Quantum Field Theory II

UNIT 1 Interactions

When you include interactions, the plot thickens. Until now, we have only considered free fields, described by a Hamiltonian H = H 0 , which we were able to diagonalize (find all eigenvalues and corresponding eigenstates). But that is no physics! It is time to do some serious physics. When you include interactions, the Hamiltonian is modified to

where H I describes interactions. It is no longer possible to solve the eigenvalue problem for H, except in very few special cases (mostly in two dimensions). The only thing we can do is perturbation theory, assuming H I is small.

This does not answer deep questions, such as "what is a proton?", but it provides a method for very accurate calculations (e.g., the magnetic moment of the electron is known to about 10 significant digits both theoretically and experimentally, and they agree with each other!). Experimental results are obtained primarily through scattering. To compare with them, we need to develop scattering theory. It turns out that this type of processes (scattering) holds all the information of a quantum field theory.

Scattering theory

Recall in quantum mechanics,

where H 0 is the kinetic energy and V the potential. If V has compact support, then at times t → ±∞, H = H 0 , because V = 0. Thus, we may define incoming and outgoing states, |in (at t → −∞) and |out (at t → +∞), respectively, which are eigenstates of H. We shall attempt to do the same in quantum field theory. Consider a state |in . It evolves in time as e −iHt |in (1.1.2)

As t → −∞, H → H 0 (no interactions), so asymptotically, our state approaches a state in the Hilbert space constructed from H 0 . Call that state |ψ . It evolves in time as e −iH0t |ψ (1.1.3)

Then the statement that this is asymptotic to the state |in amounts to Notice that if H and H 0 commute ([H, H 0 ] = 0), then we may write U (t) = e −iHI t , where we used (1.0.1). But this rarely happens. Thus, in general, U (t) is a very complicated object. Similarly, in the infinite future, we may map

where |out (|χ ) is in the Hilbert space of H (H 0 ). We wish to calculate the amplitude of the process |in −→ |out (1.1.9)

i.e., out|in = lim t±→±∞ χ|U (t + )U † (t − )|ψ = χ|S|ψ (1.1.10)

where

is the S-matrix. S maps in-asymptotes to out-asymptotes (these two Hilbert spaces may, in general, be distinct, but not here). In fact, S contains all the information of the quantum field theory. It is an important object to study. Let us bring it into a more convenient form. To this end, we need to establish some properties of U (t) first. We have dU (t) dt = iH 0 e iH0t e −iHt + e iH0t (−iH)e −iHt = −ie iH0t (H − H 0 )e −iHt = −ie iH0t H I e −iHt = −ie iH0t H I e −iH0t e iH0t e −iHt = −iH I (t)U (t) (1.1.12)

Thus U (t) is a true evolution operator with (time-dependent) Hamiltonian H I (t). The latter is obtained from H I by evolving with H 0 . The first-order ODE (1.1.12) together with the initial condition U (0) = I uniquely determine U (t).

To solve (1.1.12), convert it into an integral equation,

and then use iteration,

1.1 Scattering theory

We obtain

etc.. This may be brought into a more compact form. Look at the second-order term. In it, t ≥ t 1 ≥ t 2 ≥ 0. The two-dimensional integral is over a triangle in the (t 1 , t 2 ) plane, which is half of the square 0 ≤ t 1 , t 2 ≤ t. The other half gives the same answer, but with t 1 and t 2 interchanged. It follows that the two-dimensional integral can be written in terms of a time-ordered product as

For the nth term, we similarly obtain

It follows that

which is Dyson's formula. Next, define the generalized evolution operator

It is easily seen from its definition that it has the following properties

The S-matrix can be written as

Various important properties of the S-matrix follow.

• The S-matrix is unitary,

This may look like a trivial statement (direct consequence of the unitarity of U (t + , t − )), but we need to take limits t ± → ±∞, and that may spoil unitarity, unless the range of S is the entire Hilbert space. The latter is a consequence of the requirement that probability be conserved: everything that comes in should go out. There are cases where this is not true and particles may be trapped into bound states. The S-matrix has a way to address these issues. For the most part, we shall assume that no trapping occurs and the S-matrix is unitary.

