Download Hydrogen and the Central Force Problem

Hydrogen and the Central Force Problem
Our treatment of quantum mechanics is getting increasingly more realistic.
In the last chapter, we made the leap from 1 to 3 dimensions. In this chapter
we solve our first problem with a physically realistic potential– the attractive
electrostatic (Coulomb) potential between two oppositely charged particles. In
the case of atomic hydrogen these particles are an electron and a proton, but the
theory can be easily extended to one electron ions such as Li++ , muonic atoms
where a muon is captured in orbit about a proton, or positronium (an electron
in orbit about an antielectron). In this chapter we will ignore the complications
of electron spin.
Heretofore we have been discussing the motion of a single particle in an
“external” potential of mysterious origin. Because the proton is about 2000 times
more massive than the electron and doesn’t move very much, we could think of
it as providing an external potential to the “orbiting” electron and still get fairly
accurate results. However positronium gives us a great excuse to discuss two
particle wave functions which interact in a potential mutually created by both
particles.
The Two Particle Reduction
Lets consider two particles in one dimension. The first particle has mass m1
and is located at x1 ; the second has mass m2 and is located at x2 . The wave function becomes a function of both x1 and x2 : ψ(x1, x2 ). It is natural to generalize
the probabilistic interpretation of the wave function PDF(x1 , x2) = |ψ(x1, x2 )|2 ,
where PDF(x1 , x2 ) gives the probability density of finding particle 1 at position
x1 and simultaneously finding particle 2 at position x2 . To clarify this last
point, the probability of finding particle 2 within a distance ∆ of particle 1 would
be written as:
+∞
P(|x1 − x2 | < ∆) =
x
1 +∆
dx1
−∞
x1 −∆
1
dx2 ψ ∗ (x1 , x2)ψ(x1 , x2 )
The wave function ψ(x1 , x2 ) obeys a Schrödinger equation which is built on
the total energy of the two particle system: E = p21 /(2m1 )+p21 /(2m1 )+V (x1 , x2 ).
−h̄2 ∂ 2
−h̄2 ∂ 2
ψ(x
,
x
)
+
ψ(x1 , x2 ) + V (x1 , x2 ) ψ(x1 , x2 ) = E ψ(x1 , x2 ) (1)
1
2
2m1 ∂x21
2m2 ∂x22
We thus have a partial differential equation which is traditionally solved
using separation of variables. The separation of variables would be trivial if
V (x1 , x2 ) = V1 (x1 ) + V2 (x2 ) since we just have a diffiQ in x1 added to a diffiQ in
x2 . This would be the situation if particle 1 and 2 moved in external potentials
but felt no mutual interaction. In the case of positronium, we usually have
the opposite problem of zero external potential but the electron and positron
(antielectron) attract each other with a potential related to their separation.
The one dimensional analogy would be V (x1 − x2 ). It is very difficult to separate
variables for this potential in the original (x1 , x2 ) coordinates but there is a classic
change of variables trick which allows for a separation of variables solutions.
The clue comes from classical mechanics. In many classical situations motion
about center of mass is essentially decoupled from motion of the center of mass.
An example would be if one were to throw a system consising of two masses
coupled with a spring from the surface of the moon. The masses will oscillate
following simple harmonic motion about their center of gravity; while the center
of mass follows the same parabolic trajectory as a point like particle in the gravitational field of the moon. Guided by the classical problem, we tranform the
variables of Eq. (1) into the position of the center of mass X and the position of
x1 relative to x2 :
X=
m1
m2
x1 +
x2 where M = m1 + m2
M
M
x = x1 − x2
2
(2)
We next transform the derivatives using the chain rule.
∂X ∂
m1 ∂
∂
∂x ∂
∂
∂
=
=
+
,
+
∂x1
∂x1 ∂X ∂x1 ∂x
∂x1
M ∂X
∂x
∂
∂X ∂
m2 ∂
∂x ∂
∂
∂
=
=
+
,
−
∂x2
∂x2 ∂X ∂x2 ∂x
∂x2
M ∂X
∂x
(3)
The kinetic energy operator Ť = p21 /(2m1 ) + p22 /(2m2 ) can be written in terms
of the derivatives of Eq. (3)
−h̄2
Ť =
2m1
−h̄2
+
2m2
m1 ∂
∂
+
M ∂X
∂x
∂
m2 ∂
−
M ∂X ∂x
m1 ∂
∂
+
M ∂X
∂x
∂
m2 ∂
−
M ∂X ∂x
(4)
We note that the cross terms proportional to ∂ 2 /(∂x∂X) cancel leaving the terms:
−h̄2
Ť =
2
Ť =
m1 + m2
M2
∂2
−h̄2
+
∂X 2
2
1
1
+
m1 m2
∂2
∂x2
1
−h̄2 ∂ 2
1
1
−h̄2 ∂ 2
=
+
where
+
2
2
2M ∂X
2µ ∂x
µ
m1 m2
(5)
We see that the kinetic energy has a contribution from the center of mass motion
X which enters with a mass M = m1 + m2 and a contribution from the relative
−1 −1
.
motion x = x1 − x2 which enters with an “reduced mass” µ = m−1
1 + m2
Lets now assume a potential of the form V (x1 , x2 ) = Vr (x1 − x2 ) + Vcg (X)
which might be applicable, for example, to two masses connected with a spring
which experience a potential Vr (x1 − x2 ) = κ(x1 − x2 )2 /2, as the mass spring
system falls in a gravitational field Vcg (X) = (m1 g x1 + m2 g x2 ) = M g X. The
3
Schrödinger equation reads:
−h̄2 ∂ 2
+ Vr (x)
2µ ∂x2
ψ(x, X) +
r
−h̄2 ∂ 2
+ Vcg (X)
2M ∂X 2
ψ(x, X) = Eψ(x, X)
cg
(6)
We now have a classic separation of variables situation. We factor the wave
function as ψ(x, X) = ψr (x) × ψcg (X). Inserting this into Eq. (6) and dividing
both sides by ψr (x) × ψcg (X) we get two separate Schrödinger equations:
−h̄2 d2
ψr (x) + Vr (x)ψr (x) = Er ψr (x)
2µ dx2
−h̄2 d2
ψcg (X) + Vg (X)ψcg (X) = Ecg ψcg (X) , Er + Ecg = E
2M dX 2
(7)
To gain insight into Eq. (7) lets consider the specific example of two masses
m1 and m2 undergoing simple harmonic motion, while their center of gravity
travels to the right with a total momentum h̄K. If the two masses are tied
together with a spring with spring constant κ, the oscillation frequency will be
−1 −1
.
