# What is this unit step function for?

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The question:

Show that $\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\frac 1 {2 \omega_k}$ where $$\theta(x)$$ is the unit step function and $$\omega_k \equiv \sqrt {\vec k^2 +m^2}$$.

My solution:

$$k^2={k^0}^2 - \vec k ^2$$

$$\omega _k ^2 = \vec k^2 +m^2$$

$$k^2 - m^2 = {k^0}^2 - \omega_k^2$$

$$dk^0= \frac {d{k^0}^2} {2k^0}$$

$$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0) = \int_{-\infty}^{\infty} \frac {d{k^0}^2} {2k^0} \delta ({k^0}^2 - \omega_k^2) \theta (k^0) = \frac 1 {2 \omega_k} \theta (\omega_k) = \frac 1 {2 \omega_k}$$

However, the point of the unit step function there is unclear to me. Wouldn't the result be the same without it?

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You want energies to be positive. As k0 (the zeroth component) of the 4-momentum is the energy component, all states must be decreed to have zero amplitude for that choice. That's achieved by the step function trick.

You missed a well-known trick for delta "functions"...

The delta function satisfies,

$\delta\left(f\left(x\right)\right)=\sum_{x_{k}\in\textrm{zeroes of }f}\frac{f\left(x-x_{k}\right)}{\left|f'\left(x_{k}\right)\right|}$

for any continuous variable $$x$$ and "any" well-behaved function $$f$$ of such variable. Taking as your corresponding function and variable both $$k^{0}$$ and,

$f\left(k^{0}\right)=\left(k^{0}\right)^{2}-\left(\omega_{\mathbf{k}}\right)^{2}$

you get,

Spoiler

$\delta\left(f\left(k^{0}\right)\right)=\frac{f\left(k^{0}-\omega_{\mathbf{k}}\right)}{2\omega_{\mathbf{k}}}+\frac{f\left(k^{0}+\omega_{\mathbf{k}}\right)}{2\omega_{\mathbf{k}}}$

And that's why you need the step function: to kill the un-physical $$k^0$$'s. Negative energies do appear again in the expansion of the space of states, but they're dealt with in a different manner. This is just to define the measure for the integrals. All kinds of bad things would happen if we let those frequencies stay.

I'm sure there are better explanations out there. But the delta identity is crucial to see the point.

6 minutes ago, joigus said:

All kinds of bad things would happen if we let those frequencies components stay.

Edited by joigus
spoiler
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33 minutes ago, joigus said:

You want energies to be positive. As k0 (the zeroth component) of the 4-momentum is the energy component, all states must be decreed to have zero amplitude for that choice. That's achieved by the step function trick.

You missed a well-known trick for delta "functions"...

The delta function satisfies,

δ(f(x))=xkzeroes of ff(xxk)|f(xk)|

for any continuous variable x and "any" well-behaved function f of such variable. Taking as your corresponding function and variable both k0 and,

f(k0)=(k0)2(ωk)2

you get,

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δ(f(k0))=f(k0ωk)2ωk+f(k0+ωk)2ωk

And that's why you need the step function: to kill the un-physical k0 's. Negative energies do appear again in the expansion of the space of states, but they're dealt with in a different manner. This is just to define the measure for the integrals. All kinds of bad things would happen if we let those frequencies stay.

I'm sure there are better explanations out there. But the delta identity is crucial to see the point.

Thank you. I got it. My mistake was that when I replaced $$k^0$$ with $$\omega_k$$ I've missed that it can be + or - $$\omega_k$$. The step function is needed to kill one of them.

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Just to answer the OP question,

Wouldn't the result be the same without it?

It would not.

Without the step function it would be $\int dk^0 \delta (k^2-m^2) =\frac 1 {\omega_k}$ rather than $$\frac 1 {2 \omega_k}$$.

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I wonder, homework at your age and skills.. A new reincarnation has taken over your account..

Edited by Sensei
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3 hours ago, Sensei said:

I wonder, homework at your age and skills.. A new reincarnation has taken over your account..

It is not technically homework, but it could've been if I were technically student. Just a new textbook to work on.

I don't anymore read books that don't have equations. 🙃

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Just to answer the OP question,

It would not.

Without the step function it would be

dk0δ(k2m2)=1ωk

rather than 12ωk .

It would.

!!!

The 1/2 factor doesn't change Lorentz invariance of the metric measure, but it's quite essential to the formalism that comes later. One would think we're done with negative energies/frequencies, and such. But no. They keep biting our buttocks later with the Fourier transform. That's where the Stueckelberg-Feynman prescription for antiparticles comes in.

Edited by joigus
minor correction
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I don't anymore read books that don't have equations. 🙃

That's pretty good advice for mortals like you, except for the Harry Potter series..

Edited by Sensei
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I don't anymore read books that don't have equations. 🙃

I don't anymore read books that don't have letters..

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