atoms & an introduction to quantum...
TRANSCRIPT
ATOMS & AN INTRODUCTION TO QUANTUM MECHANICS
Yet another quote from Cambridge physicist David Tong:
“Quantum mechanics is an outrageous departure from our classical, comforting, common-senseview of the world. It is more baffling and disturbing than anything dreamt up by science fictionwriters. And yet it is undoubtably the correct description of the Universe we inhabit and hasallowed us to understand Nature with unprecedented accuracy.”
We will build up this understanding in 3 overall steps...
7.1
ATOMS & AN INTRODUCTION TO QUANTUM MECHANICS
Yet another quote from Cambridge physicist David Tong:
“Quantum mechanics is an outrageous departure from our classical, comforting, common-senseview of the world. It is more baffling and disturbing than anything dreamt up by science fictionwriters. And yet it is undoubtably the correct description of the Universe we inhabit and hasallowed us to understand Nature with unprecedented accuracy.”
We will build up this understanding in 3 overall steps...
1. Light can be particle-like, as well as wave-like.
2. Matter can be wave-like, as well as particle-like.
3. The evolution of matter in space & time is inherentlyprobabilistic, following Schrodinger’s equation.
Main application: energy levels in the H atom.
Even though quantum mechanics is very well-developed, it’s still noteasy for scientists to wrap their heads around its implications aboutwhat is “really” happening at tiny scales. Some say not to worry,and to just “shut up and calculate.”
7.2
(1) Light is sometimes “quantized” (i.e., discrete)
In 1900, it wasn’t surprising to think about matter as being quantized, both in terms of atomicmass & charge. From Maxwell’s equations, light was firmly known to be wave-ish.
However, nobody had a theory for why hot, opaque bodies emit light as a blackbody.
Stefan & Boltzmann figured out that the energy flux emitted by “good absorbers” (i.e., objectsthat don’t reflect or scatter incoming beams of light) follow a specific set of curves, depending ontheir temperatures:
7.3
Think of a blackbody as an idealized “cavity” that may let in a bit of light, but once it’s in, thelight bounces around many times.
The light energy inside the cavity eventually comes into thermal equilibrium with both: (1)any gas in the cavity, and (2) free electrons in the walls, which oscillate when light beams bounceoff them. (That induces J, which affects the electromagnetic waves in the cavity interior.)
The radiation field is isotropic, with Iν the same in all directions. The energy density(per unit ν) “trapped” in the cavity is
Uν =
∮dΩ Iν(n)
c=
4πIνc
(for more about the factor of c, review the derivation of units for UEM & Poynting flux S).
7.4
The spectrum of blackbody intensity was observed before it was understood.In the 1860s, Gustav Kirchhoff & others worked out the functional form...
Iν =C1 ν
3
eC2ν/T − 1where C1 and C2 were measured constants.
At very low frequencies, we can expand ex ≈ 1 + x + · · ·
Iν(lo-freq) ≈ C1 ν3
1 + (C2ν/T )− 1≈ C1 ν
2 T
C2
and at high frequencies,Iν(hi-freq) ∼ e−C2ν/T
Stefan and Boltzmann realized that the integral under the curve is proportional to T 4
I =
∫ ∞
0
dν Iν =
(π4C1
15C42
)
T 4 =σSBT
4
πwhere σSB = Stefan-Boltzmann constant.
Thus, a hotter blackbody emits more light.
Also, Wien found that the peak of the curve depends on T ,
νpeak ∝ T or, equivalently, λpeak ∝ 1
T.
7.5
How can we use physics to understand (i.e., derive!) these functions?
Around 1900, Rayleigh & Jeans tried, using classical E&M theory, and failed.Also in 1900, Max Planck succeeded by going beyond the classical theory.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6
How can we use physics to understand (i.e., derive!) these functions?
Around 1900, Rayleigh & Jeans tried, using classical E&M theory, and failed.Also in 1900, Max Planck succeeded by going beyond the classical theory.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Let’s first examine how Rayleigh & Jeans decided to compute the energy density (per ν) bytallying up the number density of wave modes, then multiplying by the average energy permode:
Uν = nν E
To compute nν, they first looked at the total (“cumulative”) number of wave-modes that could fitin a cubical box with sides L, with frequencies less than some given value ν.
ntot(< ν) =Nmodes
L3and nν = lim
∆ν→0
[ntot(< ν +∆ν)− ntot(< ν)
∆ν
]
≈ ∂ntot∂ν
In other words, the desired number density per-unit-ν is the derivative of the cumulative ntot.
Thus, we need to count up Nmodes that can exist in the box, less than some maximum ν(i.e., with wavelengths greater than some specified λ).
7.7
Number of standing-wave modes in each dimension: m =2L
λ=
2Lν
c
7.8
Thus, the total number of combinations, in 3D, is
Nmodes ≈ 8L3ν3
c3
(
However, the actual rigorous answer: Nmodes =8π
3
L3ν3
c3
)
Thus,
ntot(< ν) =Nmodes
L3=
8πν3
3c3so nν =
∂ntot∂ν
=8πν2
c3.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.9
Thus, the total number of combinations, in 3D, is
Nmodes ≈ 8L3ν3
c3
(
However, the actual rigorous answer: Nmodes =8π
3
L3ν3
c3
)
Thus,
ntot(< ν) =Nmodes
L3=
8πν3
3c3so nν =
∂ntot∂ν
=8πν2
c3.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Also, we need E. We’re assuming the wave-modes are in thermal equilibrium with anyrandomly-moving particles, either in the walls or in the cavity. For those particles, each degree offreedom has E = (1/2)kBT .
However, each wave-mode behaves in 1D like a harmonic oscillator (like a ball attached to aspring). It’s got kinetic energy (like the particles do), but also potential energy. As it oscillates, itgoes between “all kinetic” & “all potential.”
Averaged over long times, both components are equal, so an oscillator essentially has 2 degrees offreedom... and E = kBT .
Thus, Rayleigh & Jeans predicted that Uν = nν E =8π ν2 kBT
c3.
7.10
Recall: at low frequencies, the observed blackbody spectrum looks like
Uν =4π Iνc
≈ 4π
c
C1 ν2 T
C2and observations verified that
4π C1
cC2=
8πkBc3
X
However, at high frequencies, the blackbody function comes back down, while theRayleigh-Jeans formula −→ ∞.
