Relation with Hilbert Spaces

In possession of the concept of energy and having constructed the states of particles within the structure of our theory, we will now examine another concept which is of central importance in the traditional formalism, the concept of the eigenstates of an observable. In the traditional formalism the observables are represented by Hermitian operators in a Hilbert space and the physical states are represented by vectors in this space. In that formalism the eigenstates of an observable are the eigenvectors of the corresponding operator. Our first problem here is to determine how to characterize the property of a state being or not being an eigenstate of an observable.

In order to build this characterization of the concept of eigenstate in our formalism, it is necessary to think in terms of expectation values, as it is in terms of the direct definition of these values that our formalism is built. Let us start by recalling that, if a state $\vert e\rangle$ is an eigenstate of an observable ${\cal O}$ with eigenvalue $\omega$, then we have in the traditional formalism

$${\cal O}_{\rm op}\vert e\rangle=\omega\vert e\rangle,$$

where ${\cal O}_{\rm op}$ is an operator and $\omega$ a number. In terms of expectation values we can write this in the form

$$\langle e\vert{\cal O}_{\rm op}\vert e\rangle=\omega,$$

where we used the fact that the states are normalized, $\langle e\vert e\rangle=1$. As a consequence of the relations above we also have

$${\cal O}_{\rm op}^{2}\vert e\rangle={\cal O}_{\rm op}\omega\... ...omega{\cal O}_{\rm op}\vert e\rangle=\omega^{2}\vert e\rangle,$$

so that we may write a relation between expectation values,

$$\langle e\vert{\cal O}_{\rm op}^{2}\vert e\rangle=\omega^{2}\langle e\vert e\rangle=\omega^{2},$$

or, in other words,

$$\langle e\vert{\cal O}_{\rm op}^{2}\vert e\rangle=\langle e\vert{\cal O}_{\rm op}\vert e\rangle^{2}.$$

This is the statement that the dispersion or width of the distribution of values of the operator ${\cal O}_{\rm op}$ on the state $\vert e\rangle$ is zero,

$$\sigma_{{\cal O}}^{2}=\langle e\vert{\cal O}_{\rm op}^{2}\vert e\rangle-\langle e\vert{\cal O}_{\rm op}\vert e\rangle^{2}=0.$$

In other words, the value of the observable ${\cal O}$ on the state $\vert e\rangle$ is completely well-defined, without fluctuations. This is a representation of the concept of eigenstate that we can translate directly to the lattice formalism,

$$\sigma_{{\cal O}}^{2}=\langle{\cal O}^{2}\rangle_{e}-\langle{\cal O}\rangle_{e}^{2}=0.$$

Of course, in the representation of the structure on the lattice, one does not expect that the dispersion of the observable on its eigenstates is necessarily zero on finite lattices, but only that it goes to zero in the continuum limit, and possibly only if we take, besides this one, the $T\rightarrow \infty$ limit as well. For example, for the action per site $s_{0}$, an observable which was examined in the problems proposed in section 4.2, we know that this is true for the vacuum state, since we have that $\langle s_{0}\rangle_{0}=1/2$ and that the dispersion goes to zero in the limit $N_{T}=N_{L}=N\rightarrow\infty$. Of course, that observable is of no direct physical interest in the context of our discussion in this section.

We have then our criterion to determine whether or not a given state is an eigenstate of a given observable: it suffices to calculate the dispersion of the observable on the state and verify whether or not the result vanishes in the continuum limit. It is possible to define very singular statistical distributions for which the dispersion of any given observable is zero, even on finite lattices (problem 5.3.1), but these distributions do not correspond to physical states and are of little interest to us in the context of the quantum theory of fields. What we should verify is whether or not the vacuum state represented by the Boltzmann distribution is an eigenstate of the modified Hamiltonian $\Delta\mathbf{H}$, in the continuum limit. The same should be done with the observable number-of-particles ${\cal N}$. We should therefore calculate the dispersions of these observables in the vacuum state.

We will start by calculating the dispersion of the observable ${\cal N}$ and, therefore, of the action, because the calculations are simpler in this case, since these observables do not depend on $\bar\pi$ and we may, therefore, use directly the usual definition for the expectation values. In addition do this, these are dimensionless observables, which makes it simpler to take the continuum limit. Since we have that ${\cal N}=S-\langle S\rangle_{0}$, we can easily show that the dispersion of ${\cal N}$ is equal to the dispersion of $S$,

$$\begin{eqnarray*} \langle{\cal N}^{2}\rangle-\langle{\cal N}\rangle^{2} & = & \l... ...ngle_{0}^{2} \\ & = & \langle S^{2}\rangle-\langle S\rangle^{2}, \end{eqnarray*}$$

for any state in which we may be measuring the dispersion. This is a general fact which is important for us: the addition of a constant to an observable, be it finite or divergent in the limit, does not change the dispersion of the observable. In order to calculate the dispersion of $S$ on the vacuum, we first obtain (problem 5.3.2) the results

$$\langle S\rangle_{0}=\frac{N_{T}N_{L}^{d-1}}{2},$$

and

$$\langle S^{2}\rangle_{0} =\left(\frac{N_{T}N_{L}^{d-1}}{2}\right)^{2}+\frac{N_{T}N_{L}^{d-1}}{2},$$

so that we have for the dispersion

$$\langle S^{2}\rangle_{0}-\langle S\rangle_{0}^{2} =\frac{N_{T}N_{L}^{d-1}}{2},$$

that is, we have

$$\sigma_{{\cal N}}=\sigma_{S}=\sqrt{\frac{N_{T}N_{L}^{d-1}}{2}},$$

which, instead of vanishing in the continuum limit, diverges, thus showing that the vacuum is not an eigenstate of the observable number-of-particles. Note that this dispersion is small by comparison with the value of $\langle S\rangle_{0}=N_{T}N_{L}^{d-1}/2$, so that we have

$$\frac{\sqrt{\langle S^{2}\rangle_{0}-\langle S\rangle_{0}^{2... ...angle_{0}}=\sqrt{\frac{2}{N_{T}N_{L}^{d-1}}}\longrightarrow 0,$$

that is, the relative dispersion goes to zero in the continuum limit, but the dispersion itself is not small by comparison to the finite value of $\langle{\cal N}\rangle_{0}$. Hence, the observable ${\cal N}$ gives us the correct number of particles in each state, including the value $0$ for the vacuum, but the vacuum is not an eigenstate of this observable.

With a little more work we can repeat these calculations for the states of $n$ particles with momentum $\vec{k}$ which we introduced in section 5.2 (problem 5.3.3). In this case we obtain the preliminary results

$$\langle S\rangle_{n,\vec{k}}=\frac{N_{T}N_{L}^{d-1}}{2}+n,$$

and

$$\langle S^{2}\rangle_{n,\vec{k}} =\left(\frac{N_{T}N_{L}^{d-... ...frac{1}{2}\right) +\left(n+\frac{1}{2}\right)^{2}-\frac{1}{4},$$

so that we have for the dispersion

$$\langle S^{2}\rangle_{n,\vec{k}}-\langle S\rangle_{n,\vec{k}}^{2} =\frac{N_{T}N_{L}^{d-1}}{2}+n,$$

which diverges in the continuum limit in the same way as before. Note that these results diverge even in the case $d=1$, which corresponds to quantum mechanics. However, this fact does not cause much preoccupation because the concept of the observable number-of-particles does not play any fundamental role in quantum mechanics. For the case $d>1$ it is also possible to calculate the dispersion of the operators ${\cal N}_{\vec{k}}$ (problem 5.3.4), which are given by

$${\cal N}_{\vec{k}}=\frac{1}{2}\left[N_{T}N_{L}^{d-1} \left(\... ...{0}\right)\vert\widetilde\varphi _{\vec{k}}\vert^{2}-1\right],$$

each one of which measures the number of particles with momentum $\vec{k}$. In this case we obtain, on the state of $n$ particles with momentum $\vec{k}$,

$$\langle{\cal N}_{\vec{k}}\rangle_{n,\vec{k}}=n,$$

and

$$\langle{\cal N}_{\vec{k}}^{2}\rangle_{n,\vec{k}}=n^{2}+n+\frac{1}{2},$$

so that we obtain for the dispersion

$$\langle{\cal N}_{\vec{k}}^{2}\rangle_{n,\vec{k}} -\langle{\cal N}_{\vec{k}}\rangle_{n,\vec{k}}^{2}=n+\frac{1}{2}.$$

In this case the result does not diverge, but we still have a value for the dispersion that does not vanish in the continuum limit, showing once more that the states of particles are not eigenstates of these observables.