• S commutes with the free Hamiltonian H 0 ,

i.e., a scattering process conserves unperturbed energy.

Proof.

e iǫH0 Se −iǫH0 = lim t±→±∞ e iǫH0 e it−H0 e −it−H e it+H e −it+H0 e −iǫH0 = lim t±→±∞ e iH0(t−+ǫ) e −iH(t−+ǫ) e iH(t++ǫ) e iH0(t−−ǫ) = lim

But as t ± → ±∞, adding an ǫ makes no difference, so

Expanding in ǫ, at first order we obtain the desired result (1.1.23).

• S is Lorentz-invariant.

Proof. The Lagrangian density (a Lorentz-invariant quantity) can be split as

If L I contains no time derivatives, then

Therefore, using (1.1.21),

which is manifestly Lorentz-invariant.

The integral is over the entire space-time. It should be noted that time-ordering does not spoil Lorentz invariance. Indeed, time-ordering is frame-dependent only for spacelike separations, and fields commute at spacelike separations (causality).

• Corollary: By Lorentz invariance, it follows from (1.1.23) that S commutes with the unperturbed four-momentum,

This implies momentum conservation in scattering.

Indeed, consider an incoming (outgoing) state of free particles with momenta

(notation as in eqs. (1.1.7) and (1.1.8)). These are eigenstates of unperturbed momentum

i.e., momentum is conserved.

1.2 Wick's theorem

Wick's theorem

Notice that in the expression for the S-matrix (1.1.27), time evolution is accomplished with the unperturbed Hamiltonian H 0 , and not with H. Thus the fields in L I are asymptotic (free) fields, and can be treated in the same way we have already discussed. Let us concentrate on the scalar field φ.

The field φ in S does not satisfy the full equation of motion, which is usually non-linear. Instead, it obeys the Klein-Gordon equation. To distinguish it from the true φ, we shall call it φ I (in interaction picture). Since we shall be always talking about φ I , we might as well drop the subscript I. Hopefully the reader will not get too confused. Therefore, we may perform the familiar expansion

and define the one-particle state

where |0 is the asymptotic vacuum (not the true vacuum of the whole system).

When calculating vacuum expectation values 0|φ(x)φ(y) • • • |0 , it is convenient to normal-order. On the other hand, S contains time-ordered products. We need to relate the two types of orderings. To this end, define the contraction of two fields as the difference between the two types of ordering,

The contraction is a number (not an operator), because each time we commute a and a † , we get a number. To calculate it, we shall take the vacuum expectation value of both sides of (1.2.3). We obtain

The time-ordered product is the Feynman propagator,

The normal-ordered product contains terms of the form a † a † , a † a and aa, all of which have vanishing vacuum expectation values. Therefore, the contraction is the Feynman propagator,

This result can be generalized to a product of n fields. To simplify the notation, denote

Wick's theorem states that

For n = 1, Wick's theorem is the trivial statement

For n = 2, it was proved above.

For n = 3, we need to show

To show this, first arrange the indices 1, 2, 3 so that times are order as

First, let us try to normal-order φ 3 . To this end, split

where φ 3+ (φ 3− ) contains all the creation (annihilation) operators. Then

The commutators are numbers. We have

and similarly for [φ 2 , φ 3+ ]. We deduce

together with (1.2.9) to arrive at the desired result (1.2.8).

Now use our result for T 3 ,

where the dots represent terms that are obtained by permutations of the indices (123).

We still need to normal order the fields in the last line. Using our result for T 2 , we arrive at the desired result for T 4 ,

The above argument can be generalized to arbitrary n by induction (assume Wick's theorem holds for all n < N ; show it holds for N ).

Proof. By momentum conservation and using p 2 i = m 2 i (i = 1, 2, 3),

At O(λ 2 ), we have

We need to re-express this in terms of normal-ordered products. This introduces contractions in the above expression. Here are the various possibilities.

