functions that describe the field state
Those were either 4-potentials for an electromagnetic field or a one-dimensional scalar function for a scalar field. For a complex field, that can be a set of two functions,
φ and φ^*, which are considered as independent, i.e. there must be one q or a few q. We’ll also study the Dirac field, where we’ll have
a bispinor and the Dirac-conjugate of it. Such will be the objects of study in the classical field theory
Further: here the equations of motion are the Lagrangian equations resulting from the principle of least action. The equations are written down in a well-known way: the time derivative of the Lagrangian,
which is dependent on the generalized coordinates and velocities, and minus the derivative with respect to the coordinate
If there were a few coordinates, there was an equation for each of them. It may be useful to write down the time derivative in the way we adopted now
That will be the time derivative of the derivative of the Lagrange function with respect to the velocity. Let’s recall what is in this place in the classical field theory. The Lagrangian density is instead of the Lagrange function
It is still a function of the coordinates, but not only of the velocities. In the classical mechanics, the priority is given to the Lagrange function, and in the classical field theory to the Lagrangian density, which also depends on the coordinates
and the generalized velocities. Further, as a generalization of this equation, we derived from the principle of least action
an equation of similar structure. Earlier, the privileged time was here, and now all the equal-in-right components are in this place. The four-dimensional gradient is instead of the time derivative
Here occurs the derivative of the Lagrangian density with respect to the generalized velocity,  and again minus the derivative of the Lagrange function with respect to the corresponding coordinate
Again, for a few coordinates, there are a few corresponding equations. Then we’ve said that
the Lagrange approach is also effective because of Noether's theorem in classical mechanics, which automates the process of finding the integrals of motion. Classical Noether's theorem sounds as follows
Suppose the following transformation: t is replaced with t', which differs by a some function δt --- the quantity δt is a function of the coordinates and time
and the coordinates are replaced with some shifted coordinates, where this δq is a function of other variables and the time
If under such transformation the mode of action is not changed up to first-order terms inclusive of these variations, i.e. the first variation of the action is equal to zero, then the following is true:
the combination of the energy of the system multiplied by this function δt, minus the momentum of the system  multiplied by the variation of the coordinate is conserved
I.e. the variation of the coordinate by its conjugate momentum minus the variation of the time by its conjugate variable energy turns out to be constant
We re-wrote it in this way. This constant, as can be seen, is differentially small and thus can be designated, as in all textbooks, with the capital letter Θ
The energy is the derivative of the Lagrange function with respect to the generalized velocity multiplied by the generalized velocity,
minus the Lagrange function itself. All this is multiplied by δt, and the momentum, which is the derivative of the Lagrange function
with respect to the generalized velocity, is subtracted. This quantity is asserted to be constant, or in other words, the time derivative 
of this object is zero. What is an analog of this in the field theory?
Without proof, I made the following statement, which is actually the Noether theorem. This statement was a modified and simplified Noether theorem, adapted for the purposes of classical mechanics,
whereas the original Noether theorem concerns namely distributed systems, like the field theory. It reduces to the following: if under variations of the coordinates, not just of t,
but namely of x (variations of the spatial coordinates) and of the field coordinates,
the mode of action does not change up to the first orders of δx and δq inclusive, then the following quantity  is conserved:
the 4-vector is naturally instead of the scalar corresponding to this combination. The 4-vector is constructed in the same manner
as this picture was created, i.e. the derivative of the Lagrangian density with respect to the generalized velocity
is multiplied by the generalized velocity. Then all this will be multiplied not by δt, but
