Download Atoms, molecules and optical transitions

Atoms, molecules and optical transitions
Identical particles and Pauli exclusion principle
•
A system of
N
ψ(r1 , . . . , rN )
ith particle would be
|ψ(r1 , . . . , rN )|2 - this
identical particles can be described in quantum mechanics by a wavefunction
which depends on the coordinates of all particles. The probability (density) that
located at
ri
in a simultaneous measurement of all particle positions is given by
provides the physical meaning to the wavefunction.
•
An example of the many-particle Schrodinger equation is:

N
N
N
2
X
X
X
∂
1
− ~
+
U (ri ) +
V (|ri − rj |) ψ(r1 , . . . , rN ) = Eψ(r1 , . . . , rN )
2m i=1 ∂r2 i=1
2 i,j=1

Here,
U (r)
2
is the external potential energy seen by particles, and
interactions between two particles separated by distance
•
If particles are non-interacting,
V (r)
is the potential energy of
r.
V (r) ≡ 0, then the whole Schrodinger equation is a sum of independent
parts which involve only a single particle set of coordinates. Hence, we can separate variables and write
the wavefunction in a product form:
ψ(r1 , . . . , rN ) = ψ1 (r1 )ψ2 (r2 ) · · · ψN (rN )
Substituting this into the many-body Schrodinger equation gives
~2 ∂ 2
−
+ U (r) ψi (r) = Eψi (r)
2m ∂r2
,
N
single-particle equations:
(∀i = 1 . . . N )
Suppose we know how to solve this single-particle equation and obtain all eigenfunctions
by single-particle quantum numbers
quantum numbers
{ni |i = 1 . . . N }.
n.
Then, the many-body states are labeled by all combinations of
A many-body state can be given by specifying the single-particle
quantum states of all individual particles:
•
ψ (n) (r) labeled
(n1 , n2 , . . . , nN ).
An important physical property of identical particles is that they are absolutely indistinguishable.
While we have no choice but to label dierent particles by numbers
there is no experiment which can distinguish which of the
2,
N
i = 1, 2, . . . , N ,
it turns out that
particles we labeled by
1,
which one by
and so on. For example, all electrons in the universe are identical, and if someone exchanged places
of any two electrons without telling us, we would never be able to nd out.
•
A physically detectable property of
|ψ(r1 , . . . , rn )|2 .
Any permutation
N
P
identical particles is their probability density
of the particle labels
i
P (r1 , . . . , rn ) =
corresponds to the same physical state:
P (r1 , . . . , rn ) = P (Pr1 , . . . , PrN )
This means that the wavefunction is allowed to change by a phase factor
are exchanged. Since particles are identical, the phase angle
θ
eiθ
when two particle labels
must not depend on which two labels
we choose to exchange. Consider a simplied problem of two particles. We must require:
ψ(r1 , r2 ) = eiθ ψ(r2 , r1 )
The overall phase
eiθ
is not physical, in the sense that it cannot be measured. The state
physically identical to the state
ψ(r1 , r2 ).
ψ(r2 , r1 )
is
If we exchange the two labels second time, we must obtain
another phase factor:
ψ(r1 , r2 ) = eiθ ψ(r2 , r1 ) = ei2θ ψ(r1 , r2 )
But now, the wavefunctions on the left and right hand sides are formally identical, and they must be
single-valued which is possible only if
ei2θ = 1.
Therefore,
θ ∈ {0, π}
ψ(r1 , r2 ) = ±ψ(r2 , r1 )
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and
•
The identical particles whose wavefunction does not change sign under label exchange are called bosons.
If one nds a wavefunction
ψ(r1 , r2 )
by solving a Schrodinger equation, one has to symmetrize before
it can properly describe the many-body bosonic system:
Ψ(r1 , r2 ) ∝ ψ(r1 , r2 ) + ψ(r2 , r1 )
This wavefunction is symmetric under the particle exchange. Assuming for simplicity that the bosons
are non-interacting, we can put two bosons in the same single particle state:
ψ1 (r) ≡ ψ2 (r).
where
•
ψ(r1 , r2 ) = ψ1 (r1 )ψ2 (r2 )
In fact, we can put any number of bosons in the same single-particle state.