In short, none of our states of particles are eigenstates of any of the observables that give, as their expectation values in these states, the corresponding numbers of particles. We will now proceed to the examination of the behavior of the observables related to the energy, which are the most important ones from the physical point of view. The observable of greatest relevance to us is the modified Hamiltonian

$$\Delta\mathbf{H}=\mathbf{H}-\langle\mathbf{H}\rangle_{0},$$

recalling that the dimensionless physical energy is given by

$$\Delta{\cal E}=\langle\Delta\mathbf{H}\rangle,$$

and that the physical energy relates to this dimensionless quantity by

$$\Delta E=\Delta{\cal E}/a=N_{T}\Delta{\cal E}/T.$$

We can calculate the expectation values of $\mathbf{H}^{2}$ and $(\Delta\mathbf{H})^{2}$ both by the canonical definition and by the usual definition. Since these observables depend on $\bar\pi$ the results of these two calculations will be different. In any case, since $\Delta\mathbf{H}$ and $\mathbf{H}$ are related by the addition of a constant quantity, we know beforehand that both will have the same dispersion. Because of this, we may calculate directly the dispersion of $\mathbf{H}$. Besides the question of using the canonical definition or the usual definition, it is also necessary to consider in detail the question of the temporal average, which we used before to facilitate the calculations. To take this average is equivalent to defining an average Hamiltonian over a temporal block, which we denote as

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

As we observed before, the invariance of the lattice model by discrete temporal translation implies that $\mathbf {H}_{b}$ and $\mathbf{H}$ have the same expectation value, but these are two conceptually different observables. The observable $\mathbf{H}$ corresponds to the measurement of the energy at a perfectly well-defined instant of time, while the observable $\mathbf {H}_{b}$ constitutes a type of block variable and we should expect that its fluctuations will be smaller than those of $\mathbf{H}$, due to the average over the temporal block. Since the dispersions of the observables are a measure of the average magnitude of the fluctuations that they undergo, the dispersions will be different. We may, in fact, predict that the dispersion of $\mathbf {H}_{b}$ will be smaller than that of $\mathbf{H}$, as it is characteristic for block variables.

We see therefore that we have several calculations to do, including two possibilities for the definition of the observable and two possibilities for the definition of the averages. We will present here the calculation of the dispersion of $\mathbf {H}_{b}$, which is simpler and sufficient for our purposes, leaving the case of $\mathbf{H}$ for the problems of this section (problems 5.3.5, 5.3.6 and 5.3.7). Besides, we will start with the canonical definition of the averages, showing, first of all, the relation of the result obtained by means of this definition with that obtained by means of the usual definition. Let us recall that the Hamiltonian density is given by

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

where summation over $i$ is implicit, so that we have the Hamiltonian

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

Calculating the temporal average of this quantity we obtain the blocked Hamiltonian

$$\mathbf{H}_{b}=\frac{1}{N_{T}}\sum_{t}\mathbf{H}=\frac{1}{N_... ...\pi ^{2}+(\Delta_{i}\varphi)^{2}+\alpha_{0}\varphi^{2}\right].$$

We will now calculate the dispersion of $\mathbf {H}_{b}$, which involves the calculation of the expectation values of $\langle\mathbf{H}_{b}\rangle$ and $\langle\mathbf{H}_{b}^{2}\rangle$. We will do the calculation starting by the canonical definition of the observables. So that it be a positive quantity, recalling that $\mathbf {H}_{b}$ is purely imaginary in our formalism, we define the dispersion $\sigma_{\mathbf{H}_{b}}$ by means of

$$-\sigma_{\mathbf{H}_{b}}^{2}=\langle\mathbf{H}_{b}^{2}\rangle-\langle\mathbf{H}_{b}\rangle^{2}.$$

As we saw in section 5.1, using the canonical definition of the expectation values we have for $\langle\mathbf{H}\rangle=\langle\mathbf{H}_{b}\rangle$ the result

$$\langle\mathbf{H}_{b}\rangle=-\frac{\imath}{2}N_{L}^{d-1} -\... ...}+\rho_{i}^{2}+\alpha_{0}}{\displaystyle \rho^{2}+\alpha_{0}},$$

with

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

so that we have for the square of this quantity

$$\hspace{-2em}\langle\mathbf{H}_{b}\rangle^{2} = -\frac{1}{4N_{T}^{2}}\left[ \left(N_{L}^{d-1}N_{T}\right)^{2} +2N... ...rho_{0}^{2}+\rho_{i}^{2}+\alpha_{0}}{\displaystyle \rho^{2}+\alpha_{0}} \right.$$

$$\left.\hspace{4em} +\left( \sum_{\vec{k}} \frac{\displaystyle -\r... ...2}+\rho_{i}^{2}+\alpha_{0}}{\displaystyle \rho^{2}+\alpha_{0}} \right) \right],$$ (5.3.1)

where it is understood that the variables within each sum have as their argument the argument of the sum. Let us now calculate $\langle\mathbf{H}_{b}^{2}\rangle$, starting by the integrals over $\bar\pi$,

$$\langle\mathbf{H}_{b}^{2}\rangle=-\frac{1}{4N_{T}^{2}}\left\... ...arphi)^{2}+\alpha_{0}\varphi^{2}\right] \right\}\right\rangle,$$

where, once more, it is understood that the variables within each sum have as their argument the argument of the sum. We may write this explicitly as

$$\begin{eqnarray*} \lefteqn{\langle\mathbf{H}_{b}^{2}\rangle= -\frac{1}{4N_{T}^{2... ...{i}\varphi)^{2}+\alpha_{0}\varphi^{2}\right]_{\vec{y}} \right\}, \end{eqnarray*}$$

where the indices $\vec{x}$ e $\vec{y}$ indicate the dependencies with each one of these two sites, and we have used our freedom to interchange $\vec{x}$ and $\vec{y}$ within the sums. Recalling that the exponential can be written as