by the variation of the corresponding x. Since the index μ is occupied, this variation will be, for example, with the index ν. Then we need to differentiate
with the same index. Then we subtracted the Lagrange function; here we subtract the Lagrangian density, which now has a pair of indices,
μ and ν. So, this must be a tensor. g^μν is such a tensor. This is the answer, minus, as before,
the derivative of the Lagrangian density with respect to the generalized velocity
with the index μ, and this is multiplied, as before, by the variation of the field. As before, if there were a few coordinates, then we would need to sum over i
Here, too, we would need to sum over i
This is an analogue of the Noether theorem in the field theory. The statement that previously reduced to the
constancy of this quantity is now reduced to the fact that this vector obeys the equation of continuity, i.e. instead of the derivative of scalar quantity with respect to the zero component
we have a four-dimensional gradient. From this relation it follows that 
if I integrate the zero component of this vector over the entire volume, in classical mechanics it will turn into δΘ,
because it is made of the Lagrange function, not the Lagrangian density. This quantity will be a constant, and thus this assertion has such an analogue, an integral law
The wording is clear enough and is a complete analogue --- well, with some generalizations --- from this point of view too
If you know the proof of this in classical mechanics, the proof of this statement is just a matter of some accuracy in algebraic calculations. They are quite long, and instead of them we’ll try to do the proof by parts
We’ll consider separately a case when the action does not change under such variations and a case when the action does not change under such variations. Hereby we’ll prove the Noether theorem with less effort,
and secondly, the combination of these two proofs will almost finish the full proof. So, suppose we have such a relation. Let’s make a little bit more specific statement:
let our action be invariant with respect to shifts in the four-dimensional space, i.e. no changes
In classical mechanics, let there be no changes when some coordinate is varied (shifted), then the invariance of action means invariance of the corresponding Lagrange function, since the time does not change at all
What does this mean? This means that the space is homogeneous. The physical essence of the statement is the experimentally verifiable fact below; we’ll do an experiment at this point
I have material points; they have some initial conditions, from which the system develops. The statement is as follows: if this experiment is done in an empty space, there is no Earth, no Sun, no Universe at all, no me,
and if the same experiment is done at a different point of the empty space, from the same initial conditions it must develop as it developed there. In fact, this is the classical principle of determinism. This statement is understandable, 
unlike its consequences and proof. In the Lagrange approach in the classical mechanics the following is proved: if the space is homogeneous in this sense, the corresponding generalized momentum is conserved
In case of usual space, the total usual momentum of the system must not change in time. We are talking now of a shift in the 4-coordinate, and thus the 4-momentum will be conserved,
which includes the relativistic 3-momentum and the relativistic energy. So, let’s consider the relation between the homogeneity of the space
and the conservation 
of the four-dimensional energy-momentum. Suppose the following transformation occurs:
the coordinate x turns into the coordinate x', which differs from x by some constant value. In general formulation, here we have some functions of the coordinates and time
In our case, it is some constant value. We assume that it is small and that the action does not change in the first approximation. What are the consequences? As you remember, the constancy of action with respect to the shift of the coordinate implied
the conservation of the momentum, and the invariance of the mode in time (time homogeneity) led to energy conservation. We can expect this to lead to the conservation of the energy-momentum
First suppose that the mode of action does not change in such shift. Then Noether's theorem asserts that if the shift is like this, 
which automatically implies no changes in the field coordinates, and the answer asserts that 
here arises such a vector quantity, with the brackets. These brackets with the indices μ and ν
are a tensor. Let’s designate it as T^μν. Thus, according to this, T^μν occurs; δx is now ε_ν.
