The Klein-Gordon Equation

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In classical physics we seek to describe motion. Using equations of motion, we calculate descriptions of trajectories followed by particles and bodies. We specify the forces acting on an object, calculate the consequent acceleration of the object, and integrate the algebraic description of the acceleration to obtain algebraic formulae that describe the location and the velocity of the object as a function of the elapse of time. The equations of motion are simply the Euler-Lagrange equations applied to a certain Lagrangian function associated with the particle in accordance with the principle of least action. The Lagrangian itself is a Lorentz invariant that we obtain by multiplying together, as a vector dot product, the distance-duration four-vector and the momentum-energy four-vector that we associate with the object.

In the quantum theory we seek to describe interactions. Using equations of field evolution, we calculate descriptions of aleatric fields and waves associated with certain particles. We then apply Born’s theorem to those descriptions, the state functions of the particles, to calculate probable values of momentum, both linear and angular, and energy associated with the particles. The equations of field evolution are simply the Euler-Lagrange equations applied to a certain Lagrangian density function associated with the wave/particle. Again, the Lagrangian function must be a Lorentz invariant, but here it cannot involve a spatio-temporal interval, because those are indeterminate in the quantum theory.

Because the dot product of any four-vector with another four-vector is Lorentz invariant, we can simply square the momentum-energy four-vector. We even know the constant that the square equals, so taking the negative of the dot product, we have

(Eq’n 1)

in which m0 represents the rest mass of the particle under consideration.

In order to devise a state function that will conform to that equation and thus give us a correct description of the particle, we must recall to mind the fact that the state function so encodes the properties of the particle that differential operators can extract proper descriptions of those properties from the state function; that is,

(Eq’ns 2)

In those equations ∂i= /∂xi and ∂t= /∂t. I have also used contravariant notation, because contravariant variables represent measured quantities while covariant variables represent derived quantities. With those equations we can modify Equation 1 and write

(Eq’n 3)

In that equation the psi with the asterisk represents the adjoint of the state function (psi), the complex conjugate of the transpose (the transpose being part of it only if the state function includes a matrix as one of its factors). If we write ∂4=∂/ic∂t =∂t/ic, then we can rewrite that equation more compactly as

(Eq’n 4)

In that equation mu takes the values 1, 2, 3, 4. That’s the Klein-Gordon Lagrangian.

Applying the Euler-Lagrange equation to that Lagrangian will give us the Klein-Gordon equation, the relativistic analogue of the Schrödinger equation. Before we do that, we must so transform one of the derivatives that both derivatives have the same tensor rank. We achieve that transformation by multiplying Equation 4 by an appropriate metric tensor. In this case we use the contravariant Minkowski tensor, which expresses the coordinate geometry of flat spacetime;

(Eq’n 5)

I’ve used the contravariant metric tensor because I want both derivatives to represent measurable quantities rather than derived (covariant) quantities. It’s a subtle point because the covariant and contravariant versions of the Minkowski metric tensor are identical to each other.

In this case the Euler-Lagrange equation takes the form

(Eq’n 6)

We have the transformed Lagrangian as

(Eq’n 7)

With regard to the derivative terms we have

(Eq’n 8)

We have two terms on the right side of that equation because the operation of differentiation commutes with the operation of taking an adjoint, which fact we take into account as we apply the product rule of differentiation. With regard to the mass term the Euler-Lagrange equation gives us

(Eq’n 9)

We then have from the Euler-Lagrange equation

(Eq’n 10)

That’s just the sum of the Klein-Gordon equation and its adjoint shadow, which have the same value, though slightly different arrangements of the variables. We then have the Klein-Gordon equation as

(Eq’n 11)

That is clearly a wave equation and it has the standard plane wave solution,

(Eq’n 12)

We can also assign to the aleatric wave a wave number and an angular frequency related to the expectation values of the particle’s linear momentum and total energy,

(Eq’ns 13)