The identical particles whose wavefunction changes sign under label exchange are called fermions. If
one nds a wavefunction
ψ(r1 , r2 ) by solving a Schrodinger equation, one has to anti-symmetrize before
it can properly describe the many-body fermionic system:
Ψ(r1 , r2 ) ∝ ψ(r1 , r2 ) − ψ(r2 , r1 )
This wavefunction is anti-symmetric under the particle exchange.
Assuming for simplicity that the
fermions are non-interacting, we can show that it is not possible to put two fermions in the same single
ψ(r1 , r2 ) = ψ1 (r1 )ψ2 (r2 ) where ψ1 (r) ≡ ψ2 (r), and substituting in the equation
Ψ(r1 , r2 ) ≡ 0. This is known as Pauli exclusion principle.
particle state. Writing
above leads to
•
It was discovered in relativistic quantum eld theory that identical particles with integer spin behave
like bosons, while identical particles with half-integer spin behave like fermions. This is a consequence
of fundamental symmetry requirements: Lorentz invariance + CPT (simultaneous charge conjugation,
parity reversal and time reversal).
... Particles with integer spin,
S 2 = ~2 s(s+1) where s ∈ {0, 1, 2 . . .}, are bosons.
Examples are photons,
Hydrogen atom 1 H, etc.
... Particles with half-integer spin,
S 2 = ~2 s(s + 1)
where
s ∈ { 21 , 32 , 52 . . .},
are fermions. Examples of
fermions are electron, proton, neutron, Lithium 6 Li atom, etc.
•
Note that the above notation
ψ(r1 , . . . , rN )
is a simplication because particles in general have a spin
quantum number. For example, any fermion denitely has a spin quantum number
even if
s
numbers,
labels
•
i,
= 21 ). The wavefunction is appropriately a
ψ(r1 , ms1 ; r2 , ms2 ; . . . ; rN , msN ). By particle exchange we mean a
is the smallest possible (s
ms = −s, . . . , s,
function of all quantum
permutation of particle
which consistently permute all quantum numbers associated with individual particles.
We say that bosons and fermions have dierent statistics.
exclusion principle is fundamental for our existence.
Fermions may seem peculiar, but their
Only one type of atom (Hydrogen-like) would
exist in the universe if electrons were bosons.
Multi-electron atoms
•
Consider building a multi-electron atom by adding electrons one by one to the initially bare nucleus
with
Z
• Z =1
protons. We will consider adding exactly
Z
electrons to form a neutral atom.
(Hydrogen). The rst electron can be placed in any single-particle state obtained by solving
the Hydrogen-like Schrodinger equation. These states are labeled by four quantum numbers: principal
n = 1, 2, 3 . . .,
orbital
l = 0, 1, 2 . . . , n − 1,
m = −l . . . , l and spin ms = ± 12 . To minimize
1
lowest s-orbital: n = 0, l = 0, m = 0, ms = ± .
2
magnetic
energy, the rst electron must be placed in the
This orbital is two-fold degenerate due to spin, and either spin state can be selected.
conguration is
• Z=2
The electron
1s1 .
(Helium). Adding the second electron is a little dierent. Now, the nucleus charge is partially
screened by the rst electron, so the second one feels smaller Coulomb force than the rst one did.
This is a consequence of the mutual interaction between electrons. Adding the second electron will also
disturb the Hydrogen-like orbit of the rst electron. The full Schrodinger equation for two interacting
electrons is too dicult to solve, but the quantum numbers of single-particle states are qualitatively
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the same
(n, l, m, ms ).
The second electron can occupy the same
s-orbital
as the rst electron, but
must have the opposite spin due to Pauli principle. In other words, the two electrons will have the
quantum numbers
• Z = 3
(0, 0, 0, 21 )
(Lithium).
states in the lowest
n = 2, l = 0.
and
(0, 0, 0, − 21 )
in the ground state. The electron conguration is
The third electron we wish to add cannot go to the lowest
s-orbital
1s2 .
s-orbital because all
s-orbitals at
are already occupied. The lowest energy states available are
The rst two electrons partially screen the nuclear charge and interact with the third
electron. This interaction pushes the
p-orbital (l = 1)
to higher energy, which was not the case in a
pure Hydrogen-like ion (nucleus without any electrons). There are two spin states to choose from. The
electron conguration is
• Z=4
1s2 2s1 .
(Beryllium). The fourth electron can be added to the remaining spin-state at
electron conguration is
• Z =5
(Boron). The fth electron can go to
n = 2, l = 1.
p-orbitals
The
Apart from spin, it has a certain freedom
to choose between dierent projections of the orbital angular momentum,
in the
n = 2, l = 0.
