Hamiltonian Field Theory

Home , Hamiltonian field theory, Poisson bracket

August 31, 2016

1 Introduction

So far we have treated classical field theory using Lagrangian and an action principle for Lagrangian. This approach is called Lagrangian field theory and is best suited for quantizing a field theory using the path- integral (aka functional integral) approach. However, a quicker route to quantizing a classical field theory is thru the “canonical quantization” program which is what you will find in most introductory treatments on quantum field theory. So for us, the Lagrangian is just a crutch to extract the Hamiltonian. However there is a price to pay when we switch from a Lagrangian approach to a Hamiltonian approach, that is manifest Lorentz invariance. Recall that a field theory is a system where we have a degree of freedom located at each point in space. If this degree of freedom is a Lorentz scalar, we will denote the degree of freedom located at position x by Φ(x) and its conjugate momenta, Π(x). These degrees of freedom of course fluctuate in time, so when we include this time-dependence, the notation becomes, Φ(x, t) and Π(x, t). The job of Hamiltonian field theory is describe time-evolution of Π and Φ in equations as follows,

Π˙ = ..., Φ˙ = ...

The dot is time derivative, i.e. we have to choose a time axis or equivalently preferred reference frame which breaks manifest Lorentz invariance. Recall that earlier, in the Lagrangian/action approach we did not have to choose a time-frame because the equation of motion looked Lorentz invariant, e.g. the Euler-Lagrange equation for scalar ﬁeld,

∂L ∂L ∂µ − = 0. ∂ (∂µΦ) ∂Φ Here all Lorentz indices are contracted and the only quantities that appear are Lagrange densityL and Φ which are both Lorentz invariants/scalars - so the whole equation is manifestly Lorentz invariant. Similarly the action functional approach leads to a equation of motion of the schematic form, δI = 0, δΦ(x) which is also manifestly Lorentz invariant as both the numerator and denominator are Lorentz invariants themselves. In this note we review Hamiltonian ﬁeld theory with an eye towards canonical quantization of classical ﬁelds. Since in most texts in classical mechanics, the dependence on time for the canonical pair of variables p and q are omitted from most formulas, we too shall choose not to display the time dependence, i.e. instead of Φ(x, t) we will use, Φ(x) and instead of Π(x, t) we will use, Π(x). Just as in case of q, p it would be implicitly understood that Φ(x), Π(x) are functions of time.

1 2 Hamilton’s principle and Hamilton’s equation for ﬁeld theory

Hamilton’s principle for classical mechanics states that the equations of motion of a physical system described by the generalized coordinate and momenta, (q, p), can be obtain by extremizing (i.e. setting the ﬁrst order variation to zero) of the following functional,

I = dt [pq˙ − H(p, q)] . (1) ˆ

The function, H(p, q) is called the Hamiltonian function. One can easily check by varying the above action that the equation of motion for this system are, ∂H q˙ = , ∂p ∂H p˙ = − . (2) ∂q These equations are the well known Hamilton’s equation. Note that in contrast with the Lagrangian approach where the Euler-Lagrange equations involve second order time-derivatives in time, the equations of motion in the Hamilton’s approach are ﬁrst order diﬀerential equations in time. Of course the price to pay is that we have twice the number of equations, one set for q and one set for p.

One can easily generalize Hamilton’s framework for classical mechanics for a single degree of freedom to field theory i.e. infinite degrees of freedom. To accomplish that first we write down the action for N number of degrees of freedom, (qi, pi), i = 1, 2,...,N. The action and Hamilton’s equation for this case involving N degrees of freedom are:

" N ! # X I = dt p q˙ − H(p , q ) (3) ˆ i i i i i=1 and, ∂H q˙i = , ∂pi ∂H p˙i = − . (4) ∂qi Now in order to go to a continuum limit, we set N → ∞, then the discrete label, i turns into a continuum label such as the position coordinates, x = (x, y, z) (which are real number/continuous valued not discrete/integer valued),

qi → Φ(x),

pi → Π(x).