But that’s for a free particle and is rather trivial. Can we solve the Klein-Gordon equation for a particle moving under a force? Of course, force is not properly defined in quantum mechanics, so we can’t devise equations of motion directly. Instead, we must describe forcefields in terms of potential momentum and potential energy. In applying the electromagnetic force, for example, we create a potentials four-vector from the electrotonic potential momentum (the magnetic vector potential, A) and the electrostatic potential (ö) in the tensor form

(Eq’n 14)

If we have a particle carrying an electric charge e coming into the forcefield from field-free space, then we have

(Eq’n’s 15)

In those equations the electrotonic potential includes only the spatial components of the potentials four-vector. Equations 2 let us use those equations to define the covariant derivative,

(Eq’n 16)

and its contravariant counterpart,

(Eq’n 17)

Those equations transform the Klein-Gordon equation for a free particle into one for a particle in an electromagnetic field, which equation is

(Eq’n 18)

Starting with the solution of Equation 12, we can modify the exponential in its argument and in its applied coefficient (only implied in Equation 12) to obtain a modified state function that will enable us to calculate expectation values for the dynamic properties of the particle as it traverses the field.

We can write Equation 11 in the form

(Eq’n 19)

If we multiply that equation from the left by the adjoint of the state function and subtract from the result the adjoint of that result, we get

(Eq’n 20)

Add the expression on the right side of that equation to both sides (negating the right side) and then multiply the result by e/(2iSm0c2). The result is a continuity equation,

(Eq’n 21)

in which we have the charge density as

(Eq’n 22)

and the current density as

(Eq’n 23)

In the last three equations I have multiplied the density functions by an electric charge. That multiplication is necessitated by the fact that rho represents a probability density (not a charge density): the electric charge (either positive or negative) turns the probability density into an effective electric charge density. It does the same for the current density, transforming a probability current density into an electric current density.

Look again at Equation 10. The Euler-Lagrange equation has converted the Lagrangian function into two equations - the Klein-Gordon equation (Equation 11) and its adjoint. The solution of the adjoint equation is simply the complex conjugate of Equation 12,

(Eq’n 24)

In quantum mechanics, if we want to treat two particles as a unit, then we must multiply their state functions together. If we have two particles at rest in our inertial reference frame, one described by Equation 12 and the other by Equation 24, then we describe them as a unit with the compound state function

(Eq’n 25)

If E1=E2 (=m0c2), then the arguments of the exponentials cancel each other out and the time derivative of the result leads to an expectation value of zero for the energy carried by the two particles. That’s clearly wrong. If, on the other hand, E1=-E2, then the arguments don’t cancel out and the expectation value that we calculate for the energy of the two particles is <E>=2m0c2, which is correct. But now we have to explain the negative energy.

When Paul Dirac presented his theory of the electron, he noted that his equation produced negative-energy states, as relativistic equations must do. He noted that there must exist a zero-energy state and that electrons don’t descend below that state because all of the available negative-energy states are already filled: the electrons are prevented from going into the negative-energy states by Pauli’s exclusion principle. If we were to pull an electron out of that cosmic sea, we would have to give it at least 511,000 electron-volts of positive energy (the rest-mass energy of the electron), leaving behind it a hole with minus 511,000 electron-volts of energy. The hole behaves as a particle carrying a positive electric charge; that is, it’s a positron. Dirac inferred that, if the state function that satisfies his equation represents a certain particle, then the complex conjugate of that state function represents the antimatter counterpart of that particle. That inference applies as well to the Klein-Gordon equation and the state function that it shapes.

In devising Equation 18 we modified the derivative operators to make the Lagrangian represent a particle moving in an externally applied forcefield. We can also modify the Lagrangian by adding or subtracting terms. Thus we get the Higgs Lagrangian,

(Eq’n 26)

which contains a description of the Higgs boson, the particle whose existence necessitates the existence of inertial mass.