• 0 contractions

Same rules as before, but be sure to normal order! Topologically, this is a disconnected diagram. If we calculate a matrix element of S, we will get 2 δ-functions conserving momentum (of 6 particles).

• 1 contraction. Denote the contraction (propagator) φ(x)φ(y) by

x r ry These are not Feynman diagrams (yet).

• 2 contractions. There are 3 possibilities.

The first diagram is proportional to φ 2 (x)φ 2 (y) × φ 3 (x)φ 3 (y), etc.

• 3 contractions. Only 1 possibility.

In general, if a diagram has a symmetry (a permutation of its vertices that does not give rise to a new contraction), we need to divide by the corresponding symmetry factor. All of the diagrams above have a symmetry: just interchange the two vertices. This explains the factor of 1 2 in the expression (1.2.34) for S (2) . At higher orders, this symmetry need not be present. For example, consider the contribution to

The vertices are distinguishable, so there is no symmetry. The symmetry factor we ought to divide by is 1.

To see this in a little more detail, recall that there is a factor of 1 3! in the definition of S (3) . On the other hand, there are 3! ways of doing the two contractions, here is one:

Thus, we obtain a factor of 1 3! × 3! = 1.

EXAMPLE 3: A non-linear quantum field theory of a single field

Suppose the 3 fields are identical, φ 1 = φ 2 = φ 3 = φ, and let the interaction Lagrangian density be

where we re-defined the coupling constant λ in order to expose a factor of 1 3! for convenience. This time, the symmetry factor is the number of permutations of vertices times the number of permutations of lines.

• ❅ ❅ has symmetry factor 3! (number of permutations of (123)).

• ❅ ❅ ❅ ❅ has symmetry factor 2 (vertices) × 2 2 (two pairs of external lines) = 8.

• ❅ ❅ ❅ ❅ has symmetry factor 3! (vertices) × 2 3 (three pairs of external lines) = 48.

• ♥ has symmetry factor 2 (vertices) × 2 (internal lines) = 4.

A general graph is disconnected and consists of lower-order connected graphs. Let S (graph) be the contribution of an individual graph to the S-matrix. It can be written as 1, 2, . . . ) is the contribution to the S-matrix of the corresponding connected subgraph. Notice that we had to divide by symmetry factors n i ! (i = 1, 2, . . . ), because we have n i identical connected sub-graphs. Also, the product has to be symmetrized, because sub-graphs represent operators which do not necessarily commute with each other. 3 The above expression includes connected graphs, if we set n 1 = 1, n i = 0 (i ≥ 2). The entire S-matrix is the sum over all possibilities,

This looks hideous, but can, in fact, be neatly organized, because the sums over n 1 , n 2 , . . . , are independent of each other. Thus S can be written as an infinite symmetrized product, (c) (1.2.38) where at the end there is no longer any need to symmetrize, and S (c) is the connected part of the S-matrix, S (c) = S The two factors can be separated because they commute with each other. The first factor is a number, in fact S (c) 2 = (−i) 2 α = −α. The second factor is a (normal-ordered!) operator identical to the expression (1.2.15) derived earlier.

EXAMPLE 1: External source

The simplest interaction we can have is of the form

where J(x) is a given function (source) representing some fixed distribution of "matter" 1 The Klein-Gordon gets modified to

which may be solved exactly with the aid of the Feynman propagator. We obtain

The S-matrix is

where

It can also be written as

Of course, all S k are the same operator, but before we integrate over time (the x 0 k s), we need to remember to time-order. Since S k is linear in φ, Wick's theorem holds for the S k s. This allows us to turn time-ordering to normal-ordering,

Since all S k s are the same object, all permutations give the same result; we just need to count them. With one contraction, the number of permutations is

With two contractions, the number of permutations is the number of ways of choosing two pairs from n objects,

where we divided by 2 because of over-counting (identical pairs). Generalizing to p contractions, the number is

The sum of terms with p contractions in (1.2.14) is

Now let us collect all the terms with p contractions in S. Such terms exist in all S (n) with n ≥ 2p.