1s2 2s2 .
m = −1, 0, 1.
These states
used to be degenerate in a naive treatment of Hydrogen-like ions. However, now the
spin-orbit coupling partially lifts this degeneracy. The electron conguration is
1s2 2s2 2p1 .
• Z = 6 (Carbon). The sixth electron has an option of lling in the remaining spin state in the n = 2,
l = 1, m = 0 orbital. However, this does not happen. The repulsive interaction between two electrons
in the same n = 2, l = 1, m = 0 orbital is large enough to make it more costly than lling up one
of the m = ±1 states. Note that the shape of p-orbitals is such that electrons can eciently avoid
each other if they live in two dierent p-orbitals. The sixth electron actually prefers to have the same
spin-orientation as the fth electron, because Pauli exclusion then works against the spatial overlap
of electron wavefunctions (which results in Coulomb repulsion). This is known as Hund's rule. The
electron conguration is
• Z=7
1s2 2s2 2p2 .
p-orbital.
n = 2, l = 1 p-orbital have quantum numbers m = −1, m = 0 and
m = +1 respectively, all with the same spin, say ms = 21 . This picture is somewhat complicated by
2 2
3
the spin-orbit interaction. The electron conguration is 1s 2s 2p .
(Nitrogen). Due to the Hund's rule, the seventh electron goes to the remaining empty
The three electrons in the
• Z = 8
(Oxygen).
The eight electron has to enter one of the three
m ∈ {−1, 0, 1} states with
1s2 2s2 2p4 .
the
opposite spin to the electron which is already there. The electron conguration is
• Z=9
(Fluorine). Another
• Z = 10
•
p-state
gets occupied. The electron conguration is
(Neon). The last remaining
p-state
1s2 2s2 2p5 .
gets occupied. The electron conguration is
1s2 2s2 2p6 .
Z proceeds generally in the same fashion, but with notable
n = 2 are occupied, so the new electrons must go into s, p and d
the s-orbitals are populated, then p. After all p-orbitals are full,
Adding more electrons and increasing
exceptions. The shells
orbitals of the
n = 3
n=1
shell.
and
First
one would naturally expect that
d-orbitals
get their turn. However, due to complicated interactions
3d orbitals have a slightly higher energy than
3d ones. This goes on until 4s2 3d3
1
5
conguration 4s 3d . There are other exceptions...
between electrons and screening of the nuclear charge, the
the
4s
orbitals. Therefore, the two
4s
states are lled before any
and then the next element has another exceptional
•
Chemical properties of elements are governed by their valence electrons, beyond the fully populated
electron shell (with given
n).
Independent particle approximation: Hartree mean-eld theory
•
Solving the many-body Schrodinger equation for a multi-electron atom is dicult and can be done
only numerically. There is, however, an approximation which can give reasonably good results.
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•
Z
Suppose one wishes to calculate the properties of an atom with
protons in the nucleus and
Z
electrons. The nucleus is static and point-like for our purposes, but we want to describe individual
electrons by using single-particle wavefunctions
one to self-consistently determine
•
ψi (r), i = 1 . . . Z .
The Hartree approximation allows
ψi (r).
First, calculate the total charge distribution as a function of coordinates:
ρ(r) = −e
X
|ψi (r)|2
i
This is indeed a static charge distribution (because the atom must be in equilibrium), and it must
be spherically symmetric (otherwise, it cannot be static according to the laws of electrodynamics).
The symmetry dictates that the electric eld vector
E(r)
due to the nucleus and the electronic charge
distribution must be a function of only the radius, and pointing radially outwards. One can apply the
Gauss's law to determine the magnitude
any electron a distance
•
U (r)
The potential
knew
U (r)
r
E(r),
and from it the Coulomb potential energy
is the mean-eld produced on average by all electrons and the nucleus.
atom states, but not the same. In the Hydrogen atom
The problem is that
on
U (r).
felt by
If we
we could solve the single-particle Schrodinger equation for each electron independently and
nd all wavefunctions of dierent orbitals and shells.
•
U (r)
away from the nucleus.
These states will be similar to the Hydrogen
U (r)
is given only by the nuclear charge.
