# I QFT operators time/space asymmetry?

1. Mar 3, 2017

### TeethWhitener

I'm slowly working through Srednicki's QFT book and I had a question about section 3 (canonical quantization of scalar fields). At one point, he shows that the creation and annihilation operators $a(\mathbf{k})$ and $a(\mathbf{k}')$ are time-independent via the equation:
$$a(\mathbf{k}) =i\int{d^3x e^{-ikx}[i\partial_0 \varphi(x) +\omega\varphi(x)]}$$
I thought that was kind of odd given that the theory was supposed to be relativistic and one should expect to see time and space treated in an equal way. In one sense, we have the $e^{ikx}$ factor, where $kx$ is the dot product of the position and momentum four-vectors, so this part is at least nominally time dependent (except maybe it's constant, since it's a dot product?). On the other hand, tracking down where $a(\mathbf{k})$ came from in the first place, we have the solution to the KG equation, which Srednicki gives as the sum of plane waves weighted with arbitrary functions of the wave vector:
$$\varphi(\mathbf{x},t) =\int{ \frac{d^3k}{f(k)} [a(\mathbf{k})+b(\mathbf{k})]e^{i\mathbf{k} \cdot \mathbf{x}-i\omega t}}$$
My question is this: since the Klein Gordon equation is symmetric in time and space, why aren't the $a(\mathbf{k})$'s functions of frequency as well? e.g., why don't we have:
$$\varphi(\mathbf{x},t) =\int{ \frac{d^4k}{f(k)} [a(\mathbf{k},\omega)+b(\mathbf{k},\omega)] e^{i\mathbf{k} \cdot \mathbf{x}-i\omega t}}$$
Does this have something to do with the on shell condition? Srednicki mentions it briefly in this section but doesn't elucidate why it's important. Thanks for any help you can give!

2. Mar 3, 2017

### TeethWhitener

...or I could just plug the general solution into the KG equation and see that $\omega$ and $\mathbf{k}$ have to be related by $m$... I feel kind of dumb now

3. Mar 7, 2017

### Demystifier

The non-covariant equations you have written above can also be written in forms which look more covariant. For instance, (3.16) in Srednicki is derived from an invariant expression with $\delta(k^2+m^2)$, which expresses the mass shell condition. But non-covariant forms are used because they are more practical for actual calculations.

4. Mar 7, 2017

### Ddddx

The canonical formalism is not manifestly relativistically covariant. This is because in the canonical formalism you have X and P at a particular time t.

So you have equal time commutation relations, rather than commutation relations for two arbitrary spacetime points.

There is a way to make the canonical formalism covariant, where the phase space is not the set of X and P at a time t but the space of solutions of the equations of motion.

You sometimes see the fact mentioned that the equal time commutation relation is equivalent to the general spacetime points commutator [φ(x), φ(y)] = iΔ(x, y).

This commutator corresponds not to the poisson bracket of the usual canonical formalism but the so-called Pierls bracket of the covariant canonical formalism mentioned above where the phase space is the space of classical solutions.

5. Mar 7, 2017

### TeethWhitener

I think part of the problem is that Srednicki introduced the mass shell condition without really explaining why it was important. It seems like a necessary consequence of the Klein Gordon equation. Wikipedia said something about how virtual particles can be off shell, but I should probably just keep going through Srednicki's book and hope that he explains it eventually.

My next question was actually going to be about this equal time requirement. It didn't really make sense that we would work so hard to find solutions to the Lorentz invariant KG equation, only to throw them into a clearly non-Lorentz invariant commutator. Are you saying that equal time commutation relations are covariant, just not manifestly so?

6. Mar 7, 2017

### Demystifier

Yes, that's explained in Eq. (3.11), but perhaps not sufficiently clearly. Maybe Srednicki is not the best book for a first introduction to QFT. I would suggest you to try with Klauber:
https://www.amazon.com/Student-Friendly-Quantum-Field-Theory/dp/0984513957
It is much more pedagogic for a first book on QFT.

Last edited by a moderator: May 8, 2017
7. Mar 7, 2017

### Ddddx

Since the hamiltonian E = √(P2 - m2) is relativistic, the theory is relativistic in the end. Its just that the language is non-relativistic because E and P enter into usual hamiltonian mechanics in an asymmetric manner, not respecting relativity.

8. Mar 7, 2017

### TeethWhitener

Thank you both for your help.

9. Mar 12, 2017

### bhobba

So would I, plus:

Before that I suggest (in the following order):
https://www.amazon.com/Quantum-Field-Theory-Demystified-McMahon/dp/0071543821