ω = κ/µ where the “reduced mass” is µ = m−1
1 + m2
The ψcg (X) portion of the wave function which describes the center of gravity
is just a free particle traveling wave solution of the Schrödinger equation of the
form: ψcg (X) = exp (iKX) ; while the ψr (x) portion of the wave function which
describes the relative coordinate x1 −x2 is constructed out of Hermite polynomials
√
and given by ψr (x) = Hn ( β x) exp (−β x2 /2). The complete wave function is:
ψ(x, X) = N Hn (
β x) e−βx
2
/2
eiKX where β =
µω
h̄
(8)
The energy of the system, or E in Eq. (6), is just equal to the sum of Er (the
harmonic oscillator energy) and Ecg (the kinetic energy of the center of mass
4
which acts as a particle of mass M traveling with a momentum h̄K.
E=
1
n+
2
h̄2 K 2
h̄ω +
2M
(9)
A brief summary might be in order. Quite often in reality, we have the
problem of describing a situation where two particles interact with each other
through a potential which depends on their separation. In such cases, one can
affect a separation of variable solution by writing the Schrödinger equation in
terms of the relative displacement of two particles and location of their center
of mass. We’ve shown that the equation describing the center of mass, becomes
totally independent of the equation describing the relative coordinates. The
Schrödinger equation for the relative displacement between the two bodies is
exactly the same as the equation for a single particle of reduced mass
µ interacting in an external potential V (x) = V (x1 − x2 ). Hence the fact
that we have two particles rather than one creates no additional mathematical
complications at all!
The Angular Reduction
Besides the reduction of a two particle system down to a one particle system
with a reduced mass µ, the conservation of angular momentum creates additional
simplifications in the central force problem in both classical mechanics and quantum mechanics. A few words about the classical orbit problem (illustrated below)
wouldn’t hurt.
r
r
V(r)
φ
Perigee
rφ
2
L= µ rφ r = µ r φ
5
Apogee
We have described the elliptical orbit in terms of the polar coordinates r and
φ. The tangential velocity is vφ = r φ̇; while the radial velocity is ṙ.
1
The
energy expression is given by:
1 2
2
E = µ ṙ + (r φ̇) + V (r)
2
(10)
The angular momentum, given by L = µ vφ r = µ r 2 φ̇, is a constant of motion.
We can express φ̇ in terms of L and insert it in Eq. (10) to obtain an expression
in r only:
φ̇ =
L
1 2
L2
,
E
=
+
+ V (r)
µ
ṙ
µr 2
2
2 µr 2
(11)
Eq. (11) is very analogous to the conservation of energy in a one dimensional
potential well: E =
1
2
m ẋ2 + V (x). To complete the analogy we combine the
angular momentum term with the potential actual potential energy to form an
“effective potential”.
L2
=
+ V (r)
2 µr 2
Veff
(12)
The L2 /(2 µr 2 ) term is often called the “centrifugal potential”. We illustrate
radial motion of the orbit for a V = −K/r potential in an effective potential well
below:
V
eff
V
eff
=
-K
r
+
L2
2
2µ r
dominated by centrifugal
potential
Perigee
r
E
Apogee
dominated by
Coulomb potential
1 In classical mechanics, one often writes ṙ to stand for dr/dt.
6
The reduced mass satellite, bound in the well with a negative total energy,
oscillates in radius between a closest “perigee” and furthest “apogee” which serve
as classical turning points. The centrifugal potential prevents the satellite from
falling into the force center. As r gets small, vφ must get large so as to conserve
L = m r vφ . When r gets below the perigee, it becomes energetically impossible
to have a large enough vφ to conserve L.
In classical mechanics we used the L constraint and center of mass transformation to effectively turn the orbit problem from a problem of 6 coordinates
describing both the bodies to just 1 coordinate (the radius between the bodies)
via an effective potential.
The angular reduction in quantum mechanics
We have workesd out the Hamiltonian in spherical coordinates in the chapter
on Dimensions.
−h̄2
Ȟ =
2µ
1 ∂
r 2 ∂r
r
2∂
∂r
+
radial
Ľ2
+ V (r)
2µr 2
.
(13)
eff pot
Because of rotation symmetry, we know that the wave functions can be chosen
to be eigenstates of both Ľ2 and Lz . As disussed in the Dimension chapter, the
wave function is of the form:
ψ = Rn (r) Y m (θ, φ)
(14)
Inserting the Eq. (14) wave function into Eq. (13), using the fact that Ľ2 Y m =
( + 1)h̄2 Y m and dividing both sides of the resulting equation by Y m we have
a differential equation for the radial wavefunction only:
−h̄2 1 d
2µ r 2 dr
2d
R
r
dr
h̄2 ( + 1)
R + V (r) R = E R
+
2µr 2
(15)
Eq. (15) simplifies considerably if we re-cast Eq. (15) in terms of a function u(r)
7
where u(r) = rR(r):
−h̄2 d2 u
+
2µ dr 2
h̄2 ( + 1)
u(r)
Y m (θ, φ) (16)
+ V (r) u = E u where ψ =
2
2µr
r
We can think of Eq. (16) as the Holy Grail of the radially symmetric two body
problem in quantum mechanics! We note that it is of the exact form as the one
dimensional Schrödinger equation with x substituted for r and with an effective
potential which includes a centrifugal barrier term. In the chapter on Bound
State States in One Dimension we built up considerable insight into the
solutions of such problems. We do have a novel boundary condition however.
Since we do not generally wish the wave function to become singular at the
origin and ψ
OC
u(r)/r we require that u(r) vanish at the origin.
