[MUSIC PLAYING] Stanford University. Group velocity and
phase velocity-- OK, good. All right, let's talk
about it a little bit. Generally, the phase
velocity of a wave doesn't have anything to do
with measurable properties of the wave of the measurable
things in quantum mechanics. It's the group
velocity, which is really the velocity
at which signals move, the velocity at which
particles move and so forth. So let me just show you what's
at stake and what's going on. The first thing is if
you have a plane wave-- a sine or cosine or
an e to the ikx-- and that's all--
just e to the ikx. Let's write to e to
the ikx minus omega t. Or just sines-- the sine of
kx minus omega t will also do. In fact, let's just write that. Sine of kx minus omega t-- looks like that. And how does that wave move? How does the top
of each crest move? The top of each crest
moves in such a way that this thing inside
the sine is constant. In other words, for example,
right over here at time t equals 0. And at x equals 0, the sine wave
is not a maximum at that point. But x equals 0 and k equals
0-- the sine wave happens to be over here. OK, but where is the sine wave
after a certain amount of time t? That point here has moved. It's moved to the place where
kx minus omega t is equal to 0. Where is that? That's at the position x is
equal to omega over k times t. So it's dividing by k. Now that's not too surprising. If you think about it for a
minute, omega is a frequency. So it's inverse to the
period of oscillation, to the time of oscillation. It's 1 over the
time of oscillation. There's a 2 pi in there, but
the 2 pis appear in both things. So omega is like 1 over
the period of the wave-- 1 over t, the amount
of time that it takes for the wave to oscillate-- and k is like 1 over lambda. Actually, there are two pis
in these equations-- two pis in the numerator. Let's put them in. And so omega over k-- what's omega over k? Omega over k is lambda over t. It's the wavelength
divided by the period. The wavelength is the distance
it moves in one oscillation. The period is the
time that it takes. The ratio is the velocity. So this is the velocity
of the peak of each wave. OK, for a plane wave that's
called a phase velocity. Now let's think about
Schrodinger waves for a minute. Schrodinger waves have a
connection between omega and k. To remember it, all
you have to remember is that omega is
really secretly energy, and k is really
secretly momentum. If we set h bar equal to
one and so using the formula that energy is equal to momentum
squared divided by mass, that just becomes omega squared
is equal to k squared divided by 2m. Did I say momentum
squared divided by mass over twice the mass? This is, as usual, with
h bar equals c equals 1. In fact, there's no
c's in this formula, but there are some h bars. OK, so that will tell
you, for example, let's compute the phase velocity-- the phase velocity of
a Schrodinger wave. For a Schrodinger wave, Omega-- sorry. Omega is equal to k squared
over 2m, not omega squared. Omega is k squared over 2m So what is the group velocity? Its omega divided by k-- omega divided by k is
equal to k over 2m. That's the phase velocity-- k divided by 2m. What's the velocity
of a particle in terms of its momentum and its mass? Momentum divided by m. There's an extra
factor of two here. What's that doing there? Peculiar thing--
the phase velocity is not the velocity of the
motion of a classical particle. OK, let's just
keep that in mind. That's true, but let's
do something else now. Supposing I add a constant to
the frequency of every wave. I simply take the
frequency of every wave-- all of them for all k-- and I add the same constant. What should I call
that constant? You got a name for it? It's mc squared, but
let's not belabor. It's not important. Let's just add a constant--
an overall constant to the frequency of all waves. What does that do for
the phase velocity? Well, it changes
it omega divided by k will now have
an extra term in it. It'll have an extra term in
it, and it won't be the same. Whatever It does,
it will be c of ak. It'll have an extra term in it. The phase velocity
will have changed. But I tell you right
now, there is no physics in just adding a constant to the
frequency of every Schrodinger wave what does it correspond to? It corresponds to
adding a constant for the energy of a particle
and adding a numerical constant to the energy-- doesn't ever change anything. Only energy differences
are ever important. Can we see what's going on in
terms of the Schrodinger wave? Well, the Schrodinger
field, psi, is some kind of
sum or integral-- either a sum or an integral--
of creation operators, rather annihilation operators,
in this case, e to the ikx, e to the minus i omega t, where
omega is the omega, which is related to k. So let's write omega of k. Now supposing I add to
omega just this constant c. What does it do? It adds in or multiplies in
each one of these waves-- sorry, e to the minus ict. All of the waves in here
that make up the Schrodinger field all get multiplied by the
same time-dependent phase here. So what it does is if you had
a solution of the Schrodinger equation and you added
a constant to the phase, it would just
change the solution. Let's say psi of x and t. It would just multiply
it by e to the minus ict. That's all it would do. It would multiply the wave-- sorry-- by psi of x and t. It would change the
wave by multiplying it by a time-dependent phase. All of the quantities
of physical interest that are made out of
the Schrodinger field involve psi times its
complex conjugate. The probability density, for
example, is psi times psi star. All possible
expectation values-- quantum mechanical
expectation values always involve psi times
its complex conjugate. So what happens to psi times
its complex conjugate when you perform this operation? It cancels out. It cancels out because
psi star gets multiplied by e to the plus ict. And when you multiply
them, they cancel out. As a consequence, there is
no physical measurability to adding a constant
to all of the phases. And nevertheless, it
changes the velocity of the Schrodinger waves. So there must be something
about this change of velocity-- which is, well, it's
change of frequency. Let's say, for the
phase velocity, the phase velocity
is not something which is ever measured
or ever appears when you multiply psi time psi star. So it's an artifact of a
particular mathematical set of conventions. On the other hand,
the group velocity really does mean something. It is the motion of the wave. If you have a wave packet, which
is centered in some place-- in other words, if you have a
bunch of waves which add up, a bunch of waves, bunch
of plane waves, which add up to something
that looks like that. Maybe it has some
oscillations in it. But what you would
call like a lump, the whole lump moves
with a velocity that's called a group velocity. But let's see if we can
see just go roughly why. If you take a whole
bunch of plane waves-- every plane wave
looks like this-- goes on and on forever. How is it possible to add
them up so that you get a concentrated lump someplace? The answer is
destructive interference. Destructive interference--
so that the waves are in phase and add up over
here, and they're out of phase and cancel over here. So the question then
comes up, if we're interested in the
motion of waves and how waves of slightly
different frequency reinforce each other and make
high points and low points, let's try to follow
how that works. Let's take two waves,
not a single wave but let's take two waves of
very neighboring close momentum and try to follow where the
reinforcing constructive interference takes place. That'll be good enough. Let's just follow where the
constructive interference takes place between the two waves. So in particular,
take e to the ikx-- we could just take sine. Sine is good enough. Sine kx minus omega of
kt and let's add to it-- this is just a way
of moving along with momentum k, energy omega. And let's add to it
another wave, which is-- let's put parentheses
around this-- sine k prime where k prime and
k are very close to each other. X minus omega of
k prime times t. Where are they in phase and
where are they out of phase? In phase means they
reinforce each other. Out of phase--
and in particular, it will mean that the two
terms here have the same sine-- and there are other places where
they'll have opposite sine. Well, one easy way to find out
where they reinforce each other is to ask when is the argument-- argument meaning the thing
in the bracket here-- when does the argument
of this wave the same as the argument of this wave? Those are the
places in that case, in that particular
place where the argument of this wave and the argument
of this wave of the same. It's clear the waves will
reinforce each other. They'll reinforce each
other, and they will be constructive interference. Where is that? That's at the place where
kx minus omega of kt is equal to k prime
x minus omega-- let's just call it omega prime. Omega prime means
omega of k prime t. But let's write this
in the following way-- k minus k prime x is equal
to omega minus mega prime t. Two neighboring waves
will reinforce each other along a trajectory where x and
t are related by this equation. OK, let's just divide
it by k minus k prime. And they'll assume
that k and k prime are very close to each other. Just for simplicity, let's
imagine that they only differ by a very small amount-- a small fractional amount. Then what is omega minus omega
prime over k minus k prime? That's the derivative of
omega with respect to k. That's all it is a derivative
of omega with respect to k if omega and omega prime
are near each other. And so what it says is the
place where they reinforce each other will travel
along an orbit, which is x is equal to the derivative
of omega with respect to k. d omega by decay times t. This is also a velocity. It's the velocity
of the place where the waves reinforce
each other, and it's called the group velocity. It's the group velocity. Let's calculate it. If omega is k squared over 2m
plus an arbitrary constant-- this is the phase
velocity, v phase. What about the group velocity? That's the derivative of
omega with respect to k. Derivative of omega
with respect to k-- that's equal to 2k divided
