I am trying to find the action associated with the Lagrangian density $$ \mathcal{L} = \frac{1}{2}\left( \frac{\partial\phi}{\partial x} \right)^2 + \frac{1}{2}m^2\phi^2. \tag{1} $$ I am supposed to use the discrete expansions $$\phi_j = \frac{1}{\sqrt{Na}}\sum_p \tilde{\phi}_pe^{ipja} = \frac{1}{\sqrt{Na}}\sum_{-p} \tilde{\phi}_{-p}e^{-ipja}. \tag{2} $$

So, first I find the Lagrangian, using $$ L = \int dx \mathcal{L} = a \sum_j \mathcal{L} = \frac{a}{2}\sum_j \left[ \left( \frac{\phi_{j+1}-\phi_j}{a} \right)^2 + m^2\phi_j^2 \right] \tag{3} $$ where $j$ labels the 1D lattice sites and $a$ is the equilibrium distance between each site.
Now I plug in the expansion for $\phi_j$ into the Lagrangian, and where $\phi_j$ is squared, I use one copy of the middle ($+p$) term in eq (2) and one copy of the right ($-p$) term in eq (2), multiplied together. This is motivated by the form of the action I am supposed to get in the end.
When I do the substitution into $L$, I end up with $$ L = \frac{1}{2} \sum_p \tilde{\phi_p}\tilde{\phi_{-p}}\left[ \frac{2}{a^2}\left( 1-\cos{pa} \right) +m^2 \right] \tag{4}. $$

Now to get the action, I know that $$ S = \int L dt, \tag{5}$$ but I have no idea where time is supposed to come into this problem at all. When integrating the Lagrangian density to get the Lagrangian, I know that I had to realize that the integral over one spatial dimension becomes, in the discrete case, a sum over the positions $x_j$ times the lattice constant $a$, or just a sum over $j$, again times $a$. In addition, the spatial derivative in the Lagrangian becomes a discrete difference, as I have shown above.

Furthermore, the expression that I obtained for the Lagrangian $L$ is exactly what my textbook says I should obtain for the action $S$! Is this somehow the result of the problem not having any obvious time-dependence? So, in total, I suppose I want to know how the action relates to the Lagrangian in the case of a problem that doesn't involve time.

Just for clarity, I am going to write the question as phrased in the textbook (QFT for the Gifted Amateur):

Exercise 17.5 (a): Consider a one-dimensional system with Lagrangian $$ \mathcal{L} = \frac{1}{2}\left( \frac{\partial \phi(x)}{\partial x} \right)^2 + \frac{m^2}{2} \left[ \phi(x) \right]^2. $$ The choice of sign makes this a Euclidean theory. Descretize this theory (that is, put it on a lattice) by defining $$ \phi_j = \frac{1}{\sqrt{Na}} \sum_p \tilde{\phi}_p e^{ipja}, $$ where $j$ labels the lattice site, $a$ is the lattice spacing, and $N$ is the number of lattice points. Using the method in exercise 17.3 show that the action may be written $$ S = \frac{1}{2} \sum_p \tilde{\phi}_{-p} \left( \frac{2}{a^2}-\frac{2}{a^2}\cos{pa} + m^2 \right) \tilde{\phi}_p, $$ and read off the propagator for this theory.

The "method in exercise 17.3" is just what I described in between eq (3) and eq (4), where you expand $\phi_j$ in terms of its Fourier transforms $\tilde{\phi}_p$ and $\tilde{\phi}_{-p}$. Problem 17.3 also is the one that shows that the free propagator is $\frac{i}{2}$ times the inverse of the quadratic term in the momentum-space action, which is why this problem is asking us to find the action in the first place.

