Genady Posted February 24 Share Posted February 24 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? Link to comment Share on other sites More sharing options...

joigus Posted February 24 Share Posted February 24 (edited) You want energies to be positive. As k^{0} (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 February 24 by joigus spoiler 1 Link to comment Share on other sites More sharing options...

Genady Posted February 24 Author Share Posted February 24 33 minutes ago, joigus said: You want energies to be positive. As k^{0} (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))=∑xk∈zeroes of ff(x−xk)|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, Hide contents δ(f(k0))=f(k0−ωk)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. Link to comment Share on other sites More sharing options...

Genady Posted February 25 Author Share Posted February 25 (edited) Just to answer the OP question, 4 hours ago, Genady said: 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 \(\frac 1 {2 \omega_k}\). Edited February 25 by Genady 1 Link to comment Share on other sites More sharing options...

Sensei Posted February 25 Share Posted February 25 (edited) I wonder, homework at your age and skills.. A new reincarnation has taken over your account.. Edited February 25 by Sensei Link to comment Share on other sites More sharing options...

Genady Posted February 25 Author Share Posted February 25 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. 🙃 Link to comment Share on other sites More sharing options...

joigus Posted February 25 Share Posted February 25 (edited) 10 hours ago, Genady said: Just to answer the OP question, It would not. Without the step function it would be ∫dk0δ(k2−m2)=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 February 25 by joigus minor correction 1 Link to comment Share on other sites More sharing options...

Sensei Posted February 25 Share Posted February 25 (edited) 1 hour ago, Genady said: 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 February 25 by Sensei Link to comment Share on other sites More sharing options...

Sensei Posted February 25 Share Posted February 25 3 hours ago, Genady said: I don't anymore read books that don't have equations. 🙃 I don't anymore read books that don't have letters.. Link to comment Share on other sites More sharing options...

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