The integral of the zero component is a constant
This means that the integral of T^0ν with respect to d^3 r (ε_ν can be taken out), i.e. this quantity integrated over the entire volume, will be a constant
From this statement, I can say that the value of this vector --- let’s designate it as P^ν ---
projected on the vector ε_ν is a constant. Strictly speaking, we cannot say that all the components of P^ν are conserved, but the following requirement arises now
All this must be true for any ε_ν provided they are small. But if this equality is valid for any components, I can, for example, take the zero component different from zero,
and all other components equal to zero, then P^0 will be conserved. Then, respectively, P^x, P^y, and P^z. So, from the arbitrariness of ε_ν it will follow that 
such a quantity is conserved. This is a kind of consequence of Noether's theorem
The theorem itself was not proved, just proclaimed, and thus this also is a kind of proclamation. Let's try to prove it. I think, last time I finished telling you about the energy conservation
Let me start with the field theory, recalling how that was done in classical mechanics. In classical mechanics the total time derivative of the Lagrange function was considered
Let me still recall how that was done in classical mechanics. In classical mechanics, we took the total time derivative of the Lagrange function
This is a trivial exercise. The derivative of L with respect to q by \dot{q}∙, plus the derivative of L with respect to \dot{q} by the time derivative of the velocity itself
This is trivial. Then using the Lagrange equation, I can replace this part
with the time derivative of the derivative with respect to velocity. What did we get? This is the momentum,
the derivative of the momentum by the generalized velocity, and further is the momentum by the derivative of the generalized velocity. That is, in fact, exactly the total derivative
of two factors. The first is the generalized momentum, and the second is the generalized velocity. If you apply differentiation to this product, these two summands will appear. It is important that this statement
is valid only with the use of the equation of motion, i.e. owing to the principle of least action. The right-hand part is the total derivative, so if I move the right-hand side to the left,
I’ll get the following: the total time derivative of what? Of these brackets, minus L
All this will equal zero, so this thing is conserved, i.e. the derivative of this value is zero, and this is exactly the energy
The same reasoning can be repeated here. Let's consider the derivative of the Lagrangian density, which depends on q and ∂_μ q,
with respect not to the time, but to the 4-coordinate. Since here I took the index μ, here must be some other index, for example, ν
How can the derivative be calculated? Exactly as before: the derivative of the Lagrangian density with respect to q by the derivative of q with respect to x^ν, plus the derivative of the Lagrangian density
with respect to the generalized velocity by the derivative of this quantity with respect to x^ν
Done. Next, we use the equations of motion. According to these equations, the derivative of the coordinate is this,
i.e. ∂ of the Lagrangian density with respect to x^μ
Slightly differently: all this is applied to the derivative
of the Lagrangian density with respect to the generalized velocity. The index μ is a summation index here; this object was a relativistic scalar and it is still a relativistic scalar
Now to the right-hand part. What is this? This is the second derivative with respect to x^ν and x^μ
In the second derivative, this quantity is
exactly the second derivative of q with respect to x^ν and x^μ. If this function is not pathological, then  
the order of the  differentiation can be changed, i.e. first μ and then ν. Let's do that
We put ν here and x^μ here. What did we get? We got a derivative with respect to x^μ
of the product of the two factors. The first is like this, and the second is like this
This is ∂L/∂(∂_μ q), multiplied by the generalized velocity
Look, this is completely analogous to this quantity: the generalized momentum by the generalized velocity, and here the generalized momentum density by the generalized velocity
Great. We had the total derivative there and the total derivative in the left; here we have the divergence and the divergence in the right
But these divergences have various indices. Well, that can be corrected
I can replace this quantity as follows: if I introduce g^νμ by x^ν ---
g^νμ is symmetrical --- nothing will change. In fact this is the derivative of the Lagrange function with respect to x^μ. Yes, it is the identity transformation
The summation over μ will give us the derivative with respect to ν, as it was
Now I can write down the following under the sign of differentiation with respect to x^μ: 
the derivative of the Lagrange density with respect to the generalized velocity with the index μ by the generalized velocity with the index ν
In addition, if this derivative is moved to the left and g^μ is put under the differentiation sign, this all will be with the minus sign. g^νμ or g^μν, it does not matter, since they are symmetrical,
multiplication by L, and all this is equal to zero.
Q: in the bracket, Lμ is next to zero, and the derivative ..?
