Transcript ModSchH

Modern Physics Ch.7:
H atom in Wave mechanics
Methods of Math. Physics, 27 Jan 2011, E.J. Zita
• Schrödinger Eqn. in spherical coordinates
• H atom wave functions and radial probability densities
• L and probability densities
• Spin
• Energy levels, Zeeman effect
• Fine structure, Bohr magneton
Recall the energy and momentum operators
E
hc

 pc
p
h

From deBroglie wavelength, construct a differential
operator for momentum:
h

2
k
2

h

2

p 
 k i
 
x
2
h
Similarly, from uncertainty principle, construct energy operator:
Et

E i
t
Energy conservation  Schrödinger eqn.
E=T+V
Ey = Ty + Vy
where y is the wavefunction and
operators depend on x, t, and momentum:

p̂  i
x

E i
t
y
y
i


U
y
2
t
2m x
2
2
Solve the Schroedinger eqn. to find the wavefunction, and
you know *everything* possible about your QM system.
Schrödinger Eqn
We saw that quantum mechanical systems can be
described by wave functions Ψ.
A general wave equation takes the form:
Ψ(x,t) = A[cos(kx-ωt) + i sin(kx-ωt)] = e i(kx-ωt)
Substitute this into the Schrodinger equation to see
if it satisfies energy conservation.
Derivation of Schrödinger Equation
i
Wave function and probability
Probability that a measurement of the system will
yield a result between x1 and x2 is x 2
2
 y ( x, t ) dx
x1
Measurement collapses the wave function
•This does not mean that the system was at X before the
measurement - it is not meaningful to say it was localized at
all before the measurement.
•Immediately after the measurement, the system is still at X.
•Time-dependent Schrödinger eqn describes evolution of y
after a measurement.
Exercises in probability: qualitative
Uncertainty and expectation values
Standard deviation s can be found from the deviation from the
average: j  j  j
But the average deviation vanishes:
j  0
So calculate the average of the square of the deviation:
s 2   j 
Last quarter we saw that we can calculate s more easily by:
s 2  j2  j
2
2
Expectation values
Most likely outcome of a measurement of position, for a system
(or particle) in state y: 
x 
 x y ( x, t )
2
dx

Most likely outcome of a measurement of position, for a system
(or particle) in state y:
d x
p m
 i
dt

 * y
 y x

 dx

Uncertainty principle
Position is well-defined for a pulse with ill-defined
wavelength. Spread in position measurements = sx
Momentum is well-defined for a wave with precise . By its
nature, a wave is not localized in space. Spread in
momentum measurements = sp
We saw last quarter that
s xs p 

Applications of Quantum mechanics
Blackbody radiation: resolve ultraviolet catastrophe, measure
star temperatures
Photoelectric effect: particle detectors and signal amplifiers
Bohr atom: predict and understand H-like spectra and energies
Structure and behavior of solids, including semiconductors
Scanning tunneling microscope
Zeeman effect: measure magnetic fields of stars from light
Electron spin: Pauli exclusion principle
Lasers, NMR, nuclear and particle physics, and much more...
Stationary states
2
Y ( x, t )
 2Y ( x, t )
i

 V ( x , t ) Y ( x, t )
2
t
2m x
If an evolving wavefunction Y(x,t) = y(x) f(t)
can be “separated”, then the time-dependent term
satisfies
i
1 df
i
E
f dt
Everyone solve for f(t)=
Separable solutions are stationary states...
Separable solutions:
Y ( x, t )  y ( x )
(1) are stationary states, because
* probability density is independent of time [2.7]
* therefore, expectation values do not change
2
2
(2) have definite total energy, since the Hamiltonian is
sharply localized: [2.13]
s 2 0
H
(3) yi = eigenfunctions corresponding to each allowed
energy eigenvalue Ei.

 i En t
Y ( x, t )   cny n e
General solution to SE is [2.14]
n 1
Show that stationary states are separable:
Guess that SE has separable solutions Y(x,t) = y(x) f(t)
Y

t
2Y

2
x
sub into SE=Schrodinger Eqn
Divide by y(x) f(t) :
Y
i

t
2

Y
2
V Y
2
x
LHS(t) = RHS(x) = constant=E. Now solve each side:
You already found solution to LHS: f(t)=_________
 2 d 2y
 Vy  Ey
2
2m dx
RHS solution depends on the form of the potential V(x).
Now solve for y(x) for various U(x)
Strategy:
* draw a diagram
* write down boundary conditions (BC)
* think about what form of y(x) will fit the potential
* find the wavenumbers kn=2 /
* find the allowed energies En
* sub k into y(x) and normalize to find the amplitude A
* Now you know everything about a QM system in this
potential, and you can calculate for any expectation
value
Infinite square well:
V(0<x<L) = 0, V= outside
What is probability of finding particle outside?
2
Inside: SE becomes
d 2y
Ey  
2m dx 2
* Solve this simple diffeq, using E=p2/2m,
* y(x) =A sin kx + B cos kx: apply BC to find A and B
* Draw wavefunctions, find wavenumbers: kn L= n
* find the allowed energies:
(n )2
2
2
En 
,A 
2
* sub k into y(x) and normalize:
2mL
L
* Finally, the wavefunction is
2
n
y n ( x) 