by 2m, otherwise known as k over m. And what do we get from here? Nothing, zilch. This is the group
velocity, v group. Notice two things-- the group
velocity does not depend on this extra additive
thing that I put into each frequency-- number one. And number two, it's
exactly the velocity of a non-relativistic particle
momentum divided by mass. So the 2 went away. Yeah, that's right. The two went away-- exactly. When I differentiated k,
I got 2k divided by 2m, and 2 went away. So the group
velocity is the thing which is most closely associated
with the classical motion of a particle. It's also the only
thing that's associated with propagation of
signals and so forth, not the phase velocity. We could go ahead a
little more with this. Let's see, that we want to? If the speed of light is c, is
that phase or group velocity? Both. So let's go through that. Let's go through that
and see why that's true. The c here is not the speed
of light, incidentally. This is not the speed of light,
but let's go through that. Let's take a relativistic wave
and calculate both the group and phase velocity. For that, we need to
know a factor too. We need to start with the
relationship between energy and momentum. Sometimes they talk
about light goes faster than the speed of light. No, no, it doesn't. It doesn't. It doesn't. It doesn't. It doesn't. It is a good question, but it
doesn't have to do with that. It's a separate issue. Could he be thinking of the
classic [INAUDIBLE] waveguide for the phase velocity
is always greater than the speed of light? The group velocity
is what was left. The product is the
speed of light. Yeah, so let me give you an
example of that happening right now, not for a waveguide,
not a waveguide, just a wave equation for
a particle with a mass. The wave equation is
nothing but the relation between omega and k. To hell with the wave
equation-- just the relation between omega and k for
a relativistic particle-- particle moving close to but
not at the speed of light. It's got a mass. It's got a mass. If it didn't have
a mass, it would move with the speed of light. OK, what's the connection? Omega is energy. Let's write this
first in the form that we might remember best. What's the relation
between energy and momentum for a particle, which
is relativistic? [INAUDIBLE] What's that? Momentum equals [INAUDIBLE]? The relation with
momentum, p, square root. OK, square root of p
squared plus m squared. But do we want to put the
speed of light into here? I think there's a c to
the fourth and a squared. But let's set c equal to 1. Yeah, we've already
set c equal to 1. Later on, we can put back
the c's if we want to. It's not necessary. Here's the energy. And if we now set
h bar equal to, 1 this is exactly the
same as writing omega as equal to square root of
k squared plus m squared. So for a relativistic wave,
a relativistic particle, the frequency is the square
root of k squared plus squared. Now if k happens
to be equal to 0, then it's omega equals
the magnitude of k. Apart from a speed
of light, which I set equal to 1, that's the
relationship for a wave moving with the speed of light
that omega is equal to k. What's the group velocity? First of all, what's
the phase velocity? The phrase velocity
is omega over k. The phase is omega
over k and that's 1. What does that mean-- 1? Speed of light-- right. What about the group
velocity v group? That's the derivative over
omega with respect to k. What is that? 1. The group velocity
and the phase velocity are the same for a wave
with mass equal to 0. Forget that this is mass. This is just a mass. Of course, it's mass. Energy equals square
root of momentum squared plus mass squared. But now let's calculate
the group velocity when the mass is not equal to 0-- well, both of them. Let's calculate both of them. Omega over k-- that's
the phase velocity. That's the square root of k
squared plus m squared over k, but let's put it all
inside the square root. Let's put it all
inside the square root. Then it becomes k squared
percent squared over k squared. Is this is bigger than
or smaller than one? Bigger than one-- k squared
plus m squared over k squared is bigger than 1. So the phase velocity omega
over k is bigger than 1-- bigger than the speed of light. OK, oh we get super scared? Waves moving faster
than the speed of light. But what about the
phase velocity. Let's work that out. Group velocity-- thank you. The group velocity
is the derivative of omega with respect to k. What's the derivative of
omega with respect to K? Whoops, not t-- d omega dk. We have to take the
derivative of a square root. That gives us 1 over
twice the square root of k squared plus m squared. And then we have to
differentiate the argument here with respect to k, So
That just gives us k. This is k over this. This is the derivative,
which is also equal to the square
root of k squared over k squared plus m squared. Isn't that nice? The group velocity is
squared of k squared over k squared plus m squared. The phase velocity
is square root of k squared plus m
squared over k squared. They're inverse to each other. This one is greater than 1. This one is always less than 1. This one is less than 1. In fact, this is the velocity
of a relativistic particle with momentum k. k squared-- thank you. Yeah, they're just
inverse to each other. And as it happens-- as was pointed out by somebody-- the product of the group
velocity, the group velocity, and the phase velocity
are equal to 1. So they're just
inverses of each other. Now that's somewhat accidental. That's not a very general fact. Buried in there somewhere
is that the wave speed must be a function of frequency. What? Wave speed must be a
function of frequency. If the wave speed is
constant to the frequency, you get no dispersion. That's right. That's this case
here where omega is equal or proportional to k. That's the case where all
waves have the same velocity. So it's only in the
case where all the waves of different frequency have the
same velocity that the group and phase velocity are the same. That's a good approximation,
not only for light. It also happens to be
a good approximation for sound and other things-- to at least lower
frequency sound waves. They all move with pretty much
the same velocity and so they-- Then if it's not true,
then the [INAUDIBLE].. That's right. That's right. So that's right. That's right. So the difference of these
two kinds of velocities is connected with wave
dispersion with the fact that waves change shape
absolutely, absolutely. OK, so that's the story about
waves, about group velocity, and phase velocity. And as I said, in quantum
mechanics, for example, here the effect of an
additive shift in omega is no physical effect whatever. It's an example of the fact that
most times, the phase velocity is irrelevant to real physical
propagation of any signals or any energy flow or
anything like that. Probability flow. You said that you can't
measure the phase velocity because these complex-- At least for the
Schrodinger wave-- yeah. Well, for a photon,
they're the same. They're the same. Now it gets a little
more complicated when you have various kinds
of dispersal of materials that the photon
can move through. But that's for these
squalid state physicists. Is all of this waves that
don't spread out over time? Waves that don't
spread out over time are the ones for which all
waves have the same velocity. If different components of the
wave have different velocities, then it's going to spread out. So that's the case where omega
is a non-linear function of k, basically. If omega is a linear function of
k, then they don't spread out, and that's the case of always
having one universal velocity. At some point, maybe if
somebody comes back and asks me this question about
light cones and things, we can come back to it. But it was a good question, but
it's not quite related to this. Let's see was there anything
else I wanted to fill in before getting to fermions? Yeah, before getting
the fermions, let's discuss the issue
of momentum conservation. We discussed the issue
of energy conservation. And I showed you an example. It was really just an example
of how the time translation and variance of
the amplitudes that go into a scattering process
in quantum field theory conserve energy. I'll just remind
you very quickly and then show you how the
momentum conservation works-- very, very similar. We will come back
to these things, but then I want to
get onto fermions. That's what I want
to get to tonight. Where were we? I was going to talk about
momentum conservation just really quickly. Remember how energy
conservation works? Energy conservation we had,
for example, some process. It doesn't matter what it is-- a bunch of psi
annihilation operators, a bunch of psi
creation operators act on a state with a bunch of
particles in it with momenta. And then we calculate the
overlap of the inner product or the probability
amplitude for a final state with a bunch of other particles,
k prime, blah, blah, blah, blah, blah. OK, these fields act
on the initial states. These fields act on
the final states. And I'll just remind you a
little bit how it worked. We said supposing all
the processes take place at the same point
of space and time at the same point or
the example that I gave was one in which
one particle came in. One particle went out. And they came in and scattered
off a point in space, but we integrated over
all possible times. The particle was there,
not this particle, but the target was
there at all times. And so a particle could
have scattered at any time-- whenever it gets there. And that's represented by
integrating this over time. If you remember, each one of
these size or each one of these sizes made up out of a bunch
of things involving a's, e to the ikx's, and e to
the minus i omega t's-- omega as being the energies,
k's being the momenta. Now if we said all of this took
place of the origin of spatial coordinates, space coordinates,
then this is just equal to 1. When the operators act and
we're finished calculating everything, the only time
dependence in this amplitude comes from these e
to the i omega t's. Let me remind you
what it looked like. It looked like e to the minus
i omega of each momentum k1, t, e to the minus 1 omega of
k2 t dot, dot, dot, dot. That's for the initial states. And then for the