We obtain the following contribution to S:

Summing over p, we obtain a simple expression for the S-matrix,

A p = e − 1 2 (S1S2) : e −iS3 : = e −α/2 :

where the coefficient e −α/2 is a c-number, and

We can now study the physical properties of the system. Evidently, the source creates particles. 2 To see this, calculate the probability amplitude that the true vacuum |0 in will evolve to an n-particle state | p 1 . . . p n out . We have

where the states on the right-hand side are free particle states.

• n = 0. We have in 0|0 out = 0|S|0 = e −α/2 so the probability that no particle will be created is

To show that this is an honest probability, i.e., P (0 → 0) ≤ 1, we need to show that β ≥ 0. Using (1.2.4), we obtain

in terms of the Fourier transform of J,

Since J(x) is real, we have J(−k) = J * (k), therefore

we deduce

Integrating over k 0 , we obtain two identical contributions at k 0 = ±ω k , and so

• n = 1. In this case only the first-order term in the expansion of the normal-ordered exponential in (1.2.15) contributes. We obtain

Using p 1 |φ(x)|0 = e ip1•x , which we showed a while ago, we deduce

The probability that a single particle will be created from the vacuum is

For a general (given) n ≥ 1, the nth-order term in the expansion of the normal-ordered exponential in (1.2.15) contributes. We obtain

Only the term with n creation operators in :

contributes. By commuting all creation operators through to the left, it is easy to see that

The probability that n particles will be created from the vacuum is

Notice that we divided by n! in the expression for the probability, because these are identical particles. Thus, we have a Poisson distribution. Notice that

Probability is conserved and the S-matrix is unitary. The expected number of emitted quanta (photons) is

If the source is localized (J(x) ∼ δ 4 (x)), then the Fourier transform is J(k) = 1, so β → ∞. The divergence comes from large k (ultraviolet (UV) divergence). Thus a source can never be perfectly localized, and no experiment can measure φ(x) precisely, in accordance with Heisenberg's uncertainty principle.

EXAMPLE 2: Static external source

As a special case, consider a static source. Well, it will not be completely static, because it has to be switched on at a certain time and then switched off again, after a long time. So let

where ρ( x) represents the static ("charge") distribution, and f (t) is a smooth function of time of compact support, so that f (t) = 0 for t > |T |, say, and f (t) = 1 for t < |T |, with a smooth transition from 1 to 0 near t = ±T . We shall assume that interactions are turned on and off adiabatically. If the cutoff time T is very large, then the Fourier transform of f (t),

will have narrow support (Riemann-Lebesgue Lemma). Without the smooth transition at t = ±T , we have

It does not have compact support, but its width is ∆ω ∼ 1/T (i.e., ∆ω∆t ∼ 1). As T → ∞, f (ω) → 2πδ(ω), which has compact support (the point ω = 0), and the frequency can be measured with infinite accuracy. If we smoothen out the above f (t), then it will have compact support of width ∼ ∆ω. Let us arrange things (choose a large enough cutoff time T ) so that f (ω) has support within the interval |ω| < m, i.e., f (ω) = 0 , ω ≥ m (1.2.23)

To find the probability of particle production (radiation), notice that

In the expression for β (1.2.18), ω = ω k = k 2 + m 2 ≥ m, therefore, ω is outside the support of f , by our assumption (1.2.23), and f (ω k ) = 0. It follows that β = 0, and no particles can be created. This makes physical sense: no radiation is emitted by static charges. Thus the S-matrix is pretty boring, but we may still talk about the vacuum-to-vacuum transition amplitude 0|S|0 = e −α/2

where α is purely imaginary (we just showed that its real part β = 0). The probability is |e −α | = 1, as expected, since this is the only possible process, but what is the physical meaning of the amplitude?

