Connection with the Canonical Formalism

The concept of energy is introduced by means of what is denominated the canonical formalism, which is the one usually employed in the traditional presentation of the theory, and which is discussed in terms of states and operators in a Hilbert space. It is therefore necessary to discuss some aspects of the relation of the formalism of the Euclidean lattice with the canonical formalism of quantum field theory. Our main objective in this section is to introduce and examine the concept of energy from the point of view of the theory formulated on the Euclidean lattice. Later on we will try to see under what conditions it is possible to establish connections between the lattice formalism and elements of the operator formalism, such as particles states, the Hamiltonian operator and its eigenstates.

In this section we will construct the canonical formalism on the Euclidean lattice, introducing the concepts of conjugate momenta and of the energy. We will use the well-known case of quantum mechanics both as a guide for our construction and as a way to verify the correction of our results. Always using the free scalar field of a single component as our basic example, we have the action

$$S=\sum_{s}{\cal L},\mbox{ }{\cal L}= \frac{1}{2}\sum_{\mu}(\Delta_{\mu}\varphi)^{2} +\frac{\alpha_{0}}{2}\varphi^{2},$$

where both $S$ and ${\cal L}$ are dimensionless, ${\cal L}$ being the Lagrangian density. In order to build the canonical formalism it is necessary to separate one of the dimensions of the space, which we will call the time, from the other dimensions, which we will call spacial dimensions. To make things definite, we may think of the $d=4$ case, but the formalism can be used in any dimension. We will denote the $d$-dimensional sums by $\sum_{\mu}$ and $\sum_{s}$, while those that do not include the time will be written as $\sum_{i}$ and $\sum_{\mathbf{x}}$, respectively, and then the sum over the temporal dimension will be denoted by $\sum_{t}$. The temporal variables will be denoted by an index $0$ or $T$. The lattice will have $N_{L}$ sites in the spacial directions and $N_{T}$ sites in the temporal direction. In general these two numbers will be equal, but the possibility remains open that they be different, if and when this becomes necessary for future discussions. We may rewrite the action in this new notation, obtaining

$$S=\sum_{t}\sum_{\mathbf{x}}{\cal L}\mbox{, }{\cal L}= \frac{... ...m_{i}(\Delta_{i}\varphi)^{2} +\frac{\alpha_{0}}{2}\varphi^{2}.$$

Classically we may define the dimensionless conjugate momentum to the field $\varphi$, which we shall call $\bar\pi$, by means of

$$\bar\pi =\imath\frac{\partial{\cal L}}{\partial(\Delta_{0}\varphi)} =\imath\Delta_{0}\varphi,$$

which is the usual relation except for the factor of $\imath$, whose introduction is due to the fact that we are doing the construction in Euclidean space, as well as to the fact that $\bar\pi$ is the temporal component of a vector. This conjugate momentum may be understood as a site variable related to the link variable $\Delta_{0}\varphi$ which is associated to the link that starts at that site and points in the positive temporal direction. We may now define a dimensionless Hamiltonian density is a way analogous to the usual definition,

$${\cal H} = \bar\pi \Delta_{0}\varphi-\imath{\cal L}\Rightarrow$$

$${\cal H} = -\frac{\imath}{2}\left[\bar\pi ^{2} +\sum_{i}(\Delta_{i}\varphi)^{2}+\alpha_{0}\varphi^{2}\right],$$ (5.1.1)

which is the usual definition except for the factor of $\imath$, due to the Euclidean nature of our development, and of the factor $-1$, chosen by mere convenience. As we shall see later, independently of the choice of sign adopted here there will always exist in the theory states with both positive energy and negative energy, just as in the traditional formalism. We may now add over the spacial part and write dimensionless versions of the Lagrangian and of the Hamiltonian,

$$\begin{eqnarray*} \mathbf{L}& = & \frac{1}{2}\sum_{\mathbf{x}} \left[(\Delta_{0}... ...}+\sum_{i}(\Delta_{i}\varphi)^{2} +\alpha_{0}\varphi^{2}\right]. \end{eqnarray*}$$

Observe that we wrote $\mathbf{H}$ in terms of $\bar\pi$ and $\mathbf{L}$ in terms of $\Delta_{0}\varphi$, in the usual way. Classically nothing changes if we write $\mathbf{H}$ in terms of $\imath\Delta_{0}\varphi$, since this and $\bar\pi$ are equal. However, as we shall see in what follows, in the quantum theory it is necessary to take seriously the distinction between $\bar\pi$ and $\Delta_{0}\varphi$.