The below figure shows the usual situation for a binding effective potential
well. We illustrate it for the case of a hydrogen atom potential. At small enough
radius, the centrifugal potential will generally dominate over the interaction potential as long as = 0.
V eff (r)
large l
r
E
small l
l=0
As increases, the centrifugal potential becomes increasingly more effective
at keeping the two interacting bodies apart by moving the classical turning point
to larger values. We expect therefore that the PDF(r) for high orbitals will
8
peak further from the origin than small orbitals. In fact this is easy to see
formally if r is sufficiently small such that ( + 1)/(2µr2 ) |V (r)| or |E|. In
such a case we can neglect V (r) and E in Eq. (16) and cancel out h̄2 /(2µ):
( + 1)
d2 u
=
u
dr 2
r2
(17)
The solution of Eq. (17) is a simple power law of the form u(r) = rp . Substituting
this form into Eq. (17) we have:
p(p − 1)r p−2 = ( + 1) r p−2 → p = + 1 or − The only non-singular solution is therefore:
(18)
2
Near origin u(r) = r+1 → R(r) =
u(r)
= r
r
(19)
The below sketch compares various R(r) = r in the low r region to try to
convince you that the high orbitals are pulled further from the origin.
R(r)
r0
r1
r2
r3
r
Of course, the radial wavefunction cannot be of the form R(r) = r at large
r or it would blow up as r → ∞. At sufficiently large r, the centrifugal potential
becomes unimportant and the R(r) → 0 in a way that depends on the interaction
potential V (r).
2 Technically this should only hold for = 0 but it works for = 0 as well
9
The Coulomb Potential
Let us consider the general one ion atom with a nuclear charge of +Ze an
electron charge of −e. We will use the symbol e to represent the absolute value
of the charge of an electron: e = 1.6 × 1019 C. The atomic nucleus consists of Z
protons each of which has a charge of +e. The attractive potential is given by
the Coulomb potential:
V (r) =
−Ze2
4π
o r
and the relevant radial equation becomes:
2 2
2
2
h̄ ( + 1)
Ze
−h̄ d u
u=E u
+
−
2
2
2µ dr
2µr
4π
o r
(20)
(21)
It would take us a bit far a field to solve Eq. (21) in general. The solutions
are of the form of a polynomial in r (called an associated Laguerre polynomial)
multiplied by a decaying exponential in r. The exponential fall off of the Coulomb
wave function is much slower than the exp (−βr2 /2) fall off for the harmonic
oscillator. This makes a great deal of sense since the harmonic well becomes
much more confining as r increases while the Coulomb potential approaches zero
at large r. In the exercises I ask you to solve for the ground state energy and β
by assuming a ground state wave function of the form u(r) = r exp (−βr) with
= 0.
The maximum power of r appearing in the Laguerre polynomial is equal
to the principle quantum number n. Interestingly enough the energy of the
orbital depends only on n and is exactly the same as predicted by Bohr using
the Correspondence Principle, namely:
En =
−Z 2 e4 µ
2(4π
0h̄)2 n2
(22)
It is quite remarkable that the eigenvalues of Eq. (21) are independent of . At
one time this was known as the “accidental” degeneracy. In fact there is a rather
subtle symmetry which leads to this degeneracy.
10
A very analogous degeneracy occurs for classical orbits:
Degenerate classical orbits with
E=-
κ
d
d
High L
Medium L
Low L
Circular orbits have the maximum possible orbital momentum for a given
orbital energy. Orbits of the same energy but with very small angular momentum
( very little tangential velocity at the apogee) collapse into highly eccentric,
thin ellipses. The figure to the right shows a high L (circular orbit) and low
L (elliptical orbit) in relationship to the force center origin (at the focus of the
ellipse). As you can see the low L orbit has a much smaller perigee and spends
much more time near the origin. This is in exact analogy with the quantum
mechanical situation (the r behavior of R(r) as r → 0).
The general solution also shows that the angular momentum quantum number can take on all integral values 0 ≤ < n. This fact coupled with the
observation that − ≤ m ≤ + implies a huge degree of degeneracy. The 9
degenerate orbitals with n = 3 are illustrated below:
11
m = +2
Z
m=+1
l=2
m=0
m= -1
m=+2
m = -2
m=+1
N=3
m = +1
m=0
m=-1
m=0
l=1
m = -1
m=-2
radius=
m=0
l=0
6
In
We also show how the = 2 orbitals have 5 possible orientations for L.
general the degeneracy of orbital n is given by n2 .
Here is a list of some of the hydrogenic wavefunctions:
ψ100
ψ210
1
=√
π
1
= √
4 2π
Z
ao
ψ300
3/2
Z
ao
−Zr/ao
e
3/2
, ψ200
1
= √
4 2π
Z
ao
3/2 Zr −Zr/2ao
1
e
cos θ , ψ21±1 = √
ao
8 π
1
= √
81 3π
Z
ao
Zr
2−
ao
Z
ao
3/2
e−Zr/2ao
Z r −Zr/2ao
e
sin θ e±iφ
ao
3/2 Z 2r2
Zr
+2 2
27 − 18
e−Zr/3ao
ao
ao
The variable ao = 4π
o h̄2 /(µ e2 ) = 0.0529 nm is known as the Bohr radius. The
ratio n2 ao /Z sets the scale for atomic size. In fact one can easily show from the
virial theorem that r−1 = Z/(n2 a0 ) where Z is the number of protons in the
nucleus which is called the atomic number. Incorporation the atomic number
allows us to apply our hydrogen formalism to one electron ions such as Li++
where Z = 3.
Its easy to confirm the expected r behavior as r → 0
12
Hydrogen Wavefunction Plots
It is useful to get a good feeling for the shape of the electron cloud for the
electron in various orbitals. Below is a crude sketch of the electon clouds with
hopefully evocative names.
z
z
z
z
(2 1 1)
(2 1 -1)
donut
(1 0 0)
(2 0 0)
(2 1 0)
marble
tootsie pop
dumbbell
The spinning arrow means that the distribution can be spun about the z axis
because of its cylindrical symmetry.
