Ip2Pi f ddxiQlixiixry Qlfaixi x OT0ai0aO

P1P2

except that now there is only one term (the square of the Ax interaction term) which can account for the scattering, so the factor of \ remains. Also, since the initial state contains two creation operators, a{pa |0), and the final state two annihilation operators, (0|alp< a^ , we will need both of the annihilation operators contained in $(ii) and $(12) and both of the creation operators contained in $t(xi) and $t(i2) to "balance" the a's and a^'s in the initial and final states and give a non-zero result. Specifically, in place of (9.24) we need the following matrix element:

P1P2) = |aipiaip'iainaim<iiiaijaip1aip3|

which displays all the pairings which are possible when the particles are identical. Note that as the creation operator a\m is moved to the left, for example, a term 6mi arising from the commutator [alrn, a|j will appear, but all such terms can be ignored because they will eventually require the momenta in the initial and final states to be equal, and we are specifically excluding forward scattering from consideration. Substituting (9.85) into (9.84) and inserting the wave functions give four terms:

Using the fact that the propagator depends only on x\ - x2 and introducing the sum and difference variables R and r as we did in Eq. (9.26) allow us to extract the M-matrix as we did before. The four terms collapse in two different terms, each multiplied by 2, giving

The first term is the same as the result we obtained when the particles were not identical, and the second term is obtained from the first by symmetrizing the final state (or the initial state, but we will adopt the convention that the final state is to be symmetrized). This illustrates a new Feynman rule, Rule 4 (as they are numbered in Appendix B):

Rule 4: symmetrize between identical bosons in the final state.

Let's briefly review the calculation and see where the two terms given in (9.87) came from. For this purpose it is best to think of the interaction unfolding in space-time, where, according to (9.84), one interaction takes place at point Xi and the other at x2, where the points x\ and x2 will be defined to be the points of arrival of the incoming particles with momentum p\ and p2, respectively. The propagator describes the propagation of the neutral particle between these two interaction points. Now, because the interactions at xi and x2 involve the same particles, the particle in the final state with momentum p[ could emerge equally well from either of these two points, as illustrated in Fig. 9.6, and this explains why both of the diagrams shown in Fig. 9.6 must be present.

^ei(p'l-Pl) xi+i(p'2-pi) x2 g«(pi-pi)-l2+«(pi-p3) H _|_g«(pi-pa)-si+»(j»2-pi),a:3 + e»(pi-p3)-*a+t(pi-pi)-xi x

^ei(p'l-Pl) xi+i(p'2-pi) x2 g«(pi-pi)-l2+«(pi-p3) H _|_g«(pi-pa)-si+»(j»2-pi),a:3 + e»(pi-p3)-*a+t(pi-pi)-xi

Pi

Direct (A)

Pi y

Exchange (B)

Annihilation

Fig. 9.7 For the symmetric </>3 theory, these three diagrams contribute to second order scattering. Note the (new) annihilation diagram C.

which could balance the initial and final operators.] Furthermore, the (¡>2(x2) term can pair with the initial state in two ways, or with the final state in two ways, and counting up all of these possibilities gives 2x2 + 2x2 = 8 possible terms of this kind, which cancels the remaining factor of 8 in Eq. (9.89). A similar factor of 8 emerges for each of the two possible diagrams shown in Fig. 9.6, so the final result for the .M-matrix for this example is

which corresponds to the three diagrams shown in Fig. 9.7.

The new process, Fig. 9.7C, describes the virtual combination of the two incoming particles into a single off-shell particle, followed by its disintegration into the original two particles. While this process is physically very different from the other two, its existence and mathematical form follows from the same Feynman rules we have already obtained, including the requirement that the momentum of the propagating particle be fixed by four-momentum conservation. This example shows that we must understand what processes are possible before we can draw all of the allowed Feynman diagrams.

The reason for inserting the additional factor of 1/3! into the interaction Hamiltonian for the pure <t>3 theory is now clear. This factor was eventually canceled by the many combinations of identical terms which led to the same final physical process. However, when these particles appear inside internal loops (see Chapter 11), this factor is not canceled completely, and loop diagrams involving such particles carry with them special suppression factors, called symmetry factors. These will be discussed in Chapters 11 and 15 and need not concern us now.

With the experience we have acquired, are now ready to study a more realistic problem.

9.8 PION-NUCLEON INTERACTIONS AND ISOSPIN

We now consider the case of two nucleons (neutron and proton) interacting with neutral and charged pseudoscalar fields (pions). The Lagrangian density is composed of four terms

where Cn is the sum of two non-interacting spin | Lagrangian densities for the proton, ipp, and the neutron, ipn, with mn — mp — m^r for simplicity [recall Eq. (7.46)]. The £+ is the Lagrangian density for a non-interacting charged pseudoscalar field, identical to the charged scalar field, Eq. (7.32) [or its operator form Eq. (7.62)]. (The difference between scalar and pseudoscalar fields does not show up until we discuss £int-) This field will be denoted and describes charged 7r+ pions. Next, Co is the Lagrangian density for a non-interacting neutral pseudoscalar field, identical to Co given in Eq. (9.3). This field is a self-conjugate (Hermitian) field, denoted by (p0, and describes neutral pions, w0 and 7f0, which are identical. For simplicity, we also assume the mass of the charged and neutral pions are equal to m,.

