Dirac Brackets
Mechanics published in Belfer Graduate School of Science based on Dirac’s Lectures Dirac’s Lectures on Quantum Mechanics published Monographs Series Number Two (1964)
Steven Avery Dirac Dirac writ writes es that that the the rout routee on onee shou should ld take take to get the the mo most st gener general al relat relativi ivist stic ic quan quantu tum m field field theo theory ry is to star startt wi with th an acti action on princ princip iple le,, find find the the Lagr Lagran angia gian, n, find find the the Ha Hamil milti tion onian ian,, and then quantize the Hamiltonian to get a first approximation of the Quantum Field Theory: classical action → Lagrangian → Hamiltonian → QFT . When one writes down the most general Lagrangians which may not be quadratic in the velocity, however, one immediately runs into complications in the Hamiltonian formulation. Dirac generalizes the Hamiltonian and Poisson brackets to handle general Lagrangians possibly with constraints.
1
Review Rev iew of the Standar Standard d Hamilto Hamiltonian nian Method Method
Following Dirac, to illustrate these ideas we will start with systems of N discrete discrete particles described by dynamical variables q j j for j = 1, . . . , N , instead of a system having infinite degrees of freedom described by fields ϕ a (x). Our action, then, is S = Ldt, where L = L(q, q ˙) is our Lagrangian. Lagrangian. Requiring Requiring that the equations of motion, q j (t) make the action stationary, we find that the dynamics of the particles satisfy Lagrange’s equations:
d dt
∂L ∂ ˙ q ˙ j
=
∂L . ∂q j j
(1)
To get to the Hamiltonian formalism, where the symmetries of the theory are more easily elucidated and a transition to quantum mechanics more readily made, we define the canonical momenta ∂L p j = (2) ∂ ˙ q ˙ j and then define the Hamiltonian H (q, p) = p j ˙q j − L,
(3)
which leads to dynamics given by q ˙ j =
∂H ∂p j
and p˙ j = −
1
∂H . ∂q j j
(4)
There is a key subtlety at this point. When we switch to the Hamiltonian formalism, we also switch (via a Legendre transformation) independent variables from q j and q ˙ j to q j and p j . In other words, we must write H as function of q ’s and p’s only, with no q ˙’s. If we want to be able to use the most general Lagrangians, however, then Hamilton’s equations will not give the correct answer, and it become necessary to generalize the Hamiltonian. Before we give an explicit example of how problems arise, let’s continue with the Hamiltonian formulation of classical mechanics by introducing the Poisson bracket of two functions of the q ’s and p ’s (and maybe t): {f (q, p), g (q, p)} =
∂f ∂g ∂f ∂g − . ∂q j ∂p j ∂p j ∂q j
Then, one can show that g˙ = { g, H } +
∂g , ∂t
(5)
(6)
when q and p satisfy the equations of motion. Note that the Poisson brackets are only well defined for functions of q ’s, p’s, and t, but not if we have a function q ˙. Because of problems illustrated below we will also have to generalize the Poisson bracket when we have certain kinds of contraints. Note that the Poisson bracket satisfies the following properties: • bilinearity {af 1 + bg1 , cf 2 + dg2 } = ac {f 1 , f 2} + ad{f 1 , g2 } + bc{g1 , f 2 } + bd{g1 , g2 } for constants a, b, c , and d . • antisymmetry{f, g } = −{ g, f } • product rule {fg, h} = { f, h}g + f {g, h} • Jacobi identity {f, {g, h}} + {h, {f, g }} + {g, {h, f }} = 0 From the Poisson bracket one generally transistions to a first approximation of a quantum theory via canonical quantization by letting the q and p become operators (and therefore H ) whose commutator is i times the Poisson bracket. In general, we can think of the Poisson bracket getting promoted to a commutator in this way.
2
Complications: A Lagrangian linear in velocity
Cosider a charged particle confined to the x -y plane in a constant magnetic field B0 in the z -direction. We may write the Lagrangian for the particle as L =
q 1 mv2 + A · v − V, 2 c
where V is some external potential and we set A =
B0
2
zˆ × r =
2
B0
2
(xyˆ − y xˆ).