final states, there was something similar e to the
plus i omega k prime one blah, blah, blah, blah times t. Blah, blah, blah
means the sum of all of the outgoing
momentum, and here it's minus the sum of all
the incoming momenta-- omega k incoming times t. The last step was to
say if we integrate this over all times
representing the fact that the scattering could
happen at any times, the integral of
this gives what-- integral over time? A delta function-- a
delta function of the sum of the final momentum minus
the initial momentum-- sorry, final energies
minus the initial energies. When we integrate
this over time, what we get is a Dirac
delta function of let's call it omega final. That means the sum of all the
outgoing energies minus the sum of all the incoming energies. In other words, the answer is
0 unless the incoming energies are equal to the
outgoing energies. And this was just
as a consequence of saying that the process could
happen with equal probability at any time. Now what about
momentum conservation? Momentum conservation is
exactly the same sort of thing, except it has to do
with space translation. If a process can happen
at any point of space with equal probability, then
it's momentum that's conserved. Let me give you
just one example-- one very simple example. This is a process in which a
particle comes in and splits into two particles. Can this really
happen in nature? Yes, it can really
happen in nature-- particles decay. Particles decay. And an incoming
particle of one momentum can decay into particles
of other momenta. But as you know,
momentum is conserved, but let's see if we
can see why that is. So the incoming
particle is described-- well, of course, it's described
by a particle coming in. The outgoing
particles-- this is k-- outgoing particles k
prime 1 and k prime 2. The operator, which eats
the initial particle-- let's call that psi of x-- misrepresents the particle
being absorbed at point X. and then two particles created-- also at point x. That's a simple example. One particle is
eaten at point x, and then it spits out
two particles also at the same point
for simplicity. But it can happen at any place. It can happen in
any place and that means we have to integrate
this overall space. Let's see what we get. Again, these size have creation
and annihilation operators. The sum annihilation
operator in here-- which can annihilate or
just work on this particle-- eat that particle. That's a minus of k, and it
has an e to the ikx in it. Let's forget the omegas. We don't need the time
dependence in here. So that's the operator, which
eats this particle over here, and then there's these two
operators, which put back the particles over here. And they give us some a plus
of k1 prime, a plus of k2 prime, and then there's
an e to the minus i k1 plus k2 prime x-- same x. These operators act. They annihilate the particles. We calculate the matrix element. That's all trivial. The only thing of
interest that's left over is this function of position. This is the amplitude that the
process happens at point x. But then since x is
not a special place, it can happen anywhere as
we integrate it over x. What do we get when we
integrate it over x? A delta function of
momentum conservation. All right, so if a
process can happen with translation and
variance in space, then it will conserve momentum. And in fact, it didn't
matter how many particles were coming in and going out in
this little demonstration here. It would work the same way-- translation and
variance or the idea that the process could happen
anywhere with equal amplitude. That's momentum conservation. The idea that it can happen with
equal probability at any time-- that's energy conservation. There was another symmetry
we talked about a little bit. The other symmetry that we
talked about a little bit was a possible symmetry. If we multiply psi by e to the
I times-- let's call it lambda-- in other words, just
change the phase of psi-- what will that do to
probability amplitudes? Well, the answer is
it will do nothing to them as long as it's the same
number of psis and psi daggers. If there's the same number
of psis inside daggers, that's the same thing as saying
that the number of particles coming in is the same as the
number of particles going out. And so symmetry with respect to
changing the phase of the wave function like that-- just
multiplying it by an overall phase-- and multiplying its
complex conjugate by the opposite phase,
that's associated with certain conservation
laws, such as conservation of electric charge. Conservation of
electric charge tells us that is contained in
symmetries of this type here. If these were charge
carrying fields-- charge carrying fields
mean fields for electrons, for example-- if they were fields
for electrons, then this symmetry
would be correct only if there were as many
electrons in the initial state as in the final state. This is, of course,
a law of nature that charge is conserved. So the law of nature
that charge is conserved is the same as the
symmetry under changing the phase of the wave function. These are some
examples of symmetries but that was a quick review. Now I want to move on to
the notion of a fermion. Yeah. When you were doing the
integration over space, that's overall three
spatial dimension? Yeah, [? say. ?] That would
mean conservation of all three components of the momentum. Exactly. Yeah. Then if you have two positive
k's and a negative k, the k's may have a distribution. In other words, you could have
one is larger, one is smaller, the other one is-- Yes, yes, yes, yes. Is there anything in here
that could point to-- Some relative probability
for different ways that the momentum is shared. Yeah, not for this case here. For this case here is
basically neutral with respect to the waves. The momentum is shared. [INAUDIBLE] Here, just the number. We'll come back to these things. We'll come back to it. We'll come back to it. I simply wanted to give some
very, very simple illustrations of the connection between
symmetries and conservation laws and how quantum
fields are used to describe physical processes
of creation or annihilation of power of particles. Yeah. A technical question. I have, in my
notes, where we were doing a sum over k of
annihilation operator times either the plus ikx
times some state. And I wrote that it
reduces to one term. The state has a
product line in it. There's only one particle, There's one particle I'm
annihilating by summing over k, and I wrote that it
reduces to 1 term. And I wanted to ask
whether that means that if the terms
which are annihilating places where there
are no particles-- 0-- gives you 0. Yeah, 0 vector-- 0 vector of 0. It's nothing at all. Let's now depart from
the theory of bosons The characteristic of
bosons is, first of all, you can have any number of
them in any quantum state. And as we described a
couple of lectures ago, they have a tendency to want to
accumulate in the same state. An example of that is if a
particle decays and produces a photon, for example, an
atom decays and produces a photon it may produce
a photon in any direction with any momentum with some
probability distribution. But if at the same
time as it decays there happens to be a
large number of photons all in the same state-- in other words, all moving the
same way, with the same energy, and the same frequency-- it will preferentially decay. So that the photon
that comes out has the same momentum as
the extra photons, which are already there. That's the square
roots of n that occur in the boson operators. Fermions are very different. Fermions don't like each other. They don't like to congregate
into the same state. In fact, they simply can't
be in the same state. This, of course, was a discovery
that was first made by Pauli, I guess. Was it Pauli? Yeah, sure, Pauli
exclusion principle in the context of atoms
that two electrons never occupy the same quantum
state within an atom and that was generalized
by Pauli himself-- Pauli Fermi, others-- to
a powerful principle that fermions of particles
that simply can never be in the same quantum state
in atoms, out of atoms, anywheres very
different than bosons. OK, let's work out
some of the mathematics or some of the bookkeeping. The mathematics of creation
and annihilation operators is really a form
of bookkeeping-- useful bookkeeping. But let's try to
work out a theory of quantum fields of fermions-- particles that can never
be in the same state. Well, we start with the
same kind of construction, as we did for bosons. Imagine a quantum
field psi of x, which is, again, exactly the
same formula as for bosons. Summation over all of
the allowed momenta-- I guess I call it k, usually. All of the allowed k, if the
allowed k are continuous, it becomes an integral. Annihilation operator for
a particle of momentum k, e to the ikx. The states of the
system are again labeled by how many
particles there are in each quantum
state, but the rule is never more than one. So the state of the system
is a bunch of 0s and 1's, never more than one
particle in the same state. That's the character
of a fermion. Now, are there other
kinds of things that can have three particles
in the same state but not more or that must have more
than seven particles in the same state or
other odd varieties? The answer is no. But to prove that the
mathematics of quantum mechanics together
with special relativity dictate that the only
possibilities are fermions and bosons-- would take us far
beyond this class so we're not going to do that. But we're are going to
describe what a fermion is. All right, so this formula is
exactly the same as for bosons. Psi dagger of x-- same thing. Incidentally, this is all
non-relativistic particles. We have not got to
relativistic particles. Eventually, we will. Same thing with
creation operators-- a plus of k t to the ikx. But now the rule is a little
different than before. Yeah, there is a minus sign. All right, so now
let's focus just on one possible momentum
of the electron, or this could be an electron. Let's focus on one possible
state, not worry about all of the states. But let's just say
some particular quantum state is either filled with
an electron, or it's empty. Those are the only
two possibilities. Let's represent that. I guess we can