The summation of course is turned into an integration over the continuum position coordinate,

N X → d3x. ˆ i=1

2 Finally one should also realize that the Hamiltonian function would change into a (global) functional of the fields, Φ(x), Π(x). H(p , q ) → H [Π(x), Φ(x)] = d3x H(Π(x), Φ(x)). i i ˆ The local functional H is quite naturally called the Hamiltonian density since its volume integral is the Hamiltonian. Making these changes to the Eq.s (3, 4), we get the the action and Hamilton’s equation for a field theory, I [Φ(x, t), Π(x, t)] = dt d3x Π(x) Φ(˙ x) − H [Π(x), Φ(x)] , (5) ˆ ˆ and, δH Φ(˙ x) = , δΠ(x) δH Π(˙ x) = − . (6) δΦ(x) However the rules of functional derivatives are different from what we had in Lorentz invariant action functional approach. The new rules of functional integration can be easily deduced by generalizing the following formula for N−degrees of freedom to continuum limit,

∂qi ∂pi = δij, = δij. ∂qj ∂pj Using the prescribed rules, we now replace the i, j labels by position coordinates, x, y. To wit, δΦ(x) δΠ(x) = δ3(x − y), = δ3(x − y). (7) δΦ(y) δΠ(y)

3 Hamiltonian from the Lagrangian

Starting with a Lagrangian for a physical system, one can extract the Hamiltonian thru a Legendre transform. Two functions, f(x) and g(x) are said to be Legendre transforms of each other if their ﬁrst derivatives are functional inverse of each other, i.e.

0 0 df(y) x = f (g (x)) = dy y=g0(x) where primes denote ﬁrst derivative of the function wrt their respective arguments. Solving this condition one gets, x g0(x) = f(y) So the Hamiltonian is deﬁned to be the Legendre transform of the Lagrangian,

H(p, q) = p q˙(p) − L(q, q˙(p)).

Now hereq ˙(p) means we have inverted the relation, ∂L(q, q˙) p = ∂q˙

3 to express,q ˙ as a function of p (and more generally a function of both p and q). For a system with N- degrees of freedom, the deﬁnition is, ! X H = pi q˙i − L(qi, q˙i) i Now we can follow the prescription mentioned in the previous section to take the continuum limit, N → ∞ whereby we obtain the Lagrangian for a ﬁeld theory, H = d3x Π(x) Φ(˙ x) − L[Φ(x), Φ(˙ x)], ˆ Now since the Lagrangian itself can be expressed as a volume integral of Lagrangian density,

L[Φ(x), Φ(˙ x)] = d3xL, L = L(Φ(x), Φ(˙ x)), ˆ we have, H = d3x Π(x) Φ(˙ x) − L , ˆ where ﬁrst we need to express, Φ(˙ x) in terms of Π(x) by inverting, δL Π(x) = . δΦ(˙ x)

3.1 Example: Real (free) scalar field theory The real scalar field theory is defined by the action, 1 1 I[Φ(x)] = d4x L, L = (∂ Φ) (∂µΦ) − m2Φ2. ˆ 2 µ 2 So the Lagrangian is, 1 1 L = d3x (∂ Φ) (∂µΦ) − m2Φ2 , ˆ 2 µ 2 1 1 1 = d3x Φ˙ 2 − ∇Φ · ∇Φ − m2Φ2 . ˆ 2 2 2 The momentum conjugate to Φ(x) is, δL Π(x) = δΦ(˙ x) δ 3 1 2 1 1 2 2 = d y So Φ˙ (y) − ∇yΦ(y) · ∇yΦ(y) − m Φ (y) δΦ(˙ x) ˆ 2 2 2 δ 1 = d3y Φ˙ 2(y) ˆ δΦ(˙ x) 2 δΦ(˙ y) = d3y Φ(˙ y) ˆ δΦ(˙ x) = d3y Φ(˙ y) δ3(x − y) ˆ = Φ(˙ x).