In order to define the quantum theory there are two paths we may follow. On the one hand, we might follow our usual definition and write a dimensionless version of the energy, as well as of any other observable depending on $\bar\pi$, assuming that $\bar\pi =\imath\Delta_{0}\varphi$, as the expectation value

$$\begin{eqnarray*} {\cal E}& = & \frac{\displaystyle \int[{\bf d}\varphi]\;\mathb... ...i^{2}\right]e^{-S}}{\displaystyle \int[{\bf d}\varphi]\;e^{-S}}, \end{eqnarray*}$$

where $\mathbf{H}'$ is $\mathbf{H}$ with $\bar\pi$ exchanged by its classical value $\imath\Delta_{0}\varphi$. The dimensionfull version of the energy is given by $E={\cal E}/a=N_{T}{\cal E}/T$, in terms of the lattice spacing $a$ or of the total temporal length $T$ of the box. On the other hand, we might start with a definition of the quantum theory in terms of independent variables $\bar\pi$ and $\Delta_{0}\varphi$, with functional integrals involving both the variables $\bar\pi$ and the variables $\varphi$. As we shall see, the results are different in each case. We will adopt as the definition of the canonical version of the theory the following expression for the observables,

  $$\langle{{\cal O}}\rangle=\frac{\displaystyle \int[{\rm d}\varphi]... ...i][{\rm d}\bar\pi ] \;e^{\imath\sum_{s}(\bar\pi \Delta_{0}\varphi-{\cal H})}},$$ (5.1.2)

where ${\cal O}$ is a functional of $\bar\pi$ and $\varphi$, and ${\cal H}$ is the expression in (5.1.1). Observe that the dependency on $\bar\pi$ is always Gaussian in this definition, independently of the model under consideration, because we have

$$\imath(\bar\pi \Delta_{0}\varphi-{\cal H})=-\frac{1}{2} \lef... ...+\sum_{i}(\Delta_{i}\varphi)^{2}+\alpha_{0}\varphi^{2}\right].$$

Note that, due to the introduction of the factor of $\imath$ in the definition of $\bar\pi$, the integrals on this variable converge in Euclidean space so long as the integrations are made over real values, exactly as in the case of $\varphi$. It is understood, therefore, that at each site we have the integrals

$$\int_{-\infty}^{\infty}{\rm d}\varphi\mbox{~~~~and~~~~}\int_{-\infty}^{\infty}{\rm d}\bar\pi ,$$

along the real axis of each one of the variables.

Hence we have a complete definition of all the observables of the theory in the canonical formalism on the Euclidean lattice. It is important to observe that when ${\cal O}$ does not depend on $\bar\pi$ this definition reduces to the previous one, since in this case we can do the integrations on $\bar\pi$ explicitly and return to the usual definition. In order to do this we start by separating the variables, writing

  $$\langle{{\cal O}}\rangle=\frac{\displaystyle \int[{\rm d}\varphi]... ...ac{1}{2}\sum_{s} \left(\bar\pi ^{2}-2\imath\bar\pi \Delta_{0}\varphi\right)}}.$$ (5.1.3)

We may then complete the square on $\bar\pi$ in the exponent of the second exponential, obtaining

$$\bar\pi ^{2}-2\imath\bar\pi \Delta_{0}\varphi = (\bar\pi -\imath\Delta_{0}\varphi)^{2}+(\Delta_{0}\varphi)^{2} \Rightarrow$$

$$\int[{\rm d}\bar\pi ]\;e^{-\frac{1}{2}\sum_{s} \left[(\bar\pi -\imath\Delta_{0}\varphi)^{2}+(\Delta_{0}\varphi)^{2}\right]} = e^{-\frac{1}{2}\sum_{s}(\Delta_{0}\varphi)^{2}}\int[{\rm d}\bar\pi ] \;e^{-\frac{1}{2}\sum_{s}(\bar\pi -\imath\Delta_{0}\varphi)^{2}}.$$ (5.1.4)

We now shift the variable $\bar\pi$, defining a new variable $\chi=\bar\pi -\imath\Delta_{0}\varphi$, and obtain for the integral on $\bar\pi$

$$\int_{-\infty}^{\infty}[{\rm d}\bar\pi ] \;e^{-\frac{1}{2}\s... ...a_{0}\varphi} [{\rm d}\chi]\;e^{-\frac{1}{2}\sum_{s}\chi^{2}}.$$

Drawing now the complex-$\chi$ plane we can identify the relevant integration paths.

Completing a closed circuit with two small arcs at infinity, at which the integral is zero, we use the residue theorem for the exponential, which has no poles, obtaining

$$\int_{-\infty-\imath\Delta_{0}\varphi}^{\infty-\imath\Delta_... ...nfty}^{\infty}[{\rm d}\chi]\;e^{-\frac{1}{2}\sum_{s}\chi^{2}},$$

which is a convergent integral that gives the usual result, which is already known. However, the important fact here is that these integrals do not depend on $\varphi$, which means that in 5.1.4 we succeeded in decoupling the integrals on $\bar\pi$ from those on $\varphi$. The factor that remains from the completion of the square in 5.1.4,

$$e^{-\frac{1}{2}\sum_{s}(\Delta_{0}\varphi)^{2}},$$

completes in 5.1.3 the expression for the action, so that we have

$$\langle{{\cal O}}\rangle=\frac{\displaystyle \int[{\rm d}\va... ...\varphi]\;e^{-S}}{\displaystyle \int[{\rm d}\varphi]\;e^{-S}},$$

that is, we recovered the usual definition. However, if the observable ${\cal O}$ depends on $\bar\pi$ then the results can only be obtained by means of the canonical definition. If we substitute $\bar\pi$ by its classical value $\imath\Delta_{0}\varphi$ and use the usual definition, the result will be, in general, different from the result obtained by means of the canonical definition.