Then our Lagrangian becomes L =
m
2
( x˙ 2 + y˙ 2 ) +
qB 0 qB 0 (xy˙ − y x˙ ) − V (x, y ), 2c 2c
where we redefine the potential term slightly for later convenience. We are free to multiply a Lagrangian by a nonzero constant since it will not affect the equations of motion; therefore, we may rewrite our Lagrangian L =
1 2 1 qB 0 qB 0 (x˙ + y˙ 2 ) + (xy˙ − y x˙ ) − V (x, y ) = (x˙ 2 + y˙ 2 ) + η (xy˙ − y x˙ ) − ηV (x, y ), 2 2mc 2mc 2
where we introduce the constant η =
qB 0 2mc
. Now, let’s find the equations of motion:
¨ − η y˙ = η y˙ − η∂ x V ⇒ x¨ = 2η y˙ − η∂ x V x
(7) (8)
y¨ + η x˙ = − η x˙ − η∂ y V ⇒ y¨ = − 2η x˙ − η∂ y V.
Let’s consider the large η limit, which corresponds to large magnetic field, low mass, or large charge. Let’s make η so large that we can neglect the usual kinetic energy term in the Lagrangian and we get a Lagrangian that is linear in the velocities. This is exactly the kind of problem that interests Dirac: Lagrangians which are not quadratic in the velocity. In this case, our Lagrangian becomes L −→ η (xy˙ − y x˙ ) − ηV (x, y ).
(9)
The Euler-Lagrange equations remain valid and give us the following equations of motion: 1 ∂V 2 ∂x 1 ∂V η x˙ = − η x˙ − η∂ y V ⇒ x˙ = − , 2 ∂y
−η y˙ = η y˙ − η∂ x V ⇒ y˙ =
(10)
(11)
which correspond exactly to the large η limit of the more general equations of motion. These equations of motion may seem a little odd since they are first order differential equations, but they should be perfectly allowable in the Lagrangian formulation and Dirac wants to consider more general Lagrangians like this one. The Lagrangian density that gives the Dirac equation is of this type. At this point, having worked through the Lagrangian formulation, let’s move into the Hamiltonian formulation and see what happens. We begin, as usual by finding the conjugate momenta: ∂L = − ηy ∂ x˙ ∂L = ηx, py = ∂ y˙
px =
(12)
(13)
and this is where our troubles begin. When we complete this step with Lagrangians that are quadratic in velocity we find our conjugate momenta as (hopefully) invertible 3
functions of the q ˙’s. Not so, in this problem. We might worry that we will not be able to write down a Hamiltonian H = H (q, p), but Dirac demonstrates that this is always possible. When problems like this arise, however, something strange will happen to the momenta terms: (14) H = p x ˙x + py ˙y − L = ηV (x, y ). The momenta completely dropped out of the Hamiltonian! This seems a little fishy, but let’s continue by trying to find the equations of motion, using Hamilton’s equations. We find ∂H =0 ∂p x ∂H y˙ = =0 ∂p y ∂H ∂V = − η p˙x = − ∂x ∂x ∂H ∂V = − η . p˙y = − ∂y ∂y x˙ =
(15)
(16)
(17)
(18)
Before we address some obvious concerns, we proceed by plugging in with the definitions of the conjugate momenta into the last two equations: ∂V ∂x ∂V x˙ = − . ∂y y˙ =
(19)
(20)
Let’s summarize what we see here. We get two sets of equations of motion, one set is trivial and wrong (from the lack of momenta in the Hamiltonian), and the second set is off by a factor of 2 . Furthermore, these two sets of equations of motion are inconsistent. From this we are led to conclude that some serious changes are in order; our formalism cannot handle this problem. One might try and make the two equations consistent by sometimes expressing the Hamiltonian as a function of momenta and sometimes as a function of coordinates or maybe half of each, depending on what you are calculating. This is not valid and certainly not the general solution we want, but it illustrates the basic problem we are dealing with: the coordinates and momenta are not independent. We have constraints between them, and we have not taken the constraints into account in finding the equations of motion. It is not surprising in that context that we get inconsistant answers, since the answers only became inconsistent when we applied our constraint equations (the definition of the momenta). If we think of the definitions for the momenta as a constraint, then we see that when we computed the equations of motion there was no way in which we used the constraint. In a Lagrangian formulation of mechanics, holonomic constraints can be handled by Lagrange multipliers, which motivates how we generalize the Hamiltonian.