represent that by 0-- that represents no
electron in that state-- or one-- that represents
one electron in that state. This isn't multiplying. This is just two
possible states-- put a comma here. What is the algebra or
the abstract mathematics of the creation and
annihilation operators? Let's see if we can
make some guesses. You'll make the guess. The creation operator on
the state with no particle-- want to guess? One particle. What else could it be? What about a creation operator
on a state with one particle? 0. Now, 0 does not mean the
state with no particles. It just means the 0 vector. I don't like that as a just 0-- It doesn't mean anything or
there's not a vector at all. It's a non-state. Hmm, what's that? Louder. Is the operator still linear? Yeah, they're all
linear operators. In quantum mechanics, all
operators are linear operators. When one speaks
about an operator-- especially in
quantum mechanics-- it's always a linear operator. I think mathematical terminology
is that operator usually means, maybe always
means linear operator. Otherwise, it's operation,
but I'm not sure. But yeah, in quantum
mechanics, operators are always linear operators. OK, let's see if we can
make some more guesses. What about a minus on o? Well, you can't take away
that which is not there. But what about a minus on
the state with one particle? No particles. No, no particles. All right, so these
are operators. They do what they do. They do what they do. Just to clarify-- the 0 there--
that means that doesn't happen and you can't do that. Yeah. Yeah. Yeah. Let me give an example. If we described harmonic
oscillators by wave functions-- ordinary wave
functions-- then there would be a wave function
for the ground state, and the wave function
for the ground state. Well, let's just discuss
this for a minute. The wave function
for the ground state would be some Gaussian wave
function like e to the minus x squared-- that would be the wave function. This would be identified
as the ground state and therefore, the state
with no excitations. On the other hand,
the wave function psi of x, which is literally
equal to 0, that's the state 0. No wave function at all. Probability is just 0. It's not a state. It's just a nothing. OK, this is the starting
point for the algebra of fermionic creation and
annihilation operators. There's nothing in
the notation that helps us distinguish between
fermions and bosons then? Well, yes, I suppose we should. Yeah, it's a good point. We should probably use
a different letter. Let's not call
fermionic operators a. Could we call them alpha? We could call them alpha. Sometimes C is used. Why don't I use C? OK. The problem is that there
are many different, not inconsistent, but many different
notations in the literature. And I guess, for
us, we probably are wise to use a notation
where bosonic and fermion operators have different
letters associated with them. So let's call it C. This is a common notation also
get that fermionic operators I'll label c. But they are the
analog for fermions of the a's for bosons-- whoops. Right, let's see if we can
figure out what C dagger C-- C+ times C- does. What does that operator do
when it acts on anything? So let's figure
out what it does. What happens when we act on O? Well, C-, when it acts
on O, just gives 0. And C+ it is not going to be
helpful in getting it back to being non-zero. 0 is 0. So this is 0. You simply can't take
away what's not there. Even if you try to put
it back afterwards, you can't take it
away to begin with. What about C- times C+ on
O. What does that give? C plus-- when it
gives O gives what? 1 and then c minus? O. Let's call it O. Let's
call this O, as opposed to 0. So this gives back the state. In other words, it gives
back exactly the same state. Let's see it gives back
exactly the same state. What about C+, C- plus C-, C+? If it acts on O-- let's see. What does it give
when it acts on O? We're just adding
those two equations. It just gives O.
The ground state-- sometimes known as the vacuum. All right, let's now do the
same thing acting on one. What is C-, C+
plus acting on one? Well, C- minus an act on
one, takes the particle away, and leaves O, the vacuum,
and then C+ puts it back. So this gives one back again. What does it give
when it acts on-- One. Four. Acting on one. So minus-- yeah, that's
correct, isn't it? Yeah. Yeah. OK. Now what happens
when we do this at 0 because we can't add another
particle to one that's already present if it's a fermion. So this is 0. What happens if we add
these two equations? We find again C+, C-,
C-, C+ on 1 gives 1. So those two are equal
to i, including that one. Yeah. Both of these operators-- while this operator, this
particular combination when it acts on any state gives
back exactly the same state. So that's written in operator
notation by saying C+, C- plus C-, C+ is equal
to the unit operator. It just gives back
the same state-- whenever it acts on. This combination-- an
operator times another-- plus the interchange. What would you call if it
there was a minus here? You would it the commutator. With a plus here, it's called
the anti-commutator-- that's what it's called,
and it's the symbol for it is a curly bracket. Sometimes they use a plus sign. Sometimes. Curly bracket of A and B is,
by definition, AB plus BA. So the first thing we learn is
the creation and annihilation operators. Anti-commute to give one
on the right-hand side. This is incidentally
the analog-- yeah, it doesn't matter
which order you write them in since they're the same-- well, now, sorry-- they're not
the same but doesn't matter which order you write them in-- C+, C-, C-, C+--
plus same thing. This is the analog for
bosons of a commutator of A- with A+ is equal to 1. Analog is not the same
thing, but it's the analog. In this case, it's
commutated for bosons-- in this case, it's
anti-commutator. Now what about the
anti-commutator of C+ with C+? The anti-commutator of C+ with
C+ is just C+, C+ plus C+, C+ is just twice C+ squared-- twice this. What does this give when
it acts on anything? Well, if it acts on the vacuum,
it can create a particle, but then the next
one is going to try to create a particle on top
of a particle which is already there. You can't do that. If it acts on a particle
which is already there, can't do that. So whatever it acts on
C+, C+ always gives 0. All right, so C+, C+ gives 0. What about C-, C-
anti-commutator, C-, C-? That's just twice C- squared. If it acts on a state
with a particle, the first operator
removes a particle, and the second one looks for
particle, can't find one, and says 0. If it acts on a state without
a particle, it also gives 0. This is the analog for harmonic
oscillators of commutators. Is that 0 or O? 0. No, I mean, on C-, C-. Is that the vector for O? This is an operator. It's not a vector. Right, it's the operator 0. Yeah, it's the operator 0. Right, it's the operator 0. The operator 0 is
such about when it acts on any vector or any state, it gives 0. It really is literally 0. Whenever it acts, it just
doesn't give anything. It gives 0. There's an operator,
which is 0, and there's a vector, which is 0, not a
vector but a state vector-- a state vector which is 0. Two different uses of
0 but they're both 0. And in fact, when
the 0 operator acts on any vector, state vector,
it gives the state equal to 0-- 0 vector. OK, so fermions and bosons
have a parallel mathematics to each other where wherever
you see commutator for bosons, you replace it by
anti-commutator for fermions, and it is a consequence
of this that you can't put two particles
in the same state. If you try-- let's try
to put C+, C+ on O-- and C+ times C+ is 0. That's what this
equation is saying. C+, C+ plus C+,
C+ is equal to 0. So this algebra incorporates and
encompasses the idea that you cannot put two particles
into the same state described by these creation and
annihilation operators. I want to prove a little
observation about this. These Cs here create
and annihilate particles with given momentum. What we prove, what we've
assumed that some combination approve and assume-- oh, incidentally,
yeah, we should write this more completely. Supposing we're now talking
about creation annihilation operators for different
momenta, k and k prime, then I will just tell you
this is delta kk prime. In other words, it's one if k
equals k prime, 0 otherwise, and these things
are always just 0. C-, C- minus I'll just equal
to 0 for all k and k prime. That's the algebra of creation
and annihilation operators, which if you like,
by definition, this defines the algebra of
creation and annihilation operators for fermions. The fact that it really does
describe fermions in nature, of course, is a
marvelous correspondence between mathematics, simple
mathematics, and some physics. But here's an
interesting question. As defined up till now, what
this says is that you cannot put two particles into
the same momentum state. You cannot make two particles
carrying the same momentum. Two fermions, two fermions of
the same kind, incidentally. Two fermions of
the same species-- you cannot put them into the
same state, momentum state. Can you put them into
the same position state? Well, what this tells us is
that if we find a particle with momentum k fermions an electron,
we cannot find another one with exactly the same quantum
state, same momentum. What about the same position? Can we find two particles at
the same place in not momentum but in position? So let's try to find out. How do we create a
particle at point x? We create a particle at
point x using psi of x. Psi of x on the
vacuum represents a particle at point x. Oh, sorry. I think we should have side
dagger creation operators. So the question is what happens
if you try to create two at the same point? Well, let's see what we have. That means we're going to have
a sum over a k and a k prime of C+ of k, C+ of k prime. Let's do it. Let's do it at the origin. Let's just do it at position
0, then ez to the ik. x's aren't there on 0. What is this? Anybody tell what this is? All terms are 0, and k
does not equal k prime. No, no, there's 0 in any case. You can rewrite
this in the form-- C+ of k, C+ of k prime 1/2 C+
of k plus 1/2 C+ of k prime C+ of k because it's under
the summation here. But this is just the