Hence, the first three Lagrangian densities describe the following four non-interacting fields:

tjjp : proton p mass — m,N + charge ipn '■ neutron n mass = m^ 0 charge

4>+ : 7r+; antiparticle n+ = tt~ mass — m,r + charge

4>o : 7r°; identical to its antiparticle mass = rnn 0 charge.

These four fields interact through £int, which has a generic 4>3 structure. We take

- ¿5+:'0n75Vy </>f+ - ig+--4'p'r5ipn-4>+ , (9-92)

where gp, gn, and g+ are three real constants. Note that:

(i) The interaction conserves charge. The neutral field $o, which describes 7r° mesons, couples to the proton and the neutron with independent coupling constants gp and gn. The p —> n + n+ and n+ + n —> p charge conserving interactions are described by the two terms with coupling constant g+.

(ii) £,nt is Hermitian if gp, gn, and g+ are all real. The gp and gn terms are individually Hermitian because 4>o = 0q. The two g+ terms must have the same coupling constant to preserve hermiticity. To prove this, recall that 7o = 1 and 75t = j5, so that

The factor of i therefore makes these terms Hermitian if g+ is real.

Isospin and SU(2) Symmetry

Now suppose that the neutron and proton are indistinguishable under irN interactions. This means that they can be transformed into each other without altering the interaction. Mathematically, it means that p and n are components of a two-vector

in some abstract two-dimensional space and that the interaction is invariant under all unitary transformations in this space. The group of all unitary transformations separates into multiplication by a common phase, the U( 1) group, and the remaining group of unitary transformations with unit determinant (a condition which fixes the overall phase of the transformation), the familiar SU(2) group. Each of these groups can be considered separately. The abstract two dimensional space is referred to as isospin space, and the SU(2) transformations in this space are called isospin transformations. The mathematics of isospin transformations is identical to ordinary spin | transformations, but the space is a different, abstract spin space; hence the name isospin.

Denoting the (now) two-component nucleon field by

the interaction Lagrangian density can be written

Ant = , where $ is now a 2 x 2 matrix with the form

$ = ffpK1 +73)00 + - T3)(f>0 + g+%(Ti + iT2)<t>+ + g+ \ (n - ir2)<t>\

= \{9p + 9n)<Po + \(gP - 9n)T30O + 0+Ti £(</>+ + <j>\) + ig+T2\(<f>

It is convenient at this point to introduce two self-conjugate fields (pi and <t>2 defined by

[Mr,t),Mr',t)] = [<h(r, t), cj>2{r', t)} = [^(r, i), 02(r', i)] = 0 , and that the CCR's for 4>i (and similarly for <p2) become

[tti (r, t), (r', i)] = I { [tt*. (r, t), J>\ (r',t)}+ [tt+ (r, t), cj>+ (r', t)}}

as expected. Hence, <t>i and <t>2 are independent self-conjugate fields similar to <po = fa, and the field matrix $ can be written

$ = \{gP + gn)4>3 + \{9p - 9n)T3<t>3 + + T2^2) •

Note that the free pion Lagrangian

£x = £+ + £0 = - m2<^+ + i {d^od^o ~ m2</>£} (9.101)

can be simplified if we use

and so that ifc)

= I {d^Wfa + d^fod^fo + - m2 (02 + cfrl + <t>l)}

where <j> is a vector in an abstract three-dimensional space (corresponding to a state with isospin 1),

<$> ■ <t> = <t>\ + <t>\ + <¡>1

Because depends only on the square of the length of the vector it is clear that it is invariant under rotations of the 4> field in the isospin (one) space.

Now we demand that the full Lagrangian be invariant under isospin rotations, i.e., under transformations of

Tpp and

We have just seen that Cn + is invariant; it only remains to see what requirements must be imposed if £int is also to be invariant. Invariance of £/v and L^ requires that the transformation of ip be unitary in two dimensions and the transformation of (f> be an orthogonal transformation (rotation) in three dimensions. The transformations correspond to different representations of the same rotation.

Using our experience with rotations and angular momentum, we therefore expect (recall Sec. 5.8)

<j>' = e~i6'L <{> = [1 - iOiL'] (t> , where 6X are three continuous parameters describing the transformation and and Ll are the generators, defined by

IS7!) 5rj itijkhTk

These are the familiar algebraic properties of SU(2) symmetry.

The requirement that the interaction Lagrangian be invariant to lowest order implies

The term proportional to 6i must be zero, which requires that

2U,3}T} 2te¿i_jTj 2iti2jTj

0 0

Post a comment