Another way of viewing the hydrogen atom wave function is in terms of the
radial probability, or the probability density for finding the electron between r
and r + dr. For a spherically symmetric wave function, this probability is given
by:
dP = 4πr 2 dr
∗
ψn00
(r) ψn00 (r) or
dP
∗
(r) ψn00 (r)
= 4πr 2 ψn00
dr
(23)
since a volume element between two spherical shells is dvol = 4πr2 dr. A sketch
of the radial probability is shown below for the first three spherically symmetric
wave functions.
13
4π r2 ψ 2
r
a
d
i
a
l
p
r
o
b
(1 0 0)
4 ao
(2 0 0)
9 ao
ao
(3 0 0)
r
It is interesting to note that there are n − 1 radial nodes in the ψnoo (r) wave
function. These nodes cause the wave PDF to peak up near r → 0. The number
of nodes radial nodes decreases with increasing according to # nodes = n−−1.
Electric Dipole Transitions
The study of the hydrogen spectral lines was a major influence on the development of quantum physics. Schrödinger quantum mechanics as we’ve developed
it thus far, very accurately predicts all the energies of the electron orbitals and
the location of all of the hydrogen spectral lines. Interestingly enough, however,
it does not describe the mechanism for transitions between the various energy levels. In our treatment of quantum mechanics, an electron placed in excited state
(n) has a wave function of the form ψ(x) exp (−iωn t) which produces a time
independent PDF. There is nothing in our description which allows the electron
in an excited state to drop down to its ground state by emitting a photon and
yet typical atomic transitions occur on times scales of ≈ 10−9 → 10−8 seconds.
The solution to this problem is that the quantum mechanics which we have
developed thus far has ignored the interaction of the electron with external electromagnetic fields. These fields are present as “zero point” fluctuations in a way
analogous to the ground state oscillations of a harmonic oscillator. We can’t
begin to do justice to this important topic, which is often called second quan14
tization, in a one semester course on quantum mechanics, but a great deal of
insight can be obtained using a semiclassical argument based on radiation from
an oscillating dipole antenna which we illustrate below:
e
a
S=
θ
4 2
2
e2 ω a
sin θ
2
3
2
ε
32 ο π c r
S
E
Radiation pattern
One gets electromagnetic radiation from an accelerating electric charge. A
field. A constant moving charge has both an E
static charge has just an E
field but no Poynting’s vector or electromagnetic intensity since
field and a B
= E
×H
= 0. 5
S
Once the charge is accelerating, one can get a non-zero
Poynting’s vector which is proportional to the square of the charge × acceleration.
A simple way of getting repetitive acceleration is though the use of an oscillating
charge executing simple harmonic motion (or its equivalent – the dipole antenna
fed with an oscillating current). The above picture shows a charge on a spring
executing SHM of the form a sin (ωt) which has an acceleration of −a ω 2 sin (ωt).
We thus anticipate a radiation intensity which is proportional to (a × e)2 × ω 4
where the charge times the amplitude is called the dipole moment.
As indicated in the figure, the intensity falls off as 1/r2 so there is the same
power subtending any sphere around the dipole. The dipole antenna pattern
shows that the maximum intensity is directed normal to the dipole. The maxi5 It would violate basic relativity principles to get radiation from a charge moving with a
constant velocity, since the absence or presence of radiation would be different for different
inertial observers.
15
mum electrical field is parallel to the dipole. The time averaged power is given
by the expression:
1
Power =
3
e2
4π
o
ω4 2
a
c3
(24)
Dividing the power by the energy of a photon Eγ = h̄ω we have the rate (or
photons/second):
1
R=
3
e2
4π
o h̄c
ω3 2 1
ω 3 a2
α
a
=
c2
3
c2
(25)
where we have used α = e2 /(4π
o h̄c) ≈ 1/137.
But how does an electron in a stationary excited state spontaneously develop
an oscillating dipole moment? Imagine that the excited electron is tickled by the
the radiation field leaving it in a mixed state which is partially excited ψ2 and
partially ground ψ1 .
6
ψ = ψ2 (x) e−iE2 t/h̄ + γ ψ1 (x) e−iE1 t/h̄
(26)
We will write the 3 component dipole operator as d ≡ er = e ( x y z ) The
for the mixed state ψ is given by
expectation value of the dipole operator (d)
d = ψ | e r |ψ which is:
d = ψ2 | e r |ψ2 + γ ∗γ ψ1 | e r |ψ1 + γ e+iωt ψ2 | e r |ψ1 + γ ∗ e−iω t ψ1 | e r |ψ2 d = ψ2 | e r |ψ2 + |γ|2ψ1 | e r |ψ1 + 2Re γ ∗ψ1 | e r |ψ2 e−iω t
(27)
where ω ≡ (E2 −E1 )/h̄. We see that this mixed state will have a (real) dipole moment which oscillates sinusoidally with a frequency of ω and a phase determined
by the relative phase of γ and and the dipole moment connecting the initial and
6 The dynamics of how spontaneous emission happens is beyond the scope of the course, but
books such as Quantum Mechanics by Eugen Merzbacher have complete descriptions.
16
final state or ψ1 | e r |ψ2 . The static dipole moments such as ψ2 | e r |ψ2 would
not contribute to the radiation if they did exist (but in fact we’ll show shortly that
they vanish). The effective oscillation displacement would be a = 2γ ∗ψ1 | r |ψ2 .