anti-commutator of two C+s. The anti-commutator of C+s are
all equal to 0, so this is 0. Why? Because the
anti-commutator of C+ is 0. Should've asked you that
before because I wrote it on the blackboard. Well, yeah. Yeah. No. Right. Right. All of this is very clear if
we're talking about only one k. It gets some will
impact nontrivial impact when we generalize
it in this fashion here to all k and k prime. So there's basically some
physics that's going in here. And what is this physics? It's the physics that
not only can't you put two particles into
the same momentum state, but you can't put them into
the same position state, for example. I'm showing you is
that with this algebra. If this algebra is
correct, then it's not just that you
can't put two particles into the same momentum state. You can't put them into
the same position state. In fact, you can't put them
into the same state altogether. But this was just an example. With this algebra, it says that
you cannot put particles either into the same momentum state or
the same position state or any other state, for that matter. So these are fermions. This is the pattern
for fermions. Yeah. I've always maybe
misunderstood that you could have two electrons
at the same momentum if there were four and not one. Yeah. Yeah. The correct statements is
you can't have two electrons in the same state. Now, what constitutes a state? What constitutes a state may
not just be its momentum. It may be its
momentum and its spin, which we haven't come to yet. This is one of the
things we're going to come to shortly is
the concept the spin-- very important concept. We haven't come to it yet. So at this stage, we're
imagining that a particle is characterized by a momentum. And if it's a
fermions, you can't have two in the same momentum. Now the correct statement is,
of course, you can't have them in the same state. And if a state requires
other ingredients besides its momentum,
for example, its spin, then you
can't have two particles with the same momentum
with the same spin. But spin-- there's only really
two parallel, anti-parallel. But what about two electrons
and two hydrogen atoms that are separated some
distance away each other. Could they possibly
have the same momentum? Momentum? Momentum-- I'm just going
to stick with momentum. You want the atoms. The hybrid-- just,
well, just an electron. I maybe used a bad
example of that. Well, electrons in
the hydrogen atom are not in momentum states. They're in atomic energy levels. Now the atomic energy levels
over here and over here for two different atoms are two quite
different wave functions. So if you have an
atom over here, an electron might
occupy a wave function that looks something like this. Another atom over here
might have an electron, which looks like this. This wave function and that
wave function are not the same. They're quite distinguishable
by the fact that one of them is over here, and one
of them is over here. So yes, you can
have two electrons in two different atoms both, for
example, in the ground state-- in some particular state. Yes? But you can't have two
electrons in the same atom with the same quantum numbers. To be more generic-- What's that? To be more generic,
instead of specifically talking about atoms, could
you just use the term systems? Is that used in this language? States is the right statement. The right nomenclature
is you can't have two electrons in
the same quantum state, in the same quantum state. A system means a collection
of degrees of freedom, but it doesn't imply
a particular state. So an electron is a system. An electron can be in
a particular state. So nomenclature is
electrons' system. The state of the
electron is perhaps some particular orbital and
spin state in a particular atom. Can I just carry
that one further? Let's suppose you have an
electron beam in a picture tube, cathode ray tube. You have a stream of electrons. So that means that no two of
those in that little glass jar-- Yeah. Yeah. I really like this--
a packet over here and another packet over here
and another packet over here and another packet over here
and then not in the same state. If they're not in the same
momentum, [INAUDIBLE]?? They're not. They're not in
the same momentum. They're not in
the same position. And in big contrast to
the idea of photons, photons march down the
axis all, for example, in the same plane wave. That's why you can't build up
a coherent beam of electrons in the same sense that you
can have a coherent beam. You just can't put more than
one of them in the same state. So a formula is, in many
respects, a bigger abstraction. Than a boson? Can you have two electrons
in the same place with different momenta? No. No, no, no, no, no, no, no. No, no. Remember you either
describe a thing by a position or a
momentum, and not both. The position and
momentum are intertwined. Mutually incompatible
[INAUDIBLE].. Right. So it's only like
spin position then. That's because spin commutes
with momentum and position, but position and
momentum don't commute. So you can't speak
about an electron at a position with
a momentum, but you can speak about an
electron at a position with a particular spin. Is there some sense in which
this found electron over here and that found the
electron are, therefore, in that sense, in a position
representation or something? No, they're not
position eigenstates because the wave function
isn't that concentrated, but they approximately are. They're approximately located
at the position of the atom. You can't put two of them into
the same atom in the same way, but you can have two
different atoms, which is like having an
electron over here and an electron over
here-- that's fine. Yes? So what is the
definition of a state? What is the
definition of a state? I think I would
have to refer you either to my own lectures or
a book on quantum mechanics. But it just means it's the
quantum mechanical analog of the configuration
of a system. A system is a collection
of degrees of freedom-- a collection of coordinates and
momentum and things like that. A state is a particular
value, in classical mechanics, of the positions and momenta. In quantum mechanics,
the state of an electron is everything you have to
specify about the electron to state all the things that can
be measured about the electron. [INAUDIBLE] It's a wave function, actually. Can an electron here
and an electron, let's say, in the Andromeda
galaxy-- a free electron-- can they be in the same state? No, obviously, not-- one of
them's here, and one of them is in the Andromeda Galaxy. That's pretty different. Suppose you have one in
Nevada and one in Connecticut. Can they be in the same state? [LAUGHTER] Can they have the same
state as momentum? [INTERPOSING VOICES] What's that? They can have the same momentum. They can both be going east at-- Well, if you know
where they are, it means you don't really
know their momentum with infinite precision. So if you know anything
about where they are, it means that you don't
completely know the momentum. But yes, to within the extent
that an electron in Nevada can have a pretty well-defined
momentum but still not so well-defined that it might
be in Connecticut, right. Well-defined, but
not so well-defined that it might be in
Connecticut, then, yes, you can have an electron in Nevada,
an electron in Connecticut, which have pretty much
the same momentum. But if you really tried to make
that momentum infinitely sharp, that electron in
Nevada would be equally likely to be in Connecticut. And the one in Connecticut
would be equally likely to be in Nevada, and
you simply can't-- --can't put them both
into the same state. Just multiply the
uncertainty together and you get [INAUDIBLE]. OK. So we now know what a
fermion is, more or less. We use the algebra
of these operators in a very similar way. We write down expressions
involving electron operators, which annihilate electrons,
which take electrons out of the initial state, creation
operators which put electrons into the final state. We can even multiply them
by operators for photons-- photons incidentally
being bosons, and we can write
mathematical expressions for the process of some
number of electrons coming in, some number of electrons going
out plus a bunch of photons. And we can write similar
kinds of bookkeeping as we use to describe just
the simple processes involving bosons, and we're not
going to go into that now. I mean it is the story
of particle physics-- how you describe
those reactions. But before we get
to that, I want to talk a little more about the
differences between fermions and bosons. Question. Yeah. From the two-state example
is a poly-exclusion principle a locally communicated
principle? [? What does it mean ?] to you? There's a globally enforced
instantly entanglement. In other words, if you had to do
a few different processes going on in Nevada and
Connecticut, they're both about creating a particle. But if one is known
to be in Nevada and the other is known
to be in Connecticut, then they're not in
the same state, period. Yeah, but let's
say it's not known. Let's just say that there's a
slight probability difference. So there is a certain
amount of fuzziness. Here's the situation. Let's imagine that
Nevada is a point, and Connecticut is a point,
instead of a big state. Here's a point. Here's Nevada, and
here's Connecticut. An electron located right at the
center of Nevada, I'll call N-- N for Nevada not n for neutron,
not n for anything else. Let's call it capital
N incidentally. N is not an integer. It's just Nevada. All right, an electron in
Connecticut, I will call C. Now there is a quantum
state of an electron, which is a
superposition of being in Nevada and Connecticut. It has some probability
of being in Nevada and some probability of
being in Connecticut Nevada plus Connecticut. This is one electron
with equal probability. You should put a one over
square root of 2 here to make the probability
come out to be one. This is a state of
an electron, which has equal probability of being
in Nevada or Connecticut. It's a quantum state. And if you look
for that electron, you'll either find it
at the center of Nevada or the center of Connecticut
with an equal probability. Incidentally, there
is another state like this, which is
orthogonal to it, which has a minus sign here. What you can't have