4 We can use this to eliminate Φ˙ in terms of Π. The Lagrangian then becomes, 1 1 1 L = d3x Π2 − ∇Φ · ∇Φ − m2Φ2 , ˆ 2 2 2 and the Hamiltonian is, H = d3x Π(x) Φ(˙ x) − L ˆ = d3x Π2(x) − L ˆ 1 1 1 = d3x Π2(x) − d3x Π2 − ∇Φ · ∇Φ − m2Φ2 ˆ ˆ 2 2 2 1 1 1 = d3x Π2 + ∇Φ · ∇Φ + m2Φ2 . (8) ˆ 2 2 2 From this we can extract the Hamilton’s equations. The ﬁrst Hamilton’s equation doesn’t give us anything but the old relation, δH Φ(˙ x) = δΠ(x) = Π(x). (9) However the second equation is non-trivial,, δH −Π(˙ x) = δΦ(x) δ 1 1 1 = d3y Π2 + ∇ Φ · ∇ Φ + m2Φ2 δΦ(x) ˆ 2 2 y y 2 δ δΦ(y) = d3y ∇ Φ · ∇ Φ(y) + m2Φ(y) ˆ y y δΦ(x) δΦ(x) = d3y ∇ Φ · ∇ δ3(x − y) + m2Φ(y) δ3(x − y) ˆ y y = d3y −∇2 Φ + m2Φ δ3(x − y) ˆ y = − ∇2 − m2 Φ(x). (10) So equations are, Φ(˙ x) = Π(x), Π(˙ x) = ∇2 − m2 Φ(x). Replacing, Π = Φ˙ in the second equation we get the second equation to look like, Φ¨ = ∇2 − m2 Φ, 2 2 2 ∂t − ∇ + m Φ = 0. This is nothing but the well-familiar Klein-Gordon equation, ∂2 + m2 Φ = 0. Thus we have recovered the correct equation of motion using the ﬁeld Hamiltonian, Eq. (8) and the Hamilton’s equations Eq. (6).

5 4 Poisson brackets, Charges and algebra of charges

In general, the time evolution of a quantity, f(p, q) can be deduced from the Poisson brackets (PB) of that quantity with the Hamiltonian, H, df(p, q) ∂f dq ∂f dp = + dt ∂q dt ∂p dt |{z} ∂f ∂H ∂f ∂H = − ∂q ∂p ∂p ∂q |{z} ≡ {f, H} .

dQ In particular if the system has conserved charges, Q, since, dt = 0, one has for conserved charge, Q, {Q, H} = 0.

Again, as done previously, we can generalize the Poisson bracket expression for single degree of freedom to ﬁeld theory thru the intermediate step of going to N degrees of freedom. For N degrees of freedom, the Poisson bracket can be easily shown to be of the form,

X ∂A ∂B ∂A ∂B {A, B} ≡ − . ∂q ∂p ∂p ∂q i i i i i

P 3 So in the continuum limit, N → ∞, we replace, qi, pi → Φ(x), Π(x) and i → d x, and get, ´ δA δB δA δB {A, B} ≡ d3x − . ˆ δΦ(x) δΠ(x) δΠ(x) δΦ(x) Here of course A, B are themselves functions or functional of Φ, Π. Now, evidently,

{A, B} = −{B,A}.

This immediately implies if we set, B = A,

{A, A} = −{A, A}, =⇒ 2{A, A} = 0, =⇒ {A, A} = 0.

4.1 Example: Scalar ﬁelds and Conserved charges We already know thru Noether’s theorem that there are charges corresponding to Poincare symmetry, namely the energy-momenta, (P 0,P i) and the boost-rotation charges (Lµν). Let’s check that these are conserved by taking the Poisson bracket with the Hamiltonian. First we need to express the charges in the canonical variables, Φ, Π i.e. by eliminating the velocity, Φ˙ in lieu of the conjugate momenta, Π(x). We start from the expression,

P µ = d3x T µ0 ˆ h i = d3x (∂µΦ) Φ(˙ x) − ηµ0L . ˆ

6 h i =⇒ P 0 = d3x Φ˙ 2(x) − L ˆ 1 m2 = d3x Φ˙ 2(x) − ∂ ⊕∂µ⊕ + Φ2 ˆ 2 µ 2 1 1 m2 = d3x Φ˙ 2(x) + ∇⊕ · ∇⊕ + Φ2 ˆ 2 2 2 1 1 m2 = d3x Π2(x) + ∇⊕ · ∇⊕ + Φ2 , (11) ˆ 2 2 2 and, h i P i = d3x ∂iΦ(x) Φ(˙ x) − ηi0L ˆ = d3x ∂iΦ(x) Φ(˙ x) ˆ = − d3x ∂ Φ(x) Π(x). (12) ˆ i

i Here, we have used that due to the fact that the metric component, ηii = −1, ∂ = −∂i.

