Lagrangian

An integral of a function is the area under a curve described by that function. Let’s say you have the following integral.

S = [integral] f(dx/dt, x, t)dt

where f is a function, and x is a function of t

What function does x have to be so that S will remain stationary for small changes in x? This type of problem is called the calculus of variations. The simplest example is what is the shortest path between two points. We know the answer is a straight line. Let’s say you define a function x’(t) that is slightly different then x(t)

x’(t) = x(t) + [delta]x(t)

where [delta]x(t) = [alpha]g(t)

x(t) makes the integral stationary if

[partial derivative with respect to alpha][integral]f(dx’/dt, x’, t)dt = 0 when [alpha] = 0

You have

[partial derivative with respect to alpha][integral]f(p + [alpha]dg/dt, x + [alpha]g(t), t)dt

= [integral] [partial derivative with respect to alpha]f(p + [alpha]dg/dt, x + [alpha]g, t)dt

= [integral](dg/dt, [partial derivative of f with respect to p] + g[partial derivative of f with respect to x])dt

You would solve this integral by integrating by parts. Let me explain what that is.

Let’s say you want to integrate the following integral.

[integral]udv

Notice the following.

d(uv) = udv + vdu

[integral]d(uv) = uv = [integral]udv + [integral]vdu

Get ∫udv on one side

[integral]udv = uv — [integral]vdu

Incidentally, the minus sign we created is why you have the minus sign in the Hamiltonian and Lagrangian.

[integral]udv = [uv] — [integral] vdu where [f] = f(b) — f(a)

So now instead of having to solve for ∫udv, you have to solve ∫vdu which might be an easier integral to solve. If not, you shouldn’t use parts. You might use various substitutions and parts several times in one problem. Now, let’s go back and use parts on the problem we were working on.

[integral] (dg/dt [partial derivative of f with respect to p] + g[partial derivative of f with respect to x])dt

[g[partial derivative of f with respect to p] + [integral]g([partial derivative of f with respect to x] — d/dt[partial derivative of f with respect to p])dt

Lets say you have 3n functions q of the coordinates. You can write

xri = xri(q1, q2,…. q3n)

q is called generalized coordinates. A dot over a variable denotes differentiations with respect to time.

[delta]xri = [partial derivative of xri with respect to q][delta]q

[x dot]ri = [partial derivative of xri with respect to q][q dot]

[partial derivative of [x dot]ri] = [partial derivative of xri with respect to q]

2T = [summation]mr ([partial derivative of xri with respect to q][q dot])2

[summation]Xridxri = [summation]Xri[partial derivative of xri with respect to q]([delta]q) = Q[delta]q

You can then write δS as

[delta]S = [integral]([delta]T + Q[delta]q)dt

where T and Q are functions of q and [q dot]. Then you have

[integral][delta]Tdt = [[partial derivative of T with respect to [q dot]] + [[partial derivative of T with respect to q] — d/dt[partial derivative of T with respect to [q dot]][delta]qdot

The condition δS = 0 gives

d/dt [partial T with respect to [q dot]] — [partial derivative of T with respect to q] = Q

This is called Lagrange’s equation. It was named after Joseph Louis Lagrange (1736–1813). He was a brilliant Italian mathematician who spent most of his career in Paris. He fled France during the French Revolution but returned under Napoleon. He was one of the primary architects of the calculus of variations.

This is Hamilton’s principle. It was named after William Rowan Hamilton (1805–1865) who also invented the quaternions.

[delta]S = [delta][integral](T + W)dt = 0

where T is the kinetic energy and W is the work. The Lagrangian L can be defined as L = T + W.

[delta][integral]Ldt = [[partial derivative of L with respect to [q dot]][[delta]q] + [integral]([partial derivative of L with q] — d/dt[partial derivative of L with respect to [q dot]][delta]qdt

If t0, t1, q are varied by Δt0, Δt1, Δq, the integrated part is

[[L — q[partial derivative of L with respect to [q dot]][delta]t + [partial derivative of L with respect to [q dot]][delta]q]

since [delta]q = q + [q dot][delta]t

We define the Hamiltonian H by

H = [q dot][partial derivative of L with respect [q dot]] — L

and the canonical momentum by

p[alpha] = [partial derivative of L with respect to [q dot]]

This is Hamilton’s principle function.

