(λ^2 −1) = 0
λ=± 1
vx=±cThus, if wemeasure the velocity component in any direction, we should either get plus or
minusc. This seems quite surprising, but we should note that a component of the velocity operator
does not commute with momentum, the Hamiltonian, or even the other components of the velocity
operator. If the electron were massless, velocity operators would commute with momentum. (In
more speculative theories of particles, electrons are actually thought to be massless, getting an
effective mass from interactions with particles present in the vacuum state.)
The states of definite momentum are not eigenstates of velocity for a massive electron. Thevelocity
eigenstatesmix positive and “negative energy” states equally.
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
a
b
c
d
=λ
a
b
c
d
d
c
b
a
=±
a
b
c
d
u=
1
0
0
1
u=
0
1
1
0
u=
1
0
0
− 1
u=
0
1
− 1
0
Thus, while momentum is a constant of the motion for a free electronand behaves as it did in NR
Quantum Mechanics, velocity behaves very strangely in the Dirac theory, even for a free electron.
Some further study of this effect is in order to see if there are physical consequences and what is
different about the Dirac theory in this regard.
We may get the differential equation for the velocity of a free electron by computing the derivative
of velocity. We attempt to write the derivative in terms of the constants of the motionEand~p.
v ̇j =
i
̄h
[H,vj] =
i
̄h(− 2 Hvj+{H,vj}) =
i
̄h(− 2 Hvj+{vjpj+mc^2 γ 4 ,vj}) =
i
̄h(− 2 Hvj+{vjpj,vj})=
i
̄h(− 2 Hvj+{vjpj,vj}) =
i
̄h(− 2 Hvj+ 2vjpj) =
i
̄h(− 2 Hvj+ 2v^2 jpj) =
i
̄h(− 2 Hvj+ 2c^2 pj)This is a differential equation for the Heisenberg operatorvjwhich we may solve.
vj(t) =c^2 pj/E+ (vj(0)−c^2 pj/E)e−^2 iEt/h ̄To check, differentiate the above
v ̇j(t) =−2 iE
̄h(vj(0)−c^2 pj/E)e−^2 iEt/ ̄h=i
̄h(− 2 Evj(0) + 2c^2 pj)e−^2 iEt/h ̄