Notice that, P 0 = H, so its PB with the Hamiltonian i.e. itself will be zero due to antisymmetric nature of the PB,

δH δH δH δH P 0,H = {H,H} = d3x − = 0. ˆ δΦ(x) δΠ(x) δΠ(x) δΦ(x)

This proves, P 0 is conserved. The next case, i.e. P i given by the expression in Eq. (12). We need to evaluate, δP i δH δP i δH P i,H = d3y − . (13) ˆ δΦ(y) δΠ(y) δΠ(y) δΦ(y) Using Eq. (12) we compute,

δP i δ = − d3x ∂ Φ(x) Π(x) δΦ(y) δΦ(y) ˆ i δ (∂ Φ(x)) = − d3x i Π(x) ˆ δΦ(y) δΦ(x) = − d3x ∂ Π(x) ˆ i δΦ(y) = − d3x ∂ δ3(x − y) Π(x) ˆ i = d3x δ3(x − y) ∂ Π(x) ˆ i ∂ = Π(y), (14) ∂yi

7 δP i δ = − d3x ∂ Φ(x) Π(x) δΠ(y) δΠ(y) ˆ i δ = − d3x ∂ Φ(x) Π(x) ˆ i δΠ(y) = − d3x ∂ Φ(x) δ3(x − y) ˆ i ∂ = − Φ(y). (15) ∂yi Now recall from Eq. (9) and (10), δH δH = − ∇2 − m2 Φ(y)., = Π(y). (16) δΦ(y) y δΠ(y)

Now plugging Eq. (14), ( 15) and (16) in the expression for the PB in RHS of Eq. (13), we obtain,

δP i δH δP i δH P i,H = d3y − ˆ δΦ(y) δΠ(y) δΠ(y) δΦ(y) ∂ ∂ = d3y Π(y) Π(y) − Φ(y) ∇2 − m2 Φ(y) ˆ ∂yi ∂yi y ∂ ∂ ∂ = d3y Π(y) Π(y) − Φ(y) ∇2 Φ − m2 Φ(y)Φ(y) ˆ ∂yi ∂yi y ∂yi ∂ 1 1 m2 = d3y Π2(y) + (∇Φ)2 + Φ2(y) + d3y ∇ (∂ Φ∇Φ) , (17) ˆ ∂yi 2 2 2 ˆ i where we have used,

2 ∂iΦ∇ Φ = ∂iΦ∂j∂jΦ

= ∂j (∂iΦ∂jΦ) − ∂j (∂iΦ) ∂jΦ

= ∂j (∂iΦ∂jΦ) − ∂i∂jΦ ∂jΦ 1 = ∂ (∂ Φ∂ Φ) − ∂ ∂ Φ∂ Φ j i j i 2 j j 1 = ∇ (∂ Φ∇Φ) − ∂ (∇Φ)2 . i i 2

Thus both terms in the expression for{P i,H} i.e. the rhs of (17) are integral of a total derivative and after integration turns into surface terms at spatial infinity where of course they are zero (we are assuming for all integrals to be finite that all fields decay to zero at spatial infinity). Thus {P i,H} = 0 which means P i i.e. the linear momentum is conserved as well.

For the boost and rotation symmetry combined i.e. Lorentz symmetry, the Noether charges are given by the expression,

Lµν = d3x M µν0 ˆ = d3x xµT ν0 − xνT µ0 , ˆ

8 In particular, these charges are, the angular momenta,

Lij = d3x xiT j0 − xjT i0 , ˆ and the less familiar “Boost charges” or “Boost generators”

L0i = d3x x0T i0 − xiT 00 ˆ = d3x t T i0 − xi H , ˆ = t P i − d3x xi H ˆ

In the center of mass(momentum) frame, of course we have, P i = 0, then the Boost charges have the intepretation of the angular energy or moments of energy,

Li0 = − d3x xi H . CM ˆ CM

Lets now check that the angular momentum is conserved using the Poission Bracket with the Hamiltonian. First we express the angular momentum in Π, Φ variables by eliminating, Φ˙ in favor or Π.

