Factorization of the Correlation Functions

One fundamental concept of the traditional theory is that the physical content of a model in quantum field theory is defined by the set of its correlation functions. Having developed the necessary calculational techniques, in this section we will discuss the properties of the correlation functions of the free scalar field on a finite periodical lattice. Through this simple model, that we use as our example, we may learn some things of general interest about the structure of correlations of the theory. The $n$-point functions are defined in position space as

$$g(\vec{x}_{1},\ldots,\vec{x}_{n})= \langle\varphi(\vec{x}_{1})\ldots\varphi(\vec{x}_{n})\rangle,$$

but we may also define corresponding functions in momentum space, doing Fourier transformations for each one of the $n$ coordinates $\vec{x}_{i}$. Doing this we obtain $n$-point functions in momentum space, given by

$$\widetilde g(\vec{k}_{1},\ldots,\vec{k}_{n})= \langle\wideti... ...phi (\vec{k}_{1})\ldots\widetilde\varphi (\vec{k}_{n})\rangle.$$

As proposed in problem 3.1.1, it is not difficult to verify that, in the free theory defined by $S_{0}$, we have for the function of a single point in position space

$$g(\vec{x}_{1})=\langle\varphi(\vec{x}_{1})\rangle=0,$$

by means of arguments of symmetry and parity applied to the functional integral, and using the fact that the action is symmetrical by the exchange of the sign of the field, that is, $S_{0}[\varphi]=S_{0}[-\varphi]$. In an analogous fashion, it is easy to verify (problem 3.4.1) that the same result is valid for any functions for which $n$ is odd, independently of some of the factors $\vec{x}_{i}$ being equal or not, that is, that for any $i=0,\ldots,\infty$ and any $\vec{x}_{i}$ we have

$$g(\vec{x}_{1},\ldots,\vec{x}_{2i+1})= \langle\varphi(\vec{x}_{1})\ldots\varphi(\vec{x}_{2i+1})\rangle=0.$$

For example, we have that $\langle\varphi(\vec{x}_{1})\varphi(\vec{x}_{2})\varphi(\vec{x}_{3})\rangle$, $\langle\varphi(\vec{x}_{1})\varphi^{2}(\vec{x}_{2})\rangle$ and $\langle\varphi^{3}(\vec{x}_{1})\rangle$ are all zero, independently of the values of the vectors $\vec{x}_{i}$. Identical arguments may be applied in momentum space for expectation values of products of the Fourier components. In order to see this it suffices to write the action in terms of these components, as we already did before, verifying that it remains invariant by changes of sign of the field, that is, that $S_{0}[\widetilde\varphi ]=S_{0}[-\widetilde\varphi ]$. Hence, all the same facts that follow from this invariance are true in this case and all the expectation values of the Fourier components of the field in momentum space involving an odd number of factors are zero, that is, for any vectors $\vec{k}_{i}$ we have

$$\widetilde g(\vec{k}_{1},\ldots,\vec{k}_{2i+1})= \langle\wid... ...\vec{k}_{1})\ldots\widetilde\varphi (\vec{k}_{2i+1})\rangle=0.$$

In the case of momentum space we may refine a little this argument, extending its reach, if we take into account the fact that in this space the modes of the model are decoupled. Due to this, one can easily show (problem 3.4.2) that the expectation value of the product of two fields corresponding to distinct modes factors into the product of the expectation values of each one of the two fields. Since the components $\vec{k}_{\mu}$ assume both positive and negative values, in this case we must worry also about modes related by a change in the sign of $\vec{k}$. For example, for the case of two factors we have that, if $\vec{k}_{1}\neq\pm\vec{k}_{2}$, then

$$\langle\widetilde\varphi (\vec{k}_{1})\widetilde\varphi (\ve... ...c{k}_{1})\rangle\langle\widetilde\varphi (\vec{k}_{2})\rangle.$$