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

we make the shift

$$\chi=\bar\pi -\imath\Delta_{0}\varphi\Longrightarrow \bar\pi... ...th\bar\pi \Delta_{0}\varphi =\chi^{2}+(\Delta_{0}\varphi)^{2},$$

so that we may write, reconstituting the complete form of the action in the exponent,

$$\begin{eqnarray*} \langle\mathbf{H}_{b}^{2}\rangle & = & -\frac{1}{4N_{T}^{2}Z}\... ...{i}\varphi)^{2}+\alpha_{0}\varphi^{2}\right]_{\vec{y}} \right\}, \end{eqnarray*}$$

where $S$ is the action and

$$Z=\int[{\rm d}\varphi][{\rm d}\bar\pi ] \;e^{\imath\sum_{s}(... ...arphi][{\rm d}\chi] \;e^{-\frac{1}{2}\sum_{s}\chi^{2}}\;e^{-S}$$

is the denominator that normalizes the expectation values. We may now expand the terms inside the brackets, collecting powers of $\chi$, making exchanges of $\vec{x}$ and $\vec{y}$, and recalling that terms with odd powers of $\chi$ at a given site vanish due to the symmetry of the integration, in order to arrive at the expression

$$\begin{eqnarray*} \langle\mathbf{H}_{b}^{2}\rangle & = & -\frac{1}{4N_{T}^{2}Z}\... ...)_{\vec{x}}^{2} -4{\cal H}_{\vec{x}}'{\cal H}_{\vec{y}}'\right], \end{eqnarray*}$$

where

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

is, as usual, the Hamiltonian density with $\bar\pi$ substituted by $\imath\Delta_{0}\varphi$. We may now use the known results

$$\frac{\displaystyle \int d\chi\;\chi^{2}\;e^{-\frac{1}{2}\ch... ...i^{2}}}{\displaystyle \int d\chi\;e^{-\frac{1}{2}\chi^{2}}}=3,$$

as well as the result, which can be easily obtained,

$$\sum_{\vec{x}}\langle(\Delta_{0}\varphi)^{2}\rangle =\sum_{\vec{k}}\frac{\rho_{0}^{2}}{\rho^{2}+\alpha_{0}},$$

in order to do the integrations and write

$$\langle\mathbf{H}_{b}^{2}\rangle=-\frac{1}{4}N_{L}^{2(d-1)} ... ...2}}{\rho^{2}+\alpha_{0}} +\langle{\mathbf{H}_{b}'}^{2}\rangle,$$

where we used the expression of the Hamiltonian with $\bar\pi$ substituted by $\imath\Delta_{0}\varphi$,

$$\mathbf{H}_{b}'=\frac{1}{N_{T}}\sum_{t}\mathbf{H}'=\frac{1}{N_{T}}\sum_{\vec{x}}{\cal H}'.$$

We now observe that the result of equation (5.3.1) can be written in terms of $\mathbf {H}_{b}'$ as

$$\langle\mathbf{H}_{b}\rangle^{2}=-\frac{1}{4}N_{L}^{2(d-1)}-... ...angle\mathbf{H}_{b}'\rangle+\langle\mathbf{H}_{b}'\rangle^{2},$$

so that we may write for the dispersion of $\mathbf {H}_{b}$

$$-\sigma_{\mathbf{H}_{b}}^{2}=\langle\mathbf{H}_{b}^{2}\rangl... ...}^{2}} \sum_{\vec{k}}\frac{\rho_{0}^{2}}{\rho^{2}+\alpha_{0}}.$$

We can manipulate the last two terms of this expression and verify that they are proportional to the expectation value of $\mathbf {H}_{b}'$,

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

so that we can write the dispersion of $\mathbf {H}_{b}$ in terms of the dispersion of $\mathbf {H}_{b}'$ as

$$-\sigma_{\mathbf{H}_{b}}^{2}=\langle\mathbf{H}_{b}^{2}\rangl... ..._{b}'}^{2} -\frac{\imath}{N_{T}}\langle\mathbf{H}_{b}'\rangle,$$

that is,

$$\sigma_{\mathbf{H}_{b}}^{2}=\sigma_{\mathbf{H}_{b}'}^{2} +\frac{\imath}{N_{T}}\langle\mathbf{H}_{b}'\rangle.$$

We see here that the dispersions of $\mathbf {H}_{b}$ and of $\mathbf {H}_{b}'$, that is, the dispersions of $\mathbf {H}_{b}$ according to the canonical definition and according to the usual definition, are not too different, since the extra term is damped by a factor of $1/N_{T}$ and should not have much importance in the continuum limit.

We must now calculate, in explicit form, the dispersion of $\mathbf {H}_{b}'$, for which it is necessary to calculate $\langle{\mathbf{H}_{b}'}^{2}\rangle$. For this end it is convenient to first write $\mathbf {H}_{b}'$ in terms of the Fourier transforms of the fields,

$$\mathbf{H}_{b}'=-\frac{\imath}{2N_{T}}\sum_{\vec{x}} \left[-... ...^{2}+\rho_{i}^{2}+\alpha_{0})\vert\widetilde\varphi \vert^{2}.$$

With this we can write for the expectation value of ${\mathbf{H}_{b}'}^{2}$

$$\langle{\mathbf{H}_{b}'}^{2}\rangle =-\frac{N_{L}^{2(d-1)}}{... ...k}}\vert^{2}\vert\widetilde\varphi _{\vec{q}}\vert^{2}\rangle.$$

In order to calculate the indicated expectation values it is necessary to consider in detail and separately the cases in which $\vec{k}=\pm\vec{q}$ and the cases in which $\vec{k}\neq\pm\vec{q}$. In addition to this it is necessary to recall that the expectation values of products of four fields have different behaviors when the Fourier components are real, in comparison to the case in which they have non-vanishing imaginary parts. For simplicity of the argument, let us consider explicitly the case in which $N$ is odd, in which the only real Fourier component is the zero mode, $\vec{k}=\vec{0}$. As usual in this type of calculation, once the answer is obtained in terms of complete sums over the modes in momentum space, we may lift the restriction that $N$ be odd without affecting the validity of the answer. Under these conditions we have the following four mutually exclusive possibilities:

$$\vec{k}=\vec{q}=\vec{0},\hspace{2em} \vec{k}=\vec{q}\neq\vec... ...\vec{k}=-\vec{q}\neq\vec{0},\hspace{2em}\vec{k}\neq\pm\vec{q}.$$

Each pair $(\vec{k},\vec{q}\,)$ which is possible is exclusively in one of these four categories, while the union of the four exhausts all the possibilities for the pairs. With this we can write for our expectation value

$$\begin{eqnarray*} \langle{\mathbf{H}_{b}'}^{2}\rangle & = & -\frac{N_{L}^{2(d-1)... ...ert^{2}\vert\widetilde\varphi _{\vec{q}}\vert^{2}\rangle\right]. \end{eqnarray*}$$

The expectation values involving different momenta can be factored and, in addition to this, we can use the known results

$$\begin{eqnarray*} \langle\vert\widetilde\varphi _{\vec{k}=\vec{0}}\vert^{4}\rang... ...& = & \frac{1}{N_{L}^{d-1}N_{T}}\;\frac{1}{\rho^{2}+\alpha_{0}}, \end{eqnarray*}$$

to write

$$\begin{eqnarray*} \langle{\mathbf{H}_{b}'}^{2}\rangle & = & -\frac{1}{4N_{T}^{2}... ...}^{2}+\alpha_{0}} {\rho^{2}+\alpha_{0}}\right)_{\vec{q}}\right]. \end{eqnarray*}$$

One of the three units in the first term and two of the four units in the second term may now be joined with the third term in order to complete the sum that appears in this last one, resulting in

$$\begin{eqnarray*} \langle{\mathbf{H}_{b}'}^{2}\rangle & = & -\frac{1}{4N_{T}^{2}... ...}^{2}+\alpha_{0}} {\rho^{2}+\alpha_{0}}\right)_{\vec{q}}\right]. \end{eqnarray*}$$

The two remaining units in the first term may now be joined with the second term in order to complete the sum that appears in this one, resulting in

$$\langle{\mathbf{H}_{b}'}^{2}\rangle=-\frac{1}{2N_{T}^{2}}\su... ...^{2}+\rho_{i}^{2}+\alpha_{0}}{\rho^{2}+\alpha_{0}} \right)^{2}$$

We observe now that the last term may be written in terms of $\langle\mathbf{H}_{b}'\rangle$ and we obtain

$$\langle{\mathbf{H}_{b}'}^{2}\rangle=-\frac{1}{2N_{T}^{2}}\su... ...{2}+\alpha_{0}} \right)^{2}+\langle\mathbf{H}_{b}'\rangle^{2}.$$

