Lectures on Quantum Field Theory. Ashok Das

      This, therefore, fixes the sign of the square root in the second relation in (1.62) for various cases.

      Matching the wave functions in (1.60) and (1.61) and their first derivatives at the boundary z = 0, we determine

      so that we determine

      Let us next determine the probability current densities associated with the different waves. From (1.55) as well as the form of the potential in (1.58) we obtain

      where we have used (1.67) as well as the fact that, while p is real and positive, p′ can be positive or negative or even imaginary depending on the strength of the potential (see (1.63) and (1.65)). We can now determine the reflection and the transmission coefficients simply as

      We see from the reflection and the transmission coefficients that

      so that the reflection and the transmission coefficients satisfy unitarity for all strengths of the potential.

      However, let us now analyze the different cases of the potential strengths individually. First, for the case, E − eΦ₀ > m (weak potential), we see that p′ is real and positive and we have

      which corresponds to the normal scenario in scattering. For the case of an intermediate potential strength, −m < E − eΦ₀ < m, we note from (1.63) that p′ is purely imaginary in this case. As a result, it follows from (1.69) that

      so that the incident beam is totally reflected and there is no transmission in this case. The third case of the strong potential, E − eΦ₀ < −m, is the most interesting. In this case, we note from (1.65) that p′ is real, but negative. As a result, from (1.69) we have

      Namely, even though unitarity is not violated, in this case the transmission coefficient is negative and the reflection coefficient exceeds unity. This is known as the Klein paradox and it contradicts our intuition from the one particle scatterings studied in non-relativistic quantum mechanics. On the other hand, if we go beyond the one particle description and assume that a sufficiently strong enough electrostatic potential can produce particle-antiparticle pairs, there is no paradox. For example, the antiparticles are attracted towards the barrier leading to a negative charged current moving to the right (remember that the particles are chosen to carry positive charge so that antiparticles carry negative charge) which explains the negative transmission coefficient. On the other hand, the particles are reflected from the barrier and add to the totally reflected incident particles (which is already seen for intermediate strength potentials) to give a reflection coefficient that exceeds unity.

1.4Dirac equation

      As we have seen, relativistic equations seem to imply the presence of both positive as well as negative energy solutions and that quantum mechanically, we need both these solutions to describe a physical system. Furthermore, as we have seen, the Klein-Gordon equation is second order in the time derivatives and this leads to the definition of the probability density which is first order in the time derivative. Together with the negative energy solutions, this implies that the probability density can become negative which is inconsistent with the definition of a probability density. It is clear, therefore, that even if we cannot avoid the negative energy solutions, we can still possibly obtain a consistent probability density provided we have a relativistic equation which is first order in the time derivative just like the time dependent Schrödinger equation. The difference, of course, is that Lorentz invariance (or covariance under Lorentz transformations) would require space and time to be treated on an equal footing and, therefore, such an equation, if we can find it, must be first order in both space and time derivatives. Clearly, this can be done provided we have a linear relation between energy and momentum operators. Let us recall that the Einstein relation gives

      The positive square root of this gives

      which is far from a linear relation.

      Although the naive square root of the Einstein relation does not lead to a linear relation between the energy and the momentum variables, a matrix square root may, in fact, lead to such a relation. This is exactly what Dirac proposed. Let us, for example, write the Einstein relation as

      Let us consider this as a matrix relation (namely, an n × n identity matrix multiplying both sides). Let us further assume that there exist four linearly independent n × n matrices γ^µ, µ = 0, 1, 2, 3, which are space-time independent such that

      represents the matrix square root of p². If this is true, then, by definition, we have

      Here denotes the identity matrix (in the appropriate matrix space, in this case, n dimensional) and we have used the fact that the matrices, γ^µ, are constant to move them past the momentum operators. For the relation (1.78) to be true, it is clear that the matrices, γ^µ, have to satisfy the algebra (µ = 0, 1, 2, 3)

      Here the brackets with a subscript “+” stand for the anti-commutator of two quantities defined in (1.79)