[delta]S = [delta][integral]Ldt = [-H[delta]t + p[alpha] [alpha]q] + [integral]([partial derivative of L with respect to [q dot]] — d/dt[partial derivative of L with respect to [q dot])[delta]qdt

You replace t1 by t, and the last integral is zero so

[partial derivative of S with respect to t) = -H

[partial derivative of S with respect to q] = p[alpha]

You can use p[alpha] = [partial derivative of L with respect to [q dot]]

to get rid of [q dot] in the Hamiltonian so

[partial derivative of S with respect to t] = -H(q, p[alpha], t)

This is called the Hamilton-Jacobi equation, first written in 1836.

L is a function of q, [q dot], and possibly t. H is a function of q, p, and possibly t. Then for variations of q and [q dot], without varying t

[delta]H = [delta]([q dot], p — L) = [q dot][delta]p[alpha] + p[alpha][delta][q dot] — [partial derivative of L with respect to q][delta]q — [partial derivative of L with respect to q dot][delta][q dot]

since

p[alpha] = [partial derivative of L with respect to q dot]

therefore

[delta]H = [q dot][delta]p[alpha] — [partial derivative of L with respect to q]q

[q dot] = [partial derivative of H with respect to p[alpha]]

[partial derivative of H with respect to q] = -[partial derivative of L with respect to q]

by Lagrange’s equation

[p[alpha] dot] = d/dt[partial derivative of L with respect to q dot] = [partial derivative of L with respect to q] = -[partial derivative of H with respect to q]

therefore

[p[alpha] dot] = -[partial derivative of H with respect to q]

[q dot] = [partial derivative of H with respect to p[alpha]]

These are called Hamilton’s equations.

L = T + W

Work is minus potential energy.

W = -V

therefore

L = T — V

The Lagrangian is usually described as kinetic minus potential energy.

Let’s look at a simple system. One of the simplest where there is any movement at all is a harmonic oscillator. You could choose any example but let’s say you have a simple pendulum that swings back and forth. What do you need to specify the state of the particle, which in this case is the bob of the pendulum? Position alone doesn’t define it completely because it won’t tell you whether it’s traveling to the right or left. Therefore you need the position and momentum. The momentum is defined as

p = mv = m (dx/dt)

So you need x, and the first derivative of x. The force is determined by Newton’s Second Law.

F = ma = m (dx2/dt2)

In a one-dimensional simple harmonic oscillator

F = -kx, so

-kx = m (d2x/dt2)

also k/m = w2 so

-w2x = (d2x/dt2)

(d2x/dt2) + w2x = 0

The solution for position is

x(t) = A cos (wt + [theta])

The solution for momentum is

p = -mAw sin (wt + [theta])

Newton’s Second Law can be written as either one second order equation or two coupled first order equations

F = dp/dt

p/m = dx/dt

For one-dimensional motion, x(t) and p(t) determine the evolution of the particle’s trajectory. If you eliminate t between them, you can write p as a function of x, and derive the mechanical energy E.

p2/2m + U(x) = E

where U(x) is the potential energy. You can rewrite it as

p = ± [squareroot of] 2m(E — U(x))

If I choose any value of x, you can tell me the value of p at that position. Imagine a graph with x on the horizontal axis, and p on the vertical axis. At each instant t, the value of x corresponds to a value of p. Plot the ordered pair (x, p) on a graph. You get an ellipse.

This is called phase space. As time marches on, the state point sweeps out a trajectory in phase space. The phase space trajectory of a simple harmonic oscillator forms an ellipse. The size of the ellipse is determined by the particle’s energy E.

The two first order equations F = dp/dt and p/m = dx/dt describe the evolution of the phase space coordinates (x, p). These equations can be written in terms of energy instead of force. E, which appears on the right hand side of the equation, denotes the numerical value of the energy. The left hand side of the equation

p2/2m + U(x) = E

is a function of the phase space coordinates (x, p). The left hand side is the Hamiltonian of the system. If the potential energy includes an explicitly time-dependent factor, such as a spring constant that weakens with time, then the Hamiltonian H is explicitly a function of this time.

H(x, p, t) = p2/2m + U(x, t)

The equation p2/2m + U(x) = E says that the energy, a number, and the Hamiltonian, a function, have the same value. H(x, p, t) = E. For conservative systems, the force F is given as the minus partial derivative of H with respect to x.

