Lectures on Quantum Field Theory. Ashok Das
general and the parameter of the transformation, θ, has been arbitrary. However, if we want the transformation to diagonalize the Hamiltonian, it is clear from (3.216) that we can choose the parameter of transformation to satisfy
In this case, we have
which, from (3.216), leads to the diagonalized Hamiltonian
We see from (3.219) that the Hamiltonian is now diagonalized in the positive and the negative energy spaces. As a result, the two components of the transformed spinor
would be decoupled in the energy eigenvalue equation and we can without any difficulty restrict ourselves to the positive energy sector where the energy eigenvalue equation takes the form
For |p|
which has a natural non-relativistic expansion in powers of
There is a second limit of the Dirac equation, namely, the ultrarelativistic limit |p|
this would lead to
As a result, in this case, the transformed Hamiltonian (3.216) will have the form
which has a natural expansion in powers of
which clearly has a natural expansion in powers of
3.11Zitterbewegung
The presence of negative energy solutions for the Dirac equation leads to various interesting consequences. For example, let us consider the free Dirac Hamiltonian (1.100)
In the Heisenberg picture, where operators carry time dependence and states are time independent, the Heisenberg equations of motion take the forms (ħ = 1)
Here a dot denotes differentiation with respect to time.
The second equation in (3.228) shows that the momentum is a constant of motion as it should be for a free particle. The first equation, on the other hand, identifies α(t) with the velocity operator. Let us recall that, by definition,
where we have denoted the operator in the Schrödinger picture by
Furthermore, using (1.101) we conclude that
As a result, it follows that
In other words, even though the momentum of a free particle is a constant of motion, the velocity is not. Secondly, since the eigenvalues of α are ±1 (see, for example, (1.101)), it follows that the eigenvalues of α(t) are ±1 as well. This is easily understood from the fact that the eigenvalues of an operator do not change under a unitary transformation. More explicitly, we note that if
where λ denotes the eigenvalue of the velocity operator α, then, it follows that
where we have identified
Equation (3.234) shows that the eigenvalues of