This was called the ULTRAVIOLET CATASTROPHE.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.11
Recall: at low frequencies, the observed blackbody spectrum looks like
Uν =4π Iνc
≈ 4π
c
C1 ν2 T
C2and observations verified that
4π C1
cC2=
8πkBc3
X
However, at high frequencies, the blackbody function comes back down, while theRayleigh-Jeans formula −→ ∞.
This was called the ULTRAVIOLET CATASTROPHE.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Then came Max Planck. He tried many things to fix this. Only one thing succeeded.
He discarded the idea that the energy of the oscillating wall-charges was a continuous variable.
When we derived the properties of an ideal gas, we treated particle energy as a continuous“sample” from the Boltzmann distribution:
P ∝ e−E/kBT .
In fact, the Rayleigh-Jeans calculation for the average oscillator energy made use of thissame kind of “moment:”
E =
∫∞0 dE e−E/kBT E∫∞0 dE e−E/kBT
= = kBT .
7.12
But Planck made the (remarkable!) assumption that the energy of an oscillator with frequency νcan only be a discrete, integer-multiple of hν, where h is some constant (now called Planck’sconstant); i.e.,
E = 0 , hν , 2hν , 3hν , 4hν , · · · only!
As before, we can keep using the Boltzmann factor to give us the probability of finding anoscillator with energy E, but instead of an integration, it’s now a discrete sum:
since E = nhν , then E =
∑∞n=0(nhν)e
−nhν/kBT∑∞
n=0 e−nhν/kBT
.
How do we work out these sums? First look at the denominator:∞∑
n=0
e−nhν/kBT =∞∑
n=0
[
e−hν/kBT]n
and let’s define x = e−hν/kBT .
Thus, the denominator is∞∑
n=0
xn = 1 + x + x2 + x3 + x4 + · · ·
Note that 0 < x < 1, so this infinite series converges to a finite value.
7.13
For example, let’s try x = 0.1.
1 + 0.1 + 0.01 + 0.001 + 0.0001 + · · · = 1.11111... =10
9=
1
0.9
and, in general, it can be shown that the series converges to1
1− x.
In the numerator, we can take out a common constant hν, and∞∑
n=0
nxn = 0 + x + 2x2 + 3x3 + 4x4 + · · ·
= x[1 + 2x + 3x2 + 4x3 + · · ·
]
The thing in [square brackets] is a well-behaved series that converges nicely (for 0 < x < 1) to avalue of
1
(1− x)2.
Thus,
E = hνx
(1− x)21− x
1=
hν x
1− x=
hν e−hν/kBT
1− e−hν/kBT=
hν
ehν/kBT − 1.
7.14
This is very different from the Rayleigh-Jeans value of E = kBT , but take note of thelow-frequency limit...
ehν/kBT ≈ 1 +hν
kBT+ · · · so in this limit, E does ≈ kBT .
Thus, multiplying by the same old value of nν, we have the Planck function energy density,
Uν =8πhν3/c3
ehν/kBT − 1which explains observed blackbody curves exactly.
Anyway, we now know that the energy in one discrete photon of light can be written as E = hν,but this was a brand-new concept to Planck.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.15
This is very different from the Rayleigh-Jeans value of E = kBT , but take note of thelow-frequency limit...
ehν/kBT ≈ 1 +hν
kBT+ · · · so in this limit, E does ≈ kBT .
Thus, multiplying by the same old value of nν, we have the Planck function energy density,
Uν =8πhν3/c3
ehν/kBT − 1which explains observed blackbody curves exactly.
Anyway, we now know that the energy in one discrete photon of light can be written as E = hν,but this was a brand-new concept to Planck.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
At the start of the 20th century, there arose 3 other observations/experiments that pointed to theidea of light behaving as discrete particles in addition to behaving as waves:
a. the photoelectric effectb. Compton scatteringc. (various) double-slit experiments
We won’t derive their math in as much detail as for the Planck function, though.
7.16
(a) The Photoelectric Effect
Experiment: shining light on a metal surface may induce it to expel free electrons... which drivesa current. The effect has several interesting features:
• Kinetic energy of escaping electrons is independent of theintensity of light.
• There’s a threshold frequency of light, below which noelectrons flow.
• There’s no time lag. If electron flow is allowed by a highenough frequency, the process starts immediately when thelight starts shining.
In 1905, Einstein realized that Planck’s conjecture (discrete photons with E = hν) was the perfectfit:
For escaping electrons, (12mv2)max =
hν − Φ , if hν ≥ Φ0 , if hν < Φ
where the “work function Φ is unique to a given type of metal. Φ = the minimum energy anelectron must receive to be able to escape.
7.17
What about the lack of any time lag?
Classical E&M theory says that the energy “delivered” to a metal surface (of area A), over sometime interval ∆t, is just
E = |〈S〉| A ∆t
[
Units:
(J
m2 s
)
m2 s = J
]
and the time-averaged Poynting flux 〈S〉 depends only on amplitude of waves, not ν or λ.
Thus, to exceed a given metal’s unique value of Φ, a classical E&M wave must shine on thesurface for a sufficiently long ∆t, in order to build up enough energy.
Experimenters turned down the beam so the critical value of ∆t ought to be minutes → hours...but the photoelectric effect always started immediately!
The “quantum” perspective makes sense: even with low power, there are always a few photonshitting the metal, and if each one has the hν energy to release one electron, it happensimmediately.
7.18
(b) Compton Scattering of Light
Experiment: light beams are scattered into new directions when they pass through transulcentmaterial containing free electrons.
Classical E&M says that an incoming wave induces a charged particle to oscillate, and thatmotion emits new outgoing waves in all directions, with νincoming = νoutgoing.
7.19
(b) Compton Scattering of Light
Experiment: light beams are scattered into new directions when they pass through transulcentmaterial containing free electrons.
Classical E&M says that an incoming wave induces a charged particle to oscillate, and thatmotion emits new outgoing waves in all directions, with νincoming = νoutgoing.
However, Compton tried this with X-rays (very high ν), anddetected scattered beams with νincoming 6= νoutgoing.
The only explanation that works is that there must beindividual photons acting like “billiard balls’ as they collidewith free electrons, which recoil...
pincoming = poutgoing + precoil e
Eincoming = Eoutgoing + Erecoil e
For a photon, E = hν. To compute its momentum, recall fromrelativity that E2 = p2c2+m2c4. Form = 0, p = E/c = hν/c.