F = dp/dt

F = -[partial derivative of H with respect to x] so

dp/dt = -[partial derivative of H with respect to x]

You can also write p/m = dx/dt in terms of H. p2/2m is the kinetic energy. Notice that

p/m = [partial derivative of H with respect to p]

p/m = dx/dt so

dx/dt = [partial derivative of H with respect to p]

Therefore you have the following two first order differential equations.

dp/dt = -[partial derivative of H with respect to x]

dx/dt = [partial derivative of H with respect to p]

These are Hamilton’s equations, named after William Rowan Hamilton. They describe how the particle’s phase space point (x, p) moves with respect to time. x and p are canonically conjugate. Every position coordinate x has a specific momentum. For a particle moving in N spatial dimensions, there are 2N canonically conjugate phase space coordinates.

You can also describe the evolution of H itself. Since H = H(x, p, t), the chain rule gives

dH/dt = [partial derivative of H with respect to x](dx/dt) + [partial derivative of H with respect to p](dp/dt) + [partial derivative of H with respect to t]

When we use Hamilton’s equations to write dx/dt and dp/dt in terms of derivatives of H, two terms cancel, leaving

dH/dt = [partial derivative of H with respect to t]

If you take the derivative of something with respect to something it doesn’t contain, the result is zero. If H does not depend on a given variable, the partial derivative of H with respect to that variable is zero. If dH/dt = ∂H/∂t, that means H depends only on t. ∂H/∂t = 0 if and only if H contains no explicit time dependence. In other words, only when the system is invariant under time translation Δt. The momentum p is constant if and only if H does not explicitly contain its conjugate coordinate x when the system is invariant under a displacement Δx.

The Hamiltonian is rigorously approached through the Lagrangian. Newton’s Second Law in generalized coordinates q, assumes the following form of the Euler-Lagrange Equation

[partial derivative of L with respect to q] = d/dt[partial derivative of L with respect to [q dot]]

where the overdot represents the time derivative. The Lagrangian L is the difference between the kinetic and potential energies, written as a function of position coordinates, velocities, and possibly the time. The Euler-Lagrange Equation can be written as

[partial derivative of L with respect to t] = -d/dt([q dot][partial derivative of L with respect to [q dot]] — L)

The last two equations follow from Hamilton’s principle.

[delta][integral from a to b] Ldt = 0

Some combinations of variables occur frequently in this subject. This is the canonical momentum. It is the conjugate of the coordinate q.

p = [partial derivative of L with respect to [q dot]]

This is the Hamiltonian.

H = [q dot][partial derivative of L with respect to [q dot]] — L

The Lagrangian is a function of position and velocity. The Hamiltonian is a function of position and canonical momentum. Which you want to use depends on whether you want to use velocity or canonical momentum. H typically works out to be kinetic energy plus potential energy but it is not always the same as the energy.

Similarly, the canonical momentum is often, but not always, the mv momentum. For example, if q happens to be the polar angle θ, its conjugate momentum is the component of angular momentum about the axis of θ rotation. If a charged particle moves through an electromagnetic field, then the canonical momentum and the mv momentum differ by a vector potential term.

Hamilton’s principle can be stated in terms of the Hamiltonian rather than the Lagrangian. By using the Hamiltonian

H = [q dot][partial derivative of L with respect to q dot] — L

we can rewrite Hamilton’s principle

[delta][integral from a to b]Ldt = 0

as

[delta][integral from a to b][x dot]p — Hdt = 0

which gives Hamilton’s equations.

Let’s say you have to find the equations of motion x(t) and p(t) in a particularly difficult problem. One strategy for solving any problem is to transform it into a problem you already know how to solve. This is why you change variables. Therefore, you could try a change of phase space variables x → x’, p → p’, which produces a change in the Hamiltonian H → H’. Hamilton’s equations hold in the new phase space as well as the old.

dx’/dt = [partial derivative of H’ with respect to p’]

dp’/dt = -[partial derivative of H’ with respect to t’]

The new Hamilton’s equations are derived from minimizing the integral

[delta][integral from a to b][partial derivative of x with respect to t]’p’ — H’dt = 0

This integral must also hold.

[delta][integral from a to b][x dot]p — Hdt = 0

For the two integrals to hold simultaneously, the integrands can differ at most by a total derivative.

([partial derivative of x with respect to t]p — Hdt) — ([partial derivative of x with respect to t]’p’ — H’dt) = dS/dt

where S must include both old and new phase coordinates. It may also depend on time. Let’s consider the case where S depends on the old x and new p’.