To answer this question, use (1.1.6) and (1.1.11) to write the amplitude as e −α/2 = lim

where we used H 0 |0 = 0. For the true vacuum of our system, we have

where E 0 is the true ground state energy. From the adiabatic theorem in quantum mechanics, we know that if interactions are turned on very slowly, the state |0 (which is the ground state in the absence of interactions) evolves into the vacuum state |0 phys without getting excited. It follows that e −α/2 = phys 0|e −iH(

This is true for large cutoff time T , more precisely, as T → ∞. We deduce

Thus the S-matrix contains information about the ground state energy of the system. For an explicit expression, use (1.2.16) together with (1.2.4). The integrals over times x 0 and y 0 are

Evidently, in the limit T → ∞, this expression has support consisting of the single point k 0 = 0. It follows that for large T , it is approximately

The constant C is found by integrating both sides over k 0 . We obtain

It follows that for large T ,

which is the energy of two static ("charge") distributions interacting via the potential

This is a Yukawa potential. As m → 0, it turns into the Coulomb potential.

EXAMPLE 3: A non-linear quantum field theory

Consider three distinct Klein Gordon fields φ 1 (i = 1, 2, 3) with interaction Lagrangian density

where λ is a coupling constant. The S-matrix can be expanded in powers of λ,

The nth term is proportional to λ n , and

We shall represent S (1) by a diagram thusly:

Each line is a field. Lines that meet at a vertex have common argument which is integrated over and multiplied by −iλ. S (1) contributes to the process

where p µ i is the momentum of the particle of type i (i = 1, 2, 3). The probability amplitude for this process is

where the dots represent contributions of higher order in λ. Using 0|φ(x)|p = e −ip•x thrice, we obtain

Therefore momentum is conserved, as expected (see discussion following eq. (1.1.28)). For the general process |i → |f , we define the scattering amplitude

where

is the total momentum of the initial (final) state |i (|f ). Notice that we subtracted the amplitude representing no scattering, f |i . Earlier, for |i = |1, p 1 , |f = |2, p 2 ; 3, p 3 , we found

Also, momentum conservation implies that the decay (1.2.30) is only allowed (to all orders in λ!) if

(1.2.33)

Feynman diagrams

So far, we have been developing tools to calculate the S-matrix, which is an operator. Now, we shall apply these rules to actual scattering processes and find the corresponding rules due to Feynman.

Continuing to work with the interaction Lagrangian density (1.2.35), which is physically useless, but good enough for the moment, let us consider the 2 → 2 interaction of particles with initial momenta k µ 1 , k µ 2 and final momenta k µ 3 , k µ 4 . The amplitude is

where the dots represent terms of higher order in λ. Thus, the lowest-order contribution is O(λ 2 ).

Recall that there are various contributions, with 0, 1, 2, or 3 contractions. Since the process of interest involves 4 particles, only the part of S (2) with 4 external legs will contribute (the one with 1 contraction, of symmetry factor 2 3 = 8). We need to match the momenta with these 4 external lines.

There are 3 possibilities:

There are 8 ways to assign momenta in each of the 3 cases, so the overall symmetry factor of a diagram including momenta (i.e., a Feynman diagram) is 8 × 1 8 = 1. This symmetry factor may also be deduced by looking at the Feynman diagram itself: it is the number of permutations of internal lines and vertices (leaving external lines unchanged).

• EXAMPLE: The diagram contributing to

has symmetry factor 2 (for the internal lines). Recall that the corresponding S-matrix diagram has symmetry factor 2 × 2 = 4, since there are 2 ways of assigning momenta k 1 and k 2 , we obtain 2 × 1 4 = 1 2 , which is the correct symmetry factor of the Feynman diagram.

Let us now compute the first diagram. It is

I omitted a symmetry factor anticipating that it will eventually be 1. Using φ − (x)| p = e −ip•x |0 , the above expression becomes

Notice that we have a factor e +ip•x (e −ip•x ) for each outgoing (incoming) momentum. Introducing the Fourier transform of the Feynman propagator, we obtain

which yields 2 δ-functions, δ 4 (k 3 + k 4 − k) and δ 4 (k − k 1 − k 2 ), allowing us to draw the diagram

where the intermediate line may be thought of as representing a particle of four-momentum k µ . Momentum is conserved at each vertex, and consequently, overall,

(1.3.5)

as expected on general grounds. The intermediate "particle" is not real. Indeed consider its decay to the two outgoing particles. For this to happen, we need m 1 ≥ m 2 + m 3 , i.e., m ≥ 2m, which is not satisfied, unless m = 0. The latter is also impossible for kinematical reasons. We say that the intermediate particle is "virtual" and its existence is allowed by the Heisenberg Uncertainty Principle (k 2 = m 2 is OK in quantum mechanics). After doing all the integrals, the diagram becomes

The other 2 diagrams are similarly obtained. We deduce the scattering amplitude

Notice that it is a Lorentz-invariant expression, as it ought to be. Generalizing to more complex diagrams is straightforward. One obtains the following general rules.