An interesting exercise to illustrate the calculation of expectation values of observables that depend on $\bar\pi$ is the calculation at a given site of the observables $\langle\bar\pi (s)-\imath\Delta_{0}\varphi(s)\rangle$ and $\langle[\bar\pi (s)-\imath\Delta_{0}\varphi(s)]^{2}\rangle$, by means of which we can examine the nature of the relation between $\bar\pi$ and $\imath\Delta_{0}\varphi$ in the quantum theory. Of course, if we calculate these observables according to the previous, non-canonical definition, we will obtain zero for both. For the calculation of the first of these two observables we start from the canonical definition given in equation (5.1.2), whose numerator can be written as

$$\begin{eqnarray*} \lefteqn{\int[{\rm d}\varphi][{\rm d}\bar\pi ](\bar\pi -\imath... ...}\varphi} [{\rm d}\chi]\;\chi\;e^{-\frac{1}{2}\sum_{s}\chi^{2}}, \end{eqnarray*}$$

where $\chi=\bar\pi -\imath\Delta_{0}\varphi$ and we used once more some of the manipulations used before. We may now modify the integration circuit as we did before, obtaining

$$\int_{-\infty}^{\infty}[{\rm d}\varphi]\;e^{-S} \int_{-\inft... ...nfty}[{\rm d}\chi] \;\chi\;e^{-\frac{1}{2}\sum_{s}\chi^{2}}=0,$$

by a simple symmetry argument. We see therefore that $\bar\pi$ and $\imath\Delta_{0}\varphi$ have the same expectation value,

$$\langle\bar\pi \rangle=\imath\langle\Delta_{0}\varphi\rangle.$$

However, we may verify that $\bar\pi \neq\imath\Delta_{0}\varphi$ in a simple way, calculating the second expectation value. Repeating the same procedures and calculations used previously, and executing some Gaussian integrations (problem 5.1.1), we obtain

$$\left\langle(\bar\pi -\imath\Delta_{0}\varphi)^{2}\right\rangle=1.$$

This means that, although $\bar\pi$ and $\imath\Delta_{0}\varphi$ have the same expectation value, the two quantities fluctuate around each other in such a way that the difference between them is, typically, a non-vanishing real number with a magnitude of the order of one.

We will now calculate the expectation value of the Hamiltonian $\mathbf{H}$ in this canonical version of the formalism. Since this corresponds to the calculation of the energy of the vacuum state, this is the first step for the determination of the role that the concept of energy plays in the theory. For starters, let us worry about the calculation of the integrals over the momenta. Starting from the definition of the expectation values in the canonical formalism, and repeating once more some of the previous operations, we may write for the expectation value of $\mathbf{H}$

$$\langle\mathbf{H}\rangle=-\frac{\imath}{2}\sum_{\mathbf{x}}\... ...\;e^{-S} \int[{\rm d}\chi]\;e^{-\frac{1}{2}\sum_{s}\chi^{2}}},$$

where $\bar\pi =\chi+\imath\Delta_{0}\varphi$, so that we may write for the expression of ${\cal H}$ in the denominator

$$\begin{eqnarray*} -2\imath{\cal H}& = & (\chi+\imath\Delta_{0}\varphi)^{2} +\sum... ...phi)^{2} +\sum_{i}(\Delta_{i}\varphi)^{2}+\alpha_{0}\varphi^{2}, \end{eqnarray*}$$

so that we have

$$\langle\mathbf{H}\rangle=-\frac{\imath}{2}\sum_{\mathbf{x}}\... ...\;e^{-S} \int[{\rm d}\chi]\;e^{-\frac{1}{2}\sum_{s}\chi^{2}}},$$

where ${\cal H}'$ is the expression of ${\cal H}$ with the variable $\bar\pi$ changed to $\imath\Delta_{0}\varphi$. The linear integral on $\chi$ vanishes by symmetry and we have then

$$\begin{eqnarray*} \langle\mathbf{H}\rangle & = & -\frac{\imath}{2}\sum_{\mathbf{... ...\;{\cal H}'e^{-S} }{\displaystyle \int[{\rm d}\varphi]\;e^{-S}}. \end{eqnarray*}$$

Doing the Gaussian integrations in the first term we obtain

$$\langle\mathbf{H}\rangle=-\frac{\imath}{2}N_{L}^{d-1}+\langle\mathbf{H}'\rangle,$$

where $\mathbf{H}'$ is the $\mathbf{H}$ that corresponds to ${\cal H}'$. In other words, we see that the expectation values of $\mathbf{H}$ and $\mathbf{H}'$, calculated respectively according to the definition of the observables in the canonical formalism and according to the usual definition of the observables, differ by a quantity which is divergent in the limit $N_{L}\rightarrow\infty$, except in the case $d=1$ of quantum mechanics. Observe however that this is a constant quantity, in the sense that it does not depend on the parameters and dynamical variables of the model, being therefore of little physical relevance.