4
3
Generalizing the Hamiltonian Formulation
Having illustrated the need for a revision of the Hamiltonian for more general Lagrangians, we now outline a procedure to rectify matters developed by Dirac. With our Lagrangian in hand, we begin as usual by finding the conjugate momenta through pi =
∂L , ∂ q ˙i
which we saw may not be functions of the q ˙’s. In this case, we get certain relations of the w form φ j (q, p) = 0 between our coordinates and momenta, exactly as we found in the above example. Dirac suggests we consider these relations as constraints on our equations of w motion. The = is used to indicate “weak equality,” by which we mean something that is true when the equations of motion are satisfied or “on-shell”, but may not be generally (note that Dirac and the literature frequently define ≈ to have this meaning). The constraints derived from the definition of conjugate momentum are called primary constraints. If our system had other constraints we throw those in and call them primary constraints too. Now, if one consults the third edition of Goldstein, Poole and Safko, one will find a statement: “In the Hamiltonian formulation there can be no constraint equations among the coordinates,” and so we realize the need to modify the standard procedure. We proceed as usual, by writing down the Hamiltionian H = q ˙i pi − L. Now, one might worry that we cannot express H as a function of coordinates and momenta only; however, if we consider the differential dH , we see dH = d q ˙ j p j + q ˙ j dp j − dL ∂L ∂L = q ˙ j dp j + p j dq ˙ j − dq j − dq ˙ j ∂q j ∂ q ˙ j ∂L = p j dq ˙ j − p j dq ˙ j + q ˙ j dp j − dq j ∂q j ∂L dq j ; = q ˙ j dp j − ∂q j
that is, the variation in H is in terms of the variation of the coordinates and the momenta only. Therefore, the Hamiltonian may always be written as a function of q and p only. In fact, this is one way to derive Hamilton’s equations. If we apply Lagrange’s equations we may rewrite the last term dH = q ˙ j dp j − p˙ j dq j , which we may set equal to dH =
∂H ∂H dq j + dp j . ∂q j ∂p j
At this point, we must ask ourselves, what went wrong with the equations of motion? Our derivation of Hamilton’s equations did not anywhere require that the Lagrangian be quadratic in the q ˙’s nor did we require that the q ˙’s be functions of the momenta. What went wrong? The φ j impose constraints on what variations are allowed and we have not enforced our constraints on the dq ’s and dp ’s. 5
Now, since in the end we hope to find equations of motion that make our constraints zero, Dirac argues that we should generalize our Hamiltonian beyond the “naive” Hamiltonian H to H = H + c j φ j , ∗
since when the equations of motion are satisfied H = H . Note that our coefficients c j are not constants but arbitrary functions of our coordinates and momenta. This is just like the method of Lagrange multipliers. If we use this new Hamiltonian to compute equations of motion we should be able to allow arbitrary variations in our coordinates and momenta. According to the method of Lagrange multipliers, then, we should vary the coefficients of the constraints separately to find ∗
∂φ m ∂φ m dH = q ˙ j dp j − p˙ j dq j + φ j dc j + cm dq j + dp j ∂q j ∂p j ∂H ∂φ m ∂H ∂φ m dq j + dp j + φ j dc j . = + cm + cm ∂q j ∂q j ∂p j ∂p j ∗
Therefore, we find equations of motion ∂H ∂φ m − cm ∂q j ∂q j ∂H ∂φ m q ˙ j = + cm ∂p j ∂p j
p˙ j = −
φ j = φ j ,
which is not quite what we get when using Lagrange multipliers in the Lagrangian formulation. In that case the constraint equation comes back as an equation of motion. Dirac writes the equations of motion as ∂H ∂φ k − uk ∂q j ∂q j ∂H ∂φ k q ˙ j = , + uk ∂p j ∂p j
p˙ j = −
where uk are unknowns, which are not necessarily functions of q and p. 3.0.1 A first extension of Poisson Brackets
These equations of motion may be written with our usual Poisson brackets as q ˙ j = { q j , H } + uk {q j , φk } p˙ j = { p j , H } + uk { p j , φk },
or more generally we can say that w
f ˙ = { f, H } + uk {f, φk }.
