(2-23′)
If we now substract these two equations, we get:
(2-24)
This further condition on ψ is necessary for consistency. Now we don’t want to have any fresh conditions on ψ. We want all the conditions on ψ to be included among (2-22). That means to say, we want to have (2-24) a consequence of (2-22) which means we require
(2-25)
If (2-25) does hold, then (2-24) is a consequence of (2-22) and is not a new condition on the wave function.
Now we know that the φ’s are all first-class in the classical theory, and that means that the Poisson bracket of any two of the φ’s is a linear combination of the φ’s in the classical theory. When we go over to the quantum theory, we must havea similar equation holding for the commutator, but it does not necessarily follow that the coefficients c are all on the left. We need to have these coefficients all on the left, because the c’s will in general be functions of the coordinates and momenta and will not commute with the φ’s in the quantum theory, and (2-24) will be a consequence of (2-22) only provided the c’s are all on the left.
When we set up the quantities φ in the quantum theory, there may be some arbitrariness coming in. The corresponding classical expressions may involve quantities which don’t commute in the quantum theory and then we have to decide on the order in which to put the factors in the quantum theory. We have to try to arrange the order of these factors so that we have (2-25) holding with all the coefficients on the left. If we can do that, then we have the supplementary conditions all consistent with each other. If we cannot do it, then we are out of luck and we cannot make an accurate quantum theory. In any case we have a first approximation to the quantum theory, because our equations would be all right if we look at them only to the order of accuracy of Planck’s constant ħ and neglect quantities of order ħ2.
I have just discussed the requirements for the supplementary conditions to be consistent with one another. There is a similar discussion needed in order to check that the supplementary conditions shall be consistent with the Schrödinger equation. If we start with a ψ satisfying the supplementary conditions (2-24) and let that ψ vary with the time in accordance with the Schrödinger equation, then after a lapse of a short interval of time will our ψ still satisfy the supplementary conditions? We can work out the requirement for that to be the case and we get
(2-26)
which means that [φj, H] must be some linear function of the φ’s:
(2-27)
if we are not to get a new supplementary condition. Again we have anequation which we know is all right in the classical theory. φ j , and H are both first-class, so their Poisson bracket vanishes weakly. The Poisson bracket is thus strongly equal to some linear function of the φ’s in the classical theory. Again we have to try to arrange things so that in the corresponding quantum equation we have all our coefficients on the left. That is necessary to get an accurate quantum theory, and we need a bit of luck, in general, in order to be able to bring it about.
Let us now consider how to quantize a Hamiltonian theory in which there are second-class constraints. Let us think of this question first in terms of a simple example. We might take as the simplest example of two second-class constraints
(2-28)
If we have these two constraints appearing in the theory, then their Poisson bracket is not zero, so they are second-class. What can we do with them when we go over to the quantum theory? We cannot impose (2-28) as supplementary conditions on the wave function as we did with the first-class constraints. If we try to put q1ψ = 0, p1ψ = 0, then we should immediately get a contradiction because we should have (q1p1 − p1q1)ψ = iħψ = 0. So that won’t do. We must adopt some different plan.
Now in this simple case it’s pretty obvious what the plan must be. The variables q1 and p1 are not of interest if they are both restricted to be zero. So the degree of freedom 1 is not of any importance. We can just discard the degree of freedom 1 and work with the other degrees of freedom. That means a different definition for a Poisson bracket. We should have to work with a definition of a Poisson bracket in the classical theory
(2-29)
This would be sufficient because it would deal with all the variables which are of physical interest.Then we could just take q1 and p1 as identically zero. There’s no contradiction involved there, and we can pass over to the quantum theory, setting it up in terms only of the degrees of freedom n = 2, ..., N.
In this simple case it is fairly obvious what we have to do to build up a quantum theory. Let us try now to generalize it. Suppose we have P1 ≈ 0, q1 ≈ f(qr, Pr), r = 2, ..., N, so f is any function of all the other q’s and p’s. We could drop out the number 1 degree of freedom if we substitute f (qr , pr) for q1 in the Hamiltonian and in all the other constraints. Again we can forget about the number 1 degree of freedom and simply work with the other degrees of freedom and pass over to a quantum theory in these other degrees of freedom. Again we should have to work with the (2-29) kind of Poisson bracket, referring only to the other degrees of freedom.
That is the idea which one uses for quantizing a theory which involves second-class constraints. The existence of second-class constraints means that there are some degrees of freedom which are not physically important. We have to pick out these degrees of freedom and set up new Poisson brackets referring only to the other degrees of freedom which are of physical importance. Then in terms of those new Poisson brackets we can pass over to the quantum theory. I would like to discuss a general procedure for carrying that out.
