2m

which works out to

x2

Ψ(x, t) ∝ exp ’ . (2.35)

2

4(σ0 + i t/2m)

By evaluating Ψ— Ψ, we see that a Gaussian density is obtained with a vari-

ance σ(t) that changes in time according to

2 t2

2 2

σ (t) = σ0 1+ . (2.36)

4

4m2 σ0

The narrower the initial package, the faster it will spread. Although this

seems counterintuitive if we think of particles, we should remember that the

wave function is related to the probability of ¬nding the particle at a certain

place at a certain time, which is all the knowledge we can possess. If the

2.3 Relativistic energy relations for a free particle 25

initial wave function is narrow in space, its momentum distribution is broad;

this implies a larger uncertainty in position when time proceeds.

Only free particles broaden beyond measure; in the presence of con¬ning

potentials the behavior is quite di¬erent: stationary states with ¬nite width

emerge.

Because the packet becomes broader in space, it seems that the Heisen-

berg uncertainty relation would predict that it therefore becomes sharper

in momentum distribution. This, however, is an erroneous conclusion: the

broadening term is imaginary, and Ψ is not a pure real Gaussian; therefore

the relation σx σk = 1/2 is not valid for t > 0. In fact, the width in k-space

remains the same.

2.3 Relativistic energy relations for a free particle

The relation between energy and momentum (for a free particle) that we

used in the previous section (2.22) is incorrect for velocities that approach

the speed of light. In non-relativistic physics we assume that the laws of

physics are invariant for a translation of the spatial and time origins of our

coordinate system and also for a rotation of the coordinate system; this

leads to fundamental conservation laws for momentum, energy, and angular

momentum, respectively.1 In the theory of special relativity the additional

basic assumption is that the laws of physics, including the velocity of light,

are also invariant if we transform our coordinate system to one moving

at a constant speed with respect to the original one. Where for normal

rotations in 3D-space we require that the square of length elements (dr)2

is invariant, the requirement of the constant speed of light implies that for

transformations to a moving frame (c d„ )2 = (c dt)2 ’ (dr)2 is invariant. For

1 + 1 dimensions where we transform from (x, t) to x , t in a frame moving

with velocity v, this leads to the Lorentz transformation

’γv/c

x γ x

= , (2.37)

’γv/c

ct γ ct

where

1

γ= . (2.38)

1 ’ v 2 /c2

In Minkovsky space of 1 + 3 dimensions (ct, x, y, z) = (ct, r) vectors are four-

vectors vμ = (v0 , v)(μ = 0, 1, 2, 3) and we de¬ne the scalar or inner product

1 Landau and Lifschitz (1982) give a lucid derivation of these laws.

26 Quantum mechanics: principles and relativistic e¬ects

of two four-vectors as

def

vμ wμ = v0 w0 ’ v1 w1 ’ v2 w2 ’ v3 w3 = v0 w0 ’ v · w. (2.39)

The notation vμ wμ uses the Einstein summation convention ( 3 over re-

μ=0

peating indices is assumed, taking the signs into account as in (2.39)).2 The

square magnitude or length of a four-vector is the scalar product with itself;

note that such a square length may be positive or negative. Lorentz trans-

formations are all transformations in Minkowski space that leave dxμ dxμ =

(c d„ )2 invariant; they of course include all space-like rotations for which

d„ = 0. Vectors that represent physical quantities are invariant for Lorentz

transformations, and hence their scalar products and square magnitudes are

constants.

Without any derivation, we list a number of relevant physical four-vectors,

as they are de¬ned in relativistic mechanics:

• coordinates: xμ = (ct, r);

• wave vector: kμ = (ω/c, k);

• velocity: uμ = (γc, γv);

• momentum: pμ = muμ = (γmc, γmv).

Here m is the (rest) mass of the particle. The ¬rst component of the mo-

mentum four-vector is identi¬ed with the energy E/c, so that E = γmc2 .

Note the following constant square lengths:

u μ u μ = c2 , (2.40)

E2

pμ pμ = 2 ’ p2 = m2 c2 , (2.41)

c

or

E 2 = m2 c4 + p2 c2 . (2.42)

This is the relation between energy and momentum that we are looking for.

From the quadratic form it is immediately clear that E will have equivalent

positive and negative solutions, one set around +mc2 and the other set

around ’mc2 . Only the ¬rst set corresponds to the solutions of the non-

relativistic equation.

2 We use subscripts exclusively and do not use general tensor notation which distinguishes co-

variant and contravariant vectors and uses a metric tensor to de¬ne vector products. We note

that the “Einstein summation convention” in non-relativistic contexts, for example in matrix

multiplication, is meant to be simply a summation over repeated indices.

2.3 Relativistic energy relations for a free particle 27

Now identifying E with i ‚/‚t and p with ’i ∇, we obtain the Klein“

Gordon equation

‚2 mc 2

’ 2 2 + ∇2 ’ Ψ = 0. (2.43)

c ‚t

This equation has the right relativistic symmetry (which the Schr¨dinger

o

equation does not have), but unfortunately no solutions with real scalar

densities Ψ— Ψ exist.

Dirac devised an ingeneous way to linearize (2.42). Let us ¬rst consider

the case of one spatial dimension, where motion is allowed only in the x-

direction, and angular momentum cannot exist. Instead of taking a square

root of (2.42), which would involve the square root of the operator p, one

ˆ

can devise a two-dimensional matrix equation which in fact equals a set of

equations with multiple solutions:

‚Ψ ˆ

i = c(±ˆ + βmc)Ψ = HΨ,

p (2.44)

‚t

where Ψ is a two-component vector, and ± and β are dimensionless Hermi-

tian 2 — 2 matrices, chosen such that (2.42) is satis¬ed for all solutions of

(2.44):

(±ˆ + βmc)2 = (ˆ2 + m2 c2 )1.

p p (2.45)

This implies that

±2 p2 + (±β + β±)mcˆ + β 2 m2 c2 = (ˆ2 + m2 c2 )1,

ˆ p p (2.46)

or

±2 = 1, β 2 = 1, ±β + β± = 0. (2.47)

In other words, ± and β are Hermitian, anticommuting, and unitary ma-

trices.3 The trivial solutions of the ¬rst two equations: ± = ±1 and/or

β = ±1 do not satisfy the third equation.

There are many solutions to all three equations (2.47). In fact, when

a matrix pair ±, β forms a solution, the matrix pair U±U† , UβU† , con-

structed by a unitary transformation U, forms a solution as well. A simple