Everything seems to be correct. This multiplied by this under the differentiation sign. Well, this is precisely what we designated as T^μν
So, we got a statement that T^μν obeys
such a conservation law. In this view,
the proof was completed. Let me recall that in accordance with Noether's theorem, this quantity obeys this law: ∂/∂x_μ (δΘ^μ) = 0
If, using the arbitrariness of ε, I substitute the explicit expression, from this it follows that ∂/∂x_μ(T^μν) = 0
Namely this was proved. So, this portion of Noether's theorem is proved. We’ll use it further. I.e. we now have a statement how expressions
for the energy-momentum can be obtained. The system is defined in classical mechanics if the Lagrangian is known and in the classical field theory if the Lagrangian density is known. Knowing the Lagrangian, we can organize this tensor, T^μν,
which is called the energy-momentum tensor, via simple operations of addition and differentiation. Then it turns out that this tensor obeys the equation of continuity, which implies conservation
of the energy-momentum 4-vector. So, the procedure for obtaining the energy-momentum concept has been automated. And if the space and time are uniform,
the statement that this object is conserved was also automated. This is a very useful thing, and we have proved it directly. The next episode is associated, again, with Noether's theorem. Let's consider another useful statement,
which regards transformation of the field coordinates
The field coordinates can be functions. We’ll prove a particular case of invariance bound with the complex field. So, let's consider the following option
Suppose we have the so-called gauge transformations of the first kind, I'll explain what they are,
and the mode of action is invariant relative to these transformations. It is tied with charge conservation
minus the derivative of the Lagrangian density with respect to the generalized coordinate; the function φ itself is instead of q_1; and we take iεφ instead of δq_1 for the first function
So, in the set of fields under consideration, the electromagnetic field was taken to pieces completely
Everything is known: expansion in plane waves, reduction to a set of oscillators, quantization, and so on. Later on we’ll have a real scalar field, which is uncharged, and thus it is uninvolved. Next, a pair of charged fields,
the scalar field and Dirac spinor field. My following reasoning is important for them. Let’s consider the simplest version. Suppose that the field coordinates are
some complex function. Then we’ll have two coordinates. One is q_1 = φ, and the second coordinate is 
Let’s consider the transformation of these coordinates. Let this be the index i, running through the two values, and we have
two field coordinates. Let these coordinates be under a specific transformation of such kind. Let the coordinate φ(x) be transformed into the coordinate φ'(x), 
which is φ(x) exp(iα), where α is a real number
This is q_1
The second coordinate, φ^*(x), is automatically transformed by another law: φ^*'(x) = φ^*(x) exp(-iα)
α can be either small or big, it does not matter
It is important that if in such finite transformation the Lagrangian density is unchanged, the mode of action is unchanged, too. Let the mode of action be unchanged in such transformation,
even with a finite α. Noether's theorem, however, concerns infinitesimal transformations. Well, let's transform an infinitesimal. Let α be an infinitesimal. Then the right-hand side
Thus, φ’(x)  is equal to φ(x) + iεφ(x). This will be exactly δq_1
The right-hand part of this will be equal to φ^*(x) – iεφ^*(x). This quantity with this sign will be δq_2
Noether's theorem states that if the mode of action does not change ---
we declared that the action does not change (the Lagrangian density does not change even with finite transformations)  and thus the action does not change even with finite and infinitesimal transformations --- then the quantity subject to the law of conservation will be…
we won’t have this, i.e. the field coordinates are transformed, unlike the coordinates themselves. We defined the field transformation, 
whereas the coordinate transformation is zero. Then Noether's theorem states that we need to organize this vector; the first summand is missing;
and the second summand is present, i.e. there is some δΘ^μ equal to minus the derivative of the Lagrangian density 
with respect to the generalized velocity multiplied by the variation of the generalized coordinate
It is important that we have several functions; the sum in this case is over 1 and 2
What about this quantity? It will be
minus the same for the second coordinate: ∂L/∂(∂x_μ φ*) multiplied by the variation of the second function,
This is (-iεφ^*). This quantity must obey the equations of continuity, from which it follows that 
that the integral of this block, i.e. the integral of this quantity --- iε and the minus are taken out,
and further goes the derivative with respect to φ by φ minus the derivative with respect to φ* by φ * ---
this quantity will be conserved, which implies that this quantity will be conserved 
after exclusion of ε. Well, this is a proclamation, a statement that such a conservation law follows from Noether's theorem
Let’s try to prove it. We said that the mode of action does not change, and since the coordinates are not affected, the Lagrangian density does not change
I.e. the change in the Lagrangian density with such variations is zero. On the other hand, what is the Lagrangian density? What is the variation of the Lagrangian density?
The Lagrangian density is a function of the coordinates and generalized velocities. So, we need
to differentiate the Lagrange function with respect to the coordinate and multiply it by the variation of  the coordinate plus the derivative of the Lagrange function with respect to velocity by the variation 
of the velocity. So, this is the variation of the Lagrangian density. And then the usual course of action. What can we do here?
