# Lorentz Invariance of Propagator for Complex Scalar Field

• eudo
In summary, the task was to show that the commutator of two complex scalar fields, \hat{\phi}(x_1) and \hat{\phi}^\dagger(x_2), is equal to zero for space-like intervals (x_1 - x_2)^2 < 0. By expressing the fields in terms of creation and annihilation operators, and using the commutation relations for these operators, the problem reduces to evaluating two integrals. These integrals can then be transformed to a frame where x_1 - x_2 = -(x_1 - x_2), and since the integrands are Lorentz invariant, the second integral cancels out the first, resulting in a commutator
eudo

## Homework Statement

Show that

$[\hat{\phi}(x_1),\hat{\phi}^\dagger(x_2)] = 0$
for $(x_1 - x_2)^2 < 0$

where $\phi$ is a complex scalar field

## Homework Equations

$\hat{\phi}=\int\frac{d^3 \mathbf{k}}{(2\pi)^3 \sqrt{2\omega}}[\hat{a}(k)e^{-ik\cdot x} + b^\dagger(k)e^{ik\cdot x}]$

with

$[\hat{a}(k),\hat{a}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})$
$[\hat{b}(k),\hat{b}^\dagger(k')]=(2\pi)^3\delta^3(\mathbf{k}-\mathbf{k'})$

and all other commutation relations vanish.

## The Attempt at a Solution

This is problem (7.3) in Aitchison and Hey's QFT book. I have a couple questions about the solutions that are posted online here. They get eventually to the difference of two integrals:

$\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{-ik\cdot (x_1-x_2)}-\int\frac{d^3\mathbf{k}}{(2\pi)^3 2\omega}e^{ik\cdot (x_1-x_2)}$

So far so good. Then they argue that for space-like intervals, x1-x2 can be transformed to -(x1-x2) by a Lorentz transformation, and thus the second term cancels the first.

Now, I agree that each of the integrals must be Lorentz invariant, so you are allowed to evaluate each one in whichever frame you want, but I don't see how one is allowed to transform just the x1-x2 vector without also transforming the k.

And a related question: In a different problem (5.6), when we come to a similar integral, they suggest transforming to a frame where t1=t2, and since the resulting integral is an odd function of k, it must be zero. And since it's Lorentz invariant, it must be zero for all space-like intervals. It seems to me that this argument works for this problem, too, so I'm not sure why they bring up this other argument about transforming x1-x2 to -(x1-x2).

Anyone want to help explain what I'm missing?

Thanks

You are integrating over $\vec{k}$. Thanks to the invariance of $\mathrm{d}^3 \vec{k}/\omega$ where $\omega=\sqrt{m^2+\vec{k}^2}$ you know that the result is a Lorentz scalar $$F(x_1-x_2)$$. That means that under Lorentz transformations
$$F'(x')=F(x)=F(\Lambda^{-1} x).$$
Thus you only need to transform $x=x_1-x_2$.

It's only crucial to show that for a spacelike $x$ you can always find a Lorentz transformation such that $\Lambda x=-x$. Since you can orient your coordinate system always such that $x=(0,\xi,0,0)$ you just need to find a Lorentz transformation which makes out of this $\Lambda x=(0,-\xi,0,0)$.

1 person
Thanks for the reply. I guess I'm missing the last part of that equality $F(x) = F(\Lambda^{-1}x)$. As a test, if I define a function

$F(x)=k\cdot x$

where $k$ is some arbitrary 4-vector, then F(x) is a Lorentz scalar. If I have some transformation such that $x'=\Lambda x$, then

$F'(x')=k'\cdot x'=k\cdot x=F(x)\not=F(\Lambda^{-1} x)$

I understand that in the original problem we're integrating over $\vec{k}$, but it still seems to me we need to transform the $\vec{k}$.

Argh. That's of course a typo. The correct rule is
$$F'(x')=F(x)=F(\Lambda^{-1} x'),$$
because
$$x'=\Lambda x \; \Leftrightarrow \; x=\Lambda^{-1} x.$$

Right after I posted that, I figured that's what you meant. And it finally struck me why we don't have to worry about the $\vec{k}$. Thanks for your help.

## 1. What is Lorentz invariance?

Lorentz invariance is a fundamental principle in physics that states that the laws of physics should be the same for all observers in uniform motion. This means that the measurements of physical quantities, such as time, length, and energy, should be the same for all observers regardless of their relative motion.

## 2. What is a propagator for a complex scalar field?

A propagator for a complex scalar field is a mathematical function that describes how a particle or field propagates through space and time. It is a key concept in quantum field theory, where it represents the probability amplitude for a particle to travel from one point in space-time to another.

## 3. Why is Lorentz invariance important for the propagator of a complex scalar field?

Lorentz invariance is important for the propagator of a complex scalar field because it ensures that the laws of physics are consistent for all observers. This means that the propagator will give the same results regardless of the reference frame used to make measurements, making it a fundamental tool for understanding and predicting physical phenomena.

## 4. How is Lorentz invariance of the propagator for a complex scalar field demonstrated?

The Lorentz invariance of the propagator for a complex scalar field can be demonstrated through mathematical calculations. Using the principles of special relativity and the equations of motion for the field, it can be shown that the propagator remains the same for all observers in different reference frames.

## 5. What are the implications of Lorentz invariance of the propagator for a complex scalar field?

The implications of Lorentz invariance of the propagator for a complex scalar field are far-reaching. It allows for the consistent description of particle interactions and the prediction of physical phenomena, such as the behavior of particles in accelerators and the properties of matter. It also plays a crucial role in the development of theories such as quantum field theory and the standard model of particle physics.

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