For the present, we are going back to the classical theory. We have a number of constraints φj ≈ 0, some of them first-class, some second-class. We can replace these constraints by independent linear combinations of them, which will do just as well as the original constraints. We try to arrange to take the linear combinations in such a way as to have as many constraints as possible brought into the first class. There may then be some left in the second class which we just cannot bring into the first class by taking linear combinations of them. Those which are left in the second class I will call χs , s = 1,..., S. S is the number of second-class constraints which are such that no linear combination of them is first-class.
We take these surviving second-class constraints and we form all their Poisson brackets with each other and arrange these Poisson brackets as a determinant Δ:
I would like now to prove a
Theorem: The determinant Δdoes not vanish, not even weakly. Proof: Assume that the determinant does vanish. I’m going to show that we get a contradiction. If the determinant vanishes, then it is of some rank T < S. Now let us set up the determinant A:
A has T + 1 rows and columns. T + 1 might equal S or might be less than S. If we expand A in terms of the elements of its first column, we will get each of these elements multiplied into one of the sub-determinants of Δ. Now I don’t want all of these sub-determinants to vanish. It might so happen that they do all vanish. And in that case, I would choose the χ’s which are referred to among the rows and columns of A in a different way. There must always be some way of choosing the χ’s which occur in A so that the sub-determinants don’t all vanish, because Δis of rank T. So we choose the χ ’s m such a way that the coefficients of the elements in the first column are not all zero.
Now I will show that A has zero Poisson brackets with any of the φ’s. If we form the Poisson bracket of φ with a determinant, we get the result by forming the Poisson bracket of φ with the first column of the determinant, adding on the result of forming the Poisson bracket of φ with the second column of the determinant, and so on. Thus
This looks rather complicated, but one can easily see that every one of these determinants vanishes. In the first place, the first determinant on the right vanishes: if φ is first class, then the first column vanishes; if φ is second class, then φ is one of the χ ’s and we have a determinant which is a part of the determinant Δwith T + 1 rows and columns. But Δis assumed to be of rank T
, so that any part of it with T + 1 rows and columns vanishes. Now, the second determinant on the right vanishes weakly because the first column vanishes weakly. Similarly all the other determinants vanish weakly. The result is that the whole right-hand side vanishes weakly. Thus A is a quantity whose Poisson bracket with every one of the φ’s vanishes weakly.
Also, we can expand the determinant A in terms of the elements of the first column, and get A as a linear combination of the χ’s. So we have the result that a certain linear combination of the χ’s has zero Poisson brackets with all the φ’s. That means that this linear combination of the χ’s is first class. That contradicts our assumption that we have put as many χ’s as possible into the first class. That proves the theorem.
Incidentally, we see that the number of surviving χ ’s, which cannot be brought into the first class, must be even, because the determinant Δis antisymmetrical. Any antisymmetrical determinant with an odd number of rows and columns vanishes. This one doesn’t vanish and therefore must have an even number of rows and columns.
Because this determinant, Δ, doesn’t vanish, we can bring in the reciprocal css′ of the matrix whose determinant is Δ. We define the matrix css′ by
(2-30)
We now define new Poisson brackets in accordance with this formalism: any two quantities ξ , η have a new Poisson bracket defined by
(2-31)
It is easy to check that new Poisson brackets defined in this way satisfy the laws which Poisson brackets usually satisfy: [ξ, η]* is antisymmetrical between ξ and η, is linear in ξ, is linear in η, satisfies the product law [ξ1ξ2, η]* = ξ1[ξ2η]* + [ξ1, η]*ξ2, and obeys the Jacobi identity [[ξ , η]*, ζ ]* + [[η, ζ ]*ξ ]* + [[ζ , ξ ]*, η]* = 0. I don’t know of any neat way of proving the Jacobi identity for the new Poisson brackets. If one just substitutes according to the definition and works it out in a complicated way, one does find that all the terms cancel out and that the left-hand side equals zero. I think there ought to be some neat way of proving it, but I haven’t been able to find it. The straightforward method I have given in the Canadian Journal of Mathematics, 2, 147 (1950). The problem has been dealt with by Bergmann, Physical Review, 98, 531 (1955).
Now let us see what we can do with these new Poisson brackets. First of all, I would like you to notice that the equations of motion are as valid for the new Poisson brackets as for the original ones.
because the terms [χs′ , HT] all vanish weakly on account of HT being first-class. Thus we can write
Now if we take any function ξ whatever of the q’s and p’s, and form its new Poisson bracket with one of the χ’s, say χs′′, we have
(2-30)
Thus we can put the χ’s equal to 0 before working out new Poisson brackets. This means that the equation
(2-32)
may be considered as a strong equation.
We modify our classical theory in this way, bringing in these new Poisson brackets, and this prepares the ground for passing to the quantum theory. We pass over to the quantum theory by taking the commutation relations to correspond to die new Poisson bracket relations and taking the strong equations (2-32) to be equations between operators in the quantum theory. The remaining weak equations, which are all first class, become again supplementary conditions on the wave functions. The situation is then reduced to the previous case where there were only first-class φ’s. We have again, therefore, a method of quantizing our general classical Hamiltonian theory. Of course, we again need a bit of luck in order to arrange that the coefficients are all on the left in the consistency conditions.