This term…
This is what we need. We need to prove
that this quantity we have formed has four-dimensional divergence equal to zero. The first thing to say is that
to have the required divergence, we need to swap these signs and then here we have the derivative of δq with respect to x_μ
Here we can use the equations of motion and say that it is the derivative with respect to x_μ of the derivative with respect to generalized velocity
Combining this and, what do we get?
We get the four-dimensional divergence of these two factors
This is exactly this equation, because this is equal to zero
Now let’s recall that in fact we have a couple of components, i.e. two fields, and thus we need to add indices,
which implies a sum. Thus this equation is exactly this equation,
because  this is identical to this expression. So, we’ve proved the second part of Noether's theorem in a particular case. You can see that the proof of Noether's theorem even by parts is cumbersome,
so I did not give the full cumbersome proof. What is for future? It was important for us that this the energy-momentum tensor was bound with the law of conservation of the energy-momentum 4-vector,
and now we have another conservation law. This integrand is
-i∙∂L/∂(∂x_μ φ)∙φ plus the complex conjugate quantity. All this is the 4-current vector
The statement that the integral of the zero component of the 4-current is a constant ---
let’s designate it with the letter Q --- will be useful to us in the future. This constant is called the charge
Not necessarily it is an electric charge; it may be a baryon charge, which was experimentally proved to be conserved even better than electric charge. Restrictions on the conservation of baryon charge are more stringent than those on the conservation of electric charge
This concludes our discussion about all aspects of Noether's theorem. Let’s now take a break, and then proceed to a new topic
So, we’ve developed a common approach to the problems
of the classical field theory, and hereby we can begin consideration of the fields of interest. First let me say that we discussed two particular cases; the homogeneity of the space-time
implied the law of conservation of energy and momentum and concrete expressions for this law. Similarly, we could say that isotropy of the space implies the law of conservation of angular momentum,
in its relativistic generalization. This thing is more cumbersome, so we will take it as a correct statement, which, moreover, looks exactly the same as in classical mechanics
The more so because we won’t actually need specific expressions for the angular momentum for the fields that will be considered
Now we need to consider this entire sequence of these fields. Let’s start from the field that is most similar to the electromagnetic one. This is a real scalar field.
To emphasize its similarity to the electromagnetic field, I'll even use the large letter Φ, for it to associate with the large letter A in case of the electromagnetic field
Sorry
So, we are interested, first of all, in the corresponding Lagrangian, and the Lagrangian density, which must depend on the field Φ and its derivative. Actually a Lagrangian,
from the point of view of physics, is just an instrument; the main thing is the equation of motion. So, if you have an idea of the required equations of motion, then you construct a Lagrangian for them. If no idea,
then we act in an opposite manner, i.e. we invent various Lagrangians
from different considerations, and then look what field equations arise. It was revealed that in case of electromagnetic field, the equation of motion is a wave equation
I’ll recall it 1/c^2  ∂^2{A}//∂t^2  - ∇^2 {A} = 0
Now, we recall that we have the 4-vector p^μ, which is iħ∂^μ
In our coordinate system, it will be just i(∂/∂t, - ∇). Then this equation can be written down as follows:  again, considering c equal to 1,
p^μ ∙ Ā = 0.This is the wave equation. From it we got the statement that for plane waves the squared 4-vector, p^μ p^μ, is equal to 0
What is our goal now? We want to obtain an equation from which the appropriate four-dimensional vector
for this wave shall obey this relation. That was for photons, the mass of which is zero; for particles with non-zero mass, the equation shall look like this. Therefore, the equation
we’ll use shall look like this: (p_μ p^μ -m^2) ∙Φ =0. Correction: these shall be operators
Such an operator equation. Let’s create a Lagrange function for the equation to be exactly like this. What else is important? It is important that the Lagrange function shall be real,
In this case it is quite simple, since our field is real
Well, we’ll  eventually sort out these i characters. They appear in the same way both here and here; so everything is good. As for the equation of motion…
Well, now we’ll see everything. The functions Φ themselves are real, so all combinations of them are also real; no problem with this. Next, the Lagrangian density must not contain large numbers of derivatives
If we want to obtain such a second-order equation, then the function can only contain the function itself and the first derivative, nothing more. And finally, we know that the Lagrangian density must be the relativistic invariant
Well, we can take the quantity ∂_μ Φ∙∂^μ Φ as an invariant. If this is quadratic in the fields, then it would be convenient
to take a quadratic combination of the fields themselves. Now we need to select coefficients such that the corresponding Lagrange equations would turn to wave equations corresponding to this dispersion law,
according to which the square of the 4-momentum is equal to the corresponding mass squared. Well, that will be automatically so if I choose a factor of ½ here and a coefficient -m^2 here
Let’s check if it really leads to a correct equation. Well, this is the Lagrange function
Now let’s try to apply these equations. I have to differentiate the Lagrangian density. What do we have here?