S = S(x, p’, t)

From the chain rule, the above equation defining dS/dt says

0 = [ x dot]((partial derivative of S with respect to x) — p) + (H — H’ + [partial derivative of S with respect to t]) + [p dot]{partial derivative of S with respect to p’] + p’([x dot]’)

This equation must be true for all times and all phase space coordinates. That will happen if the grouped terms vanish separately.

p = [partial derivative of S with respect to x]

H + [partial derivative of S with respect to t] = H’

p’([x dot])’ = -([p dot])[partial derivative of S with respect to p’]

We are trying to transform the problem into phase space where it becomes a problem we know how to solve. The easiest problem we know how to solve is H’ = 0. Then from Hamilton’s equations, it follows that x’ = constant, and p’ = constant = p0.

H + [partial derivative of S with respect to t] = H’

then becomes

H + [partial derivative of S with respect to t] = 0

This is the Hamilton-Jacobi Equation (HJE). If you plug in the following definitions

H(x, p, t) = p2/2m + U(x, t)

p = [partial derivative of S with respect to x]

it becomes

p2/2m + U + [partial derivative of S with respect to t] = 0

(1/2m)[partial derivative of S with respect to x]2 + U + [partial derivative of S with respect to t] = 0

(1/2m)[partial derivative of S with respect to x]2 + U = -[partial derivative of S with respect to t]

Notice that in this form, it resembles the time-dependent Schrodinger equation.

We still haven’t actually found x(t) and p(t) so let’s do that now. We already have p from

p = [partial derivative of S with respect to x]

How do you find x? Here you have the following equation.

p’([x dot])’ = -([p dot])[partial derivative of S with respect to p’]

If you impose

x’ = [partial derivative of S with respect to p’]

it requires p’x’ to be constant. Therefore, to find x, you evaluate this derivative, and then set x’ and p’ to be constants. That will turn x’ = ∂S/∂p’ into an algebraic equation to solve for x.

First you have to find S. In a typical case where H = E, the HJE

(1/2m)[partial derivative of S with respect to x]2 + U = -[partial derivative of S with respect to t]

says that S(t) ~ -Et for its time dependent part. This suggests an additive separation of variables for S

S(t, x, p0) = W(x, p0) — Et

which combined with the above equation, gives the time independent HJE

(1/2m)[partial derivative of W with respect to x]2 + U = E

that you have to solve for W.

Let’s say you have a free particle moving to the right with energy E and momentum p0 = [squareroot of 2mE]. The time independent HJE gives

[partial derivative of W with respect to x]2 = 2mE

so that W = p0x + Δ, with Δ being an arbitrary integration constant.

S(t, x, p0) = W(x, p0) — Et

We now have S.

S(t, x, p0) = p0x — ET — [delta]

S(t, x, p0) = p02/2m + [delta]

The particle’s momentum follows from p = ∂S/∂x. To obtain x(t), you use x’ = ∂S/∂p’

x’ = [partial derivative of S with respect to p’] = x — (p0/m)t

set x’ to a constant

x0 = x — (p0/m)t

Since there are much easier ways of doing this and other problems, the HJE is rarely used for grinding out answers to practical problems. Rather, it suggests new ways of thinking about mechanics. For instance, the HJE foreshadowed particle-wave duality.

So these are Hamilton’s equations. They are frequently used in celestial mechanics.

[q dot] = [partial derivative of H with respect to p]

[p dot] = -[partial derivative of H with respect to q]

You also have this which is related to Hamilton’s equations.

p = [partial derivative of L with respect to q dot]

H is the Hamiltonian and L is the Lagrangian.

H(q, [q dot], t) = [q dot]p — L(q, [q dot], t)

If the equations defining the generalized coordinates are not functions of time,

H = T + V = E

L = T — V

Let’s define L’ so

L’ = L + d/dt f(q, t)

L’ = L + [q dot][partial derivative of f with respect to q] + [partial derivative of f with respect to t]

[partial derivative of L’ with respect to q dot] = [partial derivative of L with respect to q dot] + [partial derivative of L with respect to q]

[partial derivative of L’ with respect to q dot] = [partial derivative of L with respect to q] + d/dt [partial derivative of f with respect to q]

d/dt[partial derivative of L’ with respect to q dot] — [partial derivative of L with respect to q] = d/dt[partial derivative of L with respect to q dot] — [partial derivative of L with respect to q]

Therefore the Lagrangians represent the same equations of motion.

p’ = [partial derivative of L’ with respect to q dot] = [partial derivative of L with respect to q dot] + [partial derivative of f with respect to q] = p + [partial derivative of f with respect to q]

q’ = -[partial derivative of L’ with respect to p dot] = [partial derivative of L with respect to p dot] = q

For the interaction of a non-relativistic particle with an electromagnetic field, the Lagrangian is

L = (1/2)mv . v — q[phi] + (q/c)A . v

The Lagrangian of the electromagnetic field itself is given by

L = — F2/4q2

where F is the derivative of the four-vector gauge potential A with respect to x.

