External Sources in the Quantum Theory

We studied before, in sections 2.10 and 2.11, the role of the external sources in classical field theory. We will examine in this section the behavior of the models when one introduces external sources in the quantum theory. We saw that, in the case of the classical theory, the effect of the introduction of external sources is a deformation of the classical solution, which depends of the specific form of the source which is introduced. This happens due to the change, in the space of configurations, of the position of the minimum of the action, which is caused by the introduction of the external source. We will see that something similar happens in the case of the quantum theory. It is clear that, in this case, it is not what happens to the position of the minimum of the action which is of immediate interest, but rather what happens with the relative statistical weights $\exp(-S)$ associated to all possible configurations.

We saw, in the example of the classical theory of the free scalar field on a periodical lattice, that the classical solution without sources was simply $\varphi\equiv 0$, and that it changed to a non-zero solution $\varphi[j]$ in the presence of the external source $j$. This is not a local point-by-point relation between $\varphi$ and $j$, but rather a global relation, so that we may say that the solution $\varphi[j]$ is a type of functional of $j$: in order to determine $\varphi(s)$ at a site $s$ it is not sufficient to know $j(s)$, instead it is necessary to know $j$ at all lattice sites. In a similar way, we have that in the quantum theory of the free scalar field on a periodical lattice the expectation value $v=\langle\varphi\rangle$ of the field is zero in the absence of external sources. In this case the effect of the introduction of the external source is to cause $v(s)$ to be no longer zero, but rather a function of position that depends on the source $j$.

In order to exemplify these facts, let us calculate the expectation value of the field in the free theory, with periodical boundary conditions. When we have an external source the action is given by equation (2.10.1),

$$S_{0}[\varphi]=\frac{1}{2}\sum_{\ell}(\Delta_{\ell}\varphi)^... ...c{\alpha_{0}}{2}\sum_{s}\varphi^{2}(s)-\sum_{s}j(s)\varphi(s).$$

We know that the external source will make $v\neq 0$, unlike what happened before, since it breaks the symmetry by the transformation $\varphi\rightarrow-\varphi$, which we used before to show that the expectation value was zero. We may then rewrite the theory in terms of a new variable, the shifted field $\varphi'$ given implicitly by $\varphi'=\varphi-v$, where $\langle\varphi'\rangle=0$ by definition. If $j$ is a function of position then the same should hold for $v$. We may then rewrite the action in terms of $\varphi'$, obtaining

$$\begin{eqnarray*} S_{0}[\varphi] & = & \frac{1}{2}\sum_{\ell}(\Delta_{\ell}\varp... ...v)^{2} +\frac{\alpha_{0}}{2}\sum_{s}v^{2}(s) -\sum_{s}j(s)v(s). \end{eqnarray*}$$

Observe that the third line of the second version of this equation contains only terms which are independent of $\varphi$. When we exponentiate $S_{0}$ these terms become a constant multiplying the functional integral and, since they exist in the integrals both in the numerator and in the denominator, these factors cancel out, in all observables. It is therefore clear that we may discard these terms without changing anything in the quantum theory. We will commit here a small abuse of language and discard the terms without, however, changing the symbol $S_{0}$ to reflect the change. The first line of the equation is the action from which we started written for the field $\varphi'$, while the second contains only terms linear in $\varphi'$. If we recall our symmetry arguments leading to the fact that $\langle\varphi\rangle=0$ in the theory without sources, added to the fact that we are defining $v$ here in such a way that $\langle\varphi'\rangle=0$, we see that this will only be possible if these linear terms, which are not invariant by the transformation $\varphi'\rightarrow-\varphi'$, are in fact zero for any $\varphi'$. Clearly, this is a condition which we can use to determine $v$,

  $$\sum_{\vec{n}}\left[ \sum_{\mu}(\Delta_{\mu}v)(\Delta_{\mu}\varph... ...{n})\varphi'(\vec{n}) -j(\vec{n})\varphi'(\vec{n})\right]=0,\;\forall\varphi',$$ (3.5.1)

where we decomposed the sum over $\ell$ in sums over $\mu$ and $\vec{n}$. Since the boundary conditions are periodical, we may write this equation in another, equivalent form, performing an integration by parts in the first term to obtain

  $$\sum_{\vec{n}}\left[-\Delta^{2}v+\alpha_{0}v(\vec{n})-j(\vec{n}) \right]\varphi'(\vec{n})=0,\;\forall\varphi'.$$ (3.5.2)

In any case the important point is that it must be valid for all the configurations $\varphi'$ that exist in the ensemble of the quantum theory. Note that this equation does not have a classical limit, because $\varphi'$ is an arbitrary configuration of the quantum theory, and therefore it is not necessarily a continuous function in the continuum limit. While $\varphi'(\vec{n})$ is a completely arbitrary function on the lattice, the quantity within brackets contains only expectation values or classical quantities, all such quantities having well-defined definite values. This is true both for the given external source $j$ and for the expectation value $v$ of the field that results from its introduction. It becomes clear therefore that the only way to satisfy this equation for all $\varphi'$ is that the quantity within brackets be zero (problem 3.5.1). This equation gives us a condition involving $j$ and $v$, which determines the relation between these two quantities.

