# How does preservation of the Lorentz algebra demonstrate Lorentz invariance of a QFT?

In his book "Quantum Field Theory of Point Particles and Strings", Brian Hatfield makes the following claim (on p. 46) after canonically quantizing the free scalar field theory:

We started with a classical theory that is relativistic. We do not want to destroy this by quantization. In order to canonically quantize, we had to specify equal-time commutators. The choice made, eq.(3.11), is not Lorentz covariant. Thus, to quantize, we must choose a specific Lorentz frame. We want to verify that we get the same quantum theory no matter which frame we choose. One way to do this is to verify that the quantum operator forms of the generators of the Lorentz algebra still satisfy the proper algebra after quantization. The specific computation appears as an exercise at the end of the chapter.

Here eq.(3.11) are just the canonical quantization conditions

$$[\phi(\mathbf{x},t), \pi(\mathbf{y},t)] = i \delta (\mathbf{x}-\mathbf{y})$$

$$[\phi(\mathbf{x},t), \phi(\mathbf{y},t)] = [\pi(\mathbf{x},t), \pi(\mathbf{y},t)] = 0.$$

At the end of the chapter, the Lorentz algebra commutation relation

$$[M^{\mu\nu}, M^{\lambda\sigma}] = i(\eta^{\mu\lambda}M^{\nu\sigma} - \eta^{\nu\lambda}M^{\mu\sigma} - \eta^{\mu\sigma}M^{\nu\lambda} + \eta^{\nu\sigma}M^{\mu\lambda})$$

is provided, and the suggestion is made to

Rewrite $$M^{\mu\nu}$$ in terms of the operators $$a(\mathbf{k})$$ and $$a^\dagger(\mathbf{k})$$, and show that the algebra above still holds after quantization.

I understand what Hatfield means about the specific choice of Lorentz frame and am familiar with the generators and their associated Lorentz algebra used here. I do not see, however, why showing that the algebra is preserved upon quantization demonstrates that the theory is Lorentz-invariant. Since Hatfield makes similar arguments later in the book (e.g. when quantizing the Dirac equation on p.76) but gives no explanation beyond what is quoted above, I would greatly appreciate if someone could clarify how this works.

Requiring that the $$M^{ab}$$ satisfy the Lorentz algebra is clearly necessary, but why would it be sufficient?

Without more information about the $$M^{ab}$$, it's not sufficient. We can easily contrive operators $$M^{ab}$$ that satisfy the Lorentz algebra but which do not act as Lorentz transformations on all of the quantum fields in the model. For example, start with a single scalar field and construct the $$M^{ab}$$ as usual, then bring a second scalar field into the model but don't modify the operators $$M^{ab}$$. The $$M^{ab}$$ still satisfy the Lorentz algebra, but they don't have the required effect on the second scalar field.

Clearly, we need to do more than just checking that some given collection of operators $$M^{ab}$$ happens to satisfy the Lorentz algebra.

I don't have a copy of the book, but I'll assume that it defines the operators $$M^{ab}$$ in terms of the stress-energy tensor: $$M^{ab}\sim\int d^3x\ \big(x^a T^{0b}(x)-x^b T^{0a}(x)\big). \tag{1}$$ For a model with only scalar fields, the stress-energy tensor is$$^\dagger$$ $$\newcommand{\pl}{\partial} T^{ab} \propto \sum_n\frac{\delta L}{\delta\, \pl_a\phi_n}\pl^b\phi_n -\eta^{ab}L \tag{2}$$ where $$L$$ is the Lagrangian density and the subscript $$n$$ labels the different scalar fields. (Equations (1)-(2) eliminate the contrived counterexample I mentioned above.) Using this general expression together with the canonical equal-time commutation relations, we can verify that $$\int d^3y\ \big[T^{0b}(y),\phi_n(x)\big]\propto \pl^b\phi_n(x) \tag{3}$$ at equal time. This is clear by inspection for $$b> 0$$, and the less-clear case $$b=0$$ is already familiar because $$\int T^{00}$$ is the Hamiltonian. (That's the subject of another question.)

Now the question can be answered with the help of an appropriate perspective. Many textbooks go to great lengths to define how different types of fields should transform and to construct models that respect those transformation rules. That's important if our goal is to engineer a model that has a specified symmetry. However, if our goal is only to discover the symmetries of a model that is given to us, then life is easier. Once we have a symmetric $$T^{ab}$$ that satisfies (3), equation (1) is guaranteed to have the correct effect on the spacetime argument $$x$$ of every field. (This is clear by inspection.) Then the scalar/spinor/vector/etc character of the field can be discovered from the effects of these same transformations. In other words, instead of deciding in advance that "this field should transform like a scalar/spinor/vector/etc," we can let the calculation tell us how the field transforms. It's a discovery, not a demand.

