# Fourier transform/propagator

#### RedX

I've got a question about Fourier transforms.

Say you have the integral:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon}$$

Can I just set this equal to:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+i\epsilon}$$

since the pole structure of the denominator will force k to be on-shell?

But I thought Fourier transforms were unique? So how can two different transform functions give the same position space function?

You don't really see this type of question when using Lorentz covariant gauges, since here k_0 and vector k are split up in non-covariant way. But if you have a non-covariant gauge, you might see this.

It's just weird to me because I thought Fourier transforms were unique. Does this mean that I can use either momentum space representation of the propagator in Feynman diagrams, and I'll get the same answer?

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#### RedX

This can't be right, but I can't figure out why.

Take: $$\frac{1}{k^2+i\epsilon}=P\frac{1}{k^2}-i\pi \delta(k^2)$$.

Then if:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon }= \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+i\epsilon }$$
this implies:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2}= \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2}$$
$$\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(\frac{\vec{k}^2}{k_{0}^2})}{k^2}= \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2}$$
where the integrals are over the principal part (so they are over the real line but avoid k2=0 ). But clearly this isn't true, or if it were, it would be amazing.

My math is terrible, so I've got to be doing something wrong, but to me the pole forces the photon to have its energy squared equal to the momentum squared, so I can't get around this.

#### Dickfore

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon}$$

$$D(x, y) = 0.5 \, \int{\frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+ i\epsilon}} + \frac{0.5}{k^{2}_{0}} \int{\frac{d^{4} k}{(2\pi)^{4}} \, e^{i k (x - y)}}$$

The second integral is:

$$\delta^{4}(x - y)$$

and the first is just the propagator you thought your original expression would reduce to.

#### RedX

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon}$$

$$D(x, y) = 0.5 \, \int{\frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+ i\epsilon}} + \frac{0.5}{k^{2}_{0}} \int{\frac{d^{4} k}{(2\pi)^{4}} \, e^{i k (x - y)}}$$

The second integral is:

$$\delta^{4}(x - y)$$

and the first is just the propagator you thought your original expression would reduce to.
If k doesn't have an arrow on it, I took it to mean a 4-vector. With an arrow on it, I meant 3-vector. So on the second integral you have, the numerator doesn't cancel with one of the factors in the denominator, and the integral over dk is the entire 4-vector space so you can't bring out the timed component ko2. I should have been clearer about notation.

I should also note that by having a ko in the denominator, that changes the pole structure, but superficially, since there is no residue at ko=0 . Indeed, if you want, you can swap $$\vec{k}^2$$ with $$k_{o}^2$$, so that the pole structure isn't changed when integrating over $$k_{o}$$ first.

#### RedX

I'm using +---.

In general, if your 4-momentum squared is the same sign as the epsilon, then that is +---. If they are opposite, then it is -+++.

#### Dickfore

So:

$$0.5 + 0.5 \frac{\vec{k}^{2}}{k^{2}_{0}} = 0.5 \left( 1 + \frac{k^{2} - k^{2}_{0}}{k^{2}_{0}}\right) = 0.5 \frac{k^{2}}{k^{2}_{0}}$$

The $k^{2}$ in the numerator cancels the one in the denominator and you get $k^{2}_{0}$ in the denominator. However, your initial integral had the pole w.r.t. $k_{0}$ at:

$$k^{2}_{0} = \vec{k}^{2} - i \epsilon$$

so you need to correct:

$$k^{2}_{0} \rightarrow k^{2}_{0} + i \epsilon$$

$$D(x, y) = 0.5 \, \int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x - y)}}{k^{2}_{0} + i \epsilon}}$$

which has a different pole structure than what you had assumed, but can be integrated by the same methods.

#### RedX

thanks. you used -+++, but the gist of what you did is correct. changing to +---:

$$0.5 + 0.5 \frac{\vec{k}^{2}}{k^{2}_{0}} = 0.5 \left( 1 + \frac{-k^{2} + k^{2}_{0}}{k^{2}_{0}}\right) = 1-0.5 \frac{k^{2}}{k^{2}_{0}}$$

$$D(x, y) =\int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x - y)}}{k^2+i\epsilon} - 0.5 \, \int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x - y)}}{k^{2}_{0} + i \epsilon}}$$

and the second integral is zero since the residue is zero.