  • $\begingroup$ @Qmechanic: His question says nothing about the continuum limit. $\endgroup$
    – Daniel
    Commented Jul 26, 2015 at 21:37
  • $\begingroup$ Wouldn't taking the limit as $a$ goes to zero just be undoing the discretization that I was told to implement? $\endgroup$ Commented Jul 26, 2015 at 21:39
  • $\begingroup$ @MarkMitchison I added an exact copy of the problem at the end of my question. The Lagrangian density given seems to indicate that $\phi$ is just a function of x, not of x,t. $\endgroup$ Commented Jul 26, 2015 at 22:07
  • 1
    $\begingroup$ The question (v5) seems to be caused by a somewhat misleading exercise formulation. Exercise 17.5 (a) in QFT for the Gifted Amateur deals with a 0+1 dimensional system with Euclidean signature (where Euclidian time is called $x$), not a 1+1 dimensional system with Euclidean signature. $\endgroup$
    – Qmechanic
    Commented Jul 26, 2015 at 22:20
  • $\begingroup$ @Qmechanic: I quoted your comment in the EDIT of my answer, I don't know how to link you for more credit. Feel free to do that, if you wish $\endgroup$
    – Daniel
    Commented Jul 26, 2015 at 22:46

2 Answers 2


The question seems okay to me. It's more or less giving you the time-independent Klein-Gordon equation for a 1D lattice, right?

I think a clarification on how to discretise the derivative would go miles on this question.

If you write $$ \frac{\partial \phi}{\partial x} = \frac{\phi_{j+\frac{1}{2}} - \phi_{j-\frac{1}{2}}}{a} $$ then you can easily just substitute the Fourier expansion they give you and use the Dirac delta function, which in discrete space is given by $$ \delta_{p,q} = \frac{1}{Na} \sum_j{e^{i(p-q)ja}} $$

After cleaning everything up I think you get the end result.

  • $\begingroup$ Why would you use $j+\frac{1}{2}$ and $j−\frac{1}{2}$ as oppose to what I used in the squared expression of (3), which is my version of the discretized derivative? $\endgroup$ Commented Sep 16, 2015 at 1:33
  • $\begingroup$ They're somewhat equivalent. From a numerical methods point of view, the difference is you've got a forward difference derivative whereas I've used a central difference which generally has an increased order of accuracy. $\endgroup$
    – Aran
    Commented Sep 23, 2015 at 19:50

I'd say the problem doesn't make much sense as it stands. Without terms that include a time derivative, the system would just immediately jump the the configuration with the smallest $L$, as the speed with which it does so is not "punished" (does not increase the action).

In other words, the solution with the smallest action is the one with $L(t)=L_\text{min}$ at all times, except perhaps at $t=t_\text{initial}$.

In a sense though, $x$ and $t$ are just labels, so you could reinterpret your problem, considering $x$ as your time coordinate. $\mathcal L$ would be your Lagrangian and $L$ your action, just as your book says.

EDIT: The problem does actually talk about the Lagrangian $\mathcal L$. So this is not the Lagrangian density. Then $L$ is actually already your action. Also note what Qmechanic wrote in his comment, which I quote below for completeness

The question (v5) seems to be caused by a somewhat misleading exercise formulation. Exercise 17.5 (a) in QFT for the Gifted Amateur deals with a 0+1 dimensional system with Euclidean signature (where Euclidian time is called x), not a 1+1 dimensional system with Euclidean signature.

  • $\begingroup$ so you're saying that in the case of a problem with just one variable, let's say $x$, the action is the integral of the Lagrangian over that one variable, even if that variable isn't time? $\endgroup$ Commented Jul 26, 2015 at 22:09
  • $\begingroup$ @Physics_Plasma: "Action" is defined as the integral of the Lagrangian over time. So to be technically correct, you would interpret $x$ as "time". Indeed, as Qmechanic points out in a comment to your question $x$ is indeed the time in your problem. $\endgroup$
    – Daniel
    Commented Jul 26, 2015 at 22:41
  • $\begingroup$ Ok, thank you. The book explicitly mentions at some point that it will "follow convention" and call both $L$ and $\mathcal{L}$ "Lagrangians", so I didn't take its word choice in this question too literally when it didn't explicitly say "Lagrangian density". $\endgroup$ Commented Jul 26, 2015 at 22:54

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