U (r) depends on the solutions ψi (r) of the Schrodinger equation, and ψi (r) depend
The problem must be solved self-consistently. One rst makes a reasonable assumption about
U (r) and picks a trial function. Then, one solves the Schrodinger equation, nds ψi (r)
U (r) and constructs ρ(r) from those solutions. The corresponding potential U 0 (r) to the
0
obtained ρ(r) is dierent than U (r). So, we put U (r) in the Schrodinger equation and calculate new
0
0
00
00
0
solutions ψi (r) and the corresponding ρ (r) and U (r). The new potential U (r) 6= U (r) 6= U (r), so
the shape of
for the trial
one keeps going and re-calculating the Schrodinger equation many times. If one is lucky, the successive
results for the potentials and wavefunctions converge, and ultimately we nd the proper
gives solutions
ψi (r)
such that their charge density gives back the same
U (r)
which
U (r).
Molecules
•
Molecules are stable congurations of electrons and more than one nucleus.
The motion of nuclei
can often be neglected because nuclei are much heavier than electrons. Then, the quantum dynamics
of a molecule can be determined by solving a many-body Schrodinger equation for electrons in the
electrostatic eld of xed nuclei.
•
The simplest molecule is
H2 ,
with two protons a distance
R
apart and two electrons. Even simpler
system is a pair of protons which share a single electron among themselves. It is instructive to solve
this problem using the Schrodinger equation.
•
R from each other is E1 = −13.6
R → ∞ (the electron electron is bound to one proton, the other proton is bare), while
E2 = E1 Z 2 = −54.4 eV in the R → 0 limit (we have a Hydrogen-like ion with Z = 2 when two
The energy of a single electron bound to two protons at a distance
eV in the limit
it is
protons are at the same location). It seems that the energy can be lowered by bringing the two protons
close together, and a singly-ionized Hydrogen molecule can be stable.
•
It is not advantageous to bring the two protons too close together, because they repel each-other via
Coulomb interaction (not included in the above analysis). There is some optimal distance
R,
which
becomes the length of the Hydrogen molecular ion. This length can be calculated from the Schrodinger
equation:
2 2
e2
e2
e2
~ ∇
−
−
+
ψ(r) = Eψ(r)
−
2m
4π0 r1
4π0 r2
4π0 R
where
r1
and
r2
are distances between the electron and the two protons respectively. This is a single-
particle equation which can be solved numerically without too much trouble. Again, we assume that
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the protons are static and assume that their separation is
the ground state energy
E,
which in turn depends on
R.
R.
Solving this Schrodinger equation yields
We ultimately choose
R
which minimizes
E,
and this is the molecule size. The ground state wavefunction is such that the probability of nding the
electron in between the two nuclei is larger than anywhere else. In this manner the molecule achieves
the lowest possible energy, the electron attracts both protons and brings them closer together.
•
The neutral Hydrogen molecule has two electrons. It takes a two-body Schrodinger equation to solve
this problem, or at least the Hartree approximation (without the benet of a spherically symmetric
potential). It turns out that the ground state of a Hydrogen molecule has the two electrons in essentially
the same orbital state, but with opposite spins, similar to the
1s2
state of the Helium atom. Then,
the two electrons are fairly likely to be found in between the two protons, and they lower the molecule
energy despite their mutual repulsion.
•
The above is an example of a covalent bond. Similar bonds can be formed between separated nuclei of
more complicated molecules.
•
Another binding mechanism is an ionic bond between two atoms, one of which has a single electron
outside of a lled shell (alkali metal) and the other one an electron missing from a lled shell (halogen
element). For example, consider the NaF molecule. This molecule is formed by electrostatic attraction
between the positively charged Sodium ion and the negatively charged Fluorine ion when the valence
electron of Sodium is taken away to ll up the hole in the outer shell of Fluorine. If the two atoms
were separated at a large distance, this transfer of an electron which ionizes both of them would cost a
positive energy, and thus would not be favored. However, the Sodium ion is smaller than the Sodium
atom, while the Fluorine ion is roughly of the same size as the Fluorine atom itself.
The two ions
can approach closer to each other than the two atoms and gain potential energy by their electrostatic
attraction which overcompensates the cost of electron transfer and leads to binding.
The optimum
distance between the two ions is shaped by the short-range repulsive interactions, ultimately due to
the Pauli exclusion. Electron shells of two ions cannot simply overlap without signicantly rearranging
electron orbitals. In the extreme case when the two nuclei join into a single nucleus (with
Z = 20),
all
electrons would have to ll 20 atomic orbitals, which costs much more energy than an ionic molecule.
