Quantum Mechanics for Engineers |
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© Leon van Dommelen |
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Subsections
A.39 The relativistic hydrogen atom
The description of the hydrogen atom given earlier in chapter
4.3 is very accurate by engineering standards. However, it
is not exact. This addendum examines various relativistic effects
that were ignored in the analysis.
The approach will be to take the results of chapter 4.3 as
the starting point. Then corrections are applied to them using
perturbation theory as described in addendum {A.38}.
A.39.1 Introduction
According to the description of the hydrogen atom given in chapter
4.3, all energy eigenfunctions with the
same value of have the same energy . Therefore they
should show up as a single line in an experimental line spectrum. But
actually, when these spectra are examined very precisely, the
energy levels for a given value of are found to consist of several
closely spaced lines, rather than a single one. That is called the
hydrogen atom fine structure.
It means that
eigenfunctions that all should have exactly the same energy,
don’t.
To explain why, the solution of chapter 4.3 must be
corrected for a variety of relativistic effects. Before doing so, it
is helpful to express the nonrelativistic energy levels of that
chapter in terms of the rest mass energy
of
the electron, as follows:
|
(A.247) |
The constant is called the “fine structure constant.” It combines the constants
from electromagnetism, from quantum
mechanics, and the speed of light from relativity into one
nondimensional number. It is without doubt the single most important
number in all of physics, [19].
Nobody knows why it has the value that it has. Still, obviously it is
a measurable value, so, following the stated ideas of quantum
mechanics, maybe the universe measured
this value
during its early formation by a process that we may never understand,
(since we do not have other measured values for to deduce any
properties of that process from.) If you have a demonstrably better
explanation, Sweden awaits you.
In any case, for engineering purposes it is a small number, less than
1%. That makes the hydrogen energy levels really small compared to
the rest mass energy of the electron, because they are proportional to
the square of , which is as small as 0.005%. In simple
terms, the electron in hydrogen stays well clear of the speed of
light.
And that in turn means that the relativistic errors in the hydrogen
energy levels are small. Still, even small errors can sometimes be
very important. The required corrections are listed below in order of
decreasing magnitude.
- Fine structure.
The electron should really be described relativistically using the
Dirac equation instead of classically. In classical terms, that
will introduce three corrections to the energy levels:
- Einstein’s relativistic correction of the classical
kinetic energy of the electron.
Spin-orbit interaction
, due to the fact that
the spin of the moving electron changes the energy levels. The
spin of the electron makes it act like a little electromagnet. It
can be seen from classical electrodynamics that a moving magnet
will interact with the electric field of the nucleus, and that
changes the energy levels. Note that the name spin-orbit interaction
is a well chosen one, a rarity in physics.
- There is a third correction for states of zero angular
momentum, the Darwin term. It is a crude fix for the fundamental
problem that the relativistic wave function is not just a modified
classical one, but also involves interaction with the
anti-particle of the electron, the positron.
Fortunately, all three of these effects are very small; they are
smaller than the uncorrected energy levels by a factor of order
, and the error they introduce is on the order of
0.001%. So the exact
solution of chapter
4.3 is, by engineering standards, pretty exact after all.
- Lamb shift. Relativistically, the electron is affected by
virtual photons and virtual electron-positron pairs. It adds a
correction of relative magnitude to the energy levels,
one or two orders of magnitude smaller still than the fine structure
corrections. To understand the correction properly requires quantum
electrodynamics.
- Hyperfine splitting. Like the electron, the proton acts
as a little electromagnet too. Therefore the energy depends on how
it aligns with the magnetic field generated by the electron. This
effect is a factor smaller still than the fine structure
corrections, making the associated energy changes about two orders
of magnitude smaller.
Hyperfine splitting couples the spins of proton and electron, and in
the ground state, they combine in the singlet state. A slightly
higher energy level occurs when they are in a spin-one triplet
state; transitions between these states radiate very low energy
photons with a wave length of 21 cm. This is the source of the
21 centimeter line” or “hydrogen line
radiation that is of great importance in cosmology. For example, it
has been used to analyze the spiral arms of the galaxy, and the hope
at the time of this writing is that it can shed light on the so
called dark ages
that the universe went through.
Since so little energy is released, the transition is very slow,
chapter 7.6.1. It takes on the order of 10 million
years, but that is a small time on the scale of the universe.
