# Lorentz transformation of annihilation operator

In Srednicki's Quantum Field Theory, chapter 4, the author claims that the Lorentz transformation for given a scalar field $$\varphi(x)$$, \begin{align} U(\Lambda)^{-1} \varphi(x) U(\Lambda) = \varphi(\Lambda^{-1}x), \end{align} "implies that the particle creation and annihilation operators transform as" \begin{align} U(\Lambda)^{-1} a(\mathbf{k}) U(\Lambda) = a(\Lambda^{-1}\mathbf{k}). \end{align}

I'm trying to prove that statement. My starting point is the expression for the $$a$$ operators given in the previous chapter: \begin{align} a(\mathbf{k}) = \int d^3x e^{-ikx} \left[ i\partial_0 \varphi(x) + \omega \varphi(x) \right]. \end{align}

I then applied $$U(\Lambda)^{-1}$$ and $$U(\Lambda)$$ to left and right of this equation and used the fact that these operators commute with the integral and derivative to obtain \begin{align} U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) = \int d^3x e^{-ikx} \left[ i\partial_0 \varphi(\Lambda^{-1}x) + \omega \varphi(\Lambda^{-1}x) \right]. \end{align}

I then want to make a variable change $$x' = \Lambda^{-1}x$$. For that I first put the integral measure in a Lorentz invariant form, in a similar manner to what the author does for the measure in $$k$$-space, by defining \begin{align} \tilde{dx} := \frac{d^3x}{2 \sqrt{s^2 + \mathbf{x}^2}}, \quad s = \sqrt{-(x^0)^2 + (\mathbf{x})^2}>0. \end{align}

By making the variable change to $$x'$$ and using the Lorentz invariance of this new integration measure I then obtain \begin{align} U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) &= \int \tilde{dx}' 2 \sqrt{s^2 + (\Lambda\mathbf{x}')^2} e^{-ik(\Lambda x')} \left[ i\partial_0 \varphi(x') + \omega_k \varphi(x') \right] \\ &= \int d^3x' \sqrt{\frac{(\Lambda^0_{~~\mu} x'^\mu)^2}{(x'^0)^2}} e^{-i(\Lambda^{-1}k)x'} \left[ i\partial_0 \varphi(x') + \omega_k \varphi(x') \right] \\ &= \int d^3x' \left(\frac{\Lambda^0_{~~\mu} x'^\mu}{x'^0}\right) e^{-ik'x'} \left[ i\left(\Lambda_0^{~~\nu}\partial'_\nu\right) \varphi(x') + \left(\omega_{k'} \frac{\Lambda^0_{~~\sigma} k'^\sigma}{k'^0}\right) \varphi(x') \right], \end{align}

where $$k' = \Lambda^{-1} k$$. At this point the whole expression is in terms of the new integration variables $$x'$$, and the new momenta $$k' =\Lambda^{-1} k$$. However, it is not clear how to put this in the same form as the definition of $$a$$ (the third equation) and I don't know where to go from here.

My question is, can this last expression be simplified to obtain Eq. 3? If so, how?

Notes:

• As the author, I'm using the "mostly plus" metric, with $$(x^\mu) = (t, \mathbf{x})$$, $$(x_\mu) = (-t, \mathbf{x})$$.
• I believe the notation "$$\Lambda^{-1}\mathbf{k}$$" means the spatial part of $$\Lambda^{-1}k$$, where $$k^0 = \sqrt{m^2 + \mathbf{k}^2}$$.
• Also, are you sure that the transformation commutes with the integral and the derivative?? Commented Jan 29, 2022 at 20:28
• That is a good question. I'm not 100% sure, but I would argue like this: $U(\Lambda) = \exp(i M^{\mu \nu} u_{\mu \nu})$, where u are the generators and M the parameters of the transformation. M is just a numerical matrix of parameters, independent of x, so it commutes with the derivatives. Then it is a matter of whether the representation of the generators of the Lorentz transformations commute with a time derivative. I guessed that they shouldn't evolve with time, but maybe I'm wrong. Do they evolve in time? Commented Jan 29, 2022 at 23:18
• I think that the transformations do not commute with the derivatives, neiter with the integration measures. The reason is that Lorentz transformation include, in their definition if you like, rotations. Having said that, one can not expect derivatives and vectors not to be transformed under such transformations... Commented Jan 30, 2022 at 15:40
• Take a look here or here (Problem set 3). Note that I haven't carefully checked the content, tho. Commented Jan 31, 2022 at 14:16

Indeed, it is true, but a bit subtle to prove directly.

