﻿ Is There a Relativistic Equivalent of Schrodinger's Equation?
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Is There a Relativistic Equivalent
of Schrodinger's Equation?

The answer is Yes, but it is not so much a different equation as a difference in the nature of the variables kinetic energy and momentum. Classsically a particle of mass m traveling at a velocity v has a kinetic energy of ½mv² and a momentum of mv. These are however just the first appoximations of the relativistic quantities and valid only for velocities small compared with the speed of light in a vacuum. The correct forms will be given later.

## Hamiltonian Dynamics and Schrödinger's Equations

Let H(p, z) be the energy of a particle expressed in terms of its momentum p and its location z. It is presumed that H is not explicitly a function of time t. H(p, z) is called the Hamiltonian function for the physical system.

The Hamiltonian operator H^(p) for the system is constructed by replacing p with i∇/ h where i is the square root of −1 and h is the reduced Planck's constant. Powers of p are replaced by orders of differentiation.

For example, Classically the energy E of a paricle of mass m in a potential of V(z) is

#### E = ½mv² + V(z)

Since momentum p is equal to mv, kinetic energy is also equal to p²/(2m). Thus

and therefore

## Schrödinger's Equations

There are two Schrödinger's Equations; the time dependent form and the time independent form. It is the time independent. form which is most useful for physical analysis. That form is

#### H^ψ(z) = Eψ(z)

ψ(z) is the quantum wave function for the system. The probability density function P(z) for the system is equal to |ψ|².

For the example of a particle of mass m in a potential field of V(z) the time independent Schrödinger equation is

#### −∇²ψ/(2mh²) + V(z)ψ(z) = Eψ(z) which reduces to ∇²ψ(z) = −2mh²(E−V(z))

This may also be expressed as

#### ∇²ψ(z) = −2mh²K(z)

where K(z) is kinetic energy expressed as a function of location.

For a one dimensional coordinate x

#### ∇²ψ(x) = (∂²ψ(x)/∂x²)

Thus if K(x)=K0 then ψ(x) is a function of K0½.

## Classical versus Relativistic Dynamics

The time independent Schrödinger equation determines a probability distribution for a particle moving in a periodic path. At low speeds relative to the speed of light, relativistic dynamics results in essentially the same paths as classical dynamics. At high energies a particle approaches the situation in which it is everywhere in its path traveling at the speed of light.

## Relativistic Momentum and Kinetic Energy

Albert Einstein in his Special Theory of Relativity found that the apparent mass of a body in motion is given by

#### m = m0/(1−β²)½

where m0 is the rest mass of the body and β is the velocity of the body relative to the speed of light c; i.e., β=v/c. The total energy is mc² and therefore the kinetic energy K is then given by

#### K = (mc²−m0c²)

The expression ½mv² is not full kinetic energy; it is just the first order approximation of the kinetic energy. For more on this point see Relativistic Mechanics.

If K is the kinetic energy of a system and V is the potential energy then the Lagrangian of the system is defined as

#### L = K − V

In general the momentum pz associated with a spatial variable z is given by

#### pz = (∂L/∂vz)

where vz=(dz/dt).

The Lagrangian for a particle of rest mass m0 traveling at velocity v in a potential field V is

#### L = (mc²−m0c²) − V and the partial derivative of L with respect to v is −½(m0c²/(1−β²)3/2)(−2β(1/c)) which reduces to (m0c²/(1−β²)3/2)(v/c)(1/c) or, equivalently (m0/(1−β²)3/2)v which can be expressed as mv/(1−β²)

In other words, the relativistic linear momentum requires not only the relativistic adjustment of mass but also division by a factor of (1−β²). If the kinetic and potential energies are independent of the spatial variable that defines velocity then it is mv/(1−β²) which is constant over time.

That is to say,

## The Maclaurin Series for Relativistic Kinetic Energy and Momentum

The Maclaurin series for f(x) is

Let

Then

#### K/(m0c²) = (γ − 1)

The first two derivatives of γ with respect to β are:

#### (∂γ/∂β) = (−½)(1/(1−β²)3/2)(−2β) which reduces to (∂γ/∂β) = β/(1−β²)3/2) = βγ³ thus (∂²γ/∂β²) = γ³ + β(3γ²)(−2β) and hence (∂²γ/∂β²) = γ³ − 6β²(γ²)

At β=0, γ=1. thus

#### (∂γ/∂β) = 0 (∂²γ/∂β²) = 1

The beginning of the Maclaurin series for γ is then

Therefore

Now consider

#### p/(m0c) = γ3β

The first derivative of the RHS are:

#### γ3 + β(3γ²)(−1/2)(γ3)(−2β) at β=0 this is equal to 1

At β=0, p/(m0c)=0 therefore the beginning of the Maclaurin series for p/(m0c) is

thus

#### p ≅ m0cβ = m0v

What is needed for quantum analysis through Schrödinger's time-independent equation is kinetic energy K as a function of momentum p. From the above

#### K ≅ ½m0v² and v ≅ p/m0 K ≅ p²/(2m0)

K is a function of γ and p is also a function of γ Note that β = (1−1/γ²)½. Therefore

#### K/(m0c²) = γ −1 and p/(m0c) = γ3(1 − 1/γ²)½or, equivalently p/(m0c) = γ2( γ2 − 1)½

These may be used to construct a Maclaurin series for K in terms of p.

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