With this we finally obtain for the dispersion of $\mathbf {H}_{b}'$

$$-\sigma_{\mathbf{H}_{b}'}^{2} =\langle{\mathbf{H}_{b}'}^{2}\... ...{2}+\rho_{i}^{2}+\alpha_{0}}{\rho^{2}+\alpha_{0}} \right)^{2},$$

and, consequently, for the dispersion of $\mathbf {H}_{b}$

$$-\sigma_{\mathbf{H}_{b}}^{2} =\langle\mathbf{H}_{b}^{2}\rang... ...c{-\rho_{0}^{2}+\rho_{i}^{2}+\alpha_{0}}{\rho^{2}+\alpha_{0}}.$$

We may join the two sums that appear in this expression into a single sum, thus obtaining

$$-\sigma_{\mathbf{H}_{b}}^{2}=\langle\mathbf{H}_{b}^{2}\rangl... ..._{0}^{2}+\rho_{i}^{2}+\alpha_{0})}{(\rho^{2}+\alpha_{0})^{2}},$$

that is,

$$\sigma_{\mathbf{H}_{b}}^{2}=\frac{1}{N_{T}^{2}}{\cal S}_{\mathbf{H}_{b}},$$

with the definition of a symbol for the sum over the momenta,

$${\cal S}_{\mathbf{H}_{b}}=\sum_{\vec{k}}\frac{(\rho_{i}^{2}+... ..._{0}^{2}+\rho_{i}^{2}+\alpha_{0})}{(\rho^{2}+\alpha_{0})^{2}}.$$

Although this sum is not manifestly positive, it is in fact positive, as can be verified numerically. If we think about the symmetrical limit $N_{L}=N_{T}$ it becomes clearer that the positive terms predominate in the only factor which is not manifestly positive, the second factor in the numerator of the sum. This is due to the fact that there is an implicit sum over $i$ in $\rho_{i}^{2}$, so that we have $d-1$ positive terms and only $1$ negative term in the sum of the $\rho$'s. Since the possible values for each $\rho_{i}$ and for each $\rho_{0}$ are the same, we arrive at the conclusion that the sums are predominantly positive for sufficiently large dimensions $d$. The only case which raises some doubt, if we recall that $\alpha_{0}\rightarrow 0$ in the continuum limit, is the case $d=2$, but in this case it is possible to rewrite the sum in a manifestly positive form (problem 5.3.8).

Figure 5.3.1: The sum ${\cal S}_{\mathbf {H}_{b}}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}$, calculated according to the canonical definition of the expectation values, in the case $d=1$.

Figure 5.3.2: The sum ${\cal S}_{\mathbf {H}_{b}}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}$, calculated according to the canonical definition of the expectation values, divided by $N^{d}$, in the case $d=2$.

Figure 5.3.3: The sum ${\cal S}_{\mathbf {H}_{b}}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}$, calculated according to the canonical definition of the expectation values, divided by $N^{d}$, in the case $d=3$.

Figure 5.3.4: The sum ${\cal S}_{\mathbf {H}_{b}}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}$, calculated according to the canonical definition of the expectation values, divided by $N^{d}$, in the case $d=4$.

Figure 5.3.5: The sum ${\cal S}_{\mathbf {H}_{b}}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}$, calculated according to the canonical definition of the expectation values, divided by $N^{d}$, in the case $d=5$.

Figure 5.3.6: The sum ${\cal S}_{\mathbf {H}_{b}'}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}'$, that is, the dispersion of $\mathbf {H}_{b}$ calculated according to the usual definition of the expectation values, divided by $N^{d}$, in the case $d=1$.

Figure 5.3.7: The sum ${\cal S}_{\mathbf {H}_{b}'}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}'$, that is, the dispersion of $\mathbf {H}_{b}$ calculated according to the usual definition of the expectation values, divided by $N^{d}$, in the case $d=2$.

Figure 5.3.8: The sum ${\cal S}_{\mathbf {H}_{b}'}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}'$, that is, the dispersion of $\mathbf {H}_{b}$ calculated according to the usual definition of the expectation values, divided by $N^{d}$, in the case $d=3$.

Figure 5.3.9: The sum ${\cal S}_{\mathbf {H}_{b}'}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}'$, that is, the dispersion of $\mathbf {H}_{b}$ calculated according to the usual definition of the expectation values, divided by $N^{d}$, in the case $d=4$.

Figure 5.3.10: The sum ${\cal S}_{\mathbf {H}_{b}'}$ that appears in the expression of the dispersion of $\mathbf {H}_{b}'$, that is, the dispersion of $\mathbf {H}_{b}$ calculated according to the usual definition of the expectation values, divided by $N^{d}$, in the case $d=5$.

The issue now is to determine how these sums behave in the continuum limit. Since the term that is being added is homogeneous of order zero on the $\rho$'s, and therefore typically of the order of $1$, it is to be expected that the sums behave as $N_{L}^{d-1}N_{T}$ or, in the symmetrical limit, as $N^{d}$, which is the number of terms in the sum. One exception should be the case $d=1$, in which all the $\rho_{i}$'s disappear and the situation changes qualitatively. We can easily evaluate these sums numerically in the symmetrical case, obtaining what is seen in the figures from 5.3.1 to 5.3.5. In all cases we have that ${\cal S}_{\mathbf{H}_{b}}=1$ for $N=1$. In figure 5.3.1 we see that the case $d=1$ differs from the others, because in this case the sum has a finite limit of the order of $1$. In all the other cases the sum behaves as $N^{d}$, so that it is the ratio ${\cal S}_{\mathbf{H}_{b}}/N^{d}$ that is plotted in the graphs of figures from 5.3.2 to 5.3.5, a ratio which approaches a finite limit of the order of $1$ in these cases.

Thus we see that the case $d=1$ of quantum mechanics is the only one in which the vacuum is an eigenstate of the time-blocked Hamiltonian $H_{b}$. It suffices to observe that in this case we have for the dimensionfull dispersion $\Sigma_{\mathbf{H}_{b}}=\sigma_{\mathbf{H}_{b}}/a$, which is the one that corresponds to the dimensionfull physical energy, the behavior in the limit of large $N_{T}$,

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim\frac{1}{T^{2}},$$

so that it is enough to make $T\rightarrow \infty$ for the dispersion to vanish. Note that we may take first the limit $N_{T}\rightarrow \infty$ and only after that make $T$ go to infinity. In this particular case we can verify that this result is still obtained if we use the non-blocked Hamiltonian $H$ instead of $H_{b}$, but is a much weaker form, since we are forced to adopt a particular way to take the limits for the dispersion to vanish in the limit (problem 5.3.9).

To complete the discussion we may also examine the corresponding results for $\mathbf {H}_{b}'$, that is, calculated according to the usual definition of the expectation values, instead of the canonical definition. In this case the result can be written as

$$\sigma_{\mathbf{H}_{b}'}^{2}=\frac{1}{N_{T}^{2}}{\cal S}_{\mathbf{H}_{b}'},$$

where the sum is defined by

$${\cal S}_{\mathbf{H}_{b}'}=\frac{1}{2}\sum_{\vec{k}} \left(\... ...{2}+\rho_{i}^{2}+\alpha_{0}} {\rho^{2}+\alpha_{0}}\right)^{2}.$$

The results of the numerical evaluation of the sums ${\cal S}_{\mathbf {H}_{b}'}$, in the case of the symmetrical limit, can be seen in the figures from 5.3.6 to 5.3.10. The only case in which there is a qualitative difference in the results is the case $d=1$, whose sum no longer has a finite value as was the case for the canonical definition. Note that figure 5.3.6 shows the dispersion per site, not simply the dispersion like figure 5.3.1 does. We see that according to the usual definition of the expectation values the sums behave as $N^{d}$ in any dimension. All that one can conclude form these results, based on what happens in the case $d=1$, is that the most sensible way to calculate the dispersion of the Hamiltonian is the canonical way. But there is no qualitative change in the situation in the case $d\geq 2$.