For a photon scattered by angle θ, this model predicts
λoutgoing = λincoming +h
mec(1− cos θ) as was observed by Compton.
7.20
(c) Double-slit Experiments
First, back up 1 century. Around 1800, Thomas Young showed that light indeed behaves like awave when shining through two narrow slits...
To see the interference fringes, the slits must be similar in width as the wavelength of light.
7.21
Once the “discrete” nature of light was realized in the early 20th century,people started to wonder what would happen if one turned down theintensity, such that one photon at a time was sent through the system.
Conjecture: since any given photon can go through either slit 1 or slit 2,there’s nothing to interfere with. Thus, light should just “pile up” in twopeaks, one under each slit.
7.22
Once the “discrete” nature of light was realized in the early 20th century,people started to wonder what would happen if one turned down theintensity, such that one photon at a time was sent through the system.
Conjecture: since any given photon can go through either slit 1 or slit 2,there’s nothing to interfere with. Thus, light should just “pile up” in twopeaks, one under each slit.
Reality: interference fringes eventually appear, though it takes longer forthe image to build up, one photon at a time.
In 1909, G. I. Taylor demonstrated this with photographic film & dimmed(“enfeebled”) candle-light.
In the 1980s, Hamamatsu et al. demonstrated it with true single-photonemitters & detectors. −→ −→
What does this mean?!?
7.23
The fact that single photons show interference patterns (after enough are collected)implies that the individual photons must also simultaneously pass through BOTH slits and theninterfere with themselves!
• We’re now forced to abandon the idea that photons are localized particles, in the classicalsense of having a well-defined (even small) size. A localized particle could not pass throughboth slits at the same time and then interfere with itself.
• We’re also forced to abandon the idea that photons take a particular path from one point toanother. In reality, one can say that a photon simultaneously “takes all possible paths” fromone point to another. Only after all of the calculations are done, for all possible paths—andthe results for the different paths are combined—does the final result look like the classicalidea of a light ray traveling from one point to the next by a single path.
7.24
The fact that single photons show interference patterns (after enough are collected)implies that the individual photons must also simultaneously pass through BOTH slits and theninterfere with themselves!
• We’re now forced to abandon the idea that photons are localized particles, in the classicalsense of having a well-defined (even small) size. A localized particle could not pass throughboth slits at the same time and then interfere with itself.
• We’re also forced to abandon the idea that photons take a particular path from one point toanother. In reality, one can say that a photon simultaneously “takes all possible paths” fromone point to another. Only after all of the calculations are done, for all possible paths—andthe results for the different paths are combined—does the final result look like the classicalidea of a light ray traveling from one point to the next by a single path.
To make matters even more mysterious... material particles (such as electrons, atoms, &molecules) also display the same interference behavior as light in a double-slit apparatus.Giorgio Merli et al. were one of the first groups to do this with electrons, in the 1970s...
— content from Curtis Mobley, http://www.oceanopticsbook.info/
7.25
(2) Matter is sometimes “wave-like”
This brings us to the other huge revolution at the start of the quantum era.
We’ll eventually get to “free” electrons that can undergo wave-like interference & diffraction.However, we’ll start (historically) with “bound” electrons in atoms.
Like the blackbody story, precise observations came first, then mathematical attempts to fit them,then physics that suggests discreteness where we didn’t expect it!
In the mid-1800s, scientists realized that each elementproduced a unique “fingerprint” pattern of spectral lines.
Johann Balmer realized that pure atomic hydrogen produceslines in the visible spectrum with frequencies following asimple pattern:
νn ∝(
0.25− 1
n2
)
∝(1
22− 1
n2
)
for n = 3, 4, 5, · · ·
These are now called the Balmer lines... n = 3: “Balmer Hα” λ = 656 nm (red)n = 4: “Balmer Hβ” λ = 486 nm (cyan)n = 5: “Balmer Hγ” λ = 434 nm (violet)n→ ∞: the “Balmer break” at λ = 365 nm (UV)
7.26
Later, additional groups of H lines were found... νn ∝(
1
m2− 1
n2
)
for n > m
m = 1 Lyman series (UV)
m = 2 Balmer series (visible)
m = 3 Paschen series (near IR)
m = 4 Brackett series (far IR)
m = 5 Pfund series (far IR)
m = 6 Humphreys series (far IR)
m = 7 Strong & Hansen series
7.27
In 1913, Niels Bohr postulated some properties of the (newly discovered) nuclear atom:
1. Like planets orbiting a star, electrons orbiting the nucleus atdifferent radii r have different energies E.
2. If an electron moves from an orbit with higher Ei to one withlower Ef , it emits light with
hν = ∆E = Ei − Ef .
3. The units of h are the same as the units of angularmomentum L. Bohr assumed an electron can only occupyorbits at “quantized” values of
|L| =h
2π× 1 , 2 , 3 , 4 , · · ·
or, if we define ~ = h/2π, then
|L| = ~ , 2~ , 3~ , 4~ , · · ·Let’s look at the two-body circular-orbit problem for bothgravitational and electric-field forces.
7.28
Gravitational Electric-field
Total energy (K + U ): E =1
2µv2 − Gm1m2
rE =
1
2µv2 − |q1q2|
(4πε0) r
In circular orbits:1
2µv2 =
Gm1m2
2 r
1
2µv2 =
|q1q2|2 (4πε0) r
so, total energy: E = −Gm1m2
2 rE = − |q1q2|
8πε0 r
Above, we have one equation that gives v as a function of r. If we also posit that
ℓ = µvr = n~ (recalling the “equivalent one-body” definition),
then we’ve got 2 equations for 2 unknowns (v & r). They can be solved for
r =n2 ~2 (4πε0)
µ |q1q2|then plugging in r into... E = − µ q21 q
22
2(4πε0)2n2 ~2.
Then we plug in numbers for hydrogen specifically...1 electron (q1 = −e, m1 = me) + 1 proton (q2 = +e, m2 = mp).
7.29
Since µ ≈ me,
E = − mee4
2(4πε0)2n2~2
= − 13.6 eV
n2
where the numerator is the“Rydberg energy.”