However, since the quantity within the bracket includes a finite-difference operator, it is not so straightforward to solve it as a stand-alone equation in its current form. Another way to obtain the same result, which makes it easy to solve the equation, is to write all the functions of position in terms of their Fourier transforms. We will do this starting from equation (3.5.1). We have for the field $\varphi'$, the expectation value $v$ and the external source $j$,

$$\begin{eqnarray*} \varphi'(\vec{n}) & = & \sum_{\vec{k}'} e^{-\imath\frac{2\pi}{... ...h\frac{2\pi}{N}\vec{k}\cdot\vec{n}}\;\widetilde\jmath (\vec{k}), \end{eqnarray*}$$

so that we may write equation (3.5.1) as

$$\begin{eqnarray*} 0 & = & \sum_{\vec{n}}\sum_{\vec{k}}\sum_{\vec{k}'} \left[ \su... ...\alpha_{0}\tilde v(\vec{k}) -\widetilde\jmath (\vec{k}) \right], \end{eqnarray*}$$

where we used the fact that the complex exponentials are eigenvalues of the finite-difference operator and also the orthogonality and completeness relations. Note that, once more, this must be true for all the configurations $\widetilde\varphi '(\vec{k})$ of the field, functions which are as arbitrary in momentum space as the functions $\varphi'(\vec{n})$ are arbitrary in position space. It follows therefore, exactly as before, that it is necessary that the contents of the bracket vanish,

$$\tilde v(\vec{k})\left[\rho^{2}(\vec{k})+\alpha_{0}\right]-\widetilde\jmath (\vec{k})=0,$$

which is the version in momentum space of the equation for $v$ that one obtains from equation (3.5.2),

$$\left[-\Delta^{2}+\alpha_{0}\right]v(\vec{n})-j(\vec{n})=0.$$

We have therefore the solution for the expectation value $v$ in the quantum theory,

  $$\tilde v(\vec{k})=\frac{\widetilde\jmath (\vec{k})}{\rho^{2}(\vec{k})+\alpha_{0}}.$$ (3.5.3)

One can verify (problem 3.5.2) that this equation may also be obtained directly from the self-consistency equation $\langle\varphi'(\vec{n})\rangle=0$, by direct calculation of the Gaussian integrals involved in this expectation value. Naturally, this is the more direct and straightforward way to obtain the result. The argument presented above is a shortcut based on symmetry arguments.

Note that the solution obtained is exactly the same solution obtained in the classical case for the field in the presence of the external source. It is important to emphasize that this fact is a characteristic exclusively of the free theory, due to its linearity, and is not valid in general. Having obtained the result in momentum space it is not difficult to write it in position space. One can show directly (problem 3.5.3), taking the inverse Fourier transform of this solution, that

  $$v(\vec{n})=\sum_{\vec{n}'}K(\vec{n},\vec{n}')j(\vec{n}'),$$ (3.5.4)

where $K(\vec{n},\vec{n}')$ is given by

$$K(\vec{n},\vec{n}')=\frac{1}{N^{d}}\sum_{\vec{k}} \frac{e^{-... ...ec{k}\cdot(\vec{n}-\vec{n}')}} {\rho^{2}(\vec{k})+\alpha_{0}}.$$

One can show also (problem 3.5.4) that $K(\vec{n},\vec{n}')$ is the propagator in position space,

$$K(\vec{n},\vec{n}')=\langle\varphi'(\vec{n})\varphi'(\vec{n}')\rangle.$$

This propagator tells us how the introduction of a source at the point $\vec{n}'$ affects the average value of the field at another point $\vec{n}$, that is, it describes the propagation of relations of cause and effect within the model.

We see in this way that, in a way analogous to what happens in the classical theory, the quantum theory also establishes a functional relation between the expectation value $v$ of the field and the external source $j$. In the free theory this functional relation is the same that appears in the classical version of the theory, but this is not true in general. The fact that the relation is the same in either case in this simple example is not very important, what really matters is that in the quantum case, in a fashion analogous to what happens in the classical case, the theory establishes a well-defined relation between the external sources and the expectation value of the field. In the classical case we can establish the physical interpretation of the theory in terms of this relation, so that we have here quite a familiar way of doing the same thing on the quantum case. In fact, the effects of the quantization process on the models, that is, the consequences of the quantum theory, can be explored by means of the examination of the functional relation between $v$ and $j$ in the quantum theory. As we shall see in what follows, this can most conveniently be done in terms of a functional that we will call the effective action of the model, which is a way to encode concisely this functional relation.

Problems

1. Show in detail that the only way to satisfy equations (3.5.1) and (3.5.2) for all possible field configurations $\varphi'(\vec{n})$ on the lattice is that the expressions in brackets that appear in these equations be zero. One way to do this, among many others, is to choose a particular set of functions $\varphi'(\vec{n})$ that can constitute a basis for the representation of any element of configuration space, which is a vector space with a large but finite dimension, $\mathbb{R}^{N^{d}}$.

2. Calculate the expectation value in the equation $\langle\varphi'(\vec{n})\rangle=0$ and use it to derive the solution for $v$ in terms of $j$ in the quantum theory, showing that the answer coincides with the one derived in the text.

3. Calculate the inverse Fourier transform of equation (3.5.3) in order to obtain the solution in position space given in equation (3.5.4).

4. Show through the direct calculation of the expectation value that $K(\vec{n},\vec{n}')$ is the propagator of the model in position space, $\langle\varphi'(\vec{n})\varphi'(\vec{n}')\rangle$.

5. Show that the width of the distribution of values of the field at a single site, in the free theory with an external source $j$, which is given by

   
   

   $$\sigma^{2}_{(j)}= \langle\varphi^{2}\rangle_{(j)}-\langle\varphi\rangle^{2}_{(j)},$$
   
   

   where the index $j$ indicates the presence of the external source, is equal to the width of the theory without the external source, which is given by

   
   

   $$\sigma^{2}=\langle\varphi^{2}\rangle-\langle\varphi\rangle^{2},$$
   
   

   and, therefore, that it is independent of both $j$ and the lattice position where the expectation values are calculated.