The message to take away from that is that even errors in the ground
state energy of hydrogen that are two million times smaller than the
energy itself can be of critical importance under the right
conditions.
The following subsections discuss each correction in more detail.
A.39.2 Fine structure
From the Dirac equation, it can be seen that three terms need to be
added to the nonrelativistic Hamiltonian of chapter 4.3 to
correct the energy levels for relativistic effects. The three terms
are worked out in derivation {D.81}. But that
mathematics really provides very little insight. It is much more
instructive to try to understand the corrections from a more physical
point of view.
The first term is relatively easy to understand. Consider
Einstein’s famous relation , where is
energy, mass, and the speed of light. According to this
relation, the kinetic energy of the electron is not
, with the velocity, as Newtonian physics
says. Instead it is the difference between the energy
based on the mass of the electron
in motion and the energy based on the mass of the
electron at rest. In terms of momentum
, chapter 1.1.2,
|
(A.248) |
Since the speed of light is large compared to the typical speed of the
electron, the square root can be expanded in a Taylor series,
[41, 22.12], to give:
The first term corresponds to the kinetic energy operator used in the
nonrelativistic quantum solution of chapter 4.3. (It may be
noted that the relativistic momentum is based on the moving mass
of the electron, not its rest mass. It is this relativistic momentum
that corresponds to the operator
. So the Hamiltonian used in chapter
4.3 was a bit relativistic already, because in replacing
by , it used the relativistic
expression.) The second term in the Taylor series expansion above is
the first of the corrections needed to fix up the hydrogen energy
levels for relativity. Rewritten in terms of the square of the
classical kinetic energy operator, the Bohr ground state energy
and the fine structure constant , it is
|
(A.249) |
The second correction that must be added to the nonrelativistic
Hamiltonian is the so-called spin-orbit interaction.
In classical terms, it is due to the spin of the electron, which makes
it into a magnetic dipole.
Think of it as a magnet of
infinitesimally small size, but with infinitely strong north and
south poles to make up for it. The product of the infinitesimal
vector from south to north pole times the infinite strength of the
poles is finite, and defines the magnetic dipole moment
. By itself, it is quite inconsequential since the
magnetic dipole does not interact directly with the electric field of
the nucleus. However, moving magnetic poles create an electric
field just like the moving electric charges in an electromagnet
create a magnetic field. The electric fields generated by the moving
magnetic poles of the electron are opposite in strength, but not quite
centered at the same position. Therefore they correspond to a
motion-induced electric dipole. And an electric dipole does
interact with the electric field of the nucleus; it wants to align
itself with it. That is just like the magnetic dipole wanted to align
itself with the external magnetic field in the Zeeman effect.
So how big is this effect? Well, the energy of an electric dipole
in an electric field is
As you might guess, the electric dipole generated by the magnetic
poles of the moving electron is proportional to the speed of the
electron and its magnetic dipole moment .
More precisely, the electric dipole moment will be
proportional to because if the vector
connecting the south and north poles is parallel to the motion, you do
not have two neighboring currents of magnetic poles, but a single
current of both negative and positive poles that completely cancel
each other out. Also, the electric field of the nucleus is
minus the gradient of its potential , so
Now the order of the vectors in this triple product can be changed,
and the dipole strength of the electron equals its spin
times the charge per unit mass , so
The expression between the parentheses is the angular momentum
save for the electron mass. The constant of proportionality
is worked out in derivation {D.82}, giving the spin-orbit
Hamiltonian as
|
(A.250) |
The final correction that must be added to the nonrelativistic
Hamiltonian is the so-called “Darwin term:”
|
(A.251) |
According to its derivation in {D.81}, it is a crude
fix-up for an interaction with a virtual positron that simply cannot
be included correctly in a nonrelativistic analysis.
