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I'm a bit confused as to whether or not the Fourier transform of the momentum and coordinate field functions need to have different signs. The notation in the book is \begin{equation} \begin{cases} \hat{\phi}_k\\ \hat{\pi}_k \end{cases} \equiv\frac{1}{\sqrt{L}}\int_0^Le^{i\{\pm kx\}}dx \begin{cases} \hat{\phi}(x)\\ \hat{\pi}(x) \end{cases} \end{equation} To my knowledge, if I have chosen a particular sign I must stick with it for the Fourier transfform of both functions, so for the canonical commutation relation in $k$ basis we have \begin{equation} [\hat{\pi}_k,\hat{\phi}_{k'}]=\frac{1}{L}\iint_{0}^{L}[\hat{\pi}(x),\hat{\phi}(x')]e^{\pm(kx+k'x')}dxdx' \end{equation} However, this does not give the desired result, instead giving $\delta(k+k')$. Solutions that I have found have the signs on the Fourier transform as opposites, but I am not entirely sure why. Does the book's notation mean that $\phi$ and $\pi$ have to be Fourier transformed with opposite signs, and if so why is that the case?

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  • $\begingroup$ Isn't $e^{i\{\pm kx}$ only $e^{-i kx}$ in the definition of the FT? $\endgroup$ Feb 12, 2022 at 12:26
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    $\begingroup$ It depends on the convention you are using @Felicia $\endgroup$
    – schris38
    Feb 12, 2022 at 12:46
  • $\begingroup$ Don't they always have opposite signs? From x to p +/- and from p to x -/+? $\endgroup$ Feb 12, 2022 at 12:49
  • $\begingroup$ Maybe try to take the commutator between $\hat{\pi}_k$ and $\hat{\phi}_{k'}^*$?? $\endgroup$
    – schris38
    Feb 12, 2022 at 12:49
  • $\begingroup$ @schris38 but isn't that a completely different commutator? I've thought about doing it but the example following this exercise would have a completely different solution if I could just change $[\hat{pi}_k,\hat{\phi}_{k'}]$ to $[\hat{pi}_k,\hat{\phi}_{k'}^*]$. $\endgroup$ Feb 12, 2022 at 13:42

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First let me note that (up to how you've chosen to normalize the Fourier transform) the "Fourier transform" with opposite sign is actually the inverse Fourier transform. It doesn't matter whether you call the plus sign the transform or the inverse as this is entirely down to preference. The only important thing is that when you flip the sign you get the inverse transformation.

With that said, let me try and first give an example wherein the cause of the sign flip is manifest, then I will try and describe how this generalizes.

First of all, we can note that in a field theory specified by a Lagrangian, the momentum is not an independent variable (more on this later). Even in point particle mechanics this is the case. After all, the momentum of a free particle governed by Lagrangian $L = \frac{1}{2}m \dot x^2$ is $p = m\dot x$. So if we were to do a transformation to $x$ and expect this relation between the momentum and position to be preserved, a transformation of $p$ would be forced upon us: we would have no choice if we wish to preserve $p = m\dot x$. For an example of this, suppose $x$ is the position of a particle in 3 dimensions and apply a rotation to $x$, $x\rightarrow Rx$. Then we are forced to either transform $p \rightarrow R p$ at the same time, lest we violate $p = m\dot x$.

So then, consider a free complex scalar with Lagrangian $$ L = \partial_\mu \phi^\dagger \partial^\mu\phi - m\phi^\dagger \phi. $$ The conjugate momenta to $\phi$ here is $\pi = \partial_0 \phi^\dagger$. Hence if we wish to Fourier transform $\phi$, the transformation of $\pi$ is forced upon us and in particular there is a complex conjugation in the dagger which would flip the sign in the Fourier factor.

Of course, this may be all well and good in the special case of a complex scalar field, but based only on this example one could very easily question why a sign flip should occur if we are working with, say, a single real scalar field. In that instance there would appear to be no natural reason to have a conjugation. I mean, since everything is real we can always throw conjugations onto our field for fun, but there's no compelling reason to do so.

So this is where I would like to read further into the observation that we are preserving the relation between the field and its conjugate. So, let me point out that transformations which preserve the relation between field and conjugate are known as point transformations and, importantly, point transformations are a special type of canonical transformation (though not all canonical transformations are point transformations). Canonical transformations are precisely those transformations which preserve the commutator (or classically, the Poisson bracket). In other words, they are transformations which send conjugate pairs to conjugate pairs. As OP noted, demanding that the "Fourier transformed" variables be conjugate to each other in the standard way is enough to demand that the sign be flipped. This is how the sign flip comes about more generally.

There are two points, in my opinion, which one might wonder about at this point. Firstly, why should we restrict ourselves to only allow canonical transformations? Secondly, does allowing only canonical transformations uniquely determine that the sign must flip? That is, are there more general possibilities?

For the first point, the answer is essentially yes. We are completely free to consider transformations which are not canonical. However somewhat anti-climactically, it is typically not very useful to do so. There are some additional things involving generating charges which make canonical transforms a little more special, but I think that would go a little too far afield for this answer and at the end of the day, "non-canonical transformations tend not to be useful" is really the truth of the matter.

For the second point, no the transformation is not uniquely fixed. It is, however, probably the most natural choice. Particularly because it's the choice which makes the Fourier transformation a point transformation when we have free complex fields. The reader might be interested in looking at the Bogoliubov transformations for a somewhat more general type of transformation which mixes the plus and minus signs (of both the field and its conjugate). These transformations are canonical and are often useful for diagonalizing quadratic Hamiltonians (any quadratic polynomial can be put into the form of a sum of harmonic oscillators by a linear transformation of the fields).

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The expressions for the field operators contain the creation and annihilation operators, respectively multiplied by Fourier kernel with opposite sign in their arguments. When you compute the commutation between these field operators, it translates to a commutation between the creation and annihilation operators. Therefore it would produce a Dirac delta in which you get the difference of the respective wave vectors. I can provide more detail if your need it.

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