We may proceed now to the complete calculation of the energy of the vacuum, by evaluating the second term of the expression above. Writing this term explicitly we have

$$\langle\mathbf{H}'\rangle=-\frac{\imath}{2}\sum_{\mathbf{x}}... ...arphi^{2}\right]}{\displaystyle \int[{\rm d}\varphi]\;e^{-S}}.$$

In order to simplify the calculation, we will use translation invariance along the time direction, which is equivalent to a kind of internal conservation of energy. This invariance exists by construction in the vacuum state of the model we are dealing with. We may define a new observable $\mathbf {H}_{b}$, related to the energy, through the average of $\mathbf{H}$ over a temporal block, which might be the whole lattice,

$$\mathbf{H}_{b}=\frac{1}{N_{T}}\sum_{t}\mathbf{H}.$$

The invariance by temporal translation implies that $\mathbf {H}_{b}$ and $\mathbf{H}$ have the same expectation value, but they are really two conceptually different observables. Observe that, in the spirit of the discussion in section 4.3, it is not possible to measure an energy at a perfectly well-defined instant, that is, on a vanishing temporal interval $\Delta t$, so that in any real situation we will always be making an average over a temporal block when we measure the energy. In our case here we simply adopted a maximal block, making an average over the whole extent of the lattice. Getting back to our calculation, due to the temporal translation invariance we may add over the temporal direction and then divide by $N_{T}$, without any change in the result, thus obtaining

$$\langle\mathbf{H}_{b}'\rangle=\langle\mathbf{H}'\rangle=-\fr... ...arphi^{2}\right]}{\displaystyle \int[{\rm d}\varphi]\;e^{-S}}.$$

We may now use Fourier transforms in $d$ dimensions to calculate this in a simple way, writing the expression in the form

$$\langle\mathbf{H}'\rangle=-\frac{\imath}{2}N_{L}^{d-1}\frac{... ...t^{2}}{\displaystyle \int[{\rm d}\widetilde\varphi ]\;e^{-S}},$$

where the action $S$ may be written in momentum space as

$$S=\frac{N_{L}^{d-1}N_{T}}{2}\sum_{p} \left(\rho^{2}+\alpha_{0}\right)\vert\widetilde\varphi \vert^{2},$$

with $\rho^{2}=\sum_{\mu}\rho_{\mu}^{2}$, that, in the general case in which we have $N_{L}\neq N_{T}$, may be written explicitly as

$$\begin{eqnarray*} \rho^{2} & = & \rho_{0}^{2}+\sum_{i}\rho_{i}^{2}+\alpha_{0} \\... ...um_{i}4\sin^{2}\!\left(\frac{k_{i}\pi}{N_{L}}\right)+\alpha_{0}, \end{eqnarray*}$$

so that the expectation value of $\vert\widetilde\varphi \vert^{2}$ is given by

$$\langle\vert\widetilde\varphi \vert^{2}\rangle =\frac{1}{N_{L}^{d-1}N_{T}}\;\frac{1}{\rho^{2}+\alpha_{0}},$$

leading therefore to the result for the term we are examining,

$$\langle\mathbf{H}'\rangle=-\frac{\imath}{2N_{T}}\sum_{p} \fr... ...i}\rho_{i}^{2}+\alpha_{0}}{\displaystyle \rho^{2}+\alpha_{0}}.$$

With this result we can assemble the final expression for the dimensionless energy,

$$\begin{eqnarray*} {\cal E}& = &-\frac{\imath}{2N_{T}}N_{L}^{d-1}N_{T} -\frac{\im... ...0}}{\displaystyle \rho_{0}^{2}+\sum_{i}\rho_{i}^{2}+\alpha_{0}}, \end{eqnarray*}$$

where we used the facts that $N_{L}^{d-1}N_{T}=\sum_{p}1$ and that $\rho^{2}=\rho_{0}^{2}+\sum_{i}\rho_{i}^{2}$. We may now write the final result for the dimensionfull form $E={\cal E}/a$ of the energy,

  $$E=-\frac{\imath}{T}\sum_{k_{0}}\sum_{k_{i}} \frac{\displaystyle \... ...ht)}{\displaystyle \rho_{0}^{2}+\left(\sum_{i}\rho_{i}^{2}+\alpha_{0}\right)}.$$ (5.1.5)

In order to better understand this result it is necessary to examine the behavior of the sums it contains. For this it is useful to examine first the case $d=1$, which corresponds to the quantum mechanics of a harmonic oscillator with angular frequency $\omega$, in which case we have $\alpha_{0}=\omega^{2}a^{2}=\omega^{2}T^{2}/N_{T}^{2}$ and we may write

$$E = -\frac{\imath}{T}\sum_{k_{0}} \frac{\displaystyle \alpha_{0}}{\displaystyle \rho_{0}^{2}+\alpha_{0}}$$

$$= -\frac{\imath}{T}\sum_{k_{0}=-k_{m}}^{k_{M}}\frac{\displaystyle \... ...left(\frac{N_{T}}{T}\right)^{2} \sin^{2}\!\left(\frac{k_{0}\pi}{N_{T}}\right)},$$ (5.1.6)

where $k_{m}$ and $k_{M}$ are the minimum and maximum values of $k_{0}$ on lattices with a given $N_{T}$. Note that the term $k_{0}=0$ of this sum, unlike all the others, is simply equal to $1$ and does not depend on $\omega$, while all the others go to zero for $\omega\rightarrow 0$. It follows that in this limit the sum goes to $1$ and we have, therefore, $E=-\imath/T$ for $\omega=0$. Since $1/T$ is the energy scale associated to the temporal size of our box, we see that this effect is due to the infrared cutoff established by the box. In the limit in which we make the box infinite in the temporal direction this effect disappears and we have $E=0$ for $\omega=0$.