If that is not very satisfactory, the following much more detailed
derivation can be found on the web. It does succeed in
explaining the Darwin term fully within the nonrelativistic picture
alone. First assume that the electric potential of the nucleus does
not really become infinite as 1 at 0, but is smoothed
out over some finite nuclear size. Also assume that the electron does
not see
this potential sharply, but perceives of its
features a bit vaguely, as diffused out symmetrically over a typical
distance equal to the so-called Compton wave length
. There are several plausible reasons why it
might: (1) the electron has illegally picked up a chunk of a negative
rest mass state, and it is trembling with fear that the uncertainty in
energy will be noted, moving rapidly back and forwards over a Compton
wave length in a so-called Zitterbewegung
; (2) the
electron has decided to move at the speed of light, which is quite
possible nonrelativistically, so its uncertainty in position is of
the order of the Compton wave length, and it just cannot figure out
where the right potential is with all that uncertainty in position and
light that fails to reach it; (3) the electron needs glasses. Further
assume that the Compton wave length is much smaller than the size over
which the nuclear potential is smoothed out. In that case, the
potential within a Compton wave length can be approximated by a second
order Taylor series, and the diffusion of it over the Compton wave
length will produce an error proportional to the Laplacian of the
potential (the only fully symmetric combination of derivatives in the
second order Taylor series.). Now if the potential is smoothed over
the nuclear region, its Laplacian, giving the charge density, is known
to produce a nonzero spike only within that smoothed nuclear region,
figure 13.7 or (13.30). Since the nuclear size
is small compared to the electron wave functions, that spike can then
be approximated as a delta function. Tell all your friends you heard
it here first.
The key question is now what are the changes in the hydrogen energy
levels due to the three perturbations discussed above. That can be
answered by perturbation theory as soon as the good eigenfunctions
have been identified. Recall that the usual hydrogen energy
eigenfunctions are made unique by the square angular
momentum operator , giving , the angular
momentum operator , giving , and the spin angular
momentum operator giving the spin quantum number
for spin up, respectively down. The decisive term
whether these are good or not is the spin-orbit interaction. If the
inner product in it is written out, it is
The radial factor is no problem; it commutes with every orbital
angular momentum component, since these are purely angular
derivatives, chapter 4.2.2. It also commutes with every
component of spin because all spatial functions and operators do,
chapter 5.5.3. As far as the dot product is concerned,
it commutes with since all the components of do, chapter
4.5.4, and since all the components of commute with
any spatial operator. But unfortunately, and do not
commute with , and and do not commute with
(chapters 4.5.4 and 5.5.3):
The quantum numbers and are bad.
Fortunately, does commute with the net angular
momentum , defined as . Indeed, using
the commutators above and the rules of chapter 4.5.4 to
take apart commutators:
and adding it all up, you get 0. The same
way of course commutes with the other components of net
angular momentum , since the -axis is arbitrary. And if
commutes with every component of , then it
commutes with their sum of squares . So, eigenfunctions
of , , and are good
eigenfunctions.
Such good eigenfunctions can be constructed from the
by forming linear combinations of them that combine different and
values. The coefficients of these good combinations are called
Clebsch-Gordan coefficients and are shown for 1 and
2 in figure 12.5. Note from this figure that the
quantum number of net square momentum can only equal
or . The half unit of electron spin is not big
enough to change the quantum number of square orbital momentum by more
than half a unit. For the rest, however, the detailed form of the
good eigenfunctions is of no interest here. They will just be
indicated in ket notation as , indicating that
they have unperturbed energy , square orbital angular
momentum , square net (orbital plus spin) angular
momentum , and net angular momentum
.
As far as the other two contributions to the fine structure are
concerned, according to chapter 4.3.1 in the
Einstein term consists of radial functions and radial derivatives plus
. These commute with the angular derivatives that make up
the components of , and as spatial functions and operators,
they commute with the components of spin. So the Einstein Hamiltonian
commutes with all components of and ,
hence with , , and . And the
delta function in the Darwin term can be assumed to be the limit of a
purely radial function and commutes in the same way. The
eigenfunctions with given values of ,
, and are good ones for the entire fine structure
Hamiltonian.
To get the energy changes, the Hamiltonian perturbation coefficients
must be found. Starting with the Einstein term, it is
Unlike what you may have read elsewhere, is indeed a
Hermitian operator, but may have a delta
function at the origin, (13.30), so watch it with blindly
applying mathematical manipulations to it. The trick is to take half
of it to the other side of the inner product, and then use the fact
that the eigenfunctions satisfy the nonrelativistic energy eigenvalue
problem:
Noting from chapter 4.3 that ,
and that the expectation values of
and are given in derivation {D.83}, you
find that
The spin-orbit energy correction is
For states with no orbital angular momentum, all components of
produce zero, so there is no contribution. Otherwise, the dot product
can be rewritten by expanding
to give
That leaves only the expectation value of to be
determined, and that can be found in derivation {D.83}. The
net result is
or zero if 0.