So it is true that:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(.5+.5\frac{\vec{k}^2}{k_{0}^2})}{k^2+i\epsilon } = \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+i\epsilon}$$

which is weird as I thought Fourier transforms were unique.

#### Dickfore

$$D(x, y) =\int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x - y)}}{k^2+i\epsilon} - 0.5 \, \int{\frac{d^{4}k}{(2 \pi)^{4}} \, \frac{e^{i k (x - y)}}{k^{2}_{0} + i \epsilon}}$$

and the second integral is zero since the residue is zero.
How is the residue zero?

#### RedX

How is the residue zero?
Because of the k02 term in the denominator, that is a pole of order 2 when integrating over k0.

So what you do is multiply by the integrand by k02, and take one derivative, and set k0=0 to get your residue. In this case this will result in zero.

Another way to see it is to note that the Laurent expansion of 1/x^2 is just 1/x^2. In a general Laurent expansion:

$$f(z)=m/z^n+...n/z^2+a/z+b+cz+...dz^n$$

the residue would just be 'a', or the coefficient of 1/z.

edit: oops, forgot about the exponential term. so maybe it's not zero.

#### RedX

The basic idea I'm trying to get at is this:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}g(k)}{k^2-m^2+i\epsilon }$$

The integral over k0 is done by taking the residues at $$k_0=\pm \sqrt{\vec{k}^2+m^2}$$ to get:

$$D(x,y)=-i\theta(x_o-y_o)\int \frac{d^3k}{2 E_k(2\pi)^3} e^{-ik(x-y)}g(k) -i\theta(y_o-x_o)\int \frac{d^3k}{2 E_k(2\pi)^3} e^{-ik(y-x)}g(k)$$

where now $$k_0= \sqrt{\vec{k}^2+m^2}$$, including the k in g(k).

Therefore from the outset you should be able to set the k in g(k) on-shell in the 4-dimensional integral representation:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}g(k)}{k^2-m^2+i\epsilon }$$

But the math ain't really working out (I think it's complicated by the fact that I set m=0).

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#### RedX

Well, say $$g(k)=\frac{k_{0}^2}{\vec{k}^2}$$, or $$g(k)=1+(k_{0}^2-{\vec{k}^2)$$. These shouldn't affect the pole when integrating just over $$k_0$$. So it seems that you should be able to set:

$$g(k)=\frac{k_{0}^2}{\vec{k}^2}=1$$

and

$$g(k)=1+(k_{0}^2-{\vec{k}^2)=1$$

from the very beginning.

But if this is possible, then Fourier transforms are not unique, since g(k)=1 gives the same result when integrated over momentum space as the above g(k)'s.

Quantum mechanically Fourier transforms are unique because the momentum basis forms a linearly independent basis.

#### Dickfore

So, how exactly is:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}g(k)}{k^2-m^2+i\epsilon }$$
the same as:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon }$$

I don't know what math is not working out for you exactly.

#### RedX

So, how exactly is:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}g(k)}{k^2-m^2+i\epsilon }$$

the same as:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon }$$

I don't know what math is not working out for you exactly.
If I choose g(k)=1, they are the same. But what if I choose $$g(k)=1+(k_{0}^2-\vec{k}^2)$$. Will the two different g(k)'s give the same result, since eventually $$(k_{0}^2=\vec{k}^2)$$ when integrating over ko?

I chose $$g(k)=.5(1+\frac{\vec{k}^2}{k_{0}^2})$$, which when you set $$(k_{0}^2=\vec{k}^2)$$, gives you 1 again. However, as you pointed out, the pole structure changes so this is a bad example, so then I switched it to:

$$g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})$$

So basically choose any g(k) such that when you put k on-shell, it forces g(k)=1.

#### Dickfore

If I choose g(k)=1, they are the same. But what if I choose $$g(k)=1+(k_{0}^2-\vec{k}^2)$$. Will the two different g(k)'s give the same result, since eventually $$(k_{0}^2=\vec{k}^2)$$ when integrating over ko?