Using Eq. (25) and Eq. (27) we anticipate a photon emission rate of:
R=
α ω3 2 α ω3
4α|γ|2 ω 3
2
∗
a
=
4|γ|
ψ
|
r
|ψ
·
ψ
|
r
|ψ
=
|ψ1 | r |ψ2 |2
1
2
1
2
3 c2
3 c2
3
c2
(28)
In order to get an intrinsically real rate which reflects all three dipole components
we replace |ψ1 | r |ψ2 |2 in Eq. (28) by:
ψ1 | r |ψ2 ∗ · ψ1 | r |ψ2 ≡
|ψ1 | x |ψ2 |2 + |ψ1 | y |ψ2 |2 + |ψ1 | z |ψ2 |2
(29)
Of course we haven’t a clue without second quantization theory as to the value
of γ. Interestingly enough γ = 1 meaning that the rate for an excited state
spontaneously decaying to the ground state via an electric dipole transition is
just:
R=
4α 3
ω ψ1 | r |ψ2 ∗ · ψ1 | r |ψ2 2
3c
(30)
It is useful to get Eq. (30) in practical units since we will work some examples
in homework. In Eq. (30), |ψ2 represents the initial (higher energy) state which
we will now write as |ψi and |ψ1 represents the final (lower energy) state which
we will now write as |ψf . We are calculating the number of spontaneous decays
per second of the form |ψf → γ |ψi .
R=
4αω 3
3c2
ψf r |ψi 2 =
4αc
3
h̄ω
h̄c
3
ψf r |ψi 2 =
4αc
3 (h̄c)3
2
∆E 3 ψf r |ψi (31)
Here ∆E = Ef − Ei . Lets work out the constant 4αc/(3h̄3 c3 ) in practical units.
4αc
3 (h̄c)
3
=
4 (1/137)(3 × 108 nm/ns)
3 ( 197eV nm)
3
17
= 0.382 ns−1 nm−2 eV−3
Hence putting it all together we have:
R = 0.382 ns−1 nm−2 eV−3
∆E 3
ψf r |ψi 2
(32)
As an example of a typical electric dipole (sometimes called E1) transition
rate lets consider the hydrogen 210 → 100 transition which (as you can easily
verify) produces a λ = 122 nm photon. As a guess for
|ψ100 | r |ψ210 | we might choose 0.75 a0 = 0.04 nm.
7
The values we need to
insert into Eq. (32) are:
1
∆E = 13.6 1 − 2 = 10.2 eV ; |ψ100 | r |ψ210 |2 ≈ 0.04 nm
2
(33)
where we used the Bohr formula to find ∆E = E210 −E100 . Inserting these values
we get
R = 0.382 ns−1 nm−2 eV−3 (10.2 eV)3 (0.04 nm)2 = 0.65 ns−1 = 6.5 × 108 s−1
(34)
The rate R means we expect 0.65 spontaneous decays per nanosecond from each
excited hydrogen atoms in the |210 state. We will relate this to the lifetime of
the 210 state in the next section.
Transition rates , lifetimes , and line shapes
The meaning of this transition rate might be somewhat unclear at this point.
In the context of a large number N (such as Avagodro’s number) of hydrogen
atoms all making 2 → 1 transitions, the total number of photons per second from
all N atoms would R × N . In the context of a single hydrogen atom in the n = 2
state, the rate R−1 is related to the mean lifetime of the excited state or the
average amount of time that the electron spends in the n = 2 orbital. To see this
7 Of course there is no need to guess since you can calculate the actual integral as I ask you
to do in the exercises.
18
connection, let us imagine starting at time t = 0 with a collection of No atoms in
the n = 2 state and follow how many remain as a function of time. Let i Ri
be the total rate for the atom to leave the n = 2 state and go to a lower orbital.
Let N (t) be the number of hydrogen atoms remaining in the n = 2 orbit after a
given time t. The change in number of atoms which remain at the n = 2 orbital
will be total decay rate i Ri times the time interval ∆t times the number of
atoms presently in the excited state N or:
dN = −N dt
i
Ri
dN
= −dt
or
N
Ri
(35)
i
Integrating both sides of this equation we get:
N (t) = No e−(
i
Ri ) t
(36)
This is the exponential decay law which describes many decay processes such as
radioactive decay.
We can consider applying it to the decay of a single, excited hydrogen atom.
We set No = 1 since we initially have just one excited atom. The probability the
atom will still be excited is then:
P = exp(−t/τ ) where τ = 1
i
Ri
(37)
We can think of τ as the “lifetime” of the excited state which is the inverse of
the total rate that allows the state to de-excite. In our example, all 210 states
decay into 100 state and hence the lifetime of the 210 state is simply τ ≈ 1/R =
1/(0.65 ns−1 ) = 1.5 ns. There is an energy-time Uncertainty Principle which says
short lived states are very uncertain in energy. Hence long-lived excited atomic
states de-excite with narrow spectral lines ; while short-lived excited atomic states
produce very broad spectral lines.
Selection Rules
19
Quite often all three components of an electric dipole matrix element given
by the integral in | < ψf | r |ψi > | vanish because of symmetry. In such cases, the
transition from the initial to final state is forbidden by the electric dipole process.
9
We can formulate a set of selection rules which can be used to determine at a
glance whether or not an electric dipole transition is allowed.
A particularly instructive and easy to derive selection rule says that only
processes with ∆m ≡ mf − mi ∈ 0, ±1 can have electric dipole (E1) transitions.
This is a necessary but insufficient condition which follows from the φ part of the
< ψnf f mf | r |ψni i mi > which we can write as:
2π
Iφ =
−imf φ
dφ e
( cos φ
+imi φ
sin φ
1) e
2π
=
o
dφ e−imf φ ( e
iφ
+e−iφ
2
eiφ −e−iφ
2i
1 ) e+imi φ
o
(38)
In order that all three components vanish for Iφ we must have the condition:
2π
−imf φ
dφ e
0=
±iφ
e
imi φ
e
2π
=
o
dφ e−i(mf −±1−mi )
o
2π
and 0 =
−imf φ
dφ e
imi φ
e
o
2π
=
dφ e−i(mf −mi )
(39)
o
One can easily show that for any integer N
2π
dφ eiN φ = 0 unless N = 0
(40)
o
9 However, there always other, suppressed processes which allow the electron to ultimately
drop to the ground state including transitions involving magnetic dipoles, electric quadrupoles,
and processes where two photons are emitted. There is a suppression hierarchy where each
level is suppressed by an additional factor of ≈ α2 ≈ 10−4 in the decay rate.