Thus, changes between 2 levelsresult in photons with
hν = Ei − Ef
or
ν =mee
4
4π(4πε0)2~3
(
1
n2f− 1
n2i
)
which is what we observe.
7.30
Notes
• For n = 1, the orbital radius is r = (4πε0)~2/mee
2 ≈ 5.29× 10−11 m. This is called theBohr radius, a fundamental constant that sets the scale for sizes of atoms.
• This also applies to “hydrogen-like” atoms (1 electron, but heavier nuclei), but for preciseanswers, one needs to keep track of the reduced mass µ 6= me.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
In the 1920s, the Bohr model started to be applied in new ways.
7.31
Notes
• For n = 1, the orbital radius is r = (4πε0)~2/mee
2 ≈ 5.29× 10−11 m. This is called theBohr radius, a fundamental constant that sets the scale for sizes of atoms.
• This also applies to “hydrogen-like” atoms (1 electron, but heavier nuclei), but for preciseanswers, one needs to keep track of the reduced mass µ 6= me.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
In the 1920s, the Bohr model started to be applied in new ways.
Louis de Broglie interpreted the angular momentum quantization as a condition for discretestanding waves around the circular orbits...
2πr = nλ
where n = 1, 2, 3, 4, · · ·
7.32
In the light of special relativity, this is the same thing:
Recall E2 = p2c2 +m2c4 , and for light, m = 0 , so E = pc .
Also, E = hν =hc
λ. Thus,
hc
λ= pc =⇒ λ =
h
p.
If this is true for electrons, too, then we can write the angular momentum of an orbiting electron as
ℓ = µvr = pr =hr
λ.
Lastly, de Broglie’s standing-wave model says
r
λ=
n
2π, so ℓ =
nh
2π= n~
the same as in Bohr’s model.
Note that we didn’t actually prove that λ = h/p ought to be true for particles with mass (forwhich E 6= pc). On the next page is a bit more of a “proof” that you can read on your own, fromMcGervey’s Introduction to Modern Physics (1983).
7.33
7.34
De Broglie’s “matter wavelength” (λ = h/p) may have been a bit of a trick to explain electrons inbound orbits, but amazingly, it was soon seen to be true for “free” electrons, as well as othertypes of particles.
• In 1927, Davisson & Germer shot beams of electrons through metal foils... They emerged withwave-like interference fringes!
• Many other experiments followed, culminating in the true electron double-slit experimentdiscussed above, in the 1970s.
7.35
De Broglie’s “matter wavelength” (λ = h/p) may have been a bit of a trick to explain electrons inbound orbits, but amazingly, it was soon seen to be true for “free” electrons, as well as othertypes of particles.
• In 1927, Davisson & Germer shot beams of electrons through metal foils... They emerged withwave-like interference fringes!
• Many other experiments followed, culminating in the true electron double-slit experimentdiscussed above, in the 1970s.
Two big questions remain:
1. For light, the oscillating “stuff’ was E and B. For matter waves, what is it that is oscillating?
2. Waves are sinusoids that either go on forever (like a beam), or are wrapped back on themselves(like a standing wave). Does it make sense to use “wave language” to talk about the physics ofone isolated particle?
We’ll get to the answer to #1 soon. The answer to #2 is yes... we want to describe compactpulses or wave packets. This requires a bit more of a math digression...
7.36
“Wave Packets”
Let’s examine phenomena that occurwhen we start summing together multiplesinusoids with different values of k & ω:
7.37
“Wave Packets”
Let’s examine phenomena that occurwhen we start summing together multiplesinusoids with different values of k & ω:
Choosing one single frequency/wavenumber produces a sinusoidal wave that spreads out toinfinity in time/space.
In many cases, the more frequencies/wavenumbers that are summed, the more “localized” thesignal becomes in time/space. We’ll look at the “beat mode” case in more detail.
7.38
Consider a “packet” consisting of 2 waves of equal amplitude and nearly equal ω & k.
f(x, t) = sin(k1x− ω1t) + sin(k2x− ω2t)
with
ωavg =ω1 + ω2
2≈ ω1 ≈ ω2
∆ω = ω1 − ω2 with |∆ω| ≪ ωavg
and similar definitions for kavg and ∆k.
We can look up a trigonometric identity... sinα + sin β = 2 cos
(α− β
2
)
sin
(α + β
2
)
to get f(x, t) = 2 cos
[(k1 − k2)x− (ω1 − ω2)t
2
]
sin
[(k1 + k2)x− (ω1 + ω2)t
2
]
= 2 cos
[∆k
2x− ∆ω
2t
]
sin[
kavgx− ωavgt]
︸ ︷︷ ︸
“envelope”︸ ︷︷ ︸
∼original wave
7.39
The “original” wave propagates along with a similar phase speed as the 2 components:
Vph =ωavg
kavg≈ ω1
k1≈ ω2
k2
but the envelope moves with what we call the group velocity
Vgr =∆ω
∆kwhich can be 6= Vph .
The envelope describes how the original sine-wave has been “modulated” (i.e., chopped or variedin some way), so the group velocity is what describes how fast true “signals” propagate.
If we take a snapshot in time, we can find the spatial extent ∆x of one “group.”
It’s the extent over which the cosine goes from one zero tothe next zero,
i.e., cos(π/2) at x1 and cos(3π/2) at x2
and thus,
∆k
2x2 − ∆k
2x1 = π so that ∆k ∆x = 2π
7.40
Similarly, we could have “frozen” x and watched the packet pass by as a function of t. The extentof one group ∆t would be related to the frequency difference,
∆ω ∆t = 2π
Thus,
a
largesmall
value of ∆x requires a
smalllarge
spread of wavenumbers
a
largesmall
value of ∆t requires a
smalllarge
spread of frequencies.
Our model of a two-mode beat wave is somewhat artificial. We originally wanted a “compact”packet to represent a wave-like particle... i.e., nonzero only in a restricted range of locations, andzero everywhere else.
We can do better by adding more waves... see the example of a seven-mode beat wave on thenext page.
7.41
Superposition of 7 sinusoids with a range of amplitudes...
Here,∆x∆k ≈ 1
To increase the distance between the packets (and reduce the waves in the space), we’d need toadd even more modes.
7.42
Like at other times in this course, we can shift from discrete sums (“Fourier series”) tocontinuous integrals (“Fourier transforms”).