Hence, even if the number of fields multiplied together is even, the expectation value is still zero unless the fields are paired up in each mode, with an even number of factors in each one. In addition to this, as we saw in the case of the propagator in section 3.3, it is necessary the fields be paired up in such a way that they organize as a set of squared absolute values. This is a general fact, valid for all the correlation function in momentum space, associated to the fact that all the correlation functions are real. It can be remembered by means of the rule of association of momenta that we saw before: momentum conservation on the periodical lattice implies that each momentum that “goes into” the expectation value, associated to $\widetilde\varphi (\vec{k})$, must be equal and opposite to the one that “exits”, associated to $\widetilde\varphi (-\vec{k})=\widetilde\varphi ^{*}(\vec{k})$. Hence, we have that each factor $\widetilde\varphi (\vec{k})$ must be paired up with another factor $\widetilde\varphi (-\vec{k})$ in order for the expectation value not to be zero. For example, the expectation value $\langle\widetilde\varphi (\vec{k}_{1})\widetilde\varphi (\vec{k}_{2})\rangle$ is zero if $\vec{k}_{1}\neq-\vec{k}_{2}$, as we saw before in section 3.3.

In this way, in momentum space we may quickly reduce the number of different possibilities for correlation functions potentially different from zero. The ones that remain to be discussed are those of the types $\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{2}\rangle$, $\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{2}\vert\widetilde\varphi (\vec{k}_{2})\vert^{2}\rangle$, $\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{4}\rangle$, $\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{4}\vert\widetilde\varphi (\vec{k}_{2})\vert^{2}\rangle$, etc, where only even powers of absolute values of the Fourier components appear. As a non-trivial example of factorization, it is not difficult to verify (problem 3.4.3) that, for $\vec{k}_{1}\neq\pm\vec{k}_{2}$,

$$\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{2}\vert\wi... ...e \langle\vert\widetilde\varphi (\vec{k}_{2})\vert^{2}\rangle.$$

It is important to emphasize that, although it is true that, for $\vec{k}_{1}\neq-\vec{k}_{2}$,

  $$\langle\widetilde\varphi (\vec{k}_{1})\widetilde\varphi (\vec{k}_... ...ilde\varphi (\vec{k}_{1})\rangle\langle\widetilde\varphi (\vec{k}_{2})\rangle,$$ (3.4.1)

the analogous relation is not true in position space, that is, even if $\vec{x}_{1}\neq\vec{x}_{2}$ we have that

  $$\langle\varphi(\vec{x}_{1})\varphi(\vec{x}_{2})\rangle\neq \langle\varphi(\vec{x}_{1})\rangle\langle\varphi(\vec{x}_{2})\rangle.$$ (3.4.2)

This is due to the fact that the degrees of freedom are decoupled only in momentum space, not in position space. Hence, in the first case the fact that $\langle\widetilde\varphi (\vec{k})\rangle$ is zero implies that $\langle\widetilde\varphi (\vec{k}_{1})\widetilde\varphi (\vec{k}_{2})\rangle=0$ for $\vec{k}_{1}\neq-\vec{k}_{2}$, but in the second case, although $\langle\varphi(\vec{x})\rangle$ is zero, we have $\langle\varphi(\vec{x}_{1})\varphi(\vec{x}_{2})\rangle\neq 0$, independently of the values of $\vec{x}_{1}$ and $\vec{x}_{2}$.

Of course the factorization relation (3.4.1) in momentum space has a counterpart in position space, but it is necessary to keep in mind that this equation is only valid for $\vec{k}_{1}\neq-\vec{k}_{2}$ and not when the two momenta are equal and opposite. Doing the Fourier transformation of the left-hand side of this equation one obtains a relation for the two-point function in position space (problem 3.4.4), but it is not a factorization relation like the one suggested by equation (3.4.2). Instead of this, what one obtains is the relation

  $$\langle\varphi(\vec{x}_{1})\varphi(\vec{x}_{2})\rangle=\sum_{\vec... ...{x}_{1}-\vec{x}_{2})} \langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle.$$ (3.4.3)

The calculations involved in this kind of manipulation are usually simple but involve a few accounting subtleties involving the accounting of the terms in the sums over the modes in momentum space. This is a skill that it is very important to acquire in order to develop good control over the theory.

