Srednicki QFT: Integration measure for KG eqn?

[TeethWhitener]

Hi, I had a quick question about something from Section 3 of Srednicki's QFT book. In it, he's discussing the solution to the Klein-Gordon equation for classical real scalar fields. He gives the general solution as:
$$\int_{-\infty}^{+\infty} \frac{d^3 k}{f(k)} \left[a(\textbf{k})e^{ikx}+a^*(\textbf{k})e^{-ikx}\right]$$
and proceeds to discuss the integration measure $\frac{d^3 k}{f(k)}$. He chooses $f(k)$ such that the overall measure is Lorentz-invariant:
$$\frac{d^3 k}{f(k)} = d^4 k \delta(k^2 + m^2) \theta(k^0)$$
where $k^2+m^2=\omega^2$, $k^2 = k^{\mu}k_{\mu}$, and $\theta(x)$ is the Heaviside step function. If you integrate this measure over $k^0$, then you get:
$$\int_{-\infty}^{+\infty} dk^0\delta(k^2+m^2)\theta(k^0) = \frac{1}{2\omega}$$
I understand all of the calculation behind this. I also understand that the choice of $\delta(k^2+m^2)$ is to enforce the on-shell condition (I don't quite understand why on-shell is important yet, but I'm sure that'll come later). What I don't understand is the presence of the Heaviside step function. I remember reading somewhere (I know PF hates when people say that, but I can't find the citation), that the reason for the step function is to choose the positive energy solutions ($k^0 > 0$ implies that $E >0$), but my question is this: Why are we allowed to arbitrarily throw out the negative energy solutions? Is it just a choice we make?

 

[ShayanJ]

In 1920s when people were first trying to find the relativistic version of the quantum theory, these negative energy solutions were a problem. Dirac solved this problem for his particles, by assuming that all the negative energy levels were filled and Pauli exclusion principle forbade the positive energy Dirac particles to go down to negative energy levels. But this could only work for Dirac particles because they were fermions and it didn't work for Klein-Gordon particles. So another solution was considered: we assume that there are only positive energy solutions but every field actually has two kinds of excitations(particles) which we call particles and antiparticles. This is implemented by assuming that every field has two kinds of creation and annihilation operators for each momentum. The only point about this particular question is that for a real scalar field, particles are the same as the antiparticles.

 

[TeethWhitener]

In 1920s when people were first trying to find the relativistic version of the quantum theory, these negative energy solutions were a problem. Dirac solved this problem for his particles, by assuming that all the negative energy levels were filled and Pauli exclusion principle forbade the positive energy Dirac particles to go down to negative energy levels. But this could only work for Dirac particles because they were fermions and it didn't work for Klein-Gordon particles. So another solution was considered: we assume that there are only positive energy solutions but every field actually has two kinds of excitations(particles) which we call particles and antiparticles. This is implemented by assuming that every field has two kinds of creation and annihilation operators for each momentum. The only point about this particular question is that for a real scalar field, particles are the same as the antiparticles.

This is a fantastic explanation! Thank you!

 

[TeethWhitener]

One more thing. I'm having a hard time seeing how the integration measure $d^4k\delta(k^2+m^2)\theta(k^0)$ is Lorentz invariant. I'm fine with the volume element and the delta function, but how is $\theta(k^0)$ invariant? I don't see how $\theta(k^0) = \theta(\Lambda^0{}_{\mu}k^{\mu})$. I get as far as $\theta(k^0) = \theta(\Lambda^0{}_0k^0)$ for orthochronous transformations but I'm not sure what to do with $\mu=1,2,3$.

 

[ShayanJ]

Actually $\theta(k^0)$ is not Lorentz invariant. But its not the the whole Lorentz group that people talk about when they say that integration measure is Lorentz invariant. What they mean is the biggest subgroup of the Lorentz group which is called the proper orthochronous Lorentz transformations. Proper means, that these transformations don't change the orientation of the coordinate system and orthochronous means that these transformations take future(past)-directed time-like vectors to future(past)-directed time-like vectors,i.e. they don't change the direction of time and this means they don't change the sign of the time component of a time-like vector field. So $\theta(k^0)$ is invariant under the proper orthochronous Lorentz transformations by definition.