6
(21) (22)
As mentioned previously, the u k are not functions of q and p and so do not have a welldefined Poisson bracket with other quantities; however, we may extend the definition of the Poisson bracket by supposing that it might have a definition for quantities which are not a function of q and p and requiring that it maintain the same four properties listed above. Then we may write w
f ˙ = { f, H + uk φk }
= { f, H } + {f, uk φk } = { f, H } + {f, uk }φk + uk {f, φk } w
= { f, H } + uk {f, φk },
where in the last step we used the fact that the φ k are weakly zero. The important thing to note is that we cannot use weak equations in the Poisson bracket, since the Poisson bracket does not respect our on-shell constraints. Later, we introduce the Dirac bracket which does respect our constraint equations. We need to compute the Poisson brackets and then at the end use our constraints, but this is not the case with the Dirac bracket. 3.0.2 Finding a Total Hamiltonian and Eliminating Arbitrariness
The obvious thing to do at this point is to define the total Hamiltonian as H T = H + uk φk , so that we can make our equations even shorter. At this point, one may be concerned that we have arbitrary functions u k in our equations of motion, but there are two things to note: there are some consistancy conditions which help determine uk , and for some problems there are unphysical degrees of freedom, in which case we should expect arbitrary functions in the solution. An example of an unphysical degree of freedom is gauge freedom in electromagnetism. As hinted above there are some consistency conditions that must be satisfied. We w require that the equations of motion for our constraints satisfy φ˙ k = 0, which forces w
{φk , H } + u j {φk , φ j } = 0. There are four different types of conditions that this can give us: 1. An equation that is inherently unsatisfiable such as 1 = 0. 2. An equation that is identically true possibly after using one of our primary constraints. 3. A new constraint on the coordinates and momenta that is independent of our arbitrary u k ’s. 4. A consistency condition that helps us determine the u k . The first case tells us that the Lagrangian we are working with gives unworkable equations of motion, such as L = q . The second case does not tell us anything new and so we can forget about it. The third case gives us a new constraint on our coordinates and 7
momenta. We will call a constraint that one arrives at in this manner a secondary constraint. The distinction between primary and secondary constraints is largely an artificial one that may depend on the choice of Lagrangian and will not affect our formalism. The last case we can use to get rid of some of the arbitrary degrees of freedom in our equations of motion. w The secondary constraints we might label as φl = 0. We add these into the total Hamiltonian with some arbitrary coefficients in exactly the same manner that we added our primary constraints. Similarly, we get some new consistency conditions when we require w φ˙ l = 0, which may lead to new constraints, and so on. We continue until we get no new constraints. For the rest of these notes, we will just call all of the constraints φ j . As for the uk , they must satisfy conditions of the form
w
{φ j , H } + uk {φ j , φk } = 0. There must be a solution to this equation if the Lagranigan is consistent, but there may be many solutions depending on the number of conditions we get. Suppose that we are in the general case, where there are many solutions, then we should find one particular solution um = U m (q, p). After which we should look for all possible solutions to the inhomogeneous equation V m {φ j , φm } = 0, then the most general solution we can write is our particular solution plus a linear com bination of the inhomogeneous solutions: um = U m + va V am ,
where a runs over all the independent inhomogeneous solutions that we found and la bels each such solution V am . The va are some new arbitrary functions of time and have no consistency conditions. What then have we gained? Well, the number of arbitrary parameters should be reduced by the number of conditions that we got on the u m ’s and while the u m ’s might be functions of q and p , the v a ’s are only functions of time. In some cases, as in our example, there is only one solution, indicating that there are no unphysical degrees of freedom in our Lagrangian. At this point, we have everything we need to find the equations of motion for a general Lagrangian, but we haven’t discussed Dirac brackets, the title of these notes. I should also mention that I am skimping on the story of these unphysical degrees of freedom. In particular, it turns out we can relate some of the constraints to generators of guage transformations. 3.0.3 Dirac Brackets
Why do we need Dirac brackets? What’s wrong with the good old Poisson brackets? Well, recall the motivation for this whole exercise was to find a general method of going from a classical Lagrangian to a quantum theory. Knowing how to find the equations of motion is not sufficient, we need to know what commutation relations to impose on our quantum operators. Traditionally, we just take Poisson brackets to i times the commutator; however, this method does not work in the general case. For instance, take a system 8
of N particles with positions and momenta q j and q k . Now suppose that we have the constraints: w
q 1 = 0 w
p1 = 0.