Finally the Darwin term,
Now a delta function at the origin has the property to pick out the
value at the origin of whatever function it is in an integral with,
compare chapter 7.9.1. Derivation {D.15},
(D.9), implies that the value of the wave functions at the
origin is zero unless 0, and then the value is given in
(D.10). So the Darwin contribution becomes
To get the total energy change due to fine structure, the three
contributions must be added together. For 0, add the Einstein and
Darwin terms. For 0, add the Einstein and spin-orbit terms;
you will need to do the two possibilities that and
separately. All three produce the same final result,
anyway:
|
(A.252) |
Since is at most , the energy change due to fine
structure is always negative. And it is the biggest fraction of
for and 2, where it is
, still no more than a sixth of a
percent of a percent change in energy.
In the ground state can only be one half, (the electron spin), so
the ground state energy does not split into two due to fine structure.
You would of course not expect so, because in empty space, both spin
directions are equivalent. The ground state does show the largest
absolute change in energy.
Woof.
A.39.3 Weak and intermediate Zeeman effect
The weak Zeeman effect is the effect of a magnetic field that is
sufficiently weak that it leaves the fine structure energy
eigenfunctions almost unchanged. The Zeeman effect is then a small
perturbation on a problem in which the unperturbed
(by
the Zeeman effect) eigenfunctions derived in the
previous subsection are degenerate with respect to and
.
The Zeeman Hamiltonian
commutes with both and , so the
eigenfunctions are good. Therefore, the energy
perturbations can be found as
To evaluate this rigorously would require that the
state be converted into the one or two states with
and
using the appropriate Clebsch-Gordan coefficients from
figure 12.5.
However, the following simplistic derivation is usually given instead,
including in this book. First get rid of by replacing it by
. The inner product with can then be
evaluated as being , giving the energy change as
For the final inner product, make a semi-classical argument that only
the component of in the direction of gives a
contribution. Don’t worry that does not exist. Just
note that the component in the direction of is constrained
by the requirement that and must add up to ,
but the component normal to can be in any direction and
presumably averages out to zero. Dismissing this component, the
component in the direction of is
and the dot product in it can be found from expanding
to give
For a given eigenfunction ,
, , and
with .
If the -component of is substituted for in the
expression for the Hamiltonian perturbation coefficients, the energy
changes are
|
(A.253) |
(Rigorous analysis using figure 12.5, or more generally item
2 in chapter 12.8, produces the same results.)
The factor within the brackets is called the “Landé -factor.” It is the factor by which the magnetic
moment of the electron in the atom is larger than for a classical
particle with the same charge and total angular momentum. It
generalizes the -factor of the electron in isolation to include the
effect of orbital angular momentum. Note that it equals 2, the Dirac
-factor, if there is no orbital momentum, and 1, the classical
value, if the orbital momentum is so large that the half unit of spin
can be ignored.
In the intermediate Zeeman effect, the fine structure and Zeeman
effects are comparable in size. The dominant perturbation Hamiltonian
is now the combination of the fine structure and Zeeman ones. Since
the Zeeman part does not commute with , the eigenfunctions
are no longer good. Eigenfunctions with the same
values of and , but different values of must be
combined into good combinations. For example, if you look at
2, the eigenfunctions and
have the same unperturbed energy and good
quantum numbers and . You will have to write a two by
two matrix of Hamiltonian perturbation coefficients for them, as in
addendum {A.38.3}, to find the good combinations and their
energy changes. And the same for the
and eigenfunctions. To obtain the
matrix coefficients, use the Clebsch-Gordan coefficients from figure
12.5 to evaluate the effect of the Zeeman part. The fine
structure contributions to the matrices are given by (A.252)
when the values are equal, and zero otherwise. This can be seen
from the fact that the energy changes must be the fine structure ones
when there is no magnetic field; note that is a good quantum
number for the fine structure part, so its perturbation coefficients
involving different values are zero.
A.39.4 Lamb shift
A famous experiment by Lamb & Retherford in 1947 showed that the
hydrogen atom state 2, 0,
, also called the 2S state, has a somewhat different
energy than the state 2, 1,
, also called the 2P state. That was
unexpected, because even allowing for the relativistic fine structure
correction, states with the same principal quantum number and same
total angular momentum quantum number should have the same energy.
The difference in orbital angular momentum quantum number should
not affect the energy.
The cause of the unexpected energy difference is called Lamb shift.