•
So far we considered only electronic degrees of freedom in molecules. While much heavier then electrons,
nuclei also participate in the dynamics of molecules. For example, molecules made of two atoms can
experience two characteristic types of internal motion, vibrational and rotational. Both are properly
studied by solving the Schrodinger equation which contains the coordinates of nuclei and takes into
account their kinetic energy.
Nuclei of a molecule also interact among themselves.
interaction is just their electrostatic repulsion.
A part of this
The other part comes from electrons (for example,
assume some positions of the nuclei, solve the electron-only Schrodinger equation and interpret the
obtained ground-state energy as a potential energy of the nuclei which depends on their assumed
positions). The Schrodinger equation for the relative motion of two nuclei involves their reduced mass
and their total interaction potential. The solutions can be somewhat similar to the linear-harmonicpotential wavefunctions, which correspond to vibrational motion. Also, rotational motion is possible
with the appropriate quantization of angular momentum.
These kinds of quantum motion lead to
additional energy levels and transitions, with smaller energy scales then the electronic ones.
Optical transitions in atoms
•
Schrodinger's theory predicts that all quantum states of any atom are stable. If an electron is placed
in any excited (or ground) state of a Hydrogen atom, for example, it will remain in that state for ever.
•
Classical electrodynamics predicts that an electron performing accelerated motion in the atom has
to radiate electromagnetic waves and gradually lose its energy until it falls on the nucleus.
This is
prevented in quantum mechanics due to energy and angular momentum quantization. What prevents
even abrupt energy changes (due to the electron relaxations to lower energy levels) is the fact that charge
distribution of an electron in any stationary quantum state does not change with time. In other words,
even though the electron wavefunction
ψ(r, t) = ψ(r)eiEt/~
has time dependence, the corresponding
charge distribution
ρ(r) = −e|ψ(r, t)|2 = −e|ψ(r)|2
is static, as if no charge is being accelerated. Even
according to classical electrodynamics there will be no radiation in these circumstances.
•
In reality, however, electron transitions occur between dierent states in the presence of an external
electromagnetic eld.
Also, an electron can relax spontaneously from an excited state and emit a
photon. This is incorporated in the quantum mechanics via perturbation theory, the details of which
are beyond this course.
•
Perturbation theory is an elaborate collection of approximations which allow us to deal with small
deviations from known solvable quantum problems.
For example, if we solve for the wavefunctions
and energy levels of an isolated atom, then we can determine the changes caused by a weak external
electromagnetic eld (the perturbation).
•
If the perturbation is static, it will introduce small changes in the energy levels and wavefunctions. If
the perturbation is time-dependent, it will cause random transitions between the stationary states of
the unperturbed system, and perturbation theory allows calculating transition probabilities (rates).
•
Photon absorption probability can be calculated using perturbation theory. Photons can be described
by an electromagnetic wave whose wavelength is much larger than the atom size. Thus, the electric eld
E = E0 cos(ωt).
U (t) = −eE0 cos(ωt), which
at the atom position can be approximated by a uniform eld with oscillates in time,
An electron experiences a Coulomb potential due to this electric eld,
is to be included in the Schrodinger equation together with the static potential due to the nucleus. As
long as
E0
is small enough, the energy levels of the atom are unaltered, but there is a nite probability
that the electron will be excited to a higher energy state (assuming that
•
~ω = E2 − E1 ).
Detailed perturbative calculations reveal selection rules for transitions.
Not only must the energy
dierence between two states match the photon frequency, but transitions are possible only between
the pairs of states whose other quantum numbers satisfy certain conditions. For example, the orbital
quantum numbers of the initial and nal state must dier by one,
quantum numbers must dier by at most one:
|m2 − m1 | ≤ 1
l2 − l1 = ±1.
(including spin).
Also, the magnetic
This implies that
photons have spin equal to one, and hence must be bosons.
•
The laws of nature are invariant under time reversal. Therefore, a process opposite of absorption is
possible and it is called stimulated emission: an incoming photon can cause a transition from a higher
to a lower energy state, and hence stimulate emission of another photon of the same energy.
The
perturbative calculation of absorption rates also yields stimulated emission rates.
•
Finally, there is also spontaneous emission which occurs without any external electromagnetic eld.
An atom prepared in an excited state will eventually relax into its ground state and emit in the
process one or more photons.