The Poisson bracket does not know about these constraints and is the usual {q 1 , p1 } = 1. If we blindly follow our procedure, then we impose the commutator relation [ˆ q1 , pˆ1 ] = i ; however, if we try to enforce our constraint we see that [ qˆ 1 , pˆ1 ] = [0, 0] = 0. Of course, these constraints are not particularly interesting, but they illustrate the point. In this example, particle one has no dynamics and is uninteresting, so we could ignore it and quantize the remaining N − 1 particles. In general, it is not obvious, however, what commutation relations one should impose. Therefore, we need a generalized Poisson bracket that in some sense “knows” about the constraints and “respects” them. Before giving the answer it is helpful to distinguish between first class and second class constraints. Following Dirac’s terminology, a function of coordinates and momenta, R, is first class if its Poisson bracket with all of the constraints is at least weakly zero, ie. w {R, φ j } = 0 ∀ j. Dirac notes that anything in our theory which is weakly zero must be strongly equal to some linear combination of our φ j ’s since they are the only quantities which are weakly zero. Therefore {R, φ j } = r j φ j for some coefficients r j . The number of independent first class constraints that we have is the number of unphysical degrees of freedom in the theory, or the number of va ’s that we have. One can also demonstrate that the Poisson bracket of two first class quantities is again first class. What is of more immediate interest is the second class constraints (those constraints who have nonvanishing Poisson bracket with at least one other constraint). Notice that in the simple example given above the two constraints were second class. In fact, it is the second class constraints that are giving us trouble, since on the one hand, we should promote the Poisson bracket to commutation relations, but on the other hand the constraints should be zero in the end. Sometimes we can take a linear combination of our second class constraints and form first class constraints. Before proceeding, we should minimize the number of independent second class constraints that we have in this way. Unfortunately, we cannot always make our entire set of independent constraints first class in this way. If we could find some generalized bracket so that the bracket of any two constraints would be zero, then that would solve the problem. This would mean that we don’t have to worry about the order of evaluating the generalized bracket and using our constraints, and there would be no inconsistencies when we promoted our bracket to a commutator. We would also like the generalized bracket to have the 4 properties of the Poisson bracket and we should still be able to write our equations of motion in terms of the bracket. The solution, of course, is the Dirac bracket, defined by {f, g }D = { f, g } − {f, φ˜k }M kl 1 {φ˜l , g }, −
9
where φ˜k are only the second class constraints and the brackets on the right are the usual Poisson brackets. The matrix M is the matrix formed by the Poisson bracket of all the second class constraints, so that M ij = {φi , φ j }. One can easily demonstrate that the Dirac bracket has all of the properties listed above. One might worry that our matrix M might be singular, but one can prove that as long as we have minimized the number of secondary consraints, the matrix M will have nonzero determinant and therefore be invertible. 3.1 Working through our example
Now that we have developed the necessary formalism, let’s work through our earlier example of a charge confined to the x-y plane with a magnetic field in the z direction. Remember that we work in the high magnetic field limit which gave us the “naive” Hamiltonian H = ηV (x, y ), and conjugate momenta px = − ηy py =
ηx.
The above equations give us two primary contraints φ1 = ηy + p x φ2 = ηx − py ,
which we say are weakly equal to zero. Now we generalize our Hamiltonian to H T = ηV (x, y ) + u1 (ηy + px ) + u2(ηx − py ) w
and apply our consistency condition that {φ j , H T } = 0. Therefore, we find that {φ1 , H T } = η { px , V (x, y )} + u2 (η { px , x} − η {y, py }) = − η
∂V + 2u2 ∂x
w
= 0.