Classical mechanics can be expressed in terms of Lagrangians for point particles, or for continuous systems. The Lagrangian is T — V, kinetic minus potential energy. The action is S = [integral from t1 to t2] Ldt. When S is minimized, the Euler-Lagrange equations result, and give you Newton’s laws. The force is the derivative of the potential. Classical electrodynamics can also be written as a Lagrangian theory. E and B are the electric and magnetic fields. A and φ are the vector and scalar potentials. J and p are the current density and charge density. Using [partial derivative]j = [partial derivative with respect to xj],

you can write the field strengths in terms of the potentials.

Ei = -[del]V — [partial derivative of A with respect to t] = [partial derivative]iA0 — [partial derivative]0Ai

B = [del] x A

Here is the four-vector for the potentials.

Au = (V, A) = (A0, A)

Here is the four-vector for the current.

Ju = (p, J)

Let’s define the following antisymmetric tensor.

Fuv = [partial derivative]uAv — [partial derivative]vAu

where the components are

F0i = [partial derivative]0Ai — [partial derivative]iA0 = -Ei

Fij = [partial derivative]iAj — [partial derivative]jAi = [epsilon]ijkBk

Fuv is explicitly invariant under a transformation where Av changes by ∂vx since Fuv changes by

[partial derivative]u[partial derivative]vx — [partial derivative]v[partial derivative]ux = 0

The Lagrangian for electromagnetism is

L = (1/2)(E2 — B2) — pV + J . A

If you write it in terms of Fuv, you have

L = -(1/4)FuvFuv — JuAu

In the above two equations, what comes before the minus sign is the kinetic energy term. The kinetic energy terms in the above two equations are equal. What comes after the minus sign is the potential energy term and is also equal. The potential energy term is also called the interaction Lagrangian.

With the Lagrangian for electromagnetism, the Euler-Lagrange equation becomes Maxwell’s equations. You can write down the physics of electromagnetic fields by writing a Lagrangian that is a function of the fields or potentials.

Whenever the Lagrangian or the action is invariant under continuous transformations, a divergenceless current arises. This leads to a conserved charge. Let’s say you have the following example.

[partial derivative]uJu = 0

Integrate over d3x

[integral][partial derivative]0J0d3x + [integral][partial derivative]0J0d3x = 0

The second term can, by Gauss’ theorem, be transformed to an integral over the surface of space, and is assumed to vanish.

The second term can, by Gauss’ theorem, be transformed to an integral over the surface of space, and is assumed to vanish. The first term gives

[partial derivative with respect to t][integral]J0d3x = 0

The charge is

Q = [integral]J0d3x

and does not change with time.

Whenever there is a conserved current, there is a conserved charge, and vice versa. Whenever there is an invariance of the theory under some transformation, a conserved current arises, and can be written in terms of appropriate derivatives of the Lagrangian. Thus, whenever there is an invariance, there is a current, and thus a charge, although it might not be what we normally think of as a charge, but some conserved quantity. This is called Noether’s Theorem. It was actually devised by a woman named Emmy Noether (1882–1935).

Noether’s theorem says when a conserved quantity exists and how to define it. If there is a conserved quantity, there must exist an associated invariance under a transformation, in other words, a symmetry. If there is a symmetry, there must be a conserved quantity. If you are trying to write the Lagrangian, knowing the symmetries to build into the Lagrangian is a great help in writing it. Thus Noether’s theorem plays a central role in relating observed symmetries and conservation laws to the structure of the theory.

Particle physics is formulated by giving the Lagrangian. From it, using the rules of quantum field theory, all aspects of particle physics can be calculated. The Lagrangian defines the theory. It is written in terms of the elementary particles of the theory. Any composite objects should appear as bound states that arise as solutions of the theory. For electromagnetism, the photon is the quantum of the electromagnetic field. It is represented by the vector potential Au. The electron is represented by the fermion field ψ. The Lagrangian contains the fundamental interactions of the theory. For electromagnetism, the interaction Hamiltonian is J . A. The relativistic expression is JuAu. Actually, it is the potential energy part of the Lagrangian that specifies the theory. The kinetic energy part depends only on the spin of the particles. The potential energy part specifies the forces and is called the interaction Lagrangian. One of the main reasons why particle physics is formulated in terms of the Lagrangian is because L is a single function that determines the dynamics, and L must be a scalar in every relevant space. L is invariant since the action is invariant. Making L invariant under Lorentz transformations means that all predictions of the theory are also Lorentz invariant.

