Jo which is the property of the ¿-function, and hence

This proves the first of the relations (1.17). To prove the others, note that

A field theory may be quantized with either the CCR's (1.17) or the commutation relations (1.14) between the operators a and a*. As we have seen, these two methods are equivalent. Should either be regarded as more fundamental than the other? Many prefer to start from the CCR's because of their close connection with the fundamental relations (1.13), but in this book the relations (1.14) between the o's will be chosen as the starting point for quantizing new field theories. The reason for this choice is that the relations (1.14) are directly related to the oscillators which describe the independent dynamical degrees of freedom associated with the field, and therefore always have the same form, while the fields themselves sometimes include degrees of freedom which are not independent (such as the vector degrees of freedom of the electromagnetic field) and in these cases the *orm of the CCR's must be modified so that these dependent degrees of freedom are removed from the commutation relations. This will be apparent in the next chapter where the quantization of the electromagnetic field is discussed.


Next, we find the eigenstates of the Hamiltonian (1.16). The first step is to find the eigenstates of the operator known as the number operator. These are easy to find from the commutation relation for the a's.

Since M = a^a is Hermitian (from now on we suppress n), it has a complete set of orthonormal eigenstates. Denote these by |m). Then

This proves the first of the relations (1.17). To prove the others, note that

+ e-i(knz~k">z')+i(u>n~u'">)t [flt s «"•"-''» ii-n=o.

At this point we know only that m is real.

Now consider the state a^m). From the commutation relations (1.14) we have

Hence a^m) = C+|m + 1) , where C+ is a number to be determined. A similar argument gives a\m) = C-\m - 1) .

The numbers C+ and C_ can be determined from the norms


The axiomatic development of quantum mechanics requires that all quantum mechanical states lie in a Hilbert space with a positive definite norm. Hence we require that m > 0, or if m = 0, a|0) = 0 .

Furthermore, since m can be lowered by integers, all positive m must be integers; otherwise, we could generate negative values for m from positive values by lowering m repeatedly by one unit.

Hence, it is possible to choose phases (signs) so that (m > 0)

= Vm + 1 |m + 1)


= y/rn |m — 1)


This means that all the states can be generated from a "ground state" |0) (sometimes called the "vacuum") by successive operations of a*:

For a mechanical system like the string, these states |m) are referred to as phonon states, and if a = an, we will show that m can be interpreted as the number of phonons of energy o)„, where the quantum of energy carried by the phonon is associated with the entire system. This justifies calling N the number operator and suggests that the operators a and have the following interpretation:

* creates a phonon with frequency uin n destroys a phonon with frequency un

This description is further supported by the Hamiltonian (1.16) which now has a simple physical interpretation. If ajja„ is an operator which gives the number of phonons of frequency (energy) wn, then (1.16) expresses the total energy (if) as a sum of the energy of each phonon (un) times the number of phonons with that energy (a^an). The most general eigenstate of the Hamiltonian can therefore be written*:

Since all creation operators commute, these states are completely symmetric and satisfy Bose-Einstein statistics. Such states, with a definite number of phonons of various frequencies, are referred to as Fock states.

It is sometimes tempting to try to relate the particles associated with the field (the phonons) to the original mass points from which the string was constructed. However, there is no connection between these two kinds of particles. The phonons are associated with frequencies, or normal modes of the string, and hence are related to the motion of the string as a whole, its collective motion.

hey are localized in "frequency," or momentum space, while the particles in the string are localized in position space. Later we will see that there are also particles associated with abstract fields which have no connection with any mechanical system.


The quanta associated with a quantum field (the phonons in this example) really are physical particles which carry both momentum and energy. In the previous section we saw how the phonons carry energy. The Hamiltonian tells us that the total energy of a state with a definite number of phonons (a Fock state) is simply the sum of the energy carried by each of the phonons in the state. To complete the description of phonons as particles, we must show that they also carry momentum. This will be done in this section by first finding the momentum operator of the field and then showing that the total momentum of a Fock state is simply the vector sum of the momentum of each of the phonons in the state.

*Of course, if more than one state has the same energy (there is a degeneracy), the most general state will be a linear combination of all the states with that energy.

Hence the classical momentum density must be


We can turn (1.22) into a quantum mechanical operator by replacing the classical fields by their quantum mechanical operator equivalents. Since the field operators do not in general commute, the order of the terms in any product is important, and it is convenient to choose this order so that (in this example) the expectation value of the momentum of the ground state |0) is zero. To this end we define the normal ordered product of two field operators as follows:

<O|0i0a|O> = 4+)<P{2+) + <t>[~]4>{2+) +

Was this article helpful?

0 0

Post a comment