is two electrons both in this quantum state. You can have one
electron in Nevada, one electron in
Connecticut, but you can't have two
electrons, which are in this state of both
being equally shared between Nevada and Connecticut. Now that has implications
that probably are not at all obviously what
those implications are. But supposing we tried to make
two electrons in this state, you can't do it. You can't have two
electrons, which are equally shared in this
way between the Nevada and Connecticut. Can one have plus sign, the
other have a minus sign? Yes, yes, one can of a
plus sign and the other have a minus sign. Yes, yes, yes, yes, yes. One electron can be in the
state with a plus sign. The other one In the state with
the minus sign. And we're talking about
at the same time here. we're talking about
at the same time. So the ambiguity in that
is Nevada and Connecticut are spatially
separated, so there's a question of what do you
mean by at the same time, simultaneity. But when you think about
it in momentum space, you don't have the
spatial separation. Remember we're doing
non-relativistic physics, so there's no
limitations of the speed of light in making
experiments and so forth. At the same time, I
mean, the same time. Somebody can jump from here
to there in no time at all. Check So what happens when you
throw in relativistic physics? Not much. Not much. There is an additional fuzziness
because of relativity about how particles are located. But for the moment,
I wanted to deal with the non-relativistic ideas. Let's come to the ground
state of a system of boson and the ground state of
a system of electrons to see how different they are. Let's imagine the
three-dimensional world. And since I can never
draw three dimensions, I will only draw two. I want to plot all of the
possible allowable momenta-- two-dimensional momentum. This is the x
component of momentum. This is the y component I
really want three dimensions, but as I said, I can't draw it. Electrons confined to be
in a certain finite volume. The finite volume is
very much like when we studied one dimension,
we studied electrons on an interval of length L with
periodic boundary conditions, but that's not so
important here. If electrons are
confined or particles are confined to be
within a given volume, then the momenta are discrete. Remember when the
electron was moving on the periodic
interval, it had momenta, which had the n over l. And there's a 2 pi in there-- 2 pi n over L is equal to k. So what do we say? We said if we drew the momentum
axis, the possible values of momenta were
discrete and that was a consequence of
the space being finite-- space available to the
particle being finite. The bigger L is, the
closer these levels get. All right, let's continue
to think about things confining than a finite volume. In that case, both the
x and y and z components of the momentum would
again be discrete so the allowable values
of the momentum vector would form a lattice. This is not space. This is momentum space-- the values of the
possible momentum vectors. Let's put 0 right over
here at the center, and then they go
on and on forever. It's not bounded-- just
goes on and on forever, just like this axis
goes on and on forever. This n can be anything. So these are possible
allowable values of the momentum of a particle
in a periodic three-dimensional box. OK, let's start with boson. Let's take a
collection of in bosons all identical to each other. They're all the same
kind of particle, and let's put them into our box. And let's look for
the ground state-- the state of lowest energy. Well, let's put the
first bosons in. Where will it go
if it's supposed to go into the state
of lowest energy? 0-- 0 what? 0 momentum, 0 momentum. In other words, we'll
put the bosons in. That 0 is kx, ky, kx is equal
to 0 right at that point. That's the single-particle
state of lowest energy. Remember the energy
is k squared over 2 m. The lowest possible
energy for that boson is to carry 0 momentum. If we want to find
the ground state, we just put in the
particle with 0 momentum. Now supposing we want to
add another particle to it-- another boson of the same kind. Where do we put it to keep
the energy as low as possible? Same place. Right, just the same place. We put two particles
in the same place. In fact, we put all end
particles in the same place. That would be the
state of lowest energy. So that is called
a Bose condensate the idea of a Bose condensate
is a large number of particles all in the same state. But in this case, all in the
lowest energy state available. Incidentally, the lowest
energy state available, since it has the lowest
momentum, fills up. It's a wave function
which fills up the box. It's a wave function
which fills up the box. And so the wave function of
each one of these particles fills up the box, and we
simply put a lot of particles in, all of them very
uncertain about where they are in the box-- a lot of them in. That's, first of all,
called the Bose condensate, and it's also the ground
state of a system of bosons. The large number
of particles builds up a kind of classical
field strength so that the shrouding wave
function is pretty classical. OK, that's very, very different. Than what would happen
if we had fermions. Let's forget spin. We're not going to be
interested in spin now. Where shall we
put the first one? We got put them
in one at a time. We want to keep the
energy as low as possible. OK, obviously, the
first one to go in, we want to put into
the lowest state. What about the next one? Whatever the next energy
level is over here, the next one would
maybe go over here. They happen to be four
states of the same energy. So we could put it here,
here, here, or here. This is in two dimensions. In three dimensions, there
would be six states altogether. So by the time we put
in five electrons, they will fill up
these states here. Now let's put in
some more electrons. We can't put them in the
same state-- forbidden. So the next one will go
in the next lowest energy level, which will be over here. Incidentally, the energy being
proportional to k squared is proportional to the distance
squared from the origin. So what we want to do to keep
the energy as low as possible is crowd the electrons
in momentum space into as small a
sphere as we can. In other words, when we
have a lot of electrons or we put them in, they will
fill up in momentum space, fill up a sphere. There might be a
little confusion about the jagged of the
spherical edge here, but that's not important. This, of course, will
have a lot more energy than the corresponding
boson system. The corresponding boson system
had all of its particles in the lowest energy. If you look for a
particle, all you will find is a particle of 0 momentum. In the ground state of
the mini fermion system, you will find the more
formulas you put in, the higher the maximum
momentum will be. There will be lots and
lots and lots of particles of large momentum And of course,
only a handful at low momentum because there aren't that
many states of low momentum. There are clearly more
states within a given shell at large momentum. So first of all, the
energy of the system will be much higher, putting
fermions into a box costs more energy, one. Number two, there's this
pattern that they fill up a sphere out to
some boundary where the boundary is determined by
how many electrons you have. Depending on the
number of electrons, it will fill a sphere
and that sphere is called the Fermi sphere. It's called the Fermi sphere. And so the history of this
is a little bit confused. It was Dirac who figured
out fermions and that's why they're called fermions. It was Einstein who
figured out bosons, and that's why
they're called bosons. They all contributed very
heavily to the subject. Anyway, this is the ground state
of a collection of fermions fills the Fermi sphere. Let's draw over here
the corresponding thing in momentum space is. This is in momentum space. I don't need to
draw these lines-- just all the particles condensed
at the center of the momentum space for bosons-- bosons, fermions. Let's put a little bit
of this lattice in, so we can see where
the next states are. OK, what's the
first excited state? We have in bosons? They're all at the center. What's the first energy
level above the ground state? Well, that's easy. We take one boson and move
it to the next energy level. We take out one boson
in the ground state and put it into the
first excited state. Which boson did we do? Doesn't matter--
they're all the same. So what we have n minus 1
bosons in the ground state and one boson in the
first excited state there are four first
excited states. That's not so important. The second excited state--
it might correspond depending on the details. It might correspond to
putting one electron in the second excited state
or two electrons in the first excited state. Not electrons, bosons-- a
variety of different things. But you can see
the way the pattern to excite the system,
to give it more energy is to take electrons
from the ground state and put them into
higher energy states. OK, let's come to
the fermion system. Well, it's also very
similar, but let's try to figure out
exactly what would correspond to the lowest-- question? OK. What would correspond maybe
not to the exact lowest energy state but to the very, very low
energy states above the ground state? We want to give it just
a little bit of energy. What's the cheapest
way to give it energy? Should we take an
electron from the center here and move it outside? We can't move an electron to
someplace inside the Fermi sphere because there's
already an electron there. I think as you
were about to say, it wouldn't be too smart to
take one deep inside the Fermi sphere and bring
it to the outside because that would
take a lot of energy. It takes how much energy? You're going to take
one out of low energy and put it back at
relatively high energy. That's pretty costly. What's the cheapest thing? Right, take one from
very near the surface of the sphere,
inside the sphere, and move it to the outside. How much energy does that cost? Not much because you've taken
a pretty high energy electron and just displaced it
a little bit in energy. So that creates an
electron out here with a little bit of
