Conserved Norm

Because of the second time derivative, the conserved norm is no longer f d3ril>*ip. To find the correct norm, consider the following two expressions:

^here a and b are the quantum numbers of any two solutions. Subtracting these two expressions gives d d

axp oxv- oxoxt1

where the arrow over the operators tells in which direction they operate. In particular, an arrow pointing to the right (—>) means that the operator operates all the way to the right, including operating on any wave functions which may eventually stand on the right. Similarly, the arrow pointing to the left («—) operates all the way to the left. [The expressions -ip^Hipa and are therefore identical.] The above expression can therefore be written as the four-divergence of a four-current j^,

dx>L

where

4.3 SOLUTIONS FOR FREE PARTICLES 95 and the double arrow is an obvious generalization of the single arrow notation:

If we integrate (4.10) over all space (i.e., over a volume L3 with periodic boundary conditions), the spatial divergence integrates to zero, and the volume integral of the time component is a constant. In general, for any vector field which is conserved, so that = dj°/dt + V • j = 0, and which satisfies periodic boundary conditions, we have

jl3 jl3

dt JL 3

For the case given in Eq. (4.10), this becomes

= constant

If a = b, this is the conserved norm for the state a, and it explicitly involves the potential (the time component of the vector part). This result, unusual from the point of view of conventional quantum mechanics, is a general feature of many manifestly covariant relativistic equations (especially two-body equations). Before discussing this norm further, we will obtain the solutions of the free particle equations.

4.3 SOLUTIONS FOR FREE PARTICLES

The solutions of the free particle KG equation (i.e., with U — 0) can be obtained by separation of variables. Using box normalization with periodic boundary conditions, the solutions are

where En = yjm2 is always positive and the superscript (±) refers to the sign of the energy in the exponential. As in the cases discussed in the previous chapters, solutions with both signs occur; solutions with the superscript (+) will be referred to as "positive" energy solutions (because the factor of —iEnt in the exponential corresponds to positive energy in nonrelativistic Schrodinger theory), and those with superscript (—) will be referred to as "negative" energy solutions. The periodic boundary conditions, imposed inside of a cube of length L on each side, require

27r kn =—(nx,ny,nz) nt = 0, ±1, ±2,... . (4-16)

The norms of the positive and negative energy solutions have different sign, i J tfr^'^V = ±2EnL3N26nm , (4.17)

where 6nm = (5„imiin,raiiin,m,- Using the zeroth component of the conserved current to define invariant scalar products, the different solutions are orthogonal,

If we choose N = (2EnL3) , then the two types of solutions, and will each be normalized and orthogonal:

y/2 EnL3

i{knr*Ent)

The positive energy solutions 4>(+s> have norm +1, and the negative energy solutions 4>have norm —1.

Because negative norm solutions exist, \ \ip\\2 cannot be a probability density, and historically this was regarded as a reason for rejecting the KG equation. This point of view is too narrow, but the existence of negative norm solutions is an indication that the quantum mechanics described by such an equation departs from the classical rules of quantum theory. One of these rules is that the states span a vector space with a norm which is positive definite, and this is certainly not the case for the KG equation. Later we will see that if the KG equation is used as the basis for a field theory (recall Prob. 1.3), then the states defined by the field theory will all have positive definite norm, and the negative energy states can be reinterpreted as positive energy states of antiparticles. Before developing the field theory, however, it is useful, and maybe even necessary, to study the properties of a quantum mechanics which is based on the use of the KG equation as the equation for single particle states (referred to as the first quantized form of the theory). This is the purpose of this chapter, and it is well to realize that even though the first quantized theory can be only partially successful, it is just as important in its own right as the study of the first quantized theory for the Dirac equation (which is taken up in the next chapter). The first quantized theories for both of these equations suffer from the same fundamental disease; they both have negative energy solutions which cannot be treated fully until they are reinterpreted as antiparticles, and this is only fully successful in the second quantized (field theoretic) form.

Keeping these comments in mind, we proceed with our study. If the KG equation is applied to the description of a charged particle, the norm will be interpreted as a charge density, with positive norm states describing + charges and negative norm states describing - charges. The conservation of charge then appears as a consequence of the invariance of the norm. If the particle has no electric charge but has some other quantum number (a generalized charge) which satisfies an additive conservation law, the norm can be interpreted as the density of this generalized charge. In either case, the existence of two states, one carrying positive charge and one carrying negative charge, is assumed.

Before we develop these ideas further, it is useful to look at a simple example which illustrates how the KG norm can be consistently interpreted as a charge density and how both particles and antiparticles are described by the equation.

4.4 PAIR CREATION FROM A HIGH COULOMB BARRIER •

Since the norm is conserved, we might suppose that if we start out at some initial time t0 with a superposition of states with only positive norm, these will evolve .t a later time t into a superposition of states with only positive norm, and that therefore in this case we could still interpret the wave function as a probability density. We shall show here that when interactions are present, states with negative norm can still appear, and hence they cannot be eliminated from consideration. Our discussion will also enable us to interpret the norm physically.