Understanding this is not possible in the pure Schrodinger theory.
However, it becomes clear in quantum electrodynamics.
Vacuum unavoidably contains zero-point
quantum uctuations of electromagnetic elds. Namely, electric and magnetic elds are in some sense
conjugate variables, like position and momentum of a particle, so that there are appropriate Heisenberg
uncertainty relationships between them. It is not possible to keep both electric and magnetic elds
strictly zero at the same time. The resulting quantum uctuations can stimulate emission just like the
regular photons. However, since we cannot easily detect these zero-point uctuations, we perceive the
emission as spontaneous.
Spontaneous emission
•
It is experimentally found that an excited atom has a constant probability per unit time
a photon and relaxing to a lower energy state. The probability rate
A
A
of emitting
depends on the initial excited
and the nal relaxed state, but does not depend on time and is the same for all atoms of the same
kind. It can be calculated in the framework of quantum electrodynamics.
•
If a collection of
N (0)
atoms is prepared in the excited state at time
remaining in the excited state after time
t
will be:
N (t) = N (0)e−At
t = 0,
then the number of atoms
This behavior is seen experimentally and it is consistent only with a constant probability rate
A.
Each
individual photon-emitting relaxation event occurs at a random time, so the above relationship holds
statistically when
N (t)
is large. There is no internal clock in an atom which decides when the atom
will relax based on some mechanism unknown to us. Instead, if an atom is found in an excited state
t0 , it has the same average lifetime
0
since t = 0 until t = t to start observing
at any time
in that state following
waited
the atom.
t0 ,
regardless of how long one
Blackbody radiation and spontaneous emission
•
Einstein postulated the existence of stimulated emission before the development of Schrodinger's quantum mechanics. His postulate allowed a more natural description of the thermal equilibrium between
matter and electromagnetic radiation, relying only on statistical mechanics and quantization of photons.
•
For simplicity, assume that the material from which the blackbody is made of (a cavity) consists of
atoms which can be in two quantum states with dierent energies. The lower energy level is
the higher level is
atoms.
E1 .
Photons of frequency
ω = (E1 − E0 )/~
E0 ,
while
can be emitted and absorbed by these
The realistic existance of other atoms and transitions at other frequencies only extends the
following analysis to a wide spectrum of frequencies.
•
Thermal equilibrium at temperature
proportional to the Boltzmann factor
T implies that the average number of atoms Ni
e−Ei /kB T . Therefore, the ratio N0 /N1 is:
at energy
Ei
is
E1 −E0
~ω
e−E0 /kB T
N0
= −E /k T = e kB T = e kB T
N1
e 1 B
where
ω
is the frequency of radiation which can interact with the atoms. However, the actual number
of atoms in any of the two states constantly changes in time as photons are absorbed and emitted.
Ni
Only the average numbers
•
are constant over extended periods of time.
E0 will get excited to the state E1 is proportional to the
ω in the blackbody cavity. The total number of transitions E0 → E1
The probability that an atom in the state
number of photons
nω
at frequency
per unit time, or the transition rate, must be also proportional to the number of atoms in the state
E0 .
Therefore, the transition rate can be written as:
R0→1 = B0→1 N0 nω
where
•
B0→1
is an unknown constant.
Atoms in the excited state
E1 can spontaneously relax to the level E0 with some unknown probability A
per unit time. Furthermore, Einstein postulated stimulated emission. The probability of a stimulated
relaxation
E1 → E0
is proportional to the number of photons
The total number of transitions
E1 → E0
nω ,
just like the absorption probability.
per unit time is contributed by both the spontaneous and
stimulated processes, but must be overall proportional to the total number
N1
of excited atoms:
R1→0 = AN1 + B1→0 N1 nω
Both
•
A
and
B1→0
are unknown constants.
N0 and N1 must on average be time independent. This is possible only if
E0 → E1 per unit time is equal to the number of transitions E1 → E0 per
In the thermal equilibrium
the number of transitions
unit time:
R0→1 = R1→0
B0→1 nω N0 = (A + B1→0 nω )N1
N0
A + B1→0 nω
=
N1
B0→1 nω
65
•
The laws of physics are invariant under time-reversal. Absorption and stimulated emission are equivalent processes, one being a time-reversed version of the other. Since the constants
should not depend on
N0
and
N1 ,
B0→1
and
B1→0
we must conclude
B0→1 = B1→0 ≡ B
•
Therefore,
~ω
A + Bnω
A
N0
=
=1+
= e kB T
N1
Bnω
Bnω
where we used the earlier result for thermal equilibrium. From this we can obtain:
B
1
=
A
nω (e~ω/kB T − 1)
•
The constants
A
and
B
are microscopic parameters which describe quantum mechanical interaction
between light and matter. They cannot depend on temperature, which simply determines the statistical
distribution of degrees of freedom in dierent energy states.