20
Hence
< ψnf f mf | x |ψni i mi >=< ψnf f mf | y|ψni i mi >= 0 unless mf − mi = ±1
< ψnf f mf | z |ψni i mi >= 0 unless mf − mi = 0
(41)
A very important set of selection rules for both atomic, nuclear, solid state,
and high energy physics processes involve the concept of parity. We have discussed the idea of even (ψ(x) = ψ(−x)) or odd (ψ(x) = −ψ(−x)) one dimensional
wave functions earlier in the chapter Bound States in One Dimension for
the case of symmetric wells (V (x) = V (−x)). In the case of symmetric potentials, wave functions can be classified according to parity, since parity becomes
a symmetry of the Hamiltonian. A central force problem also has a parity symmetry (or inversion symmetry) defined by r → −r. As you can verify using
r = ( r sin θ cos φ
r sin θ sin φ
r cos θ ), one can change r → −r in spherical
coordinates by : r → r , φ → φ + π , and θ → π − θ. Since the r coordinate
is unaffected, the parity of the wave function is determined by the parity of the
angular wave function , which is the angular wave function of a central force,
means the parity of Y m .
Y 0
o
1
4π
Y1 0
3
4π
cos θ
Y1 ±1
3
∓ 8π
e±iφ sin θ
Y2 ±1
Y2 ±2
15
15
±2iφ sin2 θ
3 cos2 θ − 1 ∓ 8π
e±iφ sin θ cos θ
32π e
Y2 0
5
16π
You can confirm for a few trial spherical harmonics
10
that the parity of a
spherical harmonic is given by parity = (−1) since:
Y m (π − θ, φ + π) = (−1) Y m (θ, φ)
(42)
.
10 Note for odd states with even m, the minus sign comes from the Legendre polynomial in
θ. For odd m, the minus sign comes from the exp (imφ) part.
21
The ith component of the dipole matrix element. < ψnf f mf | ri |ψni i mi >.
will involve a volume element integral over the function Di (r) = ψn∗ f f mf × ri ×
ψni i mi . The parity of Di (r) will essentially be the product of the parities of the
intial state, final state, and ri itself. Since under parity ri → −ri the parity of a
dipole element is just:
Di (r) = (−1)f +i+1
(43)
which will be odd (or negative ) if ∆ is even, and even if ∆ is odd.
Symmetric integrals (such as the integrals from −∞ → +∞ which occur in
quantum mechanics) over odd functions vanish because of symmetry. This is
easy to see in one dimension.
f(x)
f(x)
Even
Odd
x
x
For the case of the odd function the integral over the shaded region cancels
the clear region. For the case of a two dimensional odd function, the argument
is slightly more subtle as illustrated below:
22
For an odd function, such as illustrated on the left, the two clear quadrants
will have the opposite sign and cancel as will the two shaded regions. The same
sort of thing happens in three dimensions where the space can be divided into 8
quadrants and diagonally opposite quadrants cancel for odd functions.
We thus conclude that E1 transitions will vanish unless the initial and final
state have the opposite parity (ie ∆ is odd).
The final selection rule, which I will discuss here, says that E1 transitions
vanish unless ∆ = ±1 which is a consequence of the recurrence formula and
orthogonality of the associated Legendre polynomial which describes the θ parts
of the spherical harmonics. We illustrate the argument for the case of the z dipole
moment, < ψnf f mf | z |ψni i mi >=< ψnf f mf | r cos θ |ψni i mi >. This recurrence
formula relates cos θ times the θ part of ψni i mi to the θ parts of Yim
±1 spherical
harmonics:
m
(2i + 1) cos θ Pim (cos θ) = (i + 1 − m) Pim
+1 (cos θ) + (i + m) Pi −1 (cos θ) (44)
The orthogonality property of the associated Legendra polynomials then says
that integral over ψf × cos θ × Yim
±1 will only survive if the final state contains
Y i m
±1 :
+1
d cos θ Pm (cos θ) × Pm (cos θ) = 0 unless = (45)
−1
This rule implies the following chart for some of the allowed and forbidden E1
transitions for hydrogen.
23
l=0
l=1
l=2
n=3
E1 allowed
n=2
E1 forbidden
n=1
Some allowed and forbidden electric dipole transitions
Photon Spin and the Physics of Selection Rules
The ∆m and ∆ selection rules have a fairly simple physical explanation besides the detailed mathematical explanation for the vanishing of < ψnf f mf | r |ψni i mi >.
The basic idea is that the photon itself carries one unit (h̄) of angular momentum
which is directed either along or against its direction of motion. The two spin
states of the photon correspond to left and right circular polarized light. If a
photon is emitted when an excited hydrogen atom returns to the ground state,
there must be a change of one unit of angular momentum in the atom in order to
allow for the angular momentum carried by the photon.
11
Classical electromag-
netic fields can carry both linear and angular momentum. Below is a cartoon
sketch of the transitions ψ211 → ψ100 and ψ210 → ψ100 .
11 This is a bit oversimplified since, in principle, the photon could have carry orbital momentum
which could balance its spin. In fact in E1 transitions, the photon carries no orbital angular
momentum with respect to the force center.
24
Transitions from
z
ψ
21m
ψ
100
E
m=0
y
x
m=1
By considering the φ integral it is easy to show that the non-vanishing dipole
matrix element for the first transition is < ψ100 | x − iy |ψ211 >. This means that
the photon is radiated from an x and y dipole antenna which being run 90o out of
phase (because of the i coefficient). This will result in circularly polarized light
emitted primarily along the ±ẑ axis (perpendicular to the two dipole directions).
The inital Lz = 1 of the atom in the ψ211 is now carried by the circularly polarized
photon leaving Lz = 0 for the ψ100 final state. Similarly the transition changed
by one and this one unit of angular momentum is carried away by the photon.
We next consider the process ψ210 → ψ100 . The non-vanishing dipole matrix
element for this transition is < ψ100 | z |ψ210 >. This means that the photon is
radiated from an z dipole antennae which emits light primarily in the x-y plane.
The electrical field will tend oscillate along and against the antenna direction
which means that this photon will tend to be plane polarized in the z direction.