That gives the general method of quantization. You notice that when we have passed over to the quantum theory, the distinction between primary constraints and secondary constraints ceases to be of any importance. The distinction between primary and secondary constraints is not a very fundamental one. It depends very much on the original Lagrangian which we start off with. Once we have gone over to the Hamiltonian formalism, we can really forget about the distinction between primary and secondary constraints. The distinction between first-class and second-class constraints is very important. We must put as many as possible into the first class and bring in new Poisson brackets which enable us to treat the surviving second-class constraints as strong.
DR. DIRAC
Lecture No. 3
QUANTIZATION ON CURVED SURFACES
We started off with a classical action principle. We took our action integral to be Lorentz-invariant. This action gives us a Lagrangian. We then passed from the Lagrangian to the Hamiltonian, and then to the quantum theory by following through certain rules. The result is that, starting with a classical field theory, described by an action principle, we end up with a quantum field theory. Now you might think that that finishes our work, but there is one important problem still to be considered: whether our quantum field theory obtained in this way is a relativistic theory. For the purposes of discussion, we may confine ourselves to special relativity. We have then to consider whether our quantum theory is in agreement with special relativity.
We started from an action principle and we required that our action should be Lorentz-invariant. That is sufficient to ensure that our classical theory shall be relativistic. The equations of motion that follow from a Lorentz invariant action principle must be relativistic equations. It is true that when we put these equations of motion into the Hamiltonian form, we are disturbing the four-dimensional symmetry. We are expressing our equations in the form
(1-21)
The dot here means dg/dt and refers to one absolute time, so that the classical equations of motion in the Hamiltonian form are not manifestly relativistic, but we know that they must be relativistic in content because they follow from relativistic assumptions.
However, when we pass over to the quantum theory we are making new assumptions. The expression for HT which we have in the classical theory does not uniquely determine the quantum Hamiltonian. We have to decide questions about the order in which to put non-commuting factors in the quantum theory. We have something at our disposal in choosing this order, and so we are making new assumptions. These new assumptions may disturb the relativistic invariance of the theory, so that the quantum field theory obtained by this method is not necessarily in agreement with relativity. We now have to face the problem of seeing how we can ensure that our quantum theory shall be a relativistic theory.
For that purpose we have to go back to first principles. It is no longer sufficient to consider just one time variable referring to one particular observer; we have to consider different observers moving relatively to one another. We must set up a quantum theory which applies equally to any of these observers, that is, to any time axis. To get a theory involving all the different time axes, we should first get the corresponding classical theory and then pass from this classical theory to the quantum theory by the standard rules.
I would like to go back to the beginning of our Hamiltonian development and consider a special case. We started our development by taking a Lagrangian L, which is a function of dynamical coordinates and velocities q, , introducing the momenta, then introducing the Hamiltonian. Let us take the special case when L is homogeneous of the first degree in the ’s. Then Euler’s Theorem tells us that
(3-1)
That just tells us that pnn − L = 0. Thus we get in this special case a Hamiltonian that is zero.
We necessarily get primary constraints in this case. There must certainly be one primary constraint, because the p’s are homogeneous functions of degree zero in the velocities. The p’s are thus functions only of the ratios of the velocities. The number of p’s is equal to N, the number of degrees of freedom, and the number of ratios of the velocities is N − 1. N functions of N − 1 ratios of the velocities cannot be independent. There must be at least one function of the p’s and q’s which is equal to zero; there must be at least one primary constraint. There may very well be more than one. One can also see that, if we are to have any moti
on at all with a zero Hamiltonian, we must have at least one primary first-class constraint.
We have the expression (1-33) for the total Hamiltonian
H′ must be a first-class Hamiltonian, and as 0 is certainly a first-class quantity we may take H′ = 0. Our total Hamiltonian is now built up entirely from the primary first-class constraints with arbitrary coefficients:
(3-2)
showing that there must be at least one primary first-class constraint if we are to have any motion at all.
Our equations of motion now read like this:
We can see that the ġ’s may all be multiplied by a factor because, since the coefficients ν are arbitrary, we may multiply them all by a factor. If we multiply all the dg/dt’s by a factor, it means that we have a different time scale. So we have now Hamiltonian equations of motion in which the time scale is arbitrary. We could introduce another time variable τ instead of t and use τ to give us equations of motion
(3-3)
So we have now a Hamiltonian scheme of equations of motion in which there is no absolute time variable. Any variable increasing monotonically with t could be used as time and the equations of motion would be of the same form. Thus the characteristic of a Hamiltonian theory where the Hamiltonian H′ is zero and where every Hamiltonian is weakly equal to zero, is that there is no absolute time.
The Dreams That Stuff is Made of Page 93