The last two terms in that Lagrangian are analogous to the potential energy in the classical Lagrangian. Particles emerge as quantized vibrations of the ψ-field from the general ground state of that field, the state in which the potential, U, has its minimum value. If we split the state function into its real and imaginary parts (ψ=ψ1+iψ2), then we can lay out an Argand diagram and represent U(ψ) as an altitude above the complex plane on which we plot ψ1 on one axis and ψ2 on the other. Calculating the minimum values of U through the usual dU/dψ=0, we find that the minimum points form a circle

(Eq’n 27)

centered on the origin of the complex plane.

Shifting the coordinate frame along the ψ1-axis by μ/λ transforms the Lagrangian in a way that splits it into two versions of the Klein-Gordon Lagrangian plus some interactive terms. The first of those Klein-Gordon Lagrangians is a function of only ψ1 and describes a massive particle moving freely and the second is a function of only ψ2 and describes a massless Goldstone boson, which doesn’t exist. We assert that there is a point on the minimum potential circle where the Goldstone boson does not exist, but it’s not one of the points where ψ2=0, because that would change the value of the Lagrangian. The Goldstone boson comes from the part of the Lagrangian that’s purely imaginary, so we want to eliminate that part without changing the value of the Lagrangian.

If we draw a straight line across the ψ1-ψ2 plane and it makes an angle θ with the ψ1-axis, then we have a transformation of the state function,

(Eq’n 28)

The value of the Lagrangian remains invariant under that transformation (ψ*ψ=ψ*ψ), so we are free to pick a value of θ that makes the Lagrangian wholly real. We get that value with

(Eq’n 29)

That transformation changes the Lagrangian into the sum of the Klein-Gordon Lagrangian describing the Higgs field and its associated boson and something that looks like a Proca Lagrangian for a particle of indeterminate mass (depending on the value of lambda). Thus we see how the existence of the Higgs field necessitates the existence of mass in other particles.

Except for the metric tensor, the Klein-Gordon Lagrangian does not involve any matrices. That fact means that the state function is a single number; thus, the state function does not specify a direction in space. That fact necessarily implies that the particles described by the Klein-Gordon equation have zero spin, that the Klein-Gordon equation applies to pseudoscalar mesons, such as the pi meson.

As with all relativistic particle/wave equations, we can present the Klein-Gordon equation in a form identical to that of the Dirac equation. We can factor Equation 1 into a product of two linear operators if we split the metric tensor into the product of the Lorentz Transformation matrix, [Λ], and its transpose, so we get

(Eq’n 30)

We can pick one of those factors in the third line to act on the state function, which is still a single number, to get, for example,

(Eq’n 31)

Again, that equation does not describe a spinning particle, even though we have a four-vector involved.

If the metric tensor involves functions of the coordinates, bringing General Relativity into play, then we can generalize the first line of Equation 30 to produce a General Relativistic form of the Klein-Gordon equation,

(Eq’n 32)

In that equation g is a single number that represents the determinant of the metric tensor (treated as a matrix) and R represents the scalar curvature of spacetime (the Ricci scalar). When m0=0, that equation becomes conformally invariant.

There exists one other way to bring matrices into play in regard to the Klein-Gordon equation. We can use the Dirac-like Duffin-Kemmer-Petiau equation,

(Eq’n 33)

In that equation beta represents the Duffin-Kemmer-Petiau matrices, which mathematicians define through the commutation relation,

(Eq’n 34)

Eta represents a constant diagonal matrix,

(Eq’n 35)

For a spin-0 particle the Duffin-Kemmer-Petiau matrices consist of four 5x5 arrays of elements, all of which equal zero except the following;

(Eq’ns 36)

The state function of Equation 33 is not a single number, but is a column matrix,

(Eq’n 37)

In this case we are using the Minkowski signature gμν=diag(+1,-1,-1,-1), with time being the zeroth coordinate rather than the fourth. Equation 33 is then equivalent to

(Eq’n 38)

Thus we have all that we need to know about the Klein-Gordon equation.

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