C Q wn 6n 6n

to obtain finally

This is a straightforward generalization of the result for the string. Now there are number operators for photons with momenta in all three spatial directions and with two polarization states.

We leave it as an exercise, Prob. 2.1, to show that the momentum operator

This expression shows that the total momentum of the field is a vector sum of the momenta of each photon, as expected from our study of the string.


In the final section of this chapter we show that the particles which emerge from the quantization of the EM field (the photons) have spin one. Spin can be regarded as an internal degree of freedom of the quanta which is closely connected to the structure of the field from which they emerge. In particular, the quantization of vector fields always gives rise to quanta with spin one, while the quantization of scalar fields gives quanta with spin zero.

To obtain these results, it is necessary to discuss the behavior of the field under rotations. Just as the energy of the quanta is displayed by the Hamiltonian (the generator of time translation) and the momenta are displayed by the momentum operator (the generator of space translations), so it is that the spin will emerge from a discussion of the angular momentum operator, the generator of rotations. This section therefore begins with a brief discussion of the rotations of vector fields. A deeper discussion of these topics is postponed until Chapter 8.


When transforming vector fields, which are continuous functions of space and time, both the components of the vector and the arguments of the function must be transformed. For example, a scalar function under rotation transforms in a non-trivial way, as illustrated in Fig. 2.2. In this book, all transformations will be interpreted as active transformations, i.e., they transform the state functions, leaving the coordinate system fixed. From examination of the figure, we see that, under an active rotation R, the transformed function <pR(r,t), where r is a shorthand notation for the three spatial coordinates (x, y, z), is related to the untransformed function <p(r, t) by is n,a n,a

Sri = -60e%jtre = -i66 (/¿)jí rl

Similarly, the change in a vector field A under the same infinitesimal transformation, using (2.52), is

= Aj(r, t) + i66i \—IjeAt(r, t) + I^mrmdeA3 {r, t)

Introducing the familiar orbital angular momentum operator, Ll = —i f-im?rmdt, this can be written

Now we are ready to apply these ideas to the EM field.

Angular Momentum Operator

The angular momentum operator for the EM field is n1 = J d3r[rx:{E_L xB):f

This can be reduced to a more tractable form by expanding out the double cross product, fi* = J d3reijer:ieekrnEk1ernabdaAb = J d3reljer3 (6la6kb - 6ka6tb) Ek±daAb

= j d3rei]er>{EtldeAb-E±-VAe} , and simplifying the last term by integrating by parts and recalling that Ej_ is transverse,

- J d3r eijtrjE± ■ VAe = J d3r elje (rjV ■ExAe + 6jbEb±Ae)

d rtibiE±A

Substituting for Ex and using the orbital angular momentum operator L V) give a more compact form for fi:


The extra term in both commutators is required to make each expression consistent with the Coulomb gauge.

From the vector nature of the field we expect it to have spin one, but our task is to see how this comes about naturally in the particle picture. To do this we will express fispin in terms of the a's and a^'s, using real polarization vectors for simplicity,

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