Here's the catch: For the discovery-not-demand approach to work, we still need to check that the operators (1) really do satisfy the Lorentz algebra, because that's what ensures that all of the model's fields do transform in some representation of the (covering group of the) Lorentz group. Equation (3) ensures this for the spacetime argument $$x$$, and then the Lorentz-algebra condition ensures this for the "spin" degrees of freedom.$$^{\dagger\dagger}$$

This whole argument rests on knowing, a priori, that equation (1) properly accounts for angular momentum when $$T^{ab}$$ is symmetric, regardless of the model's details. I neglected to explain how we know that, but the answer https://physics.stackexchange.com/a/69578 addresses this issue beautifully. For some related insights in a non-relativistic context, see Why is the (non-relativistic) stress tensor linear and symmetric?.

Footnotes:

$$^\dagger$$ We should use the symmetric version of $$T^{ab}$$. This makes an important difference for spinor fields, so we get the correct "spin" term in the angular momentum operators (1). Recall that Noether's theorem does not determine the conserved current uniquely. Any expression for $$T^{ab}$$ consistent with Noether's theorem will satisfy (3), but we need to use the symmetric version to get the correct Lorentz transformation properties. If we define $$T^{ab}$$ by varying the action with respect to the metric field, then $$T^{ab}$$ is automatically symmetric.

$$^{\dagger\dagger}$$ I've never seen this perspective in a textbook. To help explain why it's valid, let's play a game. I'll step into another room where you can't see what I'm doing, and while I'm in there, I'll engineer a model to be either Lorentz-symmetric or not. Then I'll mix up the notation to obscure which components of which fields belong together. After that, I'll hand the model to you, with the mixed-up notation, and your job is to determine whether or not the model is Lorentz-symmetric. How can you do that? You can do it by constructing $$M^{ab}$$ as described above (using the symmetric stress-energy tensor) and then checking whether or not the $$M^{ab}$$ satisfy the Lorentz algebra. That tells you whether or not the model is Lorentz-symmetric, and if it is, is also tells you how to organize the fields' components into all of the right Lorentz representations.

• This is a very clear answer, and your assumption about how the book defines $M^{ab}$ is correct — thank you! Before I accept it, could you please clarify a final point related to your first footnote? As I understand, the logic is that the angular momentum operators give the right infinitesimal coordinate transformations, and that by checking they form the right Lie algebra we can use the exponential map to get a representation of the connected component of our desired group. Commented Apr 20, 2020 at 10:40
• My question is, how do we know beforehand that the action/equations of motion will be invariant under this representation? Your first footnote seems to suggest that by choosing a non-symmetric stress-energy tensor we can get "wrong" Lorentz transformation properties, which I assume means a wrong representation (or do you mean that the algebra will just fail to hold?) Why is this construction with the angular momentum (and a correctly chosen $T$ tensor) capable of discovering the right representation for the given model? Commented Apr 20, 2020 at 11:13
• I guess the classical Noether current $\mathcal{M}^{\mu a b} = \sum_n \pi^\mu_n \left.\frac{\partial \phi_n}{\partial \epsilon_{ab}}\right|_{\epsilon = 0} - \mathcal{F}^{\mu a b}$, where $\left.\frac{\partial\mathcal{L}}{\partial\epsilon_{ab}}\right|_{\epsilon=0} = \partial_\mu\mathcal{F}^{\mu a b}$, means that that upon quantisation the symmetry generators should act as $[M^{ab}, \mathcal{L}] \sim \partial_\mu\mathcal{F}^{\mu a b}$ and the action should thus be invariant under the representation they give. I still don't see quite why the symmetric $T$ tensor is the desired one though. Commented Apr 20, 2020 at 15:13
• @Quasihorse You're asking exactly the right question! The whole argument rests on knowing, a priori, that equation (1) properly accounts for angular momentum when $T$ is symmetric, regardless of the model's details. I neglected to explain how we know that. The answer physics.stackexchange.com/a/69578 addresses this issue beautifully in the relativistic context. (The answers to physics.stackexchange.com/q/62963 give some related insights in a non-relativistic context.) Commented Apr 20, 2020 at 16:11
• Thank you very much! I believe that, thanks to your latest comments and the links given therein, I now fully understand the argument. I am accepting your answer, and a big thank you again. Commented Apr 20, 2020 at 16:59