Note that in the case of quantum mechanics, although the sum now diverges as $N_{T}$, this still does not prevent us from making the dimensional dispersion go to zero in the limit, since in this case we have for $\Sigma_{\mathbf{H}_{b}'}=\sigma_{\mathbf{H}_{b}'}/a$

$$\Sigma_{\mathbf{H}_{b}'}^{2}\sim\frac{N_{T}}{T^{2}},$$

so that we can make the dispersion go to zero in limits in which we make $T$ increase with $N_{T}$ is a sufficiently fast way. This type of limit is the same that we are forced to use if we try to employ the non-blocked Hamiltonian $H$ with the canonical definition of the expectation values (problem 5.3.9). Limits of this type will be discussed in detail in a little while, for the case of the quantum theory of fields. On the other hand, if we try to use the non-blocked Hamiltonian $H$ and the non-canonical definition of the expectation values, then we verify that it is not possible to make the dispersion go to zero in the limit even in the case of quantum mechanics (problem 5.3.10).

In order to discuss in a more direct way the physical significance of these results, it is necessary to first translate these dimensionless results in terms of the dimensionfull physical energy, as we did above for the case of quantum mechanics. We will use in this discussion the results obtained according to the canonical definition of the expectation values. The dispersion $\Sigma_{\mathbf{H}_{b}}^{2}=\sigma_{\mathbf{H}_{b}}^{2}/a^{2}$ of the dimensionfull energy is given by

$$\Sigma_{\mathbf{H}_{b}}^{2}=\frac{1}{T^{2}}{\cal S}_{\mathbf{H}_{b}},$$

where we recall that $T$ is the temporal size of the box. We see that in the case $d=1$, since ${\cal S}_{\mathbf {H}_{b}}$ tends to a constant, the dispersion goes indeed to zero when we make $T\rightarrow \infty$, so that in this case the vacuum state is indeed an eigenstate of the blocked Hamiltonian. However, in all other cases the fact that the sum ${\cal S}_{\mathbf {H}_{b}}$ diverges as $N^{d}$ means that we have $\Sigma_{\mathbf{H}_{b}}^{2}\sim N^{d}/T^{2}$, so that it is not possible to make the dispersion go to zero in the limit, and therefore in all these cases the vacuum state is not an eigenstate of the Hamiltonian. The borderline case is the case $d=2$, in which we have

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim\frac{N^{2}}{T^{2}}.$$

We see here that, if we make $T$ increase in the continuum limit in a sufficiently fast way, in order to compensate the increase of $N$, we end up preventing the limit from being in fact a continuum limit, since in order to cause the dimensionfull width to vanish it is necessary to make $a\rightarrow\infty$ instead of $a\rightarrow 0$. We can see this if we recall that $T=Na$, so that the expression above can be written as

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim\frac{1}{a^{2}}.$$

For larger dimensions the situation gets progressively worse. So long as we limit ourselves to taking the continuum limit in the symmetrical way, the situation seems to be that the concept of eigenstate and, ultimately, the concept of Hilbert space, only apply to the case of quantum mechanics, and not to the case of the quantum theory of fields.

We will therefore examine what happens if we take the continuum limit in a non-symmetrical way, which obviously only makes sense for $d\geq 2$. The most extreme case in this context is to take first the limit $N_{T}\rightarrow \infty$ with fixed $N_{L}$, as in the case of quantum mechanics, and only after that take the limit $N_{L}\rightarrow\infty$. The effect of this procedure is to first reduce the system to the quantum mechanics of an arbitrary but finite number of degrees of freedom, and only after that make the number of degrees of freedom increase without limit. In this case all the sums ${\cal S}_{\mathbf {H}_{b}}$ and ${\cal S}_{\mathbf {H}_{b}'}$ for $d\geq 2$ behave simply as $N_{T}$ when we take the first limit. In fact, examining the behavior of the terms of the sums in the limit we can see that the sums tend to the value $N_{L}^{d-1}N_{T}$. Writing explicitly the general term $t(\vec{k})$ of the sum for the case of ${\cal S}_{\mathbf {H}_{b}'}$ we have

$$t(\vec{k})=\left[\frac{-4\sin^{2}\!\left(\frac{k_{0}\pi}{N_{... ...\!\left(\frac{k_{i}\pi}{N_{L}}\right) +\alpha_{0}}\right]^{2}.$$

Recalling that $\alpha_{0}=m_{0}^{2}a^{2}$ for some finite mass $m_{0}$, which implies that $N_{T}^{2}\alpha_{0}=m_{0}^{2}T^{2}$, as well as that for finite $k_{0}$ and $N_{T}\rightarrow \infty$ the argument of the sine function goes to zero, so that we may approximate it by its argument in the first terms in the numerator and in the denominator, we may multiply numerator and denominator by $N_{T}^{2}$ and write, for most terms in the sum, that

$$t(\vec{k})\longrightarrow\left[\frac{-(2\pi k_{0})^{2} +N_{T... ...i}{N_{L}}\right) +(N_{T}/N_{L})^{2}m_{0}^{2}L^{2}}\right]^{2}.$$

Now, since $1/N_{L}$ does not go to zero, the sine function that appears in the second terms in the numerator and in the denominator does not become small, so that these terms diverge like $N_{T}^{2}$, as do the third terms. Hence, the only difference between the numerator and the denominator, which is the sign of the first term, tends to disappear, so that we obtain, in the limit $N_{T}\rightarrow \infty$ with fixed $N_{L}$ and finite $T$,

$${\cal S}_{\mathbf{H}_{b}'}\longrightarrow\sum_{\vec{k}}1=N_{L}^{d-1}N_{T}.$$

The same is true if we make $T\rightarrow \infty$ together with $N_{T}\rightarrow \infty$, since $T$ has to increase slower than $N_{T}$ in order to guarantee that $a\rightarrow 0$, that is, that we have in fact a continuum limit. It follows that the term involving $m_{0}$ always becomes negligible in the limit, besides the fact that its presence would not change, in any case, the fact that the terms of the sum ${\cal S}_{\mathbf {H}_{b}'}$ tend to $1$. As one can see, in this type of asymmetrical limit all the sums tend to $N_{L}^{d-1}N_{T}$ in the limit, diverging, therefore, as $N_{T}$ in the first limit involved. The same type of behavior can be verified for the sum ${\cal S}_{\mathbf {H}_{b}}$ (problem 5.3.14).

Note that this argument for the evaluation of the sums is not completely rigorous, because it is clear that there are always some terms of the sums for which $k_{0}$ is of the order of $N_{T}$ and for which we cannot approximate the sine function by its argument. If one examines the behavior of these terms one realizes that we may have over-evaluated the sums. However, with basis on the fact that these terms were not enough to avoid the divergent behavior of the sums as $N_{L}^{d-1}N_{T}$ even in the case of the symmetrical limit, in which they are relatively more important, we may expect that they do not change the divergent behavior of the sums in our limit here. At most, we may expect a change in the multiplicative constant, to the effect that the sums behave as

$${\cal S}_{\mathbf{H}_{b}},{\cal S}_{\mathbf{H}_{b}'}\longrightarrow C(d)N_{L}^{d-1}N_{T},$$

with, in each dimension, some positive constant $C(d)$, smaller than and of the order of $1$. However, in order to check these facts beyond any doubt, it is necessary to evaluate numerically these sums in this type of asymmetrical limit (problem 5.3.15).