Let us start from $$a(k) = i\int_{\Sigma}e^{-ik x} n_\Sigma \cdot \partial \phi(x) - \phi(x)n_\Sigma \cdot \partial e^{-ik x} d \Sigma(x)$$ where $$\Sigma$$ is any 3-rest space of any Minkowskian reference frame, the left hand side being independent of that essentially due to the fact that the integral is the charge of a conserved current and using Stokes’ theorem. Notice that $$a(k)$$ is labelled by four vectors $$k$$ which stay on the mass shell. It does not make much sense to me to use the invariant measure and next to label these operators by the spatial components of the four momenta. Let us go on. From the transformation property of the field, $$U(\Lambda)^{-1} \phi(x) U(\Lambda) = \phi(\Lambda^{-1}x)$$ the transformation of the conjugate momentum, $$\pi_n(x) := n\cdot \partial \phi(x)$$ where $$n$$ is the chosen temporal axis, reads $$U(\Lambda)^{-1} n\cdot \partial \phi(x) U(\Lambda) = \Lambda^{-1} n \cdot \partial \phi (\Lambda^{-1}x).$$ Therefore, assuming the temporal axis $$n_\Sigma$$ normal to $$\Sigma$$,

$$U^{-1}_\Lambda a(k) U_\Lambda = i\int_{x\in \Sigma}e^{-ik x}\Lambda^{-1} n_\Sigma \cdot \partial \phi(\Lambda^{-1} x) - \phi(\Lambda^{-1} x)n_\Sigma \cdot \partial e^{-ikx} d\Sigma(x).$$ Notice that $$\Lambda^{-1}n_\Sigma = n_{\Lambda^{-1}\Sigma}.$$ Due to the fact that $$\Lambda$$ is an isometry, thus preserves the measure, the identity above an be rewritten as $$U^{-1}_\Lambda a(k) U_\Lambda = i\int_{\Lambda x\in \Lambda \Sigma}e^{-ik \Lambda x} n_{\Sigma} \cdot \partial \phi(\Lambda^{-1} \Lambda x) - \phi(\Lambda^{-1}\Lambda x)n_{\Sigma} \cdot \partial e^{-ik \Lambda x} d\Lambda \Sigma(\Lambda x).$$ In other words, $$U^{-1}_\Lambda a(k) U_\Lambda = i\int_{\Lambda x\in \Lambda \Sigma}e^{-i\Lambda^{-1} k x} n_{\Sigma} \cdot \partial \phi(x) - \phi(x)n_{\Sigma} \cdot \partial e^{-i\Lambda^{-1}kx} d \Lambda \Sigma(\Lambda x).$$ Using again the fact that isometries preserve the measure, $$U^{-1}_\Lambda a(k) U_\Lambda = i\int_{x\in \Sigma}e^{-i\Lambda^{-1} k x} n_{\Sigma} \cdot \partial \phi(x) - \phi(x)n_{\Sigma} \cdot \partial e^{-i\Lambda^{-1}kx} d \Sigma(x) =a_{\Lambda^{-1}k}.$$

What Srednicki is trying to say with the first equation you have written (well, maybe the with the second too) is that both the field $$\varphi(x)$$ and the creation/annihilation operators do not have a vector or a tensor nature. Rather, they are scalar quantities and as scalar quantities they should transform under Lorentz transformations. What you are trying to do is redundant I think. The essence of a scalar quantity is that it transforms trivially under Lorentz tranformations. You can think of it as a defining property. Also, the same way Lorentz transformations act on spatial vectors, they also act on momentum vectors, so there is no reason why the creation/annihilation operators do not obey the same "transformation rules" with the scalar field $$\varphi(x)$$...

• What I am struggling to understand is how the fact that the field $\phi(x)$ transforms as a scalar implies that the annihilation and creation operators must transform as a scalar as well. Conceptually it seems reasonable, but I would like to prove it mathematically. Commented Jan 29, 2022 at 23:32
• Also, your last sentence seems aimed towards proving that $\tilde{\phi}(\mathbf{k}, t)$ should transform the same way as $\phi(x)$, but the annihilation operators are a little bit more complicated than that... Commented Jan 30, 2022 at 0:10
• The Lorentz transformation does not see how each function depends on its (vector) arguments. The only thing the Lorentz transformation cares about is whether or not it is dealing with functions, vectors, more general tensors etc... I realize that what I am suggesting might appear as a "believe without proof" sort of thing, but I think it is not. Commented Jan 30, 2022 at 15:44
• What you are trying to do (prove it mathematically) seems redundant to me, because you are somehow going from some sort of function to some other and then using the definition (i.e. that scalar functions are invariant under Lorentz transformations), rather than using the definition directly in the first place... Commented Jan 30, 2022 at 15:49
• So you're saying annihilation operators are defined as scalars under Lorentz transformations? But aren't they defined from $\phi$? Commented Jan 30, 2022 at 19:39