In order to determine the exact value of this sum in the opposite case, when instead of $\omega=0$ we have $\omega\gg 1/T$, it is necessary to make a numerical evaluation. It is possible, however, to establish analytically upper and lower bounds to the sum (problems 5.1.2, 5.1.3 and 5.1.4), showing that, for even $N_{T}$,

$$\frac{\omega}{2}\sqrt{\frac{(2N_{T})^{2}}{(\omega T)^{2}+(2N... ...ga}{2}\sqrt{\frac{(2N_{T})^{2}}{(\omega T)^{2}+(2N_{T})^{2}}},$$

a relation that, in the limit $N_{T}\rightarrow \infty$, results in

$$\frac{\omega}{2}\leq\imath E\leq\frac{\omega}{2}+\frac{1}{T}.$$

In order to examine the case $\omega\gg 1/T$ we may make an approximation of the sum by an integral, which is a good approximation for large $N_{T}$. In this case we have for the minimum variation of the momentum $dp=2\pi/T$, so that we get

$$\begin{eqnarray*} E & = & -\frac{\imath}{T}\frac{T}{2\pi} \int_{-N_{T}\pi/T}^{N_... ...math\omega}{\pi}\arctan\!\left(\frac{N_{T}\pi}{\omega T}\right). \end{eqnarray*}$$

Figure 5.1.1: The energy as a function of the variable $\omega T$. The curved line corresponds to the numerical result for $N_{T}=1000$. The central straight line corresponds to the continuum limit in an infinite temporal box. The other straight lines correspond to the upper and lower bounds which are mentioned in the text and proposed as problems to the reader.

In the limit $N_{T}\rightarrow \infty$ with finite $\omega$ and $T$ the arc-tangent tends to $\pi/2$ and hence we obtain the expected result for the harmonic oscillator,

$$\imath E=\frac{\omega}{2}.$$

Note that this is the exact result only in the case in which the temporal box is infinite, with $T\rightarrow \infty$, so that the approximation by an integral is not sufficient to show the infrared effects due to the finite temporal box.

Figure 5.1.2: The energy as a function of the variable $\omega T$. The straight line corresponds to the continuum limit in an infinite temporal box, that is, to the case $N_{T}\rightarrow \infty$ and $T\rightarrow \infty$. The curved lines correspond to the numerical results in progressively larger lattices, from $N_{T}=10$ to $N_{T}=1000$, showing how they approach the continuum result in the limit $N_{T}\rightarrow \infty$.

For a more detailed examination of the complete behavior of the energy as a function of $N_{T}$ and $\omega T$ it is necessary to calculate the sum numerically (problem 5.1.5). The results of such a calculation appear in figures 5.1.1 and 5.1.2. The graph in figure 5.1.1 shows the energy in the continuum limit, as well as the upper and lower bounds that it is possible to establish analytically for it, in this limit. Besides the result in an infinite temporal box, the result in a finite box is also shown, to illustrate the infrared effects that exist in this case. In this graph the central straight line corresponds to the continuum limit in an infinite temporal box, that is, to the case $N_{T}\rightarrow \infty$ and $T\rightarrow \infty$. The other straight lines correspond to the upper and lower bounds which are proposed as problems to the reader. The lower straight line corresponds to the lower bound to which problem 5.1.2 makes reference, while the improved lower bound proposed in problem 5.1.3 coincides with the central straight line. The curved line corresponds to the numerical result for $N_{T}=1000$.

The graph in figure 5.1.2 shows the energy in the continuum limit and on various finite lattices, illustrating the way in which these results approach their limit when $N_{T}\rightarrow \infty$. Note that, for each finite lattice, the lattice result is above the continuum limit for $\omega T$ sufficiently small, but falls below it above a certain value of this variable. We are seeing here an ultraviolet effect: if we increase sufficiently the frequency $\omega$, decreasing therefore the corresponding wavelength until it is of the order of the lattice spacing $a$, we start to see clearly the distortions cause by the discrete character of the lattice. The numerical evidence indicates clearly that $\imath E=\omega/2$ is the exact result for the energy in the continuum limit within an infinite temporal box, but it seems that it is rather difficult to obtain this result analytically.

We may now return to the discussion of the $d$-dimensional case for $d\geq 2$, whose result for the energy is given in equation (5.1.5), which we may rewrite as

  $$E=-\frac{\imath}{T}\sum_{k_{i}} \left(\sum_{k_{0}}\frac{\displaystyle A^{2}}{\displaystyle \rho_{0}^{2}+A^{2}}\right),$$ (5.1.7)

where $A^{2}=\sum_{i}\rho_{i}^{2}+\alpha_{0}$. Note that the sum in $k_{0}$ has the same form of the sum that we just discussed in the case of quantum mechanics. However, the continuum limit of this sum behaves in a way that is very different from what happens in the case $d=1$, due to the way in which the quantity $A$, which contains the spacial components $\rho_{i}$ of the dimensionless momentum, scales in the limit, for $d\geq 2$. In fact, the quantity $\imath ET$ diverges in this case, when we make $N_{T}\rightarrow \infty$ and $N_{L}\rightarrow\infty$.