sin 
x
L
 L 
Square well: homework
Repeat the process above, but center the infinite square
well of width L about the point x=0.
Preview: discuss similarities and differences
Infinite square well application:
Ex.6-2 Electron in a wire (p.256)
Summary:
• Time-independent Schrodinger equation
has stationary states y(x)
• k, y(x), and E depend on V(x) (shape & BC)
• wavefunctions oscillate as eiwt
• wavefunctions can spill out of potential
wells and tunnel through barriers
That was mostly review from last quarter.
Moving on to the H atom in terms of Schrödinger’s wave
equation…
Review energy and momentum operators

p̂  i
x

E i
t
Apply to the Schrödinger eqn:
EY(x,t) = T Y(x,t) + V Y(x,t)
y
y
i

 Vy
2
t
2m x
2
Find the wavefunction
for a given potential V(x)
2

Y ( x, t )   cny n e
n 1
 i En t
Expectation values

f   y * f y dx

Most likely outcome of a measurement of position, for a system
(or particle) in state yx,t:

x 

x y ( x, t ) dx
2
where
y  y *y
2

Order matters for operators like momentum – differentiate y(x,t):
d x
p m
 i
dt

 * y
 y x

 dx

H-atom: quantization of energy for V= - kZe2/r
Solve the radial part of the spherical Schrödinger equation (next
quarter):
Do these energy values look familiar?
Continuing Modern Physics Ch.7:
H atom in Wave mechanics
Methods of Math. Physics, 10 Feb. 2011, E.J. Zita
• Schrödinger Eqn. in spherical coordinates
• H atom wave functions and radial probability densities
• L and probability densities
• Spin
• Energy levels, Zeeman effect
• Fine structure, Bohr magneton
Spherical harmonics solve spherical Schrödinger
equation for any V(r)
You showed that Y210 and Y200 satisfy Schrödinger’s equation.
H-atom: wavefunctions Y(r,q,f) for V= - kZe2/r
R(r) ~ Laguerre Polynomials, and the angular parts of the wavefunctions for any
radial potential in the spherical Schrödinger equation are
Y (r ,q ,  )  R(r )Ylm (q ,  ) where
Ylm (q ,  )  spherical harmonics
Radial probability density
P(r )  r Rn, (r )
2
2
Look at Fig.7.4. Predict the probability (without calculating)
that the electron in the (n,l) = (2,0) state is found inside the Bohr
radius.
Then calculate it – Ex. 7.3. HW: 11-14 (p.233)
H-atom wavefunctions ↔
electron probability distributions:
l = angular momentum wavenumber
Discussion: compare Bohr model to Schrödinger model for H atom.
ml denotes possible orientations of L and Lz (l=2)
Wave-mechanics L ≠ Bohr’s n
HW: Draw this for l=1, l=3
QM H-atom energy levels: degeneracy for states
with different Y and same energy
Selections rules for allowed
transitions: n = anything
(changes in energy level)
l must change by one, since
energy hops are mediated by
a photon of spin-one:
l = ±1
m = ±1 or 0 (orientation)
DO #21, HW #23
Stern-Gerlach showed line splitting, even when l=0. Why?
l = 1, m=0,±1 ✓
l = 0, m=0 !?
Normal Zeeman effect
“Anomalous”
A fourth quantum number: intrinsic spin
Since L 
l (l  1), let S 
s(s  1)
If there are 2s+1 possible values of ms,
and only 2 orientations of ms = z-component of s (Pauli),
What values can s and ms have?
HW #24
Wavelengths due to energy shifts
E
hc

dE  _________ d 
  ________ E
Spinning particles shift energies in B fields
Cyclotron frequency: An electron moving with speed v
perpendicular to an external magnetic field feels a Lorentz
force:
F=ma
(solve for w=v/r)
Solve for Bohr magneton…
Magnetic moments
shift energies in B fields
Spin S and orbit L couple to total angular momentum
J = L+ S
Spin-orbit coupling: spin of e- in magnetic field of p
Fine-structure splitting (e.g. 21-cm line)
(Interaction of nuclear spin with electron spin (in an atom) →
Hyper-fine splitting)
Total J + external magnetic field → Zeeman effect
Total J + external magnetic field → Zeeman effect
Total J + external magnetic field → Zeeman effect
History of atomic models:
• Thomson discovered electron, invented plum-pudding model
• Rutherford observed nuclear scattering, invented orbital atom
• Bohr quantized angular momentum, improved H atom model.
• Bohr model explained observed H spectra, derived En = E/n2
and phenomenological Rydberg constant
• Quantum numbers n, l, ml (Zeeman effect)
• Solution to Schrodinger equation shows that En = E/l(l+1)
• Pauli proposed spin (ms= ±1/2), and Dirac derived it
• Fine-structure splitting reveals spin quantum number