Lij = d3x xiT j0 − xjT i0 ˆ = d3x −xi∂ Φ(x) Π(x) + xj∂ Φ(x) Π(x) . ˆ j i Then second step, we evaluate the functional derivaties,

δLij δ ∂ ∂ = d3y −yi Φ(y) Π(y) + yj Φ(y) Π(y) δΦ(x) δΦ(x) ˆ ∂yj ∂yi ∂ δ ∂ δ = d3y −yi Φ(y) Π(y) + yj Φ(y) Π(y) ˆ ∂yj δΦ(x) ∂yi δΦ(x) ∂ ∂ = d3y −yi δ3(x − y) Π(y) + yj δ3(x − y) Π(y) ˆ ∂yj ∂yi ∂ ∂ = d3y δ3(x − y) yi Π(y) − yj Π(y) ˆ ∂yj ∂yi i j = ∂j x Π(x) − ∂i x Π(x) i j = x ∂jΠ(x) − x ∂iΠ(x). (18)

9 and,

δLij δ ∂ ∂ = d3y −yi Φ(y) Π(y) + yj Φ(y) Π(y) δΠ(x) δΠ(x) ˆ ∂yj ∂yi ∂ δ ∂ δ = d3y −yi Φ(y) Π(y) + yj Φ(y) Π(y) ˆ ∂yj δΠ(x) ∂yi δΠ(x) ∂ ∂ = d3y −yi Φ(y) δ3(x − y) + yj Φ(y) δ3(x − y) ˆ ∂yj ∂yi i j = −x ∂jΦ(x) + x ∂iΦ(x). (19)

Finally we compute the Poisson brackets by using Eq.s (16), (18) and (19)

δLij δH δLij δH Lij,H = d3x − ˆ δΦ(x) δΠ(x) δΠ(x) δΦ(x) = d3x xi∂ Π(x) − xj∂ Π(x) Π(x) − xi∂ Φ(x) − xj∂ Φ(x) ∇2 − m2 Φ(x) ˆ j i j i Π2(x) m2Φ2(x) = d3x xi∂ − xj∂ + − xi∂ Φ(x) − xj∂ Φ(x) ∇2Φ(x) ˆ j j 2 2 j i Π2(x) m2Φ2(x) Π2(x) m2Φ2(x) = d3x ∂ xi + xi − ∂ xj + xj ˆ j 2 2 i 2 2 i j 2 − x ∂jΦ(x) − x ∂iΦ(x) ∇ Φ(x) . (20)

The ﬁrst two terms are of course integrals of total derivatives which would lead to surface terms at spatial inﬁnity which in turn vanish. The last term needs to be massaged a littel bit . We have alrady seen that,

1 ∂ Φ∇2Φ = ∇ (∂ Φ∇Φ) − ∂ (∇Φ)2 . i i i 2

So then,

1 1 xi∂ Φ − xj∂ Φ ∇2Φ = xi∇ (∂ Φ∇Φ) − xj∇ (∂ Φ∇Φ) − xi∂ (∇Φ)2 + xj∂ (∇Φ)2 j i j i j 2 i 2 1 1 = xi∂ (∂ Φ∂ Φ) − xj∂ (∂ Φ∂ Φ) − xi∂ (∇Φ)2 + xj∂ (∇Φ)2 k j k k i k j 2 i 2 1 1 = ∂ xi ∂ Φ∂ Φ − xj∂ Φ∂ Φ − ∂ xi (∇Φ)2 + ∂ xj (∇Φ)2 . k j k i k j 2 i 2

This is also a total derivative. Thus all three terms in the {Lij,H} are integrals of total derivatives which vanish on integration, dLij {Lij,H} = 0 =⇒ = 0. dt