ω
independent, the average number of photons with frequency
to:
B/A to be
T must be
temperature
proportional
1
e~ω/kB T − 1
nω ∝
Note that this is not the same as the Boltzmann factor
E = ~ω .
In order for
at temperature
e−E/kB T
appropriate for the photon energy
Instead, this expression reects the Bose quantum statistics of many identical photons, and
can be derived from fundamental statistical principles, without considering how photons couple to
matter (which is shown below).
•
n photons can be given by
n bosons in arbitrary single-particle states. If we concentrate on a single frequency ω ,
dierent allowed quantum states can have n = 0, 1, 2, 3 . . . photons at that frequency. The probability
−n~ω/kB T
that there will be n photons around is proportional to the Boltzmann weight e
- this is a
The statistical derivation : Photons are bosons, so that a quantum state of
any collection of
fundamental statistical principle merely dening the thermal equilibrium.
must be normalized:
p(n) =
where
z=
∞
X
e−n~ω/kB T =
n=0
1 −n~ω/kB T
e
z
∞ n
X
e−~ω/kB T
=
n=0
The average number of photons with frequency
nω ≡ hni = N (ω)
ω
at
•
ω,
N (ω)
∞
X
np(n) =
N (ω)
e~ω/kB T − 1
is the degeneracy of photon modes with frequency
having dierent wavevectors
k
such that
1
1 − e−~ω/kB T
is:
n=0
where
However, the probability
ω
(the total number of dierent modes
ω = c|k|).
We obtained an expression of this kind when we discussed blackbody radiation from the Planck's
postulate, and we used it to express the energy density of equilibrium electromagnetic radiation as a
function of frequency. The derivation based on the Planck's postulate alone did not have any basis in
the knowledge of any details about how radiation couples to matter. We only now see that radiation
and matter must interact in such a way that absorption, spontaneous emission and stimulated emission
take place. At the time of Einstein's derivation, stimulated emission was not known, but it's existence
was required in order to consistently explain the blackbody radiation.
•
Finally, we can determine the coecient
B
for the absorption and stimulated emission rates:
B=
This implies that the spontaneous emission
Re ,
A
N (ω)
stimulated emission
Rs
and absorption
Ra
rates are:
Re = AN1
AN1
e~ω/kB T − 1
AN0
Ra = BN0 nω = ~ω/k T
B
e
−1
Rs = BN1 nω =
Again,
A
is the spontaneous emission probability per unit time, which is a microscopic parameter
characterizing atoms and their coupling to the zero-point quantum uctuations of electromagnetic
eld in vacuum.
Lasers
•
N0 > N1
In normal circumstances
- more atoms are found in the lower energy state than in the higher
energy state, especially at low temperatures. This can be formally seen from the Boltzmann weights.
Therefore, absorption will occur at a higher rate than spontaneous emission. This is what keeps the
number of photons constant.
equilibrium and achieve
•
However, it is possible to drive the system of atoms out of thermal
N1 > N0 .
Then, the stimulated emission rate exceeds the absorption rate.
Each spontaneous emission event is caused by a photon incoming on an atom in the excited state. This
photon continues propagating, but stimulates the emission of another photon of the same kind, with
the same frequency. Therefore, one photon comes in, two come out. This kind of photo-multiplication
can carry on, leading to an exponential proliferation of photons. All of the created photons have the
same frequency, so the light is monochromatic. Using mirrors, the photons can also be focused into a
beam. This is the basic principle of laser operation.
•
Each stimulated emission is accompanied by the relaxation of one atom. A relaxed atom is incapable
of undergoing another stimulated emission. So, in order to maintain a continuous beam of laser light it
is necessary to provide an independent mechanism which promotes relaxed atoms back to their excited
state. We do not want this mechanism to be based on the absorption of the same photons we wish to
create. There are many known ways to accomplish this eect, the so called pumping which maintains
the inverted population
N1 > N0 .
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