Let us observe that in this type of limit we have for the dimensionfull dispersions, both for $\Sigma_{\mathbf{H}_{b}}$ and for $\Sigma_{\mathbf{H}_{b}'}$, in dimensions $d\geq 2$, the behavior

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim N_{L}^{d-1}\frac{N_{T}}{T^{2}}.$$

This allows us to make the dispersion vanish in the limit, it suffices to make $T$ go to infinity sufficiently fast as a function of $N_{T}$, in order to compensate the increase of $N_{T}$. We can do this by relating $T$ to some finite temporal length ${\cal T}$ (something like a mean life) by means of

$$T=N_{T}^{p}{\cal T}.$$

There are limits for the possible values of the power $p$. In order for $T$ to go to infinity in the limit, we must have $p>0$. On the other hand, we must remember that $T=N_{T}a$ and that there is also the need to make $a$ go to zero in the limit, so that it be in fact a continuum limit. Combining this condition with the equation above we obtain

$$T=N_{T}a=N_{T}^{p}{\cal T}\Longrightarrow{\cal T}=N_{T}^{1-p}a,$$

so that in order that we have ${\cal T}$ finite with $a\rightarrow 0$ it is necessary that $1-p>0$, that is, that $p<1$. Joining these two conditions we see that $p$ must be inside the open interval $(0,1)$. Writing now the dispersion $\Sigma_{\mathbf{H}_{b}}$ in this type of limit we obtain

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim N_{L}^{d-1}\frac{N_{T}}{{\ca... ...^{2}N_{T}^{2p}}= \frac{N_{L}^{d-1}}{{\cal T}^{2}}N_{T}^{1-2p}.$$

In order for this to vanish in the limit we must have $1-2p<0$, that is, $p>1/2$. This set of conditions over $p$ can be satisfied by values of $p$ in the open interval $(1/2,1)$, for example $p=3/4$.

As another way to define asymmetrical limits, we can also generalize the symmetrical limits to the case in which both $N_{L}$ and $N_{T}$ increase in the limit, but with $N_{T}$ increasing faster than $N_{L}$, thus establishing an asymmetry. We will call this type of limit the “simultaneous asymmetrical limits”. It suffices to establish between these two quantities a relation of the type

$$N_{L}=N_{T}^{q}.$$

In order for $N_{L}$ to increase slower than $N_{T}$, but so that both still increase simultaneously, we must have $0<q<1$. Observe that the terms of the sums still have the same type of behavior that we saw before in the case of the fully asymmetrical limits. In this case, since both $N_{T}$ and $N_{L}$ increase in the limit, all the sine functions that appear in the terms of the sums can be approximated by their arguments. For example, in the case of ${\cal S}_{\mathbf {H}_{b}'}$, which we examined before, we now have

$$t(\vec{k})\longrightarrow\left[\frac{-(2\pi k_{0})^{2} +(N_{... ...(2\pi k_{i})^{2} +(N_{T}/N_{L})^{2}m_{0}^{2}L^{2}}\right]^{2}.$$

Due to the factor $(N_{T}/N_{L})^{2}=N_{T}^{2(1-q)}$, which still diverges because $q<1$ implies that the exponent is strictly positive, it is still true that, as before, the second and third terms of the numerator and of the denominator diverge with respect to the first, so that the terms approach $1$ for finite $\vec{k}$. Combining these results with the increase of $T$ in the limit, as was discussed before, we obtain for the dimensionfull dispersion $\Sigma_{\mathbf{H}_{b}}$, for example, the behavior

$$\Sigma_{\mathbf{H}_{b}}^{2}\sim\frac{1}{T^{2}}N_{L}^{d-1}N_{... ..._{T}^{q(d-1)}N_{T} =\frac{1}{{\cal T}^{2}}N_{T}^{1+q(d-1)-2p},$$

so that in order for $\Sigma_{\mathbf{H}_{b}}$ to vanish in the limit we must have $2p>1+(d-1)q$. Since $0<p<1$ this condition results in

$$\frac{1+(d-1)q}{2}<p<1.$$

We may satisfy all the conditions over $p$ and $q$ with, for example, the choice $q=C/(d-1)$, with some constant $C$ in the open interval $(0,1)$ and $p$ chosen in the open interval $((C+1)/2,1)$. The conclusion is that there is no qualitative change in the results when we include this type of simultaneous asymmetrical limit. It is always possible to find limits in which the dispersion of the energy goes to zero, so long as we make $N_{L}$ increase slower than $N_{T}$ in the limit, and so long as we also make $T$ increase without limit in the limit.

However, none of these asymmetrical limits helps us to solve completely the problem of how to make the vacuum state become an eigenstate of the Hamiltonian and at the same time keep intact all the fundamental physical characteristics of the theory. The reason for this is that any limit that is not symmetrical, that is, any limit in which one has $N_{L}=N_{T}^{q}$ with $q\neq 1$, destroys the on-shell condition and causes the theory not to contain any states of particles with energy different from zero, in the continuum limit. We can see this writing once more the expression, in Minkowski space, of the energy of the state of one particle with momentum $\vec{k}$ which we discussed in section 5.2,

$$\Delta E_{1,\vec{k}}=\frac{-1}{T}\; \frac{\rho_{0}^{2}(\vec{... ...ho_{0}^{2}(\vec{k})+\sum_{i}\rho_{i}^{2}(\vec{k})+\alpha_{0}}.$$

Let us recall that, since $T\rightarrow \infty$, this energy does not go to zero only if one of the two factors in which the denominator can be factored vanishes as $1/T$ in the limit. Writing explicitly the $\rho$'s we have

$$\Delta E_{1,\vec{k}}=\frac{-1}{T}\; \frac{4\sin^{2}\!\left(\... ...{i}4\sin^{2}\!\left(\frac{k_{i}\pi}{N_{L}}\right)+\alpha_{0}}.$$

We may now multiply the numerator and the denominator by $N_{T}^{2}$ and take the limit $N_{T}\rightarrow \infty$ with $N_{L}=N_{T}^{q}$ and $q>0$. Note that, since in this case the momentum $\vec{k}$ is fixed and finite, we can approximate all the sine functions by their arguments without introducing any imprecision of thought. When we do this we obtain

$$\Delta E_{1,\vec{k}}\sim\frac{-1}{T}\;\frac{(2\pi k_{0})^{2}... ...(1-q)}\sum_{i}(2\pi k_{i})^{2} +N_{T}^{2(1-q)}m_{0}^{2}L^{2}}.$$

Since $k_{0}$, $k_{i}$ and $m_{0}$ are finite and $q\neq 1$, in the limit in which $N_{T}\rightarrow \infty$ with $T\rightarrow \infty$ the first term becomes negligible by comparison with the other two, both in the numerator and in the denominator, so that we are left with the relation

$$\Delta E_{1,\vec{k}}=\lim_{T\rightarrow\infty}\frac{1}{T},$$

which goes to zero. Since the energy of the corresponding state of $n$ particles is $n$ times this result, we see that the energies of all the particle states collapse to zero in this type of limit. In other words, none of the states has energy different from the energy of the vacuum state in the continuum limit, whatever the momentum-space mode it is related to. Another way to say this is that, in this type of limit, there are no physical states left except the vacuum, once the limit is taken. One is left with an empty theory.