We will now calculate some of the examples that remain of non-zero correlation functions in momentum space, to illustrate the important phenomenon of the factorization of all higher-order correlation functions in terms of the propagator, which is characteristic of free field theories. A we already saw before, from equation (3.3.3) we have the following fundamental result for the two-point function, which cannot be factored in terms of the one-point function,

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle= \frac{1}{N^{d}[\rho^{2}(\vec{k})+\alpha_{0}]}.$$

Besides this, we have the result of equation (3.3.4) for the basic functional integral in momentum space,

$$\int[{\bf d}\widetilde\varphi ]e^{-S_{0}[\widetilde\varphi ]... ...c{k}} \sqrt{\frac{2\pi}{N^{d}[\rho^{2}(\vec{k})+\alpha_{0}]}},$$

starting from which it is easy to calculate all the others by means of differentiation with respect to the quantity $-N^{d}[\rho^{2}(\vec{k})+\alpha_{0}]/2$. However, it is necessary to treat separately the cases in which $\widetilde\varphi (\vec{k})$ is real (such as, for example, the case $\vec{k}=\vec{0}$) and the cases in which $\widetilde\varphi (\vec{k})$ has a non-zero imaginary component, because there exists in the sum that defines $S_{0}[\widetilde\varphi ]$ only one term containing a real component such as $\widetilde\varphi (\vec{0})$, but two identical terms containing a Fourier component that has non-zero imaginary part. As an example of this kind of calculation, let us consider the quantity

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{4}\rangle= \fr... ...t[{\bf d}\widetilde\varphi ]\;e^{-S_{0}[\widetilde\varphi ]}},$$

for which we may write, in the case in which $\widetilde\varphi (\vec{k})$ is real,

$$\begin{eqnarray*} \langle\vert\widetilde\varphi (\vec{k})\vert^{4}\rangle & = & ... ...\\ & = & 3\frac{1}{\{N^{d}[\rho^{2}(\vec{k})+\alpha_{0}]\}^{2}}. \end{eqnarray*}$$

In this way, comparing this result with the fundamental result of equation (3.3.3), we obtain the factorization relation

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{4}\rangle= 3\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{2},$$

for the case $\vec{k}=\vec{0}$ and other purely real modes, showing that the four-point function factors into two two-point functions. The same is true for higher-order functions, as one can verify without difficulty. For example, one can obtain for the purely real modes (problem 3.4.5) a more general factorization formula,

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{2n}\rangle= (2... ...!!\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{n},$$

for any integer $n$, involving a double factorial $(2n-1)!!=(2n-1)(2n-3)(2n-5)\ldots 1$.

In the case in which $\widetilde\varphi (\vec{k})$ has a non-zero imaginary part we have

$$\begin{eqnarray*} \langle\vert\widetilde\varphi (\vec{k})\vert^{4}\rangle & = & ... ...\\ & = & 3\frac{1}{\{N^{d}[\rho^{2}(\vec{k})+\alpha_{0}]\}^{2}}. \end{eqnarray*}$$

so that we obtain, in a way analogous to the previous one, the relation

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{4}\rangle= 2\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{2},$$

showing once more that the four-point function factors into two two-point functions, but with a different coefficient. For higher-order functions one can obtain (problem 3.4.5) the general formula

$$\langle\vert\widetilde\varphi (\vec{k})\vert^{2n}\rangle= n!\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{n},$$

for any integer $n$. In fact, in future volumes we will see that it will be convenient to build a small table of such relations between the functional integrals of the free theory, because they will show up repeatedly in the development of the perturbative theory for interacting theories, like the non-linear models that we will examine in the future.

