Integral with delta and unit step functions in the integrand

[Hill]

Homework Statement

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 $\omega_k \equiv \sqrt {\vec k^2 +m^2}$.

Relevant Equations

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

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

Wouldn't the result be the same without the unit step function?

 

[TSny]

Homework Statement: 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 $\omega_k \equiv \sqrt {\vec k^2 +m^2}$.
Relevant Equations: $k^2={k^0}^2 - \vec k ^2$

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

Wouldn't the result be the same without the unit step function?

With the theta function, $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\int_{0}^{\infty} dk^0 \delta (k^2-m^2) $$

Without the theta function, we have $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2)$$ The integrand is an even function of $k^0$, so $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) = 2\int_{0}^{\infty} dk^0 \delta (k^2-m^2)$$ This is twice the result for the case with the theta function.

I good approach to evaluating the integral is to use property #7 of the delta function listed here.
Property #6 in the list is a special case of #7 and is directly relevant to your integral.

 

[Hill]

With the theta function, $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) \theta (k^0)=\int_{0}^{\infty} dk^0 \delta (k^2-m^2) $$

Without the theta function, we have $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2)$$ The integrand is an even function of $k^0$, so $$\int_{-\infty}^{\infty} dk^0 \delta (k^2-m^2) = 2\int_{0}^{\infty} dk^0 \delta (k^2-m^2)$$ This is twice the result for the case with the theta function.

I good approach to evaluating the integral is to use property #7 of the delta function listed here.
Property #6 in the list is a special case of #7 and is directly relevant to your integral.

Thank you very much. I see my mistake now: when replacing $k^0$ with $\omega_k$ I've missed that it can be $+\omega_k$ and $-\omega_k$. The unit step function eliminates one of them.

 

[Hill]

The last part of that exercise is to show that $$\int \frac {d^3k} {2 \omega_k}$$ is Lorentz invariant.
Using the equation above, I get $$\int \frac {d^3k} {2 \omega_k}=\int {d^3k} \int dk^0 \delta (k^2-m^2) \theta (k^0)= \int d^4k \delta (k^2-m^2) \theta (k^0)$$
Now, $d^4k$ and $k^2-m^2$ are Lorentz invariant, but $k^0$ isn't. What do I miss?

 

[renormalize]

The last part of that exercise is to show that $$\int \frac {d^3k} {2 \omega_k}$$ is Lorentz invariant.
Using the equation above, I get $$\int \frac {d^3k} {2 \omega_k}=\int {d^3k} \int dk^0 \delta (k^2-m^2) \theta (k^0)= \int d^4k \delta (k^2-m^2) \theta (k^0)$$
Now, $d^4k$ and $k^2-m^2$ are Lorentz invariant, but $k^0$ isn't. What do I miss?

The step-function $\theta (k^0)$ depends only on the sign of $k^0$ and so is invariant under any Lorentz transform that doesn't include time-reversal.