The role of q is played by the function itself; the role of the generalized velocity is played by the derivative of this function. Hence, this quantity will be the derivative of the Lagrange function with respect to…
well, let it be the derivative of the Lagrange function with respect to ∂_μ Φ
We need to differentiate this with respect to x^μ,
which yields this quantity. So, I’ve done this part
Since differentiation here may be done with respect to both the first factor and the second one, the factor ½ disappeared. That was the first part; I’ve got the first summand
Now I need to subtract the derivative of the Lagrange function with respect to Φ, i.e. this quantity. This will give us +m^2 Φ, and all this should be equal to zero
Now, if I turn this value in ∂_μ and multiply this all by i…
For nothing to change, I’ll write the minus sign and -i^2 --- I’ve changed nothing  ---  and we get the desired equation,   (\hat{p}_μ \hat{p}^μ - m^2) Φ(x) = 0
However, in future, it will be more convenient to use the initial equation, without any i
Both this type and the other one are equations for a scalar field. This equation is called the Klein-Fock-Gordon equation, although Vladimir Fock is not mentioned in foreign literature,
whereas he wrote and published his work earlier than Gordon did, and that work was much more substantial than the work by Klein
That was a large work concerning a certain five-dimensional generalization of usual space-time, and this was a small piece in this work. The work was so complicated that most scientists somehow missed it
and only later on, the tribute was paid to this remarkable scientist, including the Soviet literature (in particular, in a book by Berestetsky, Lifshitz, and Pitaevsky). He was the head of the chair of theoretical physics at the University of Leningrad. I was a student there and when we started
the course of quantum mechanics, the first chapters were delivered by Vladimir Fock. It was an interesting sight. The introduction itself was unemotional and somewhat dry, 
whereas the communication with that person was unforgettable. Immediately after a ring for a break or the end, he was surrounded by students. I remember a strong impression from one episode
An advanced student, Victor Cogan, who had already read the beginning of quantum mechanics and its interpretation by Bohr, Infield, Born, and especially de Broglie, would attack Fock:
"You say one thing, while de Broglie is reasoning differently in his book." Fock listened for some time and then said, "de Broglie is wrong." That turned out to be the absolute truth; de Broglie was a terribly ignorant man. Moreover, when he wrote a really excellent work,
in which he suggested considering particles as having wave properties, they in the quantum capital of the world, in Goettingen, did not pay attention to this work, for they already knew de Broglie as a real muddle-head. He had written a few articles on optics,
and Max Born, the head of Gottingen physicists --- you are probably familiar with the thick book "Optics", by Born and Wolf --- was a specialist in optics and knew that de Broglie was really a muddle-head and did not pay attention to this work. Einstein was the first to note this work and initiated 
the Schrödinger research. Well, back to our equation. What’s next? We’ll act as we did in the case of electromagnetic field,
with the following difference: the potentials of electromagnetic field are related to well understandable  quantities. Indeed, the corresponding derivatives of the potential are the intensities of the electric and magnetic fields, which can be directly measured with classical instruments,
whereas the potential of the real scalar field has no such classic, measurable analog. Only when we quantize it, quanta of this field will be interpreted as particles with nonzero mass
Their coordinate and momentum can be measured. We'll start with the old scheme. Let’s recall the scheme in case of electromagnetic field. In a special gauge, where only the vector potential was left,
with two independent components, we started from expanding it into plane waves. Of course, the coefficients of the expansion were vectors
Naturally, they had to be attributed the index k, over which summation was performed. The coefficients themselves were functions of time. Since the field was real,
we needed to take this value and its complex conjugate, and then this expansion automatically turned out to be real