I have described the Lagrangian for an electromagnetic field. Often however, you use a much simpler Lagrangian of a scalar field. This is the spinless field φ(x). The field φ(x) can be thought of as arising from a source the same way as the electromagnetic field arises from charged particles. Like electromagnetism, you can talk about the fields without concerning yourself with the source. Let’s say you want to write the Lagrangian analogous to the one for electromagnetism, for a real scalar field φ(x) with mass m. We derive the Lagrangian like we did before. We assume L depends explicitly on the fields and their derivatives but not on xu. L = ([phi], [partial derivative]u[phi]).

You begin with a small variation in φ(x).

[phi]’(x) = [phi](x) + [delta][phi](x)

[delta]S = [integral]d4x[delta]L

[delta]S = [integral]d4x[[partial derivative of L with respect to phi][delta][phi] — [partial derivative of L with respect to [partial derivativeuphi][delta] ([partial derivative]uphi)]

Using

[delta]([partial derivative]u[phi]) = [partial derivative]u[delta][phi]

and integrating by parts gives

[delta]S = [integral]d4x[[partial derivative of L with respect to phi] — [partial derivative]u[partial derivative of L with respect to [partial derivative]([partial derivative]u[phi])]]

The condition for the action to be stationary (δS = 0) is

[partial derivative of L with respect to phi] — [partial derivative]u[partial derivative of L with respect to [partial derivative]([partial derivative]u[phi])] = 0

This is the Euler-Lagrange equation for φ. A real scalar field has the Lagrangian

L = (1/2)[[partial derivative]u[phi][partial derivative]u[phi] — m2[phi]2]

The factor 1/2 is a convention.

The Euler-Lagrange equation gives the wave equation or Klein-Gordon equation

[partial derivative]u[partial derivative]u[phi] + m2[phi] = 0

That is what you would expect since

E2 = p2 + m2

E = i[partial derivative]0 and p = i[del] so

E2 — p2 = -[partial derivative]02 + [del]2 = -[partial derivative]u[partial derivative]2

Therefore every field that describes a particle of mass m should satisfy

-[partial derivative]u[partial derivative]u[phi] = m2[phi]

Here is the Lagrangian

L = (1/2)[[partial derivative]u[phi][partial derivative]u[phi] — m2[phi]2]

The first term is the kinetic energy term since it arises from p2.The second term is called the mass term since it’s proportional to m2. In this case, there is no potential term, or interaction Lagrangian, since it describes a non-interacting field.

Some people say that what I’ve been calling the Lagrangian L, is actually the Lagrangian density, and the true Lagrangian is

[integral]d4L(x, t)

However, most people just call L(x, t) the Lagrangian.

Let’s say you have a field normalized to a single quantum of energy w, and momentum k. If w2 = k2 + m2, then a solution to the wavefunction is

[phi] = A cos (k . r — wt)

The Hamiltonian is

H = [phi dot][partial derivative of L with respect to phi dot] — L

so

H = (1/2)([phi dot]2 + ([del][phi}2 + m2[phi]2)

E = [integral]dx3H so

E = [integral](1/2)d3x[[phi dot]2 + ([del][phi}2 + m2[phi]2]

We know that E = [h bar]w for a single quantum of energy. We’ll use natural units so [h bar] = 1. We can solve for A, and then plug it into the original equation for φ.

E = [integral](1/2)d3x[[phi dot]2 + ([del][phi}2 + m2[phi]2]

since [phi] = A cos (k . r — wt)

E = (1/2) A2[integral](w2 + k2) sin2 (k . r — wt) + m2cos2 (k . r — wt)]d3x

[integral]sin2(k . r — wt) = [integral]cos2(k . r — wt) = 1/2 so

E = (1/2) A2 [(1/2) (w2 + k2) + (1/2)m2]

w2 + k2 + m2 so

E = (1/2)A2[(1/2)(k2 + m2 + k2) + (1/2)m2]

E = (1/2)A2[(1/2)(2k2 + m2) + (1/2)m2]

E = (1/2)A2[k2 + (1/2)m2 + (1/2)m2]

E = (1/2)A2[k2 + m2]