The fundamental reason causing this behavior can be identified as the underlying relativistic invariance of the theory. It is this invariance that implies the form of the action and therefore the form of the terms in the results for the energy, with the sum of $\rho_{0}^{2}$ and of the $\rho_{i}^{2}$, all with coefficients $1$. The terms $(\Delta_{0}\varphi)^{2}$, $(\Delta_{i}\varphi)^{2}$ and $\alpha_{0}\varphi^{2}$ in the action lead directly to the terms $\rho_{0}^{2}$, $\rho_{i}^{2}$ and $\alpha_{0}$ contained in our results. In the ultimate analysis, the relativistic invariance requires that the continuum limit be taken in a symmetrical way.

Note that this does not mean that the box inside which we are defining our model has to be exactly cubical. We may have a fixed proportionality relation between $N_{T}$ and $N_{L}$, such as $N_{T}=CN_{L}$ with some constant $C$, meaning that the temporal size $T$ and the spacial size $L$ of the box may not be the same. However, it is necessary that the continuum limit be taken in a symmetrical way, that is, that $N_{T}$ and $N_{L}$ increase with the same speed in the limit. So long as the lattice spacing $a$ remains the same in all the directions of the lattice there is no change in the form of the action and therefore no change in our results here. In other words, the requirement of symmetry in the continuum limit is a characteristic related to the ultraviolet regime, not to the infrared regime of the theory.

We can improve to some extent our understanding of the difficulties that we face in the definition of the quantum theory of fields if we once again turn our attention to the interpretation of our lattice structure in the terms of usual quantum mechanics. If we take the limit in the completely asymmetrical form, keeping $N_{L}$ fixed, our structure is reduced to the quantum mechanics of a certain number of degrees of freedom associated to the sites. It becomes in fact a set of coupled harmonic oscillators with mass $M$, located at the sites, with frequency $\omega=m_{0}=\sqrt{K/M}$, whose elastic constant $K$ is associates to the term $\alpha_{0}\varphi^{2}$ of the action by

$$\alpha_{0}=\omega^{2}a^{2}=\frac{K}{M}a^{2} =\frac{K}{M}\left(\frac{T}{N_{T}}\right)^{2},$$

where $T/N_{T}$ is the lattice spacing $a$ in the temporal direction. Note that finite $M$ and $K$ imply that $\alpha_{0}$ goes to zero in the limit, as usual. On the other hand, in the asymmetrical limit we should look at the term $\beta_{0}(\Delta_{i}\varphi)^{2}$ of the action, with $\beta_{0}=1$, as a coupling term between two oscillators at neighboring sites, a term which naturally depends on the difference of position (or rather of elastic elongation) $\Delta\phi=\sqrt{M}\Delta x$ between the two neighboring oscillators. This is an elastic interaction with spring constant $K'$ between these two neighbors, but the coefficient involved is simply

$$\beta_{0}=\frac{K'}{M}a^{2} =\frac{K'}{M}\left(\frac{T}{N_{T}}\right)^{2}=1,$$

which does not go to zero in the limit like $\alpha_{0}$ does, so that these interactions between neighbors are infinitely strong from the point of view of quantum mechanics, corresponding to $K'\rightarrow\infty$ when $N_{T}\rightarrow \infty$.

In order to compensate for this fact, making the elastic interactions between the sites finite in the limit, it would be necessary to make $K'$ finite, which implies making $\beta_{0}\rightarrow 0$ in the limit, which is equivalent to violating completely the relativistic symmetries of the action of the corresponding quantum theory of fields. In order to understand the significance of all this from the point of view of field theory, let us observe that from the point of view of that theory the introduction of $\beta_{0}\neq 1$ is equivalent to the introduction into the system of a velocity $\nu$ different from the velocity of light $c=1$, through the relation $\beta_{0}=\nu^{2}$, so that we can now write the action as

$$\begin{eqnarray*} S & = & \frac{1}{2}\sum_{s}\left[(\Delta_{0}\varphi)^{2} +\bet... ...+\nu^{2}\sum_{i}(\partial_{i}\phi)^{2}+m_{0}^{2}\phi^{2}\right]. \end{eqnarray*}$$

If we consider the case of the massless theory $\alpha_{0}=0$, then this parameter is indeed the velocity of propagation of waves in the system (in its de-Euclideanized version, of course), which ceases to be $c=1$ and becomes equal to $\nu$. We now see that our condition for the regularization of the asymmetrical limit, thus leading to a quantum-mechanical system of coupled harmonic oscillators with finite couplings, which is to make $\beta_{0}\rightarrow 0$, also implies that $\nu\rightarrow 0$, that is, leads to the total absence of wave propagation in the continuum limit in Minkowski space.

Note that recovering the balance between the three terms of the sums in our results for the energy, which leads to the on-shell condition, also implies making $\beta_{0}\rightarrow 0$ in this asymmetrical limit. This analysis can be extended to the case of the simultaneous asymmetrical limits, with the same basic results. Any trial at recovering the on-shell condition in these limits implies the absence of wave propagation in the continuous limit in Minkowski space. With the introduction of $\beta_{0}$ the numerator and the denominator that appear in our results acquire the form

$$\pm 4\sin^{2}\!\left(\frac{k_{0}\pi}{N_{T}}\right) +\beta_{0... ...{i}4\sin^{2}\!\left(\frac{k_{i}\pi}{N_{L}}\right) +\alpha_{0},$$

so that in order to compensate the factor $(N_{T}/N_{L})^{2(1-q)}$ that appears in the second terms when we multiply both the numerator and the denominator by $N_{T}^{2}$ we must make $\beta_{0}\rightarrow 0$ in the limit, which is equivalent, from the point of view of quantum mechanics, to making finite the couplings between sites and, from the point of view of quantum field theory, to the absence of wave propagation in the continuum limit in Minkowski space.

We end this section with the interesting historical observation that this is not the first time that the existence of a useful Hilbert space for quantum field theory is submitted for discussion. In a very interesting (but difficult to find) little book titled “Lectures on Quantum Field Theory” [4] no lesser a figure than Dirac gave us his views about the state of the subject. In this book one finds the following two statements, that we take the liberty of quoting here:

The interactions that are physically important in quantum field theory are so violent that they will knock any Schrödinger state vector out of Hilbert space in the shortest possible time interval.

[The Schrödinger picture] is thrown out by the interactions which physicists are interested in being so violent in the high frequencies, and it doesn't seem to be possible to get interactions satisfying relativity which do not have this violent behaviour in the high frequencies.

These statements are by no means exactly the same that we are led by our results to make here, since Dirac is talking about interactions between fields and the Schrödinger picture of quantum mechanics, but the reference to the lack of usefulness of the Hilbert space, because the dynamics of the theory does not allow it to permanently contain the states, and the reference to relativistic invariance as being in conflict with the usual Hilbert space structure, are both, at the very least, extremely interesting and suggestive.

Problems

1. Consider the delta-functional type of state, defined by a statistical distribution that attributes to a given configuration $\varphi_{0}$ of the fields the probability $1$ and to all other configurations the probability $0$. In other words, we can represent such a state by the distribution

   
   

   $$\vert\varphi_{0}\rangle\sim \prod_{\vec{x}}\delta[\varphi(\vec{x})-\varphi_{0}(\vec{x})].$$
   
   

   Show that this state is an eigenstate of all the observables of the theory, including the Hamiltonian. In terms of the traditional formalism you will be showing, for example, that

   
   

   $$\varphi_{\rm op}\vert\varphi_{0}\rangle =\varphi_{0}(\vec{x})\vert\varphi_{0}\rangle,$$
   
   

   where $\varphi_{\rm op}$ represents the field operator, as well as

   
   

   $$\mathbf{H}_{\rm op}\vert\varphi_{0}\rangle =\mathbf{H}[\varphi_{0}(\vec{x})]\vert\varphi_{0}\rangle.$$
   
   

   Observe that a state like this corresponds to a situation in which the field does not fluctuate at all and is, therefore, devoid of any physical meaning in the quantum theory.