Consider the reflection of positively charged mesons (7r+ mesons for example) from a high Coulomb barrier. The KG equation (4.7) for this case is where V is zero to the left of the barrier and a constant to the right, as shown in Fig. 4.1 A. We seek a solution of the form e~xEt, corresponding, in region I, to a free particle with positive energy. We guess the solutions in regions I and II to be of the form

'Much of the material in this section is discussed in an interesting paper by Winter [Wi 59],

= Aei(pz~Et) + Be-'(pz+Et) 4>n(z, t) = C e«Pz-Etï + D e^Pz+Et>

To see how these solutions develop in time, we smear in E to make wave packets. Smearing around E — Eq we have, for ordinary solutions like those in region I, rE0+AE __-+A E .

JEO-AE J-AE

where v0 - po/EQ, and the envelope function is f(rj) = r dEe^= .

Note that these packets travel in the direction of p with the classical relativistic velocity v — p/E.

The packets in region II behave differently, however. We have

E0+AB dE¿W('V-W-m>z-Et) a ei{Poz-Eot) fAE dEe-iE('-^z+t) Eo-AE J-AE

where u0 =

Note that this packet propagates to the left; its group velocity is negative even though its phase velocity is positive. It travels with the classical relativistic velocity of a free particle with kinetic energy corresponding to eV - Eq. Since a 7r+ cannot have a positive kinetic energy in this region, the packet must be describing something else. If it were negatively charged, then it would see the potential barrier as a deep hole, and it could have positive kinetic energy. In that case it would have total energy fitotal = (eV - E) - eV = -E , as shown in Fig. 4.IB. A consistent picture of a particle of mass m and charge —e emerges; it is a tt~ meson!

To complete the description, we compute the coefficients for a state which is initially a pure ir+ traveling to the right toward the barrier. This means that any 7T~ meson must be produced by the interaction and will travel toward the right into region OL Hence the boundary condition is that C = 0, and it is convenient to choose D = 1. Then the continuity of the wave function and its derivative at z — 0 require

which gives

and the solution becomes (renormalizing so A = 1)

To interpret this solution, smear in E and assume that the coefficients are slowly varying functions of E which can be approximated by their value at the central energy E0. We then get

Using the fact that the envelope functions are non-zero only when their arguments are small, a moving picture of this state can be constructed as shown in Fig. 4.2.

Note that the norm of the state is a constant of the motion. At t —♦ — oo, only the incoming packet on the left-hand side exists, and if we take its norm to be one, then the norm of the reflected packet (traveling to the left in region I) is and, recalling Eq. (4.23), the norm of the "transmitted" packet traveling to the right in region II is where the ratio (eV — Eq)/Eq comes from the energy factor in (4.23) divided by a similar factor which appears in the norm of the states in region I (which must be divided out because the incoming state is normalized to unity), and the ratio uo/vQ is the effect of the fact that the z dependence of the envelope functions is scaled by the velocities in the two regions (see Fig. 4.2). Note that T < 0, corresponding to the negative charge of the particle in region I, so that even though R > 1,

as required by the conservation of the norm (the charge).

We interpret this result by saying that the incident pion not only scatters from the barrier but also stimulates the barrier to produce 7r+7r~ pairs, and the 7r" particles travel into the barrier, which is negative to them, while the 7r+ particles produced by the scattering join the scattered 7r+ particles. If the norm is interpreted as a charge density, the interpretation is consistent and makes good physical sense. Energy and charge are conserved.

4.5 TWO-COMPONENT FORM *

In order to display these two states explicitly, and to further develop our understanding of the KG equation, we will now discuss how the equation can be cast into a "two-component" form. This will help to understand the equation and to study its nonrelativistic limit.

Any second order differential equation can be transformed into two coupled first order differential equations. If this transformation is applied to the time dependence of the KG equation, we emerge with a set of coupled equations of the form where 0 is a vector in a complex two-dimensional space and H is a 2 x 2 matrix. We gain several advantages from this reduction. First, the equation is now first order in the time, so that the time dependence of the two-component wave function is uniquely determined by its initial value, in agreement with the rules of quantum mechanics. This means that the perturbation theory we developed in the preceding chapters, which implicitly assumed an equation which is first order in time, can be used with the two-component KG equation. Finally, study of the matrix structure of the two-component equation is good preparation for the study of the Dirac equation, which has a similar matrix structure. However, the two-component KG equation is no longer manifestly covariant, and for this reason it will be discarded after this chapter is concluded. When we encounter the KG theory again in Chapter 7, we will use the original version presented in the preceding sections.

The transformation to two-component form can be carried out by introducing ip and dip/dt as independent functions. However, instead of ip and dip/dt, it is helpful to take a more symmetric linear combination. Introduce two functions cp+ and 4>_ determined from ip and dip/dt by

'Much of the material in this section is drawn from the interesting review by Feshbach and Villars [FV 58],

'Much of the material in this section is drawn from the interesting review by Feshbach and Villars [FV 58],

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