This is not a secondary constraint but a consistency condition for our arbitrary coefficient u2 . Similarly, {φ2 , H T } = η
∂V + 2 u1 ∂y
w
= 0,
another consistency condition, but no new constraints, so we only have our two primary constraints to deal with. Therefore, we need not add any new terms to our Hamiltonian. We need to find the Poisson brackets between our constraints to determine whether they are first-class or second-class constraints: {φ1 , φ2 } = −{φ2, φ1 } = − η − η = − 2η, and obviously the Poisson bracket of the contraints with themselves is zero. Note that for Grassmannian constraints this may not be true. Since we see that our two primary 10
constraints are in fact second-class, we need to define Dirac brackets. The first step is to find our matrix 0 −1 M = 2η , 1 0
whose inverse is easy to compute, namely −1
M
Note that we may write M ij 1 = −
1 = 2η
0 1 . −1 0
1 2η ij
. Onto the Dirac brackets:
{A, B }D = { A, B } −
ij {A, φi }{φ j , B }. 2η
Finally, let’s compute our equations of motion using the Dirac brackets. We find x˙ = { x, H T }D
1 {x, φ1 }{φ2 , H T } 2η 1 = u 1 {x, φ1 } − {x, φ1 }η (∂ y V + 2u1 ) 2η 1 1 ∂V = − {x, φ1 }∂ y V = − , 2 2 ∂y = u 1 {x, φ1 } −
exactly the result we found through Lagrange’s equations. Similarly, we find 1 2
y˙ = { y, H T }D = − u2 + (∂ x V + 2u2 )
=
1 ∂V , 2 ∂x
and for the momenta, p˙ x = { px , H T }D = − η η
1 ∂V − ηu 2 + { px , φ2}{φ1 , H T } 2η ∂x
= − η∂ x V + ∂ x V − ηu 2 + ηu 2 2 1 ∂V = − η . 2 ∂x
Finally, for p y : p˙y = { py , H T }D = − η
∂V ∂V 1 + 2u1 ) − ηu 1 + η 2( 2η ∂y ∂y
1 ∂V = − η . 2 ∂y
11
At this point, if we have done everything correctly, then we should find that the equations of motion for our momenta are consistent with our constraints φ j . Let’s check this: 1 ∂V 2 ∂x
p˙x = − η y˙ = − η
1 ∂V , 2 ∂y
p˙ y = η x˙ = − η
exactly the equations of motion we found from the Dirac bracket, using our constraint equation. Therefore, we have found equations of motion that are consistent and match up with the Lagrange equations. If we wish to quantize this theory we should compute the Dirac brackets between all of our momenta and coordinates so that these may be promoted to commutation relations. 1 1 {x, φ1}{φ2, y } = − 2η 2η 1 1 {x, px }D = 1 − (1)(η ) = 2η 2 {y, px }D = 0 {x, py }D = 0 1 1 {y, py }D = 1 + (−1)(η ) = , 2η 2
{x, y }D = { x, y } −
and we find something very interesting. The Dirac bracket between our two coordinates is nonzero, so when we move to a quantum thoery we find that our coordinates no longer commute. This describes a noncommutative geometry. Note that, naturally, we also get an uncertainty principle between x and y ! As mentioned above, with the Dirac bracket, we do not need to be careful about applying our constraints last. For instance, having computed {x, px }D we may find {x, y }D by using our constraint: 1 1 {x, y }D = − {x, px }D = − , 2η η and we get exactly the result we found directly.
4
What I Left Out
I left out some of the information about dealing with unphysical degrees of freedom and how we can find generators of transformations in the unphyical directions. Dirac works through the example of the electromagnetic 4-potential. The only other thing I left out was Dirac’s discussion of quantization on curved surfaces and some of the proofs.
12
5
More Examples
5.1 An example with Grassmann coordinates: Peskin and Schroeder 9.2d
The motivation for using Lagrangians which are linear in velocity is that those are the Lagrangians which give the correct equations of motion for fermions. Classically, there is not a good analog of fermions, but it is useful to be able to write down classical fermionic Lagrangians to be used in Feynman path integrals. This requires the introduction of Grassmann numbers. For example, if we have grassmann numbers α and β , then we require that αβ = −βα , or [α, β ]+ = 0. This implies that α2 = β 2 = 0. Fermions are described by grassmann-valued coordinates ψ (t), such that ψ (t) is an independent Grassmann number at every time t. These strange Grassmann numbers will require the introduction of slightly different definitions, usually just changing a sign or ordering. The problem gives us the Euclidean Lagrangian LE = ψ¯ψ˙ + ω ¯ ψψ,
where ψ¯ and ψ are independent real grassmann-valued coordinates. The Euclidean Lagrangian means that it uses imaginary time τ = it. First, we would like to find the corresponding Minkowski Lagrangian and then find the corresponding quantized Hamiltonian with appropriate (anti-)commutation relations. First, we need to know how the Euclidean Lagrangian is related to the regular Lagrangian. We know iS = − S E , (23) from which we can conclude that L = i ψ¯ψ˙ − ω ¯ ψψ.