Figure 5.1.3: The energy per site as a function of $N$ in the symmetrical case $N_{T}=N_{L}=N$. Each pair of curves corresponds to a different dimension $d$, converging to the value $(d-1)/d$ in the limit $N\rightarrow \infty$. In each pair the curve that converges faster is the one that corresponds to $m_{0}=1$.

One can understand this fact by observing that the result above is a sum, of something similar to the energy of the ground state of a harmonic oscillator, over all the $N_{L}^{d-1}$ degrees of freedom of a $(d-1)$-dimensional section of the lattice, so that this sum is certainly divergent at least as $N_{L}^{d-1}$. In addition to this, the harmonic oscillators over which we are adding have all the possible frequencies in the temporal direction of the lattice, so that their ground state energies vary from values of the order of $1$ to values of the order of $N_{T}$. For this reason, the quantity $\imath ET$ diverges in fact as $N_{L}^{d-1}N_{T}$. We can verify this fact doing a numerical evaluation of the sums involved (problem 5.1.6). Since the sums diverge, it is more convenient to evaluate the energy per site $e=E/(N_{L}^{d-1}N_{T})$,

  $$\imath eT=\frac{1}{N_{L}^{d-1}N_{T}}\sum_{k_{0}}\sum_{k_{i}} \fra... ...}^{2}+\alpha_{0}}{\displaystyle \rho_{0}^{2}+\sum_{i}\rho_{i}^{2}+\alpha_{0}},$$ (5.1.8)

The graph in figure 5.1.3 shows the result of such a calculation, done in the symmetrical case $N_{T}=N_{L}=N$. As one can see, the quantity $\imath eT$ converges quite rapidly to the value $(d-1)/d$, in each dimension, when we make $N\rightarrow \infty$, for any value of the mass $m_{0}$. In fact, one can demonstrate that this quantity can be written (problem 5.1.7) as

$$\imath eT =\frac{d-1}{d}+\frac{\alpha_{0}}{d}\sigma^{2}(N_{T},d,\alpha_{0}),$$

where $\sigma ^{2}$ is the local width of the field, a quantity that was extensively discussed in section 4.1. Since for $d\geq 2$ this quantity does not diverge faster than or as fast as $N_{T}^{2}$ in the $N_{T}\rightarrow \infty$ limit, while $\alpha_{0}=m_{0}^{2}/N_{T}^{2}$, the second term goes to zero because $\alpha_{0}$ goes to zero for finite $m_{0}$. The convergence is progressively faster for progressively smaller values of $m_{0}$.

Observe that, if we make $T\rightarrow \infty$ while we take the continuum limit, thus eliminating the infrared effects due to the finite size of the temporal box, the energy ceases to diverge as $N_{T}^{d}$ and becomes divergent as $N_{T}^{p}$, with a power $p$ in the range $d-1<p<d$. This is due to the fact that we can make $T$ go to infinity as $N_{T}^{q}$ only with $0<q<1$, since $q=0$ would of course correspond to keeping $T$ finite, while $q=1$ would imply that the lattice spacing $a$ would be kept finite, in which case the limit $N_{T}\rightarrow \infty$ would would no longer be a continuum limit.

In the traditional formalism of quantum field theory the result for the energy of the vacuum is constructed as the sum of a collection of quantum-mechanical ground-state results, one for each mode. We can try to obtain, from the result for the energy shown in equation (5.1.7), a somewhat clearer relation between the results of quantum mechanics and of the quantum theory of fields. However, we will see that this relation cannot be established in a completely exact and precise form, for reasons related to the order in which the limits involved are to be taken, a subject that will turn out to be very important later on. In order that we be able to sketch an argument to this effect, it is necessary that we consider the symmetrical limit in which $N_{T}=N_{L}=N$. The result obtained in the case of quantum mechanics for the sum over $k_{0}$ that appears in equation (5.1.7), in the $N_{T}\rightarrow \infty$ limit, for a large temporal box, implies that we have

$$\sum_{k_{0}}\frac{\displaystyle A^{2}}{\displaystyle \rho_{0}^{2}+A^{2}}\sim\frac{N_{T}A}{2},$$

so that we may write for the energy of the vacuum

$$E\sim-\frac{\imath N_{T}}{2T}\sum_{k_{i}} \sqrt{\sum_{i}\rho... ...rac{-\imath}{2}\sum_{k_{i}}\sqrt{\sum_{i}p_{i}^{2}+m_{0}^{2}},$$

where we used the fact that $\alpha_{0}N_{T}^{2}/T^{2}=m_{0}^{2}$ and that, for large $N_{T}=N_{L}=N$, $\rho_{i}N_{T}/T\approx p_{i}=2\pi k_{i}/T$, the dimensionfull linear momentum. In short, we may write that

$$\imath E\sim\frac{1}{2}\sum_{k_{i}}\sqrt{\mathbf{p}^{2}+m_{0}^{2}},$$

which is a sum of the relativistic energies of free particles with rest mass $m_{0}$ and linear momentum $\mathbf{p}$, with an overall factor of $1/2$, there being one term in the sum for each Fourier mode existing within a $(d-1)$-dimensional box. This is the sum of the so-called zero-point energies, the energies of the ground states of each one of the $N_{L}^{d-1}$ uncoupled harmonic oscillators that are associated to each one of these $(d-1)$-dimensional modes.