We see therefore that the complete solution of the theory of the free scalar field, that is, the calculation of all its correlation functions, is reducible to the calculation of the propagator. It follows that all the physics of the theory is contained in the structure of this propagator. The factorization of the higher-order functions in terms of the propagator means that there are no physical interactions between the objects that propagate in this model. One way to understand this using our classical intuition is to remember that the theory is linear, being characterized classically by a linear equation of motion for which there is a principle of linear superposition, that is, the waves that propagate in space-time in the non-Euclidean version of the theory superpose linearly, going right through one another as in classical electrodynamics, without interacting with one another. The factorization of the correlation functions means that this linearity is preserved in the quantum version of the theory, that is, that the quantum fluctuations superpose linearly and in this way do not affect the linearity of the theory. The propagator itself, the sole correlation function in momentum space that does not factor our in terms of other lower-order functions, describes how objects propagate in this model, which ends up being the only physics that it contains.

Problems

1. Show, using the same ideas that were suggested for problem 3.1.1, that the correlation function of an odd number of points in position space is zero, that is, that for any $i=0,\ldots,\infty$ and any $\vec{x}_{i}$,

   
   

   $$g(\vec{x}_{1},\ldots,\vec{x}_{2i+1})= \langle\varphi(\vec{x}_{1})\ldots\varphi(\vec{x}_{2i+1})\rangle=0.$$
   
   

2. Show, using the explicit form of the action $S_{0}[\widetilde\varphi ]$ and the definition of the functional integral in momentum space, that the expectation value of two Fourier components of the field with different momenta factors into the product of the expectation values of each component, that is, that for $\vec{k}_{1}\neq\pm\vec{k}_{2}$ we have

   
   

   $$\langle\widetilde\varphi (\vec{k}_{1})\widetilde\varphi (\ve... ...c{k}_{1})\rangle\langle\widetilde\varphi (\vec{k}_{2})\rangle.$$
   
   

3. Show, using the same ideas used in problem 3.4.2, that for $\vec{k}_{1}\neq\pm\vec{k}_{2}$ we have the factorization

   
   

   $$\langle\vert\widetilde\varphi (\vec{k}_{1})\vert^{2}\vert\wi... ...e \langle\vert\widetilde\varphi (\vec{k}_{2})\vert^{2}\rangle.$$
   
   

4. Show that equation (3.4.2) holds. In other words show that, unlike what happens in momentum space as described by equation (3.4.1), this type of factorization does not happen in position space. In order to do this, first show that $\langle\varphi(\vec{x})\rangle=0$, so that the right-hand side of equation (3.4.2) is zero. Then apply Fourier transformations to the left-hand side of equation (3.4.1) and obtain the result given in equation (3.4.3) for the two-point function in position space. Finally, show that the right-hand side of this last equation is never zero.

5. Using the techniques given in the text for the calculation of the functional integrals, demonstrate the factorization formula

   
   

   $$\langle\vert\widetilde\varphi (\vec{k})\vert^{2n}\rangle= (2... ...!!\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{n},$$
   
   

   for the case of purely real modes $\widetilde\varphi (\vec{k})$, while for the case of modes in which $\widetilde\varphi (\vec{k})$ has a non-zero imaginary part, demonstrate the factorization relation

   
   

   $$\langle\vert\widetilde\varphi (\vec{k})\vert^{2n}\rangle= n!\langle\vert\widetilde\varphi (\vec{k})\vert^{2}\rangle^{n},$$
   
   

   for an arbitrary integer $n$ in either case.

6. Using previous results already known and/or Fourier transforms, calculate explicitly, at an arbitrary site $\vec{n}$, the quantities $\langle\varphi^{4}(\vec{n})\rangle$ and $\langle\varphi^{2}(\vec{n})\rangle$, showing in this way that the following factorization formula is valid for them:

   
   

   $$\langle\varphi^{4}(\vec{n})\rangle=3\langle\varphi^{2}(\vec{n})\rangle^{2}.$$