For a continuous distribution of k values, where theamplitudes of each mode are distributed like a Gaussian,
P (k) ∝ exp
[
−(k − k0)2
2∆k2
]
and ∆k is the standard deviation, the corresponding spatialwave-pattern is
f(x) ∝ exp(ik0x) exp
(
− x2
2∆x2
)
where
∆x∆k = 12 and similarly, in time, ∆t∆ω = 1
2 .
Note that the Gaussian shape is special: 12 is the minimum
value one can ever get. For all other shapes (like our beats),
∆x∆k ≥ 12
and ∆t∆ω ≥ 12
.
7.43
In general, the group velocity is defined as
Vgr =∆ω
∆k(for 2 modes) −→ Vgr =
∂ω
∂k(for ≫ 2 modes)
and we have Vgr = Vph only when
ω = Ck , i.e.,ω
k= C and
∂ω
∂k= C .
That case is called “non-dispersive.”
When Vgr 6= Vph, that requires ω(k) to not be a linear function; these waves are “dispersive.”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.44
In general, the group velocity is defined as
Vgr =∆ω
∆k(for 2 modes) −→ Vgr =
∂ω
∂k(for ≫ 2 modes)
and we have Vgr = Vph only when
ω = Ck , i.e.,ω
k= C and
∂ω
∂k= C .
That case is called “non-dispersive.”
When Vgr 6= Vph, that requires ω(k) to not be a linear function; these waves are “dispersive.”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Back to Physics...
We’d like to represent a “freely moving” particle (i.e., no forces on it) as a wave packet. Let’s saythe particle is moving along the x-axis with velocity v ≪ c. We know
p = mv and E =p2
2m.
However, from de Broglie and Planck, we also know
p =h
λ=
h
2π
2π
λ= ~k and E = hν =
h
2π2πν = ~ω .
7.45
Thus,
E =p2
2mbecomes ω =
~k2
2m.
If this is a packet containing multiple modes, they’re dispersive (with Vgr 6= Vph). What are thesevelocities?
Vgr =∂ω
∂k=
~
2m(2k) =
~k
m=
p
m= v
so the wave packet’s group velocity = the “classical” particle velocity!
What about the phase velocity?
Vph =ω
k=
~k
2m=
p
2m=
v
2? ! ?
From a physics standpoint, Vph doesn’t represent anything “real.”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.46
Thus,
E =p2
2mbecomes ω =
~k2
2m.
If this is a packet containing multiple modes, they’re dispersive (with Vgr 6= Vph). What are thesevelocities?
Vgr =∂ω
∂k=
~
2m(2k) =
~k
m=
p
m= v
so the wave packet’s group velocity = the “classical” particle velocity!
What about the phase velocity?
Vph =ω
k=
~k
2m=
p
2m=
v
2? ! ?
From a physics standpoint, Vph doesn’t represent anything “real.”. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Aside: What about special relativity?
If we used E =√
p2c2 +m2c4 instead of E =p2
2m, we would get:
Vgr = = v X and Vph = =c2
v> c (similarly meaningless).
7.47
We still haven’t answered a key question: For these waves, what is it that’s oscillating?
In the 1920s, de Broglie, Schrodinger, Heisenberg, & others gradually realized the wave“amplitude” is related to the probability of finding a particle at given values of x & t.
Recall: for photons in the double-slit experiment, we measure energy flux (proportional to |E|2).For particles in the double-slit experiment, we detect them individually at different locations, andthe signals pile up.
Thus, if we express the particle wavefunction as ψ(x, t) similarly, the number of particles wefind should be proportional to |ψ|2.
Since ψ is complex, the all-real, positive-definite (physically measurable) “energy-like” quantity isψ∗ψ. And, if this is interpreted as a probability, then it should be normalized:
∫ +∞
−∞dx ψ∗ψ = 1 .
7.48
If these wavefunction packets are real things, they must obey
∆x∆k ≥ 12 and ∆ω∆t ≥ 1
2
but since p = ~k and E = ~ω, these turn into
∆x∆px ≥ ~/2 and ∆E∆t ≥ ~/2
(note that we should really use px = ~kx for the component of motion along the x-axis).
These are Heisenberg’s uncertaintyprinciples, first worked out in 1927.
The ∆’s tell us about the spread ofuncertainty in the probability of findinga particle at given values of x, p, E, or t.
Note that as some of these go down,others must go up.
7.49
Example: “Particle in a Box”
The uncertainty principle gives strange answers at tiny/atomic scales, but it is more reassuring &intuitive on large/human scales.
Consider a particle of mass m inside a cubical box of length L. If we know for sure it’s inside thebox, the uncertainty on its position in x must be
∆x ≤ L .
But Heisenberg says ∆x∆px ≥ ~/2 , so ∆px ≥ ~
2L.
For Gaussian wave packets, ∆px is the standard deviation of a distribution of possible valuesof px one could measure. If the average value of px is zero (symmetry), the most probablevalue of |p| would be ≈ ∆px itself. Thus, a box of size L has minimum values of
|p|min ≈ ~
2Lor also |v|min =
|p|min
m≈ ~
2mLor also Emin =
|p|2min
2m≈ ~
2
8mL2.
Electron in an atom (m = 9× 10−31 kg, L = 5× 10−11 m): Emin ≈ 3 eV (appropriate!)
Marble in your hand (m = 1 gram, L = 10 cm): |v|min ≈ 10−31 m/s (unobservably small!)
The reason we don’t “see” the uncertainty principle in the everyday world is because ~ is so tiny.
7.50
More about the Uncertainty Principle
If ψ = Aeikx, then the particle it describes has one, definitive value of momentum p = ~k.
However, ψ is an infinite sine-wave, and ψ∗ψ is constant everywhere. Thus, the position of theparticle is completely unpredictable.
On the other hand, if the wavefunction is infinitessimally narrow (i.e., ψ 6= 0 at one value of x),that requires superposition of an infinite number of modes... i.e., we’ve lost all informationabout its k (or momentum).
The act of measuring either x or p removes all uncertainty, and will “collapse” the wavefunction.
This isn’t an actual physical process. It’s just a change in our knowledge... we didn’t knowsomething before (i.e., the probability distribution was broad) but now we do know it.
If two particles are close together, theirwavefunctions may overlap. Their propertiesare entangled with one another.
