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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

H(p, z) = E = p²/(2m) + V(z)

and therefore

H^(∇, z) = (i)²∇²/(2mh²) + V(z)
which reduces to
H^(∇, z) = −∇²/(2mh²) + V(z)

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,

d(mv/(1−β²))/dt = 0

The Maclaurin Series for
Relativistic Kinetic Energy
and Momentum

The Maclaurin series for f(x) is

f(x) = f(0)/0! + f'(0)x/1! + f"(0)x²/2! + f"'(0)x³/3! + ...

Let

1/(1−β²)½ = γ

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

γ ≅ 1 + (0)β + ½β² +

Therefore

K/(m0c²) = γ − 1 ≅ ½β²
and hence
K ≅ ½m0

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

p/(m0c) = 0 + (1)β + ...

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 ≅ ½m0
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)½
or, 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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