E = (1/2)A2w2

since E = [h bar]w, so with [h bar]= 1, then E = w, which means A = [squareroot of](2/w)

E = (1/2)([squareroot of](2/w)]2w2 = (1/2)(2/w)w2 = w2/w = w

so A = [squareroot of](2/w) so putting that in the original equation

[phi] = [squareroot of 2/w] cos (k . r — wt)

One definition of the cosine function is

cos z = (1/2)(eiz + e-iz)

Also, the hyperbolic cosine is defined by

cosh z = (1/2)(ez + e-z)

The sine function is defined by

sin z = (1/2i)(eiz — e-iz)

Anyway so

[phi] = [squareroot of 2/w]((1/2)(ei(k . r — wt) + e-i(k . r — wt)))

[squareroot of 2/w](1/2) = [squareroot of 2]/[squareroot of 2] x 1/2 = [squareroot of 2]/2[squareroot of w] = [squareroot of 2]/(squareroot of 2)(squareroot of 2)(squareroot of w) = 1/(squareroot of 2)(squareroot of w) = 1/[squareroot of 2w] so

[phi] = 1/[squareroot of 2w][ ei(k . r — wt) + e-i(k . r — wt)]

In a quantized theory, we want to speak of the creation and destruction of particles. If nk> is a state with n particles, all of the same momentum k and energy w at time t, you have

|nk> is proportional to einwt

Then a state with one less particle will be

|nk — 1> is proportional to ei(n-1)wt

You would then expect the operator a that creates or destroys quanta to be a multiple of ei(n-1)wti(k . r — wt). We define a† as the operator that creates quanta associated with the field φ and a destroys them.

[phi] = 1/[squareroot of 2w][ aei(k . r — wt) + [a dagger]e-i(k . r — wt)]

So far we’ve discussed only free fields without considering their source or how they interact. Here is the Lagrangian of a scalar field.

L = (1/2)[[partial derivative]u[phi][partial derivative]u[phi] — m2[phi]2]

Let’s say you add to it a part describing interactions, such as

Lint = [phi]p(x, t)

Then the wave equation

[partial derivative]u[partial derivative]u[phi] + m2[phi] = 0

then has a source term

[partial derivative]u[partial derivative]u[phi] + m2[phi] = p

In terms of the analogy with electromagnetism, you can think of p as a source of the field φ. Let’s say you have the simple case.

p = g[delta](x)

Here you have a time independent point source at the origin of strength g. Then you can solve the problem by using a Fourier transform.

Since p is not time dependent, you have

(-[del]2 + m2)[phi] = g[delta](x)

Here you have

[phi](x) = (1/(2[pi])3/2))[integral]d3keik . x[phi tilde](k)

[phi tilde](k) = (1/(2[pi])3/2))[integral]d3xe-ik . x[phi](x)

[del]2 = -k2 so

(k2 + m2)[phi tilde](k) = g/(2[pi])3/2

You can then divide to get [φ tilde] and substitute back into the original equation to get φ.

[phi](x) = g/(2[pi])3[integral]d3k (eik . x)/(k2 + m2)

If we had not taken a time independent source, the denominator would have been -k02 + k2 + m2 — k2, where k2 = kuku. The denominator is the propagator whenever a particle is exchanged in the interaction.

Putting k . x = k r cos θ you have

[integral from 0 to infinity]k2dk/k2 + m2 [integral from 0 to 2 pi]d[phi] [integral from -1 to 1] d(cos [theta])eikr cos [theta] =

= 2[pi]/r [integral from 0 to infinity] (dk2/k2 + m2) sin kr

= [pi]/ir [integral from 0 to infinity] (dk2/k2 + m2) (eikr — e-kr)

= [pi]/ir [[integral from 0 to infinity] (dk2/k2 + m2) (eikr) — [integral from 0 to minus infinity] (dk2/k2 + m2) (eikr)]

= [pi]/ir [integral from minus infinity to infinity] (dk2/k2 + m2) (eikr)

This can be evaluated as a contour integral. Closing the contour where Im k > 0 so the integral is convergent, and picking up a contribution from the residue at k = im, gives finally

[phi] = (g/4[pi]) (e-mr/r)

Hideki Yukawa (1907–1981) identified φ as a meson field with the nucleon as a source, analogous to how any charged particle is the source of an electromagnetic field. Just as the effects of the electromagnetic field are transmitted by photons, the effects of the meson field should be transmitted by mesons. When the particles have a mass m, the field is significant in size only out to a range of force r ~ 1/m.