I chose $$g(k)=.5(1+\frac{\vec{k}^2}{(k_{0}^2})$$, which when you set $$(k_{0}^2=\vec{k}^2)$$, gives you 1 again. However, as you pointed out, the pole structure changes so this is a bad example, so then I switched it to:

$$g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})$$

So basically choose any g(k) such that when you put k on-shell, it forces g(k)=1.
Clearly you have some misunderstandings on how to apply the residue theorem in complex integration problems. The g(k) you chose, namely:

$$g(k)=.5(1+\frac{\vec{k}^2}{(k_{0}^2})$$

has extra poles in the complex $k_{0}$-plane, so your logic from post #11 does not apply here.

#### RedX

Clearly you have some misunderstandings on how to apply the residue theorem in complex integration problems. The g(k) you chose, namely:

$$g(k)=.5(1+\frac{\vec{k}^2}{(k_{0}^2})$$

has extra poles in the complex $k_{0}$-plane, so your logic from post #11 does not apply here.
Right, I messed up on that one. But what if I had this one instead:

$$g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})$$

This does not introduce a pole in the complex $k_{0}$-plane.

The idea is that two transform functions, one with g(k)=1 and one with $$g(k)=.5(1+\frac{k_{0}^2}{\vec{k}^2})$$, give the same result. This shouldn't be possible if the momentum basis is linearly independent.

#### Dickfore

Do you have a non-zero mass or no?

#### Dickfore

I just thought of something. All these "g(k)" you had presented are not Lorentz invariant, so I don't think your expressions make physical sense, because they have different value in different reference frames.

#### RedX

Do you have a non-zero mass or no?
Since there are computers that can do simple Fourier transforms, I thought I'd experiment with some simple ones ( http://www.wolframalpha.com/input/?i=fourierx[(1)/(x^2+1)] and http://www.wolframalpha.com/input/?i=fourier[(-x^2)/(x^2+1)] ):

$$\frac{1}{\sqrt{2\pi}} \int \frac{e^{-ikx}}{x^2+1} dx=\sqrt{\frac{\pi}{2}}e^{|k|}$$

$$\frac{1}{\sqrt{2\pi}} \int \frac{e^{-ikx}(-x^2)}{x^2+1} dx=\sqrt{\frac{\pi}{2}}e^{|k|}-\sqrt{2 \pi} \delta(k)$$

Since the pole is at $$x=\pm i$$, one would think that the second transform should be the same as the first.

And they almost are the same except for a funny delta function. First of all the second integral doesn't even satisfy Jordan's lemma as the integrand doesn't go to zero as x goes to infinity. But I guess that's okay.

The only difference between the two Fourier transforms are at a single point, k=0. This would correspond to (x0-y0)=0 in the previous examples (I switched up k and x in this post).

Now when (x0-y0)=0, I think the propagator, defined as the time-ordered product, kind of screws up, because what it the fields are taken at the same time? Actually you can evaluate the propagator at equal times.

As for Lorentz invariance, it can be broken so long as it's part of the gauge condition. If the theory is gauge invariant, you can transform to a non-covariant gauge, thereby breaking Lorentz invariance, but your theory should still give you the same answer since your theory was gauge invariant.

I guess you're right:

$$D(x,y)=\int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(\frac{k_{0}^2}{\vec{k}^2})}{k^2+i\epsilon } = \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}(\frac{k^2+\vec{k}^2}{\vec{k}^2})}{k^2+i\epsilon}= \int \frac{d^4k}{2 \pi^4} \frac{e^{ik(x-y)}}{k^2+i\epsilon}+ \delta(x_0-y_0)\int \frac{d^3k}{2 \pi^3} \frac{e^{i\vec{k}(\vec{x}-\vec{y})}}{\vec{k}^2+i\epsilon}$$

So the Fourier transforms aren't the same when the times are equal. The last term in fact is a Yukawa term and proportional to:

$$\delta(x_0-y_0) \frac{1}{|\vec{x}-\vec{y}|}$$

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