Quantum computers utilize entanglement tosimulate a continuum of states between “0”and “1.”
7.51
More about Quantum “Measurement”
1. At the quantum level, all measurement activities that can give usinformation about a system must also disturb or modify it in someway.
2. If two possible measurement results have equal values of ψ∗ψ, theoutcome of 1 measurement is truly random. (i.e., it seems that“God really does play dice!”)
3. Although ψ itself isn’t measurable, one can “map out” ψ∗ψ by repeatinga measurement many times. Just like rolling many dice to see theprobabilities of different outcomes.
7.52
(3) How do Wavefunctions Evolve? The Schrodinger Equation
We know that ψ(x, t) behaves like a wave packet, so ideally we would like a wave equation totell us how it evolves in space & time.
However, the classical wave equation∂2f
∂t2= V 2 ∂
2f
∂x2doesn’t give us the right physics.
Its solutions are non-dispersive (i.e., a constant V gives Vph = Vgr, but freely traveling particlesobey a dispersive relation,
E =p2
2m⇒ ⇒ ~ω =
~2k2
2m.
7.53
(3) How do Wavefunctions Evolve? The Schrodinger Equation
We know that ψ(x, t) behaves like a wave packet, so ideally we would like a wave equation totell us how it evolves in space & time.
However, the classical wave equation∂2f
∂t2= V 2 ∂
2f
∂x2doesn’t give us the right physics.
Its solutions are non-dispersive (i.e., a constant V gives Vph = Vgr, but freely traveling particlesobey a dispersive relation,
E =p2
2m⇒ ⇒ ~ω =
~2k2
2m.
If we assume a single sinusoid (i.e., localized in p but not in x), let’s figure out what differentialequation gives ω ∝ k2. For ψ(x, t) = ψ0e
i(kx−ωt), we know
∂ψ
∂t= −iωψ ,
∂ψ
∂x= +ikψ ,
∂2ψ
∂x2= −k2ψ
so...
~ω =~2k2
2mis consistent with i~
∂ψ
∂t= − ~
2
2m
∂2ψ
∂x2
and that’s essentially the evolution equation that Schrodinger came up with in the 1920s.
7.54
However, it’s a good idea to generalize that result in 2 ways:
• 1D ⇒ 3D: For a wave traveling along k with ψ ∝ ei(k·r−ωt),
the appropriate spatial derivative goes from∂2
∂x2to ∇2 .
7.55
However, it’s a good idea to generalize that result in 2 ways:
• 1D ⇒ 3D: For a wave traveling along k with ψ ∝ ei(k·r−ωt),
the appropriate spatial derivative goes from∂2
∂x2to ∇2 .
• Non-free particles: If the particle is acted upon by a conservative force (recall F = −∇U ),then its total energy is
E =p2
2m+ U (r)
and we define the “Hamiltonian operator” as H = − ~2
2m∇2 + U
so the full Schrodinger equation is
i~∂ψ
∂t= Hψ i.e., i~
∂ψ
∂t= − ~
2
2m∇2ψ + Uψ .
For non-relativistic motions, this equation describes how particle wavefunctions evolve inspace & time... for any kind of wave packet or sum of multiple modes.
7.56
The last part of our survey of quantum mechanics will examine solutions to the Schrodingerequation for a few example cases:
We’ll illustrate We’ll derivethe solutions the solutions
(a) A “free” Gaussian wave packet X
(b) A particle in a 1D finite box X
(c) A beam of particles striking a barrier X
(d) The hydrogen atom X
(a) A Free Gaussian Wave Packet
For a particle feeling no forces, U = 0. That simplifies the Schrodinger equation, but the solutioninvolves an integral over a continuous spectrum of modes. Math is nasty, so we’ll just show theinitial condition and the time-dependent solution.
At t = 0 , ψ(x, 0) ∝ eik0x e−x2/(2a2) where k0 = p0/~ (initial momentum)
and a is the initial spatial “width” of the packet in the x direction.
7.57
The solution at future times looks like
ψ(x, t) ∝ 1
σ2exp
[
ik0
(
x− p0t
2m
)]
exp
[
−(x− p0t/m)2
2a2 σ
]
where σ = 1 +i~t
ma2.
The centroid of the packet moves with x = p0t/mi.e., x = v0t. (group velocity) X
The crests & troughs of the sinusoidal part travelat a speed of p0/2m = v0/2, which is the same“meaningless” phase velocity we saw earlier.
What’s the effect of σ changing in time? Let’s lookinstead at how the shape of the “probability cloud”evolves...
ψ∗ψ (x, t) =1
ξ√πexp
[
−(x− p0t/m)2
ξ2
]
where ξ(t) is the spatial “width” of the packet as it evolves. It spreads over time:
ξ = a√σ∗σ = a
√
1 +~2t2
m2a4=
√
a2 +~2t2
m2a2=√
a2 + a2Heisenberg .
7.58
The “Heisenberg” term in ξ comes from the uncertainty in momentum (i.e., H’s uncertaintyprinciple). Recall
∆x∆p ∼ ~ so that ∆p ∼ ~
∆x∼ ~
a.
If two classical particles started out together at t = 0, but with momenta that differed by ∆p,after time t their x positions would differ by
∆v t =∆p
mt =
~t
ma= aHeisenberg .
How long does it take for the spreading to double? (i.e., at what time does ξ = 2a)?
Solving for that time, one gets t ≈ ma2
~.
Examples:
Electron in an atom (m = 9× 10−31 kg, a = 5× 10−11 m): t ≈ 10−17 seconds.On quantum scales, a free particle’s uncertainty spreads rapidly!
Grain of sand (m = 10−5 kg, a = 1 mm): t ≈ 1023 seconds ≫ age of the universe!
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.59
(b) A Particle in a 1D Finite Box
Despite being somewhat artificial, this problem illustrates someinteresting things that happen to all “confined” quantum systems.
Here, U (x) =
+∞ , x ≤ 00 , 0 < x < L+∞ , x ≥ L
Outside the box, setting U to infinity is a sure-fire way of demanding the particle always remaininside – i.e., it can never have enough energy to “climb out of the potential well.”
Mathematically, this provides boundary conditions for ψ(x, t) of 0 at x = 0 and x = L(and also everywhere else outside the box).