(24)
Let’s confirm that: iS
¯ ˙ ¯ = d ( ) ( )− () () ¯ ˙ ¯ = (− ) d ( ) ( )− ( ) ( ) ¯ ˙ ¯ = − d + i i
t iψ t ψ t
i
ωψ t ψ t
τ i2ψ τ ψ τ
t ψψ
ω ψ τ ψ τ
ω ψψ .
Note that in the first line ψ˙ means ddψt and in the second and third lines ψ˙ = ddψτ . This accounts for the extra factor of i . At this point, we need to introduce a differentiation convention. We choose left differentiation, which means that ∂ ¯ ∂ ¯ ψ ) = − ψ. (ψψ ) = − (ψ ¯ ∂ψ ∂ψ
13
Using this definition, let’s find the conjugate momenta: π =
¯ = π
∂L ∂ ψ˙ ∂L ∂ ψ¯˙
= − iψ¯
= 0.
(25)
(26)
This gives us two primary constraints ¯ φ1 = π + iψ,
and φ2 = π¯.
(27)
Now, we write down the naive Hamiltonian. When we have Grassmann variables there is an ordering ambiguity. If we are using left-differentiation to define the momenta, then we need to put the q ˙’s before the p’s. Therefore, we find ˙ + ψ¯˙ π¯ − L H naive = ψπ ˙ ¯ = − iψ ψ + (0) − iψ¯ψ˙ + ω ¯ ψψ = ω ¯ ψψ,
(28)
and our total Hamiltonian becomes ¯, H total = ω ¯ ψψ + α (π + iψ¯) + β π
(29)
where α and β are our arbitrary coefficients which must be Grassmann odd since the Hamiltonian and Lagrangian must be Grassmann even. At this point, we need to general define the Poisson bracket for Grassmann variables: {f, g } = −
∂f ∂g ∂g ∂f + (−1)f g , ∂p ∂q ∂p ∂q
where f is the Grassmann parity of f (ie. is 0 for g-even f and 1 for g-odd f ). This means, for example that {π, ψ } = { ψ, π } = − 1. Now we can apply our consistency conditions, being careful about various signs and order of differentiation: ¯ β π¯ } {φ1, H T } = { π, ω ¯ ψψ } + i{ψ, = ω ¯ ψ − iβ = 0 ⇒ β = − iω ¯ ψ
(30)
¯, ω ¯ ¯, αψ¯} {φ2 , H T } = { π ψψ } + i{π = − ωψ + iα = 0 ⇒ α = − iωψ.
(31)
and
14
Therefore, our consistency conditions determined α and β and we have no secondary constraints. We can plug in with α and β to rewrite the total Hamiltonian: ¯ H T = ω ¯ ψψ − iωψ (π + iψ¯) − iω ¯ ψπ = iω (πψ + π ¯ ¯ ψ)
(32)
Ultimately we want to find the appropriate anticommutation relations to apply, so we need to evaluate all of the Dirac brackets. First, we calculate the Poisson brackets between our constraints: {φ1 , φ1 } = 0 {φ2 , φ1 } = − i −1
which gives us M
{φ1 , φ2 } = − i {φ2 , φ2 } = 0,
(33)
= iσ 1 . Our Dirac bracket, then, is
{A, B }D = { A, B } − i{A, φ1 }{φ2, B } − i{A, φ2}{φ1, B }.
(34)
Now, let’s evaluate our Dirac brackets: {ψ, ψ }D = 0 {ψ, ψ¯}D = − i {ψ, π }D = − 1 ¯ }D = 0 {ψ, π ¯ ψ¯}D = 0 {ψ, ¯ π }D = 0 {ψ, ¯ π {ψ, ¯ }D = 0 {π, π }D = 0 {π, π¯ }D = 0 ¯, π¯ }D = 0. {π w
As expected the Dirac bracket of anything with ¯π = 0 vanishes. These suggest we enforce the following commutation relations: ˆ] = , and [ ˆ ψ, ψ¯ +
[ ˆ ˆ ]+ = − i , ψ, π
(35)
and all other operators anticommute. Note that we take the Dirac bracket between two g-variables into an anticommutator of quantum operators. The hats denote operators. Since our commutation relations respect our constraint, we are free to use them to rewrite the total Hamiltonian as ˆ T = ω ˆ H ψ¯ ˆ ψ. This describes a two-state system.
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