FEYNMAN RULES (in momentum space)

• To each external line of a diagram assign an incoming or outgoing momentum.

• To each internal line assign the most general momentum consistent with momentum conservation at each vertex.

• Include a factor (−iλ) for each vertex.

• Include a factor i k 2 −m 2 +iǫ for each internal line of momentum k µ .

• Integrate with

2π) 4 for each undetermined momentum. 4 • Divide by the appropriate symmetry factor. These rules yield the scattering amplitude iA after removing the overall factor (2π) 4 δ 4 (k f − k i ). EXAMPLE: Consider the propagation of a single particle of momentum k µ

There is 1 arbitrary momentum, and the symmetry factor is 1 2 , so

We managed to derive an expression for this amplitude without much effort, but alas, it is an infinite integral. We will come back to this later. GRAPH TOPOLOGY: The topology of a graph is related to the number of undetermined momenta, so it is important when one tries to determine the complexity of the integral(s), as well as the order in the quantum mechanical expansion of matrix elements of the S-matrix. EXAMPLES

has no undetermined momenta.

• ♥ has one undetermined momentum.

In general, consider a connected graph with V vertices and I internal lines.

• Each vertex gives a δ-function conserving momentum. One of them is the overall factor (2π) 4 δ 4 (k f − k i ) (for connected graphs), so we have V − 1 δ-functions constraining momenta.

• Each internal line provides an arbitrary momentum. Therefore, we have L integrals to do, where L = I − (V − 1) = I − V + 1 (1.3.9)

The following theorem, relating the integrals to the topology of the graph, holds.

Theorem. L is the number of loops in the graph.

Proof. The proof is by induction on L. When L = 0, we have a tree graph and I = V − 1 (because each time you add a vertex, you add an internal line, so V − I =const. in a tree; if I = 0, then obviously V = 1, therefore V − I = 1).

To add a loop, join two external lines. This increases I → I + 1 and V is unchanged. So if we start with I − V = L − 1, we end up with (I + 1) − V = L = (L + 1) − 1, which completes the inductive step.

PHYSICAL CONSIDERATIONS. Let us go back to the 2 → 2 scattering process. In the center of mass frame, k 1 = − k 2 , k 3 = − k 4 , E 1 + E 2 = E 3 + E 4 therefore, E 1 = E 2 = E 3 = E 4 = E , | k

showing that we have an elastic collision. E is the beam energy in an accelerator. The total centerof-mass energy (a Lorentz-invariant quantity) is E cm = 2E, because E 2 cm = (k 1 + k 2 ) 2 = (E 1 + E 2 ) 2 = 3E 2

It is also convenient to define the momentum transfers

Their norms are Lorentz-invariant quantities, and found to be q 2 = (k 1 − k 3 ) 2 = 2k 2 (1 + cos θ) , q ′2 = (k 1 − k 4 ) 2 = 2k 2 (1 − cos θ)

where θ is the angle between the vectors k 1 and k 3 . The scattering amplitude (1.3.7) reads

(1.3.10)

The second term can be easily understood. It is the Fourier transform of a Yukawa potential (eq. (1.2.26)). This matches the result from non-relativistic quantum mechanics (Born approximation) for scattering off of a potential,

The third term has a similar interpretation. Its presence is possible quantum mechanically because we have identical particles. The first term has poles at E cm = ±m. It represents a relativistic effect. Note that the poles are at E cm = ±mc 2 , so in the non-relativistic limit, c → ∞, and the poles go away to infinity.

1.4 Reaction rates