However, this argument cannot do more than to give us a general but imprecise idea about the relation between the case $d=1$ and the case $d\geq 2$ since, in order to make the argument rigorous, it would be necessary to first take the limit $N_{T}\rightarrow \infty$, thus reducing the problem to the quantum mechanics of a system with a finite number of degrees of freedom in its $(d-1)$-dimensional section, taking only after that the limit $N_{L}\rightarrow\infty$. However, the fact is that the results of the $d=1$ case that were used above are not valid if we take the limits over $N_{T}$ and $N_{L}$ separately in this order. This is due to the fact that the quantity $A^{2}=\sum_{i}\rho_{i}^{2}+\alpha_{0}$ which appears in the denominator of the sum over $k_{0}$ in equation (5.1.7) behaves in a way that is different from the behavior of the corresponding quantity in the case of quantum mechanics, $\alpha_{0}=(wT/N_{T})^{2}$. We should recall that the complete expression of $A^{2}$ is

$$A^{2}=\alpha_{0} +\sum_{i=1}^{d-1}4\sin^{2}\!\left(\frac{k_{i}\pi}{N_{L}}\right),$$

where we have $N_{L}$ within the argument of the sine function, not $N_{T}$. While $N_{T}^{2}\alpha_{0}$ has a finite limit when we make $N_{T}\rightarrow \infty$, the quantity

$$N_{T}^{2}\sin^{2}\!\left(\frac{k_{i}\pi}{N_{L}}\right)$$

does not have a finite limit, but instead of that diverges as $N_{T}^{2}$. If we had $N_{T}=N_{L}$, that is, the symmetrical limit, then the sine function would go to zero as $N_{T}^{-1}$, compensating for this divergence, but it is not possible to take the limit $N_{T}\rightarrow \infty$ while $N_{L}$ is kept finite and still obtain finite results. In addition to this, in this case we cannot say that $\rho_{i}N_{T}/T$ approaches $p_{i}=2\pi k_{i}/T$ as we did above.

As an exercise to illustrate the difference between the results of the two formalisms, starting from the result of the traditional formalism for the energy of the vacuum,

$$\imath E=\frac{1}{2}\sum_{k_{i}}\sqrt{\mathbf{p}^{2}+m_{0}^{2}},$$

we may translate it to the lattice, writing a corresponding result for the energy per site, valid for the case $N_{T}=N_{L}=N$ in the limit $N\rightarrow \infty$,

$$\imath eT=\frac{1}{2N^{d-1}} \sum_{k_{i}}\sqrt{\sum_{i}\rho_{i}^{2}+\alpha_{0}^{2}},$$

and then calculate numerically, for large values of $N$, the sum which appears in the resulting expression (problem 5.1.8). One verifies that the result obtained in this way is always larger that the result of the lattice formalism, for any value $d\geq 2$ of the dimension. In $d=4$, for example, we obtain approximately $1.1938$ for this result, to be compared to the lattice result $(d-1)/d=3/4=0.75$.

We see in this way that the results of the quantum theory of fields have the potential to depend in a significant way on the order of the limits over $N_{T}$ and $N_{L}$. In our case here this fact is of only secondary importance, because the energy of the vacuum diverges anyway, whatever the order of the limits, having therefore no direct physical relevance. We will see later on that, in order to define an energy that makes physical sense, it will be necessary to consider only the variations of the energy with respect to the energy of the vacuum, not the absolute value of the energy, a procedure that corresponds to what is called a subtractive “renormalization” of all the energies, exactly as is done in the traditional presentation of the theory. However, we see here that the usual argument of the traditional presentation, that this divergence is due to the sum of an infinite number of zero-point energies of harmonic oscillators, cannot be taken as more than an approximate intuitive argument, without exact mathematical validity.

Problems

1. Using the calculational techniques illustrated in the text, calculate the value at a give site $s$ of the observable $\langle[\bar\pi (s)-\imath\Delta_{0}\varphi(s)]^{2}\rangle$, obtaining the result shown in the text. You will have to perform some Gaussian integrations. Remember that the extremes of integration of the integrals on $\chi=\bar\pi -\imath\Delta_{0}\varphi$ depend on $\varphi$ until one makes the deformation of the integration contour in the complex $\chi$ plane.