Further, using the wave equation, we prove that this quantity obeys the equation for the oscillator, and therefore its time dependence is the same as that of the oscillator, but the frequency
is strictly tied to the wave vector. Well, this quantity was, respectively, proportional to exp(+iω_k t). Next we expanded this quantity in a series over two vectors,
mutually orthogonal and in a plane perpendicular to the vector k, but that was associated with the peculiarities of the electromagnetic field. The main thing we managed to do was the transformation of the electromagnetic field in a set of oscillators
Let’s to do the same here, i.e. let’s try turn this field into a set of oscillators. To do this we need to
likewise represent Ф(x) as a function of x containing both  t and the spatial coordinates, in the form of plane waves
It does not matter in which the function is expanded, but the variables k are convenient for the wave vector, the square of the 4-vector of which is zero. Using momenta is habitual for the 4-momentum, and so we choose expansion in momenta
running through the same set of values as those wave vectors k do. No need to act on the corresponding coefficient in accordance with its vector nature; it will be just some time-dependent coefficient
a_k. Since this field is real, we need to take the respective conjugate value. Sorry, this must be a_p
Let me remind that the choice of the normalization coefficient was an important part of our program. Let’s take
some coefficient factor C_p such that our final Hamiltonian would be a sum of Hamiltonians. As you remember, here we selected the corresponding coefficients such that
the expression for the field energy would be such a quantity: the electric field strength, the magnetic field strength, by 8π and integrated over all the space. All this reduced to
a set of oscillators. Since their dependencies were mutually supplementary, all this turned into
a time-independent sum corresponding to individual oscillators, and a and a* within a factor were the coordinate and momentum of these oscillators. Here we want to do the same, expressed in the following way: we want the expression for the energy
to turn into a sum over the momenta of a*_p(t) ∙ a_p(t)), and want the energy to be here
What is the energy? There the energy was such a quantity, and in this case, since we hope that the particle has mass, the energy should be
such a relativistic dependence. This is our global aim. Let’s choose the coefficient C such that
the expression for the energy would look like just a set of oscillators, each with its own energy, which depends on the variable momentum p. To this end, 
a from here and a from here have to explicitly depend on the time. How can we get this? From the equations of motion. Let’s apply these equations,
act with the operator (p _μ p^μ - m^2)∙Φ and equate this to zero. The summands with this and this exponentials are independent
Let’s consider the summands at the coefficient exp (i p r). What do they look like? The common factor C_ p does not matter; it will remain the common factor. When I do this action,
what will I get? Let me remind that p_μ p^μ = p_0^2 - p^2. How does p_0^2 act
on this structure? It acts only on a, right? p_0^2 is exactly -∂^2/∂t^2 and its action on a yields (- ä_p (t)), the exponential, and C_ p
I’ll take them out as common factors. Further goes minus this vector. This vector is -∇^2
Action of each gradient yields the factor (i p), and in the second time we get the factor (i p)^2, multiplied by a
In addition, we’ll have the common factor -m^2, also multiplied by a. For this equation to be valid, all this has to be zero
Let’s collect all in one. Let’s change the signs; we’ll have pluses here, here and here
Minus (i p)^2…What is wrong here?
We had -p^2 and -∇^2. Each gradient yielded ip, and we also had this common minus,
which we’ve missed. The result is such an equation…
If the square root of this quantity 
with the plus sign is called the energy, this will be an oscillator with a frequency ε_p. Therefore,
according to the equation of motion, we get that this factor will obey the law exp(-iε_p t) and this one the law exp(+iε_p t)
So, we automatically obtained the expansion in plane waves with the correct dependence of energy on momentum. The coefficient C_ p still remains a problem we’ll sort out next time.
Good bye