We can also assume the particle in the box has one specific value for its energy E. Since E = ~ω,this assumes there’s just one frequency in the solution. This means we can write it as
ψ(x, t) = Ψ(x) e−iωt (“separation of variables”)
and our goal is now only to solve for the spatial variation Ψ(x).
7.60
With this, the Schrodinger equation turns into an ordinary differential equation. In 1D,
i~∂ψ
∂t= − ~
2
2m
∂2ψ
∂x2+ Uψ
ւ ց
LHS = (i~)(−iω)ψ = ~ωψ = Eψ RHS = − ~2
2m
∂2ψ
∂x2(since U = 0 inside).
There’s a factor of e−iωt on both sides. If we cancel it, our “time-independent Schrodingerequation” is
EΨ(x) = − ~2
2m
d2Ψ
dx2=⇒ d2Ψ
dx2+ K2Ψ = 0
where K2 =2mE
~2and we’ve seen before that the solutions are sinusoids:
Ψ(x) =
A sinKxB cosKx
but our x = 0 boundary condition demands B = 0.
7.61
Thus, we can only have the sine solutions... but only when sinKx = 0 at x = L.This demands only a discrete collection of solutions...
or their sum...
Ψ(x) =
∞∑
n=1
An sinKnx
where
Kn =nπ
L
like standing waves in a cavity:
7.62
The boundary conditions have quantized our system, and only certain discrete values of theenergy are allowed:
En =~2
2mK2n =
~2
2m
n2π2
L2=
(~2π2
2mL2
)
n2 where n = 1, 2, 3, 4, · · ·
reminiscent of energy levels in an atom!
Note also the lowest energy level (n = 1) is not zero, similar to our results from thinking aboutthe uncertainty principle. A finite-sized system has a finite “ground-state” energy.
7.63
The boundary conditions have quantized our system, and only certain discrete values of theenergy are allowed:
En =~2
2mK2n =
~2
2m
n2π2
L2=
(~2π2
2mL2
)
n2 where n = 1, 2, 3, 4, · · ·
reminiscent of energy levels in an atom!
Note also the lowest energy level (n = 1) is not zero, similar to our results from thinking aboutthe uncertainty principle. A finite-sized system has a finite “ground-state” energy.
It’s useful to compare to the classical solution for a particle bouncing around in a 1D box.
If there’s no force on it, and it has some initial px = mvx, it just bounces back and forth betweenthe walls. The normalized “probability density” of finding it at any place between 0 and L is
P(x) =1
Lso that ∫ L
0
dxP(x) = 1
equal to the average of the quantum probability Ψ∗Ψ.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.64
(c) A Beam of Particles Striking a Barrier
Consider a finite-valued potential that is non-zero everywhereexcept between 0 < x < L.
If a free particle travels left to right, what happens when it reachesx = 0? It sees a sharp gradient (∇U ), so it feels a force!
Classically, the result depends on the value of E = 12mv2:
• If E < U0, the particle is bounced back to the left.
• If E > U0, the particle keeps going to the right, but is slowed down to
v =√
2(E − U0)/m
then at x = L, it is accelerated back up to its original speed.
7.65
(c) A Beam of Particles Striking a Barrier
Consider a finite-valued potential that is non-zero everywhereexcept between 0 < x < L.
If a free particle travels left to right, what happens when it reachesx = 0? It sees a sharp gradient (∇U ), so it feels a force!
Classically, the result depends on the value of E = 12mv2:
• If E < U0, the particle is bounced back to the left.
• If E > U0, the particle keeps going to the right, but is slowed down to
v =√
2(E − U0)/m
then at x = L, it is accelerated back up to its original speed.
For quantum particles, an incoming “beam” is represented by Ψ(x) ∝ eikx.
When computing the proper solution to the Schrodinger equation, it turns out that even forE < U0, there can be Ψ 6= 0 for x > 0.
Quantum particles always have some probability to undergo “tunneling” through barriers thatclassical physics says are impassable.
7.66
As the potential barrier gets higher (U0 ↑) or the particle energy gets lower (E ↓), theprobability of getting through is lower...
We’ll see in a few weeks how George Gamow used the probability of getting through (which goesroughly as e−U0/E) to learn some fundamental things about nuclear fusion & fission.
Also, as the potential barrier gets thicker (L ↑), the probability of getting through is lower...
Barriers as small as a particle’s own de Broglie wavelength are relatively “easy” to getthrough. Thicker ones... not so much.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.67
(d) The Hydrogen Atom
This is kind of the “crown jewel” of mathematical solutions to the Schrodinger equation.So much modern-day atomic physics rests on it.
As before, let’s treat the two-body H atom as an “equivalent one-body” system... an electron withreduced mass µ and kinetic energy p2/2µ, moving in a spherically symmetric electrostaticpotential
U (r) =1
4πε0
q1q2r
= − Ze2
4πε0 rand we generalize to hydrogen-like ions, with positive charge Ze in the nucleus.
Assuming constant total energy E = ~ω, the time-independent Schrodinger equation is
EΨ = −~2
2µ∇2Ψ + UΨ
which we rearrange as ∇2Ψ =2µ
~2(U − E)Ψ and write in spherical coordinates as:
1
r2∂
∂r
(
r2∂Ψ
∂r
)
+1
r2
[1
sin θ
∂
∂θ
(
sin θ∂Ψ
∂θ
)
+1
sin2 θ
∂2Ψ
∂φ2
]
=2µ
~2(U − E)Ψ .
7.68
The key step is to use the “separation of variables” trick to assume Ψ is the product of 3 functionsof each spatial variable:
Ψ(r, θ, φ) = R(r) Θ(θ) Φ(φ) .
If we substitute this back into the Schrodinger equation, we can separate it into 3 ordinarydifferential equations.
(Aside: How do we know separation of variables will work? Can we assure all solutions are of theabove form? In general, the answer to both is “we can’t!” But for problems like this that involve∇2Ψ, this trick does give us a very general set of functions that can be summed together to formALL possible solutions!)