2. Consider the expression of the energy derived in the text for the case of quantum mechanics, that is for $d=1$, which is shown in equation (5.1.6). It contains the sum

   
   

   $$\Sigma=\sum_{k_{0}=-k_{m}}^{k_{M}}\frac{\displaystyle 1}{\di... ...a T}\right)^{2}\sin^{2}\!\left(\frac{k_{0}\pi}{N_{T}}\right)}.$$
   
   

   Consider the case in which $N_{T}$ is even, for which the limits of $k_{0}$ are $k_{m}=1-N_{T}/2$ and $k_{M}=N_{T}/2$. Show that this sum satisfies the inequalities

   $$\begin{eqnarray*} \Sigma & \geq & -1+\frac{(\omega T)^{2}}{(\omega T)^{2}+(2N_{T... ...ga T}{2}\sqrt{\frac{(2N_{T})^{2}}{(\omega T)^{2}+(2N_{T})^{2}}}. \end{eqnarray*}$$
   
   
   
   

   In order to do this you should consider in separate the terms $k_{0}=0$ and $k_{0}=N_{T}/2$ of the sum and observe that all the others can be organized in pairs. In this way, rewrite the sum in terms of a sum of a monotonically decreasing function. After that, compare the sum with the integral of this function taken in an appropriate interval. Making a graph of the function, including the points of the sum and the integral, may help a lot. You will have to look up in a table of integrals the value of the definite integral

   
   

   $$\int_{0}^{\pi/2}\frac{1}{1+A^{2}\sin^{2}x}.$$
   
   

   Take the $N_{T}\rightarrow \infty$ limit of these relations and show that

   
   

   $$\frac{\omega T}{2}-1\leq\Sigma\leq\frac{\omega T}{2}+1.$$
   
   

3. The lower bound derived in problem 5.1.2 can be improved. Re-examine the comparison between the sum and the integral in this case and take into consideration the collection of triangles that can be fitted between the graph of the integral and the graph of the sum. Once more, drawing the graph with some care may help a lot. Calculate exactly the sum of the areas of these triangles and show that sum $\Sigma$ satisfies the inequality

   
   

   $$\frac{\omega T}{2}\sqrt{\frac{(2N_{T})^{2}}{(\omega T)^{2}+(2N_{T})^{2}}}\leq\Sigma,$$
   
   

   which is tighter than the previous one. You will have to show that the graph of the monotonically decreasing function that appears inside the sum has its concavity turned upwards along all the domain of the integral. You can do this calculating the second derivative of the function and showing that the first derivative is also a monotonic function, in this case an increasing function. After that take the limit $N_{T}\rightarrow \infty$ and show that

   
   

   $$\frac{\omega T}{2}\leq\Sigma.$$
   
   

   Show also that the value obtained for the lower bound of the sum in the limit $N_{T}\rightarrow \infty$ is larger than the value of the lower bound of the sum for finite $N_{T}$, that is, show that taking the $N_{T}\rightarrow \infty$ limit tightens the lower bound.

4. Repeat the analysis made in problems 5.1.2 and 5.1.3 and determine upper and lower bounds for the sum $\Sigma$ in the case in which $N_{T}$ is odd, in which the limits of $k_{0}$ are $k_{m}=-(N_{T}-1)/2$ and $k_{M}=(N_{T}-1)/2$. Write the corresponding inequalities for $\imath E$ and verify that they are compatible with the inequalities obtained for the case of even $N_{T}$.

5. Write a program to calculate numerically the sum that appears in the expression of the energy in the case of quantum mechanics, given in equation (5.1.6). Use your program to reproduce the data that are shown in the graphs of figures 5.1.1 and 5.1.2. Use lattices with sizes between $10$ and $1000$ and calculate $\imath ET$ for $\omega T$ from $0$ to $20$.

6. Write programs to calculate numerically the sum that appears in the expression of the energy per site for the cases $d\geq 2$, given in equation (5.1.8). Remember that in the quantum theory of fields we have $\alpha_{0}=m_{0}^{2}/N^{2}$ and consider only the symmetrical case $N_{T}=N_{L}=N$. Use your programs to reproduce the data that are shown in the graph of figure 5.1.3.

7. Show analytically that the sum that you calculated in problem 5.1.6 tends to the value $(d-1)/d$ in the limit $N\rightarrow \infty$. Your demonstration does not have to be strictly rigorous and you may consider the case $m_{0}\rightarrow 0$ it this simplifies things. However, be careful, because there is a term of the sum (the zero mode) which diverges for $m_{0}=0$. You can demonstrate this result by first showing that the energy per site may be written as

   
   

   $$\imath eT=\frac{d-1}{d}+\frac{\alpha_{0}}{d}\sigma^{2}(N,d,\alpha_{0}),$$
   
   

   where $\sigma ^{2}$ is the local width defined in section 4.1, which has a finite limit in the continuum limit for $d\geq 3$. The case $d=2$ has to be examined in separate, refer to the section mentioned in order to verify the behavior of $\sigma ^{2}$ in this case.

8. Write a program to calculate numerically the sum that appears in the result of the traditional formalism for the energy per site,

   
   

   $$\imath eT=\frac{1}{2N^{d-1}} \sum_{k_{i}}\sqrt{\sum_{i}\rho_{i}^{2}+\alpha_{0}^{2}},$$
   
   

   where $\rho_{i}=2\sin(k_{i}\pi/N)$. Remember that in the quantum theory of fields we have $\alpha_{0}=m_{0}^{2}/N^{2}$. Use your program for progressively larger values of $N$ and try to show that, in $d=4$, the limiting value for this result is approximately $1.1938$. Try to relate the fact that this result is larger than the corresponding result in the lattice formalism ($0.75$) with the behavior in the limit of the sum that appears in the lattice formalism in the case $d=1$, which is shown in the graph of figure 5.1.2.