotational level
There are many other
complications, such as spin-orbit coupling,
nuclear spin, etc.,
which split levels.
E = Eelectonic + Evibrational + Erotational
As a result, molecules generally have very complex spectra.
Generalized Uncertainty Principle
Define the Commutator of two operators, A and B:
[ ],A B AB BA≡ −
Then the uncertainty relation between the two corresponding
observables will be:
[ ]*12
,A B A B∆ ∆ ≥ Ψ Ψ∫So if A and B commute, the two observables can be measured
simultaneously. If not, they can’t.
Example: [ ] ( ) ( ),p x px xp i x x ix x
xi i x x i i
x x x
∂ ∂ Ψ ≡ − Ψ = − Ψ − − Ψ
∂ ∂
∂ ∂Ψ ∂Ψ = − Ψ − − − = − Ψ
∂ ∂ ∂
h h
h h h h
[ ],p x i= − hSo: / 2p x∆ ∆ ≥ hand
Two Types of Uncertainty in Quantum Mechanics
We’ve seen that some quantities (e.g., energy levels) can be
computed precisely, and some not (Lx).
Whatever the case, the accuracy of their measured values is
limited by the Uncertainty Principle. For example, energies can
only be measured to an accuracy of ħ /∆t, where ∆t is how long we spent doing the measurement.
And there is another type of uncertainty: we often simply don’t
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 is excited,
but which one? We might say that each atom has a 1% chance of being in an excited state and a 99% chance of being in the
ground state. This is called a superposition state.
Superpositions of states
Stationary states are stationary. But an atom can be in a
superposition of two stationary states, and this state moves.
1 1 1 2 2 2( , ) ( ) exp( / ) ( ) exp( / )r t a r iE t a r iE tψ ψΨ = − + −r r r
h h
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
ψ ψ
ψ ψ
Ψ = + +
−
r r r
r rh
where |ai|2 is the probability that the atom is in state i.
Interestingly, this lack of knowledge means that the
atom is vibrating:
Superpositions of states
⇔
Energ
y
Ground level, E1
Excited level, E2
∆E = hν
The atom is at least partially in
an excited state.
The atom is vibrating
at frequency, ν.
Vibrations occur at the frequency difference between the two levels.
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
ψ ψ
ψ ψ
Ψ = + +
−
r r r
r rh
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Calculations in Physics: Semi-classical
physics
The most precise computations are performed fully quantum-
mechanically by calculating the potential precisely and solving Schrodinger’s Equation. But they can be very difficult.
The least precise calculations are performed classically,
neglecting quantization and using Newton’s Laws.
An intermediate case is semi-classical computations, in
which an atom’s energy levels are computed quantum-mechanically, but additional effects, such as light waves, are
treated classically. Our Thomson Scattering calculation of αand n was an example of this.