Here’s what we get by substituting in Ψ = RΘΦ:
ΘΦ
r2d
dr
(
r2dR
dr
)
+RΦ
r2 sin θ
d
dθ
(
sin θdΘ
dθ
)
+RΘ
r2 sin2 θ
d2Φ
dφ2=
2µ
~2(U − E)RΘΦ
and notice that we can change all ∂ to d. If we divide every term by Ψ/r2 and rearrange a bit,
1
R
d
dr
(
r2dR
dr
)
+2µr2
~2[E − U (r)] = − 1
Θ sin θ
d
dθ
(
sin θdΘ
dθ
)
− 1
Φ sin2 θ
d2Φ
dφ2
where the LHS = function of r only, RHS = function of θ & φ only.
7.69
If the LHS is a function of r only, and the RHS is a function of θ & φ only... then:
• Any variations in r cannot change the RHS, and
• Any variations in θ or φ cannot change the LHS.
Thus, the only way for LHS=RHS (in general) is for both sides to be equal to the sameconstant. With some foresight, we call that constant ℓ(ℓ + 1).
Preview: ℓ is one of three integer-valued “quantum numbers” that will characterize solutions forhydrogen. The 1D potential well has one number like this (n = 1, 2, 3, · · · ) and 3D potentials likethis have three (n, ℓ, m).
The “angular equation” is
ℓ(ℓ + 1) = − 1
Θ sin θ
d
dθ
(dΘ
dθ
)
− 1
Φ sin2 θ
d2Φ
dφ2
and we can rearrange it as1
Φ
d2Φ
dφ2= −ℓ(ℓ + 1) sin2 θ − sin θ
Θ
d
dθ
(
sin θdΘ
dθ
)
.
Once again, we see that the LHS is a function of φ only, and the RHS is a function of θ only.Thus, both sides must be equal to another constant, which we will call −m2.
7.70
The left side of the above equation is then
1
Φ
d2Φ
dφ2= −m2 i.e.,
d2Φ
dφ2+ m2Φ = 0
which we’ve seen before. Solutions are sinusoids: Φ(φ) ∝ eimφ.
Because φ is an angle, we require Φ(φ) to always be equal to Φ(φ + 2π).
Like de Broglie’s standing waves, this requires m = 0 , ±1 , ±2 , ±3 , ±4 , · · · .
The RHS of the angular equation is an O.D.E. for Θ(θ) that contains 2 parameters: ℓ and m.
We won’t go through the process of solving that equation in detail. The solutions are well-knownto mathematicians as Pm
ℓ (θ), the associated Legendre polynomials.
Symmetry for θ demands that ℓ is a positive integer (or zero), and that each value of ℓ restrictsthe available values of m:
ℓ = 0 m = 0 onlyℓ = 1 m = −1, 0,+1ℓ = 2 m = −2,−1, 0,+1,+2ℓ = 3 m = −3,−2,−1, 0,+1,+2,+3general ℓ −ℓ ≤ m ≤ +ℓ
7.71
Illustration of the associated Legendre polynomials:
7.72
The combined angular solutions Y mℓ (θ, φ) are called spherical harmonics, and they’re used in
many fields of study to decompose complex functions on a sphere into their component“multipoles:”
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Lastly, we need to solve the radial component of the Schrodinger equation...
E R(r) = − ~2
2µr2d
dr
(
r2dR
dr
)
+ℓ(ℓ + 1)~2
2µr2R(r) + U (r)R(r)
and it was written this way to bring out the commonalities with the classical one-body energy...
E =1
2µr2 +
L2
2µr2+ U (r)
especially since we’ve rewritten the particle momentum as r =prµ
=~krµ
= −i~µ
d
dr.
This commonality means that this system has the following value of the angular momentummagnitude:
L = ~
√
ℓ(ℓ + 1)
and the condition ℓ = integer means that angular momentum isquantized, like in the Bohr model.
One can also show that the z-component of the angular momentumis also quantized, with Lz = m~ .
Putting both quantizations together puts a constraint on the possibledirections of the atom’s vector angular momentum. −→
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The radial equation is solved by... surprise, surprise... a set of esoteric functions tabulated bymathematicians: the Laguerre polynomials. There’s a discrete set of them, because the“potential well” in the r direction has a finite extent.
Like the 1D “box,” these solutions correspond to quantized values of the energy E, and there’s aset of principal quantum numbers n. Summing up:
n = 1 , 2 , 3 , . . .ℓ = 0 , 1 , 2 , . . . (n− 1)m = −ℓ , −ℓ + 1 , . . . − 1 , 0 , +1 , . . . + ℓ .
Each choice of n, ℓ,m produces a unique Ψ(r, θ, φ), and there’s a unique value of E that goesalong with it. In our idealized hydrogen atom, E depends only on n, with
E = −(
µZ2e4
2(4πε0)2~2
)1
n2same as in the Bohr model!
In the radial direction, the probability density Ψ∗Ψ usually has one dominant peak. This tends tooccur occur near the radii predicted in the Bohr model.
(In this case, however, there is always a chance of finding an electron anywhere....)
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Illustrations of the Laguerre functions & 3D snapshots of Ψ∗Ψ:
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Last quantum topic: Multi-electron atoms
Many of the insights about energy levels (n) & discrete angular momentum states (ℓ,m) extendsimilarly to more complex atoms, too.
However, there are additional quantum numbers that describe electron spin. One can think of itlike another vector component of the
total angular momentum vector: J = L + S
The number of m values depends on ℓ. The number of possible ms values depends on s, which isis always 1
2. Thus, ms = −12 or +
12.
Electrons can “spin” either parallel, or anti-parallel to the z-axis defined by ℓ & m.
Multi-electron atoms are governed by Pauli’s exclusion principle: “No 2 electrons in an
atom can have the exact same set of quantum numbers n, ℓ,m,ms.”Thus, each primary electron “shell” (i.e., configuration of electrons of specific n & ℓ) can contain,at most:
(for ℓ = 0, “s”) 2 electrons i.e., 1 value of m × 2 values of ms
(for ℓ = 1, “p”) 6 electrons i.e., 3 values of m × 2 values of ms
(for ℓ = 2, “d”) 10 electrons i.e., 5 values of m × 2 values of ms
(for ℓ = 3, “f”) 14 electrons i.e., 7 values of m × 2 values of ms
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Pauli’s exclusion principle thus gives us the structure of the periodic table of elements...
The hydrogen-atom model for E depends onlyon n. In real atoms, E depends mostly on n,but also a bit on ℓ.
In chemistry, the Aufbau rule (German:“building-up”) says that as more electrons areadded to a nucleus, they fill up in order ofincreasing energy...
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