In general, a plane polarized photon consists of equal amplitudes for left and right
circularly polarized light. For the ψ210 → ψ100 transition there was no change
in Lz in the atom and the emitted photon carries no Lz . There was however
a change in the total angular momentum ∆ = 1 which represents the angular
momemtum carried by the the photon.
We will show in the next chapter, that a free electron traveling through space
carries intrinsic spin as well.
25
Lasers and Masers
As a practical illustration of some of the concepts presented in this chapter, we
give a brief, thumbnail sketch of the operaton of the LASER (Light Amplification
by Stimulated Emission of Radiation) and MASER (Microwave Amplification by
Stimulated Emission of Radiation). Of course, lasers are ubiquitous in science and
engineering. In the future, minature solid state lasers driving optical fibers might
ultimately replace most wires in electrical circuits. Masers are less common, but
still provide the lowest noise method for amplifying microwaves.
One of the primary virues of a laser which is a light amplifier as opposed
to just a light source, is that it creates or amplifies a nearly perfectly coherent
light beam with a well defined, single , frequency and phase. It is the coherence
property which allows its light beam to be nearly perfectly unidirectional and
monochromatic.
The operant part of the laser/maser acronym is the term stimulated (as oppose to spontaneous) emission. Stimulated emission means that excited atom is
tickled by an external electromagnetic field, rather than the zero point fluctuations of the radiation field responsible for spontaneous emission. Like a high
quality electrical or mechanical resonator, the maximum energy absorption or
emission occurs when the incident radiation is very close to a natural resonant
frequency, which for an atom is ω = ∆E/h̄ ,where ∆E is the energy difference
between two atomic energy levels.
As you probably know, an electrically or mechanically driven oscillator can
either provide or absorb power depending on the phase of the driver relative to
the oscillator. The same is true of the atom, which we illustrate for an atomic
electron making transitions between the ground state and an excited state.
26
E2
n2 A
n2
2 1
n2 B
s
p
o
n
t
e
m
i
t
s
t
i
m
a
b
s
o
r
b
s
t
i
m
2 1 ρ(ω)
e
m
i
t
E1
n1
n1 B 1
2 ρ(ω)
According to the classic analysis of Einstein, there are three processes comprising the dynamics of the two level transition: spontaneous emission, stimulated
emission and stimulated absorption. Since stimulated emission or absorption is
due to the interaction of atomic electrons with the external radiation field, the
transition rate per atom is proportional to the density of photons with the resonant frequency or ρ(ω). The rate per atom for spontaneous emission, given by
Eq. (32) , is of course independent of the external spectral density ρ(ω). Total
rates will be proportional to the rate per atom times the number of atoms at
each level. In terms of the famous Einstein A and B coefficients the total rate for
1 → 2 transitions is given by n1 ρ(ω) B1→2 where n1 is the number of atoms at
level E1 and B is related to the rate for stimulated emission. The rate for 2 → 1
transitions is fed both by spontaneous emission and stimulated emission and is
given by n2 (A2→1 + ρ(ω) B2→1 ) where A2→1 describes spontaneous emission.
All three coefficients B1→2 , B2→1 and A2→1 depend on the matrix elements between the initial and final state and can be computed using the theory of second
quantization.
The name of the game for a laser or maser is to arrange the total rate for
stimulated emission to be greater than the rate for stimulated absorption, since
the photons produced via stimulated emission will be in phase with those from
the external radiation field leading to coherant amplification. Light produced via
27
spontaneous emission will of course contribute to the final radiant output of the
device but in a totally uncorrelated (incoherent) manner.
As early as 1917, Einstein was able to relate spontaneous to stimulated emission and demonstrate that the coefficient for stimulated emission equals the rate
for stimulated absorption using an ingeneous thermodynamic argument. We
will briefly summarize this argument since it gives considerable insight into laser
dynamics. Einstein considered the case where the two level are in thermal equilibrium with the external radiation field. If the two level system is in thermal
equilibrium , the overall rate for 1 → 2 transitions will equal the rate for 2 → 1
transitions:
n1 ρ(ω) B1→2 = n2 (A2→1 + ρ(ω) B2→1 )
(46)
The relative n1 and n2 populations of the two orbitals in equilibrium will be given
by the temperature dependent Boltzman factor.
n2
= e−∆E/KT
n1
(47)
Eq. (46) and Eq. (47) can be combined to give a prediction for the equilibrium
spectral density ρ(ω):
ρ(ω) =
A2→1 /B2→1
B1→2 /B2→1 eh̄ω/(KT ) − 1
(48)
Much earlier, Planck made the first use photon quantization arguments to compute the spectral density for photons in equilibrium with the walls of a “black
cavity” and derived the Planck distribution
2h̄ ω 3
ρ(ω) =
πc3
12
1
eh̄ω/(KT ) − 1
(49)
Einstein argued that spectral density Eq. (48) describing the spectral density for
photons in equilibrium with a collection of atoms must be the same as Planck’s
12 This describes the spectral density of photons found in a hot “black body” such as the
filament of an incandescent bulb. The term “black” means that the surface is as efficient
as possible for radiating or absorbing photons.
28
expression Eq.(49) for photons in equilibrium with the walls in a black cavity.
One consequence is that the rate for stimulated emission equals the rate for
stimulated absorption.
I find this very analogous to driving a black car in the summer. During the
day the car gets hot quickly since the black surface is a good radiation absorber.
At night, the black car cools off quickly since the black surface is a good radiation
emitter.
How can one make an amplifier or strong light source using atomic transitions? The idea is to somhow enhance the total rate for stimulated emission of
light over the rate for stimulated absorption: This cannot be done in thermal
equilibrium since in equilibrium n1 ρ(ω) B1→2 = n2 ρ(ω) B2→1 + n2 A2→1 which
means that stimulated absorption rate actually exceeds the rate for stimulated
emission. Since B2→1 = B1→2 and the spectral density factors cancel , the only
way to meet the condition that stimulated emission rate exceeds the stimulated
absorption rate to achieve a non-equilibrium situation with n2 > n1 which is
called a population inversion. We note that for typical atoms, at reasonable temperatures we usually have the reverse situation where the bulk of the atoms are
in their ground state n1 >> n2 and absorption will dominate over stimulated
emission.