2. Calculate the dispersion of the action $S$ on the vacuum state. Consider carefully and in separate the terms in which $\vec{q}\neq\pm\vec{k}$ and those in which $\vec{q}=\vec{k}$. Among these last ones, consider in separate the cases in which the momentum vector corresponds to a real mode and those in which the mode has a non-vanishing imaginary part.

3. Calculate the dispersion of the observable number of particles ${\cal N}$ on the state of $n$ particles with momentum $\vec{k}$. Consider carefully and in separate the terms in which $\vec{q}\neq\pm\vec{k}$ and those in which $\vec{q}=\vec{k}$. Among these last ones, consider in separate the cases in which the momentum vector corresponds to a real mode and those in which the mode has a non-vanishing imaginary part.

4. Calculate the dispersion of the projector ${\cal N}_{\vec{k}}$ of the number of particles with momentum $\vec{k}$ on the state of $n$ particles with momentum $\vec{k}$. Consider carefully and in separate the terms in which $\vec{q}\neq\pm\vec{k}$ and those in which $\vec{q}=\vec{k}$. Among these last ones, consider in separate the cases in which the momentum vector corresponds to a real mode and those in which the mode has a non-vanishing imaginary part.

5. Relate, in the case of the Hamiltonian without the average over the temporal block, the dispersion $\sigma_{H}^{2}=\langle\mathbf{H}\rangle^{2}-\langle\mathbf{H}^{2}\rangle$ with the dispersion $\sigma_{H'}^{2}=\langle\mathbf{H}'\rangle^{2}-\langle{\mathbf{H}'}^{2}\rangle$, doing the integration over the variable $\bar\pi$ and thus showing that

   
   

   $$\sigma_{H}^{2}=\sigma_{H'}^{2}+\imath\langle\mathbf{H}'\rangle.$$
   
   

6. Calculate, according to the usual (non-canonical) definition of the expectation values, using the Fourier components of the fields, the dimensionless dispersion $\sigma_{H'}^{2}=\langle\mathbf{H}'\rangle^{2}-\langle{\mathbf{H}'}^{2}\rangle$, of the non-blocked Hamiltonian, obtaining
   $$\begin{eqnarray*} \sigma_{H'}^{2} & = & \frac{1}{2N_{T}^{2}}+\frac{1}{N_{T}^{2}}... ...^{2}(q_{0})+\sum_{i}\rho_{i}^{2}(\mathbf{k})+\alpha_{0}\right]}. \end{eqnarray*}$$
   
   
   
   

7. Combine the results of problems 5.3.5 and 5.3.6 and calculate completely the dispersion $\sigma_{H}^{2}=\langle\mathbf{H}\rangle^{2}-\langle\mathbf{H}^{2}\rangle$, obtaining
   $$\begin{eqnarray*} \sigma_{H}^{2} & = & \frac{1}{2N_{T}^{2}}+\frac{1}{2N_{T}}\sum... ...^{2}(q_{0})+\sum_{i}\rho_{i}^{2}(\mathbf{k})+\alpha_{0}\right]}. \end{eqnarray*}$$
   
   
   
   

   Compare your result with the dispersion of the blocked Hamiltonian $\mathbf {H}_{b}$ and show that $\sigma_{H}^{2}>\sigma_{H_{b}}^{2}$, as expected.

8. Show that the sum ${\cal S}_{H_{b}}^{2}$ in dimension $d=2$,

   
   

   $${\cal S}_{H_{b}}=\sum_{\vec{k}}\frac{(\rho_{1}^{2}+\alpha_{0... ...{2}+\alpha_{0})} {(\rho_{0}^{2}+\rho_{1}^{2}+\alpha_{0})^{2}},$$
   
   

   can be written as a manifestly positive quantity in the case of the symmetrical limit, in which $N_{L}=N_{T}=N$. Use the possibility of interchanging, in this symmetrical case, the variables $k_{0}$ and $k_{1}$ within the sum.

9. Write the dispersion $\sigma_{H}$ of the non-blocked Hamiltonian in the $d=1$ case of quantum mechanics. Evaluate the behavior in the $N_{T}\rightarrow \infty$ limit of the sums that appear in this result. In order to do this, consider the expressions in the limit $\alpha_{0}\rightarrow 0$, which is the most relevant for us because $\alpha_{0}$ in fact vanishes in the continuum limit. Note that in this limit you must treat separately the terms with $k_{0}=0$ and/or $q_{0}=0$. Use your results to show that the corresponding dimensionfull dispersion behaves in the limit as

   
   

   $$\Sigma_{H}^{2}\sim\frac{-N_{T}}{T^{2}}.$$
   
   

   Find out how to define limits in which $N_{T}\rightarrow \infty$ and $T\rightarrow \infty$ simultaneously so as to guarantee that this dimensionfull dispersion vanishes in the limit.

10. Write the dispersion $\sigma_{H'}$ of the non-blocked Hamiltonian in the $d=1$ case of quantum mechanics, using the usual, non-canonical, definition of the expectation values. Evaluate the behavior of the sums that appear in this result in the $N_{T}\rightarrow \infty$ limit, just as was done in problem 5.3.9. Use your results to show that the corresponding dimensionfull dispersion behaves in the limit as

    
    

    $$\Sigma_{H'}^{2}\sim\frac{N_{T}^{2}}{T^{2}},$$
    
    

    showing in this way that, in this case, it is not possible to take the limits $N_{T}\rightarrow \infty$ and $T\rightarrow \infty$ in such a way that we have both that $a\rightarrow 0$ and that this dimensionfull dispersion vanishes in the limit.

11. Write a collection of programs to calculate, in the symmetrical case $N_{L}=N_{T}=N$, as a function of $N$, in dimensions from $d=1$ to $d=5$, the sums ${\cal S}_{H_{b}}$ that appear in the expression of $\sigma_{H_{b}}^{2}$. Also do the same for the sums ${\cal S}_{H_{b}'}$ that appear in the expression of $\sigma_{H_{b}'}^{2}$. Use your programs to reproduce the graphs which are shown in the text.

12. Calculate in detail the quantity $\langle{H_{b}'}^{2}\rangle$ in the case in which $N$ is even, thus completing the argument presented in the text, where the calculation was presented in the case in which $N$ is odd. Be mindful of the correct identification and counting of all the terms of the sums over the momenta, and remember that in this case both the value $0$ and the value $N/2$ of the components $k_{\mu}$ are associated to real Fourier components of the fields.

13. Calculate the dispersion of the blocked Hamiltonian $\mathbf {H}_{b}$ on the state of one particle with momentum $\vec{k}$.

14. Show, examining the behavior of its terms, that the sums ${\cal S}_{H_{b}}$ tend to $N_{L}^{d-1}N_{T}$ in the asymmetrical limit, in which we make $N_{T}\rightarrow \infty$ with fixed $N_{L}$. Do the same in the case of the simultaneous asymmetrical limit, in which we make $N_{L}=N_{T}^{q}$ with $0<q<1$. In either case assume that we can limit the discussion to terms with finite $k_{0}$.

15. Write programs to calculate, in the asymmetrical case $N_{L}\neq N_{T}$, as functions of $N_{T}$ and for a fixed value of $N_{L}$, in dimensions from $d=2$ to $d=5$, the sums ${\cal S}_{H_{b}}$ that appear in the expression of $\sigma_{H_{b}}^{2}$. Also do the same for the sums ${\cal S}_{H_{b}'}$ that appear in the expression $\sigma_{H_{b}'}^{2}$. Use your programs for values of $N_{L}$ from $4$ to $20$ and values of $N_{T}$ from $1$ to $200$, thus showing that these sums do in fact diverge as $N_{T}$ in this type of asymmetrical limit.