Now we want to extend this to see how one nucleon would interact with another by sensing its meson field. The interaction Hamiltonian between two nucleons, the second one described by p2(x), is

H = -[integral]d3x[phi](x) p2(x)

[phi](x) = (1/4[pi]) [integral]d3x’ p1(x’) ((e-m|x — x’|)/(|x — x’|))

H12 = -(1/4[pi]) [integral]d3x d3x’ p1((x) p2((x’)((e-m|x — x’|)/(|x — x’|))

Therefore the potential can be written as

V(r) = -(1/4[pi]) (e-mr/r)

Notice that since photons are massless, in that case, the equation reduces to the familiar case from electrostatics.

V(r) = -(1/4[pi]) (e-(0)r/r) = -(1/4[pi]) (e0/r) = -(1/4[pi]) (1/r) = -1/4[pi]r

This result leads to the general interpretation in quantum field theory that all interactions are due to the exchange of field quanta. Force and interaction become interchangeable. The equation for H12 gives the interaction in position space. Usually, you write matrix elements in momentum space. The momentum space quantity representing the exchanged particle of mass m is

1/(k2 — m2)

This is called the propagator. The complete propagator also has a phase factor and a numerator that depends on the spin of the exchanged particle.

Here are some Lagrangians used in particle physics.

Real spin-zero field of mass m (scalar or pseudoscalar)

L = (1/2) [([partial derivative]u [phi])( [partial derivative]u [phi]) — m2 [phi]2]

where ([partial derivative]u[partial derivative]u + m2)[phi] = 0

Complex scalar or psuedoscalar field of mass m, or two real scalars of the same mass

L = (1/2) [([partial derivative]u [phi]1)( [partial derivative]u [phi]1) — m2 [phi]12] + (1/2) [([partial derivative]u [phi]2)( [partial derivative]u [phi]2) — m2 [phi]22]

L = ( [partial derivative]u [phi])* )( [partial derivative]u [phi]) — m2[phi]*

where [phi](x) = ([phi]1(x) + i[phi}2(x))/[squareroot of 2]

and ([partial derivative}u [partial derivative}u + m2)[phi] = ([partial derivative}u [partial derivative}u + m2)[phi]* = 0

Spin — 1/2 fermion field of mass m

L = [psi bar](i[gamma]u[partial derivative]u — m)[psi]

where ψ satisfies the Dirac equation

(i[gamma]u[partial derivative]u — m)[psi] = 0

Massive Abelian vector field

If there were an Abelian field Bu, like the electromagnetic field except massive, the following term is added to the Lagrangian

(1/2)m2BuBu

If you see the term BuBu, it’s coefficient is (mass)2

Non-Abelian vector field

There are non-Abelian gauge fields such as the gluon or W fields. The vector potential Au of electromagnetism is generalized to a non-Abelian field that has an internal index a so that the non-Abelian vector potential can be written as Wau. For SU(2), a = 1, 2, 3. For SU(3), a = 1, 2, 3, …8. Then we define

Wauv = [partial derivative]u Wa v — [partial derivative]v Wa u + gfabc Wbu Wcv

The fabc are structure constants. For SU(2), fabc = εabc. The extra term is necessary to the make the vector, in internal space, transform correctly under rotations. The form of Wauv is gauge invariant.

L = -(1/4) Wauv Wuva + (1/2)m2 Wau Wua

Electromagnetism

L = (1/2)(E2 — B2) — pV + J . A

L = -(1/4)FuvFuv — JuAu

Quantum Chromodynamics

L = (1/4g2)Guva Guva + [summation of j][q bar]j(i[gamma]uDu + mj)qj

where Guva = [partial derivative]uAva — [partial derivative]vAua + ifbca Aub Avc

and Du = [partial derivative]u + it Aua

Here mj and qj are the mass and quantum field of the quark of jth flavor. A is the gluon field with spacetime indices u and v, and color indices a, b, and c. The numerical coefficients f and t guarantee SU(3) symmetry. The quark masses and coupling constant g are free parameters.

The Yang-Mills equations are four dimensional equations for the gluon fields in strong interactions, meaning quantum chromodynamics. In a formalism where time is imaginary, which is Euclidean space, these equations have a soliton-like solution called the instanton. The instanton is a solution of the classical Yang-Mills equations in Euclidean space. It is the basis of a lot of semiclassical calculations which are based on classical solutions. The main property of instantons is that they carry the topology of the vacuum.