There are many clever ways of achieving the population inversion which insures that stimulated emission will dominate over stimulated absorption. A particularly simple method (called optical pumping) was used in the first laser,
invented in 1960, which is constructed out of a ruby rod. Ruby is a transparent
crystal of consisting primarily of Al2 O2 with a small percentage of chromium
Cr+++ ions which are responsible for the laser properties. A rough schematic of
the atomic transitions responsible for the ruby laser:
29
blue
n2
excitation
n1
1.96 eV
18.7 eV
n2 metastable
green
collisional
1.79 eV
n1
Helium
chromium
Neon
The red appearance of the ruby crystal is due to the fact that Cr+++ strongly
absorbs blue and green light. In a ruby laser a flash tube is used to optically pump
electrons up to the green and blue energy levels and thereby de-populate the n1
level. The excited Cr+++ can then de-excite (through somewhat complicated,
non-radiative transitions which transfer mechanical energy to the ruby crystal)
and feed a metastable state (no E1 transitions) which populates the n2 level
and sets up the population inversion. A few of the n2 atoms will spontaneously
emit photons which will then be coherently amplified since stimulated emission
will occur at a larger rate than stimulated absorption owing to the population
imbalance.
The very common, helium-neon laser, invented in 1967, uses a 4 level optical
pumping scheme and can produce a continuous beam. The helium-neon laser
consists of a roughly 5 to 1 mix of neon to helium in a gas discharge tube. A gas
discharge puts a large number of helium atoms in their first excited state. This
first excited state is nearly degenerate with a particular excited state of neon and
therefore is very likely to excite this level through resonant collisional excitation.
The transition rates are such that the collisionally excited state drops to a state
lying 1.96 eV beneath it, rather than de-exciting to the ground state. Because
the n1 state lies so much above the ground state it is virtually unoccupied initally according to the Bolzman factor. The n1 population is continually depleted
30
through spontaneous emission. The n2 population is continually fed from collisions with the optically pumped helium. We thus have an automatic, highly
efficient population inversion. The lasing action amplifies the familiar 1.96 eV
red photons. Clearly lasers are a fairly inefficient way of generating light, but
they are highly efficient at concentrating light. One can burn holes though metal
plates by concentrating a power equivalent to a 60 Watt light bulb into a very
small area.
Masers are similar to lasers but involve transitions of more like 1/1000 of
an electron volt rather than 2 eV. One of the first masers used the double well
ammonia molecular transition which we discussed extensively in the chapter on
Bound States in One Dimension.
31
Important Points
1. The hydrogen atom consists two particles which interact via a mutual potential of the form V (x1 − x2 ). In close analogy with classical mechanics,
it is possible to decouple the two particle Schrödinger equation by working in relative coordinates x = x1 − x2 and center of mass coordinates
= (m1 x1 + m2 x2 )/(m1 + m2 ). The relative coordinate Schrödinger
X
equation is identical to the Schrödinger equation for a particle of mass
−1 −1
µ = m−1
in an external potential of the form V (x). The re1 + m2
sultant wave function is the product of a wave function in x times a wave
funtion in X.
2. Because of rotational symmetry, the solutions of the central force problem
are eigenstates of angular momentum of the form ψ = R(r) Ym . Guided
by classical physics, one can write a radial Schrödinger Eq. in terms of an
“effective” potential which includes the true interaction potential as well as
a “centrifugal” potential of the form Ľ2 /(2µr 2).
−h̄2 d2 u
+
2µ dr 2
h̄2 ( + 1)
u(r)
Y m (θ, φ)
+ V (r) u = E u where ψ =
2
2µr
r
At short distances, where the centrifugal potential dominates, one can show
that R ∝ r which means large states tend to stay away from the origin
more than low states.
3. For the Coulomb potential, relevant to the one electron ion like the hydrogen atom, the energy levels are described by a principle quantum number
n in exactly the same as the Bohr model:
1
1
−13.6 eV
En = − α2 µc2 2 =
2
n
n2
The radial wave function, on the other hand, depends on both n and .
The fact that the energy levels do not depend on is surprising but has
32
an analog for classical, gravitational orbits. The Coulomb potential wave
function is described by three integer quantum numbers {n, , m} which
obey the relationship < n and the general relation − ≤ m ≤ +. These
rules imply a degeneracy of n2 ,
4. We discussed spontaneous electic dipole emission which describes the dominent process by which excited atoms return to their ground state. The rate
for this process per atom is given by:
R=
4α 3
ω < ψ1 | r |ψ2 >∗ · < ψ1 | r |ψ2 >
3 c2
or R = 0.382 ns−1 eV −3 nm−2 (∆E)3 |ψf |r|ψi |2
5. The rate for spontaneous emission is related to the lifetime of an excited
electron orbital as well as the width of its spectral line according to the
Uncertainty Principle.
τ=
1
i
Ri
, Γ=
h̄
τ
Γ is the full width at half maximum for the excited state’s uncertain energy.
6. We developed several selection rules which describe when electric dipole operator vanishes because of symmetry. The most general rule is that the parity of the initial state must differ from the parity of the final state. The parity is given by (−1) . We also have the rules ∆ = 1 and ∆m = −1, 0, +1.
Often the components of electric dipole matrix element | < ψf | r |ψi > |
are relatively imaginary which implies that their dipole antenna radiate 90o
out of phase. This means that circularly polarized light is emitted from the
atom.
7. Ultimately the selection rules imply that the photon carries one unit of angular momentum which can be directed either along its axis (left circularly
polarized light) or against its axis (right circularly polarized light).
33
8. Finally we discussed the role of stimulated emission in the production or
amplification of coherent light in a laser or maser. Stimulated emission
occurs when an electron is tickled by an external field and emits a photon
which is in phase with the external field. Unfortunately the amplifying
effects for stimulated emission are countered by the reverse process called
stimulated absorption which occurs at the same rate. In order to achieve
laser action, it is necessary to achieve population inversions which means
that there are more electrons at a higher energy orbital than at a lower
energy orbital. This inversion is often accomplished by optical pumping.
34