i tr (M*) = è tr ([&, + i^Titi] [4>a - i^Titt]) = = |<|>|2 ,

If the meson masses are physical, so that m2 > 0, this equation has only one solution, and the minimum is at |0| = 0 [see Fig. 13.2A], Recalling that


Fig. 13.2 The Hamiltonian H^ as a function of <j> for two signs of the mass term.

the requirement that |())| = 0 automatically implies that the value of all the classical fields in the minimum energy state (the vacuum) is zero:

4><r = <f>„ =0 (for the vacuum) . (13.100)

This is in agreement with our intuition that the vacuum should have no a or 7r particles present.

Consider now the interesting case when m2 < 0, so that -m2 > 0. This appears to correspond to choosing an unphysical meson mass, but as we shall soon see, this is not really the case. In this case, the H$ curve now has a minimum for a finite value of |(j)| = v, determined from the solution to Eq. (13.99):

This case is illustrated in Fig. 13.2B. The Lagrangian is still mathematically invariant under chiral transformations, but the vacuum no longer is. This is because |<j>| = v implies that at least one of the four fields (<t>„, must be non-zero, and hence, in general, we cannot treat all four components of (j) in a symmetric way. The choice m2 < 0 forces the vacuum to break the chiral symmetry spontaneously, and this is the origin of the term spontaneous symmetry breaking. We choose 4>c to be non-zero and introduce a new field which is s = 0a-v . (13.102)


The linear sigma model requires both a pion and a scalar (sigma) field in order to maintain chiral symmetry. The presence of the scalar field makes the model less than satisfactory for low energy applications, because, at low energy, scalar particles exist only as very broad resonances with masses in the neighborhood of 1 GeV, very much larger that the pion mass. It does not seem natural for such particles to play the fundamental role suggested by the linear sigma model, and in this section we will discuss how the sigma model can be reformulated so that it contains no sigma mesons.

The most obvious idea on how to eliminate the sigma meson is to let the a mass become infinite (since it is a free parameter); we would then hope that all contributions from a exchange forces would be vanishingly small. For example, the Lagrangian (13.105) predicts a a exchange force in NN scattering which goes like q2

However, this idea does not work because the sigma does not decouple from the pion. This is because the nns (or a) coupling also is proportional to ml, so the a exchange force between pions cannot be ignored:

Since the sigma couples to the pion, and the pion to the nucleón, it cannot be eliminated from the Lagrangian (13.105).

We can eliminate the sigma, however, if we are willing to consider a meson matrix (|) which is non-linear in the pion fields. This approach was developed by Weinberg in the late 1960's [We 68]. The idea is to exploit the fact that the length |<()| is invariant under chiral rotations, and hence if we set it equal to a constant (which is denoted /„■), we still preserve chiral symmetry, but <pa is no longer an independent field but is related to <j>v by fl = |(t>|2 = <t>l + cfi . (13.108)

With this constraint, we can construct a Lagrangian which depends on the pion field only, but the replacement of by a function of cp„ will give a meson matrix non-linear in the pion field. The non-linearities are quite severe; the Lagrangian will include all powers of the pion field.

Theories based on a non-linear Lagrangian of the type discussed in this section cannot be quantized using the methods we have discussed in the preceding chapters and require the techniques we will discuss in Chapter 14. Generally, these Lagrangians give simple reliable results when used in tree approximation, but the calculation of loops with such a Lagrangian leads to many infinities which can only be removed by introducing many renormalization constants. Recently a systematic method for absorbing these renormalization constants into undetermined parameters (known as chiral perturbation theory [DW 69]) has been developed. It can be used to calculate the interactions of very low energy pions and nucleons with considerable reliability.

Once the idea of a non-linear Lagrangian is accepted, we have great freedom, and a new pion field 7r can be chosen so as to give a convenient form for the Lagrangian. Using the notation r ■ 7T

7 2U

two forms which appear frequently in the literature are

where y2 = y y = y2. Both of these forms satisfy the constraint (13.108) and, up to second order in the pion field, are equal to

in agreement with Eq. (13.87) and the constraint (13.108) if we identify 0„. = ir. Hence the two fields </>,, and it differ only in higher order.

Starting from the linear sigma model Lagrangian (13.91), we consider the Lagrangian

Note that this Lagrangian is identical to (13.91) except that terms involving |<|)2|2 have been dropped because |<J>2|2 = /2 is a constant and is no longer a dynamical variable and cannot be used to construct the Lagrangian.

Next we simplify this Lagrangian by reducing the meson kinetic energy term. Note that

0 0

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