To explain why it occurs would require quantum electrodynamics, and
that is well beyond the scope of this book. Roughly speaking, the
effect is due to a variety of interactions with virtual photons and
electron/positron pairs. A good qualitative discussion on a
nontechnical level is given by Feynman [19].
Here it must suffice to list the approximate energy corrections
involved. For states with zero orbital angular momentum, the energy
change due to Lamb shift is
|
(A.254) |
where is a numerical factor that varies a bit with from
about 12.7 to 13.2. For states with nonzero orbital angular momentum,
|
(A.255) |
where is less than 0.05 and varies somewhat with and
.
It follows that the energy change is really small for states with
nonzero orbital angular momentum, which includes the 2P
state. The change is biggest for the 2S state, the
other state in the Lamb & Retherford experiment. (True, the
correction would be bigger still for the ground state 1, but
since there are no states with nonzero angular momentum in the ground
state, there is no splitting of spectral lines involved there.)
Qualitatively, the reason that the Lamb shift is small for states with
nonzero angular momentum has to do with distance from the nucleus.
The nontrivial effects of the cloud of virtual particles around the
electron are most pronounced in the strong electric field very close
to the nucleus. In states of nonzero angular momentum, the wave
function is zero at the nucleus, (D.9). So in those states
the electron is unlikely to be found very close to the nucleus. In
states of zero angular momentum, the square magnitude of the wave
function is 1 at the nucleus, reflected in both the
much larger Lamb shift as well as its approximate 1 dependence
on the principal quantum number .
A.39.5 Hyperfine splitting
Hyperfine splitting of the hydrogen atom energy levels is due to the
fact that the nucleus acts as a little magnet just like the electron.
The single-proton nucleus and electron have magnetic dipole moments
due to their spin equal to
in which the -factor of the proton is about 5.59 and that of the
electron 2. The magnetic moment of the nucleus is much less than the
one of the electron, since the much greater proton mass appears in the
denominator. That makes the energy changes associated with hyperfine
splitting really small compared to other effects such as fine
structure.
This discussion will restrict itself to the ground state, which is by
far the most important case. For the ground state, there is no
orbital contribution to the magnetic field of the electron. There is
only a spin-spin coupling
between the magnetic moments
of the electron and proton, The energy involved can be thought of most
simply as the energy of the electron in the
magnetic field of the nucleus. If the nucleus is modelled
as an infinitesimally small electromagnet, its magnetic field is that
of an ideal current dipole as given in table 13.2. The
perturbation Hamiltonian then becomes
The good states are not immediately self-evident, so the four
unperturbed ground states will just be taken to be the ones which the
electron and proton spins combine into the triplet or singlet states
of chapter 5.5.6:
or for short, where
and are the quantum numbers of net spin
and its -component. The next step is to evaluate the four by four
matrix of Hamiltonian perturbation coefficients
using these states.
Now the first term in the spin-spin Hamiltonian does not produce a
contribution to the perturbation coefficients. The reason is that the
inner product of the perturbation coefficients written in spherical
coordinates involves an integration over the surfaces of constant
. The ground state eigenfunction is constant on
these surfaces. So there will be terms like
in the integration, and those
are zero because is just as much negative as positive on these
spherical surfaces, (as is ). There will also be terms like
in the integration. These will be zero too because by symmetry the
averages of , , and are equal on the
spherical surfaces, each equal to one third the average of
.
So only the second term in the Hamiltonian survives, and
the Hamiltonian perturbation coefficients become
The spatial integration in this inner product merely picks out the
value 1 at the origin, as delta functions
do. That leaves the sum over the spin states. According to addendum
{A.10},
Since the triplet and singlet spin states are orthonormal, only the
Hamiltonian perturbation coefficients for which
and
survive, and these then give the leading order changes in the energy.
Plugging it all in and rewriting in terms of the Bohr energy and fine
structure constant, the energy changes are:
|
(A.256) |
The energy of the triplet states is raised and that of the singlet
state is lowered. Therefore, in the true ground state, the electron
and proton spins combine into the singlet state. If they somehow get
kicked into a triplet state, they will eventually transition back to
the ground state, say after 10 million years or so, and release a
photon. Since the difference between the two energies is so tiny on
account of the very small values of both and
, this will be a very low energy photon. Its wave
length is as long as 0.21 m, producing the 21 cm hydrogen line.