 

[TeethWhitener]

Thanks @ShayanJ for your response. My question isn't about what's intuitively going on, but rather the math of it. If I Lorentz transform $k^0$ in the step function, I get $\theta(k'^0) = \theta(\Lambda^0{}_0 k^0 +\Lambda^0{}_1 k^1 +\Lambda^0{}_2 k^2 +\Lambda^0{}_3 k^3)$. For orthochronous transformations, I have $\Lambda^0{}_0 \geq 1$, but I don't see how that alone gives me $\theta(k^0) = \theta(k'^0)$.

 

[ShayanJ]

To simplify the calculations, I consider the 1+1 dimensional case. The Lorentz transformations can be written as $\left( \begin{array}{cc} \cosh\chi &\sinh\chi \\ \sinh\chi &\cosh\chi \end{array} \right)$. I implement $\Lambda_0^0>1$ by assuming that $\cosh\chi=b+1$ with b a positive real constant. So the transformation given becomes $\left( \begin{array}{cc} b+1 &\sqrt{b(b+2)} \\ \sqrt{b(b+2)} &b+1 \end{array} \right)$. A future directed time-like vector can be written like $\left( \begin{array}{c} \sqrt{x+y} \\ \sqrt y \end{array} \right)$ with x and y both positive real constants. The time-component of the transformed vector is equal to $(b+1)\sqrt{x+y}+\sqrt{by(b+2)}$. Now considering that

$(b+1)\sqrt{x+y}=\sqrt{(b+1)^2(x+y)}=\sqrt{(b^2+2b+1)(x+y)}=\sqrt{b^2y+2by+b^2x+2bx+x+y}=\sqrt{by(b+2)+b^2x+2bx+x+y}>\sqrt{by(b+2)}$

So the transformed vector will also be future directed regardless of the sign of the space-component.

 

[PeterDonis]

by assuming that $\cos\chi=b+1$

I think you mean to use $\cosh$ and $\sinh$, not $\cos$ and $\sin$, correct? This formula will work for $\cosh \chi$, but not for $\cos \chi$, since the cosine function can't give values greater than 1.

 

[TeethWhitener]

Ok thanks so much for your help. I've been working on the (3+1)D case and I just want to make sure I'm clear on everything. First, even though, in principle, $k^{\mu}$ is an arbitrary 4-vector, orthochronous transformations only matter when $k^{\mu}$ is timelike, because if it were spacelike, the time direction would be irrelevant. So assuming $k^{\mu}$ is timelike and future directed, we have that $k^0 > \sqrt{(k^1)^2+(k^2)^2+(k^3)^2} = |\textbf{k}|$, or equivalently, $k^0/|\mathbf{k}|>1$. For a general Lorentz transformation, we have
$$k'^0=\gamma(k^0-\mathbf{\beta} \cdot \textbf{k})$$
where $\gamma \geq 1$ for an orthochronous transformation and $\mathbf{\beta}$ is a vector such that $0\leq |\mathbf{\beta}|<1$. Since gamma is positive, $k'^0$ will have the same sign as $k^0$ as long as $k^0>\mathbf{\beta} \cdot \textbf{k}$. Dividing throughout by the norm of $\mathbf{k}$, and writing $\mathbf{\beta} \cdot \mathbf{k} = |\mathbf{\beta}||\mathbf{k}|\cos{\theta}$, we get:
$$\frac{k^0}{|\mathbf{k}|} - |\mathbf{\beta}| \cos{\theta}$$
and since the first term is greater than 1 and the second term is less than 1, $k'^0$ has the same sign as $k^0$. Is that about right?

 

[ShayanJ]

Yeah, that's right, and really nice!