energy above the edge of the Fermi sphere. But it also leaves something--
it leaves a hole in the Fermi sphere behind it. Let's suppose these
are really electrons. OK, let's say let's consider
a real-world situation. We have a box. And that box is full
of positive charge. The positive charge is
protons, of course, or nuclei. But for simplicity, let's just
imagine the positive charge is smeared over the box. And then we put electrons
in on top of it. And we put enough electrons
in to fill up a Fermi sphere like this. It's electrically
neutral because there are as many electrons as
there are protons in the box, so it's electrically neutral. In the ground
state, the fermions fill up this Fermi circle,
this Fermi sphere here. Now we take an electron out
from just below the Fermi sphere and put it in just
above the Fermi sphere. We have not changed the charge. It's still electrically neutral. But we have created
one extra particle with a little bit of extra
energy above the Fermi surface and left a hole in the Fermi
surface and left a hole where a missing particle,
a missing particle-- just think of it
literally as a hole. It's an absence of a
bit of negative charge-- a bit of negative charge with
a certain momentum over here. It can be thought of as the
presence of a positive charge-- whole. A whole can be thought
of as a kind of particle. In this context, the whole
behaves as a particle. Is it a particle with
negative energy because it's below the Fermi sea? Now let's see how
much energy it takes. How much energy does it take
to move a particle from here to here? Got a square in it. Let's just say we took
out a particle of energy. Let's call a particle
that we took out. Let's call it energy epsilon. And we put in another particle
of energy, epsilon prime. , When we took out the particle
of momentum of energy. how much energy does that cost? Difference-- the difference. [INAUDIBLE] It's both. Is it the sum or the difference? I think the difference. You think the difference. I think the sum. OK. [LAUGHTER] I'm sorry. I'm sorry. Let me be more precise. It is the difference,
but I'm getting tired. I'm reaching a point of-- the point is a custom energy-- a custom energy. I'll tell you how
much energy later. Would you have to push
energy into the system in order to pop an electron
off for the next energy level? Well, the point is how much
energy do you have to put in? You have to put in
the amount of energy that it would take to
remove this particle and bring it exactly to
the Fermi surface and then an additional amount of energy
to bring it from the Fermi surface off the Fermi surface. So here's the picture. Here's the Fermi sphere. The Fermi surface is the
edge of the Fermi sphere. Let's take a particle just
below the Fermi surface, bring it from here to here. How much energy does it take? Well, it takes two bits
of positive energy. We can count it this way. The first bit of positive
energy is the energy that it would take to
remove this particle and bring it right
to the Fermi surface. That's a little bit
of positive energy to bring it to here and then
a little more positive energy to bring it to here. So it's actually the sum
of two positive terms. The two positive terms
being the absolute value of the difference
of the energies from the Fermi surface. You have to add two
little bits of energy. One of them can be thought of
as the energy of the electron over here measured relative
to the Fermi surface and the other one can be thought
of as the energy of the whole. And they're both positive. This one has a bit
of positive charge because it's an electron. This one has a bit of negative
charge because it's missing-- no sorry. This one has a bit of negative
charge because it's a positron. No, I did that on purpose. I didn't. This one has a bit
of negative charge because it's an
electron, and this one has a bit of positive
charge because it's a hole. Thank you. Now this electron
could be moving around and then suddenly
pop into that hole. The hole is available now. The hole is available for the
electron to drop down into. Just as an electron can
drop from an excited level of an atom down to an
unoccupied quantum state-- if there's an unoccupied
quantum state-- this electron can drop
down into this hole. What will happen when an
electron drops its energy and drops down into the hole? Well, energy has
to be conserved. So the energy of the
electron has decreased. A photon is emitted. You can say it another way. You can say an electron
comes together with a hole and annihilates. The electron annihilates
the hole and a photon or more than one photon. Some number of photons goes out. So you can speak about
this in the language that a hole is a
kind of particle because it's a fake particle. It's not a real particle in
a real sense of the word. It's a kind of
fake particle that has the opposite charge
of the true electron. It can be counted as having
some positive energy. And when the electron
drops down into the hole, you can think of it
as the electron hole coming together,
annihilating, and producing some heat, radiation, whatever--
the process of annihilation. Question. Yeah. When you say you're
moving these electrons, you're moving from one lattice
point to another lattice point-- Lattice in momentum space. Yeah, no, no, no. You're moving them from one
energy level to another. So it's the analog
of having an atom. All right, let's speak
about the analog of what this means in atomic physics. In atomic physics, the box could
be not the momentum space box but the box that confines
the electron could just be the atom. The proton at the center
is some positive charge, and we fill it up with some
negative charged electrons. Of course, in this
case, the proton charge is not smeared over
the atom, so it's concentrated in the center. It's a little bit different. But we take an atom, which has
the same number of electrons as protons, and we start
with all of the electrons in the lowest available states. The lowest available states if
the ground state of the atom-- that's the analog of filling
up the Fermi sphere here. Now a photon comes along
and hits an electron and kicks it up into
an excited state. Photon comes along,
hits an electron just below the Fermi
surface, and kicks it up just above the Fermi surface. It's a low energy electron,
a low energy photon, so it doesn't have much energy. It can't take an
electron from the center here and bring it
to the outside. That would take a lot of energy. It's a very low energy
photon, and so it comes in and all it can do
is hit an electron but be absorbed by an electron
near the Fermi surface and kick it just above
the Fermi surface. Another language is the
photon has annihilated and been replaced by a particle
or an electron and a hole. So it's called
particle, hole, creation by taking the energy
from a photon. In the same way that a
photon hitting an atom excites the atom, of course,
when it excites the atom, it leaves a vacancy. The vacancy in this
language is called a hole. Holes has gone So across the whole sphere,
if the photon hits anything in that sphere, will it
kick off something nearly-- Well, if it's a
low-energy photon, let's say some
radio wave photon-- very low energy-- and
it tries to get absorbed by an electron near
the center, it just can't give that
electron enough energy to get out of the Fermi sphere. It can't jump to any place
inside the Fermi sphere because it's already occupied. So the answer is a
low energy photon can't be absorbed by an
electron in the interior. So the cross section
doesn't get bigger as the sphere gets bigger? It's only into the
surface, not the volume-- It all comes off the surface. That's right-- from
a high-energy photon. Of course, that can kick an
electron from deep inside. The language is that things
inside the Fermi sphere are in the Fermi Sea, S-E-A. This is deep in the Fermi Sea. This is near the surface
of the Fermi Sea. Only a high energy
photon or a collection of high-energy photons
coming together can kick an electron from
deep inside the Fermi Sea to the outside. A low-energy photon can take
a shallow electron or one just beneath the surface
and put it just above. Yeah? Is there anything that
keeps the electrons from interacting with electrons
directly without using photons? Nothing, except
we're ignoring it. No, interactions
between electrons make the story somewhat
more complicated, but interactions between
electrons are pretty weak. They're pretty weak. They're not very strong. And for the most part,
they can be ignored inside of metal for example. If a photon has excess
energy, it has enough to kick multiple electrons-- Say it again. If a photon has a
lot of energy, enough to the excite multiple
elections, then what happens? Does one electron go get extra
excited and then fall back? Well, anything
can happen as long as it's consistent with
the conservation laws. But typically, it's a much
lower probability process to create two
electrons in two holes. If it kicked out two electrons,
it would make two holes. So that would be
a process in which four particles, four particles
were created altogether, and that's much less likely. [INAUDIBLE] super excited one. Yeah, that's right. It's more likely to
super excite one. Another question. Photo-electric
effect here, is sort of what we're talking about. Is it ever the case two
photons simultaneously can be absorbed by the electron? Yeah. One of them wouldn't do
it, but both of them will. Yeah. OK. Often, you need both of
them to conserve both energy and momentum. It's just a
probability situation. It' not prohibition. If you have a low-energy photon
that's doesn't have enough energy to excite any of the
electrons, does it absorb-- Well, it just takes a
tiny, tiny bit of energy to kick it from here to here. So yeah, there is some
energy, which is so low that it can't do anything. It's true. Does it then not interact
or does it absorb it? Yeah, pretty much just
passes through the system. Well, that's not really true. Yeah, very low energy
photon will find other ways to interact. It'll start the whole collection
of electrons sloshing, and that's more complicated. You mean that they're
actually moving on the grid back and forth? They actually would be moving on
the grid back and forth-- yeah. Space. From space, well, that's
a little bit late, but that's more
complicated question. No. Why did you say
electrons interact weakly with each other? Wait-- say it again. You said that electrons will
not interact with each other very strongly. But don't they all have
the same negative charges? Yeah, but they're also
in a background sea of positive charges. So the whole thing is
electrically neutral. The electrons are
interacting with each other, but the electron
interactions are pretty weak. Partly, they're weak because
the electron charge is screened by the proton
charges, but I really didn't want to get into
the electron interaction. That's a separate issue that-- Will we talk about
that in screening? Yeah, we can talk about it. No, I don't want to
talk about that now. Yeah, screening is an important
phenomenon in particle physics, and it has its analog here. But let's come back
to it another time. What I really wanted to get
to was the Dirac equation, and I have 15 minutes. To give you the very,
very simplest version of the Dirac equation, which we
will come back to repeatedly-- but the very, very simplest
version of the Dirac equation just sort of
complete a story here. You know what story I'm
starting to tell you. I'm telling you the story
about antiparticles. Holes are antiparticles. But what does this have to
do with not a metal in a box but with relativistic
quantum field theory and the vacuum and
particles and antiparticles? So I want to spend just a
couple of minutes giving you the very, very simplest
version of Dirac's logic. And to keep it
really simple, let's stay with one
dimension of space-- one dimension of space
on only one dimension, x. We'll come back to more
dimensions next time. And let s write a wave equation. This wave equation is
going to be for a field that I'm going to call psi. OK? Ah. This electron-- there's
not an electron. It's a neutrino. I don't know what it is. It's a particle. It's a fermion. It's a fermion, but
it happens to move with the speed of light. It moves with the speed of light
and that means omega equals k. Omega equals k is
the relationship between frequency
and wave number or between energy and momentum-- C equal 1. Omega equals k. There is another possibility-- omega is equal to k. And what's the
other possibility? Omega equals minus k, OK? But let's take omega equals k. What kind of wave equation
does that correspond to? Well, that corresponds to the
wave equation, d psi by dt is equal-- I think it's minus d psi by
dx-- this is partial derivative. This is even simpler than
the Schrodinger equation. Why do I say that this
is omega equals k? Let's imagine the psi is equal
to e to the ikx minus omega t. d psi by dt t brings down
a factor minus i omega. d psi by dx brings
down a factor, ik. If the relation
between omega and k is just omega equals k
corresponding to c equals 1, then that just says d psi by dt
is equal to minus d psi by dx. So this is a very, very
simple wave equation, and it describes waves which
only move in one direction. They cannot move
in both directions. They move only is it to
the right or to the left? Yeah, this these waves
move to the right. What would it take to make
waves that move to the left? You want d psi by dt
equals plus d psi by dx. With either of
those two equations, you have particles which
move in one direction-- let's say, in this
case, to the right. And let's suppose
these are fermions, and let's even imagine
they carry electric charge. So they're electrically
charged fermions. OK, now there's something a
little bit crazy about this. Omega equals k--
energy equals momentum. Can k be negative? Sure, why not? I can write e to the
ikx, and I can also write e to the minus ikx. I can write both possibilities. But there's something
a little ugly-- not ugly, a little
bit dangerous-- about negative values of k. They carry negative values
of omega and negative values will make omega mean what? Negative energy-- right. So this equation
describes electrons of positive and negative energy. Let's just draw a picture. Omega equals k. Here's omega equals
k, omega versus k. It's just a straight line. For k positive, the
energy is positive. For k negative, the
energy is negative. This equation simply
describes electrons, which can have either
positive or negative energy. That sounds like trouble. Why is it trouble? Why is electrons with
negative energy trouble? Well, if electrons can have
arbitrarily low momentum-- remember what happens in an
atom to an electron, which is excited. It radiates and drops
down to a lower state. If there were unboundedly low
energies arbitrarily negative, then let's say positive
energy electrons could emit photons and drop down energy. But they wouldn't necessarily
have to stop at 0 energy. They could keep dropping down
and dropping down and dropping down and make arbitrarily
negatively energy electrons while at the same
time pumping up the energy of the
electromagnetic field. This sounds very dangerous,
and it is dangerous-- something wrong with this. You can't have
arbitrarily low energies. Well, what is the
lowest energy state of the multi-electron system? [INTERPOSING VOICES] What's that? The Fermi sphere? What is the Fermi
sphere in this case? The Fermi sphere in this
case is [INAUDIBLE]?? Let's suppose we have a zillion
electrons-- as many as we like. And we put the first
electron into the system. Where does it go? To the hole. It goes down into the
lowest energy state, but the lowest energy
state is way the hell down there somewhere. It's off at minus
infinity somewhere. The next one off
at minus infinity. This is very dangerous. This is a bad idea. This is the simplest
version of Dirac equation. This is a Dirac equation
in one dimension. What Dirac said is let's pretend
that all of the negative energy states were filled. Just like filling
the Fermi sphere, fill all the negative energy
states all the ways up to 0 energy. So all of these
states are filled. There's an electron in
every negative energy state. You can't make a state of lower
energy than that incidentally. That's the lowest
possible energy you have. You filled up all the
negative energy states. Every time you put a
negative energy thing in, that lowers the energy. You've put as many negative
energy states in as you can. Now if I start putting in
positive energy electrons, that will increase the energy. So the very lowest
energy state-- the thing that you could
call a ground state-- is to fill it with as many
negative energy electrons as there are negative
energy states. And that's your starting point. Instead of a starting
point with no electrons, your starting point is fill up
completely the negative energy sea. That's the vacuum. That's empty space,
according to Dirac's theory, Dirac's simple version. Start with that. Since it's the state
of lowest energy, let's call it the vacuum. And then what can you do? You can take the
only thing you can do is to take an
electron out of the sea-- the negative energy sea-- and give it positive energy. So you can take an
electron from over here and put it someplace else. what are the properties
of a state with one electron over here with positive
energy and positive momentum and one missing
electron over here? It's a missing electron. How much momentum does
it have [INAUDIBLE]?? [INTERPOSING VOICES] k is negative. k negative. Well, we removed the
negative momentum electrons. Yeah, so it's 0. No, no no, no,
no, no-- the hole. How much momentum
does the hole have? Is it positive or negative? Positive, positive, positive. Why is it positive? Because it's a missing
deficit of negative momentum. So what we've done is
create a positive energy electron and positive momentum. We've created twp
positive momentum objects both moving to the right-- one with minus charge, the
electron with minus charge, and the missing electron
having positive charge-- negative charge and
positive charge. That in summary-- and
we're going to do this more completely-- but that was
Dirac's theory of the electron and the positron-- that the positron
was simply a hole in the infinite
negative energy sea. You wrote down this equation
or the generalization of it, looked at it. And said I love that
equation, but it's got this terrible disease of
having negative energy states. No problem-- these are fermions. You can't put more than one
electron in a negative energy state. So just fill it up completely
and call that empty space. Having called that empty
space, what can you do to it? You can remove one of the
negative energy electrons and put it back as a
positive energy electrons-- yes, put it back as
positive energy electron. That leaves a hole. And Dirac side that hole is a
particle of positive charge. Now at first, he
was very excited because he thought
it was the proton-- wasn't the proton. It's easy to prove that
the mass of the particle here has to be the same
as the mass of the-- these are particle
and antiparticle. This was this discovery of
particles and antiparticles. But this was the basic
logic, and it was equations like this which led to it. Notice what a disaster you'd
be in if you tried to describe a boson by this equation. That would be a madhouse. You would just keep
putting particles into the negative energy
lowest for as many as you like, and there would be
no ground state. There would be no ground state. You could just keep putting in
more and more and more negative energy bosons. So this wave equation only
makes sense for fermions, would not make sense for bosons. And that was Dirac's
great discovery-- that fermions can be
described by Dirac equations like this that would not
make sense for bosons. Why? Well, just the
reasons I explained and the idea of antiparticles. We'll come to antiparticles
for bosons next time, and we'll discuss the Dirac
equation in more completeness-- the real Dirac equation. This is an elementary
version of it, but we'll come to it next time. There's a number of things
wrong with this electron. For one thing, it only
moves to the right-- that's crazy. For another thing, it moves
at the speed of light-- that's crazy. So is a number of things that
are wrong with this electron. No, it's a neutrino. No, it's a neutrino. Neutrino is a boson. This neutrino is a boson. This is a neutrino, but
it's not quite a neutrino because the neutrino
also has mass. For more, please visit
us at Stanford.edu.
