# Lorentz group representations in QFT: what's the vector space?

In QFT, a representation of the Lorentz group is specified as follows: $$U^\dagger(\Lambda)\phi(x) U(\Lambda)= R(\Lambda)~\phi(\Lambda^{-1}x)$$ Where $\Lambda$ is an element of the Lorentz group, $\phi(x)$ is a quantum field with possibly many components, $U$ is unitary, and $R$ an element in a representation of the Lorentz group.

We know that a representation is a map from a Lie group on to the group of linear operators on some vector space. My question is, for the representation specified as above, what is the vector space that the representation acts on?

Naively it may look like this representation act on the set of field operators, for $R$ maps some operator $\phi(x)$ to some other field operator $\phi(\Lambda^{-1}x)$, and if we loosely define field operators as things you get from canonically quantizing classical fields, we can possibly convince ourselves that this is indeed a vector space.

But then we recall that the dimension of a representation is simply the dimension of the space that it acts on. This means if we take $R$ to be in the $(1,1)$ singlet representation, this is a rep of dimension 1, hence its target space is of dimension 1. Then if we take the target space to the space of fields, this means $\phi(x)$ and $\phi(\Lambda x)$ are related by linear factors, which I am certainly not convinced of. EDIT: This can work if we view the set of all $\phi$ as a field over which we define the vector space, see the added section below.

I guess another way to state the question is the following: we all know that scalar fields and vector fields in QFT get their names from the fact that under Lorentz transformations, scalars transform as scalars, and vectors transform like vectors. I would like to state the statement "a scalar field transforms like a scalar" by precisely describing the target vector space of a scalar representation of the Lorentz group, how can this be done?

Let me give an explicit example of what I'm trying to get at: Let's take the left handed spinor representation, $(2,1)$. This is a 2 dimensional representation. We know that acts on things like $(\phi_1,\phi_2)$.

Let's call the space consisting of things of the form $(\phi_1,\phi_2)$ $V$. Is $V$ 2 dimensional?

Viewed as a classical field theory, yes, because each $\phi_i$ is just a scalar. As a quantum field theory, each $\phi_i$ is an operator.

We see that in order for $V$ space to be 2 dimensional after quantization, we need to able to view the scalar quantum fields as scalar multipliers of vectors in $V$. i.e. we need to view $V$ as a vector space defined over a (mathematical) field of (quantum) fields.

We therefore have to check whether the set of (quantum) fields satisfy (mathematical) field axioms. Can someone check this? Commutativity seem to hold if we, as in quantum mechanics, take fields and their complex conjugates to live in adjoint vector spaces, rather than the same one. Checking for closure under multiplication would require some axiomatic definition of what a quantum field is.

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And it's me again, nagging at your question ;) What is $U$? Are you certain that the definiton of a rep is not: "Under a Lorentz transformation $x \mapsto \Lambda x$, the fields transform as $\phi(x) \mapsto R(\Lambda)\phi(\Lambda^{-1}x)$." ? (Also, of course are $\phi(x)$ and $\phi(\Lambda^{-1}x)$ related by a linear factor, the factor is $1$ - a scalar field does not change under Lorentz trafos) – ACuriousMind Jul 24 '14 at 23:36
The way I understand it: $U$ transforms $\phi$ as if it's an operator on a space of states, $R$ transforms $\phi$ as if it's something in $R$'s target vector space. So $U$ is a representation onto the space of states in the field theory, where $R$ is a representation onto some structure involving the fields, and precisely what that structure is, is the content of my question. – bianchira Jul 24 '14 at 23:40
No $\phi(x)$ and $\phi(\Lambda^{-1}x)$, in general, are different operators, scalars do not transform in the sense that the form is invariant, not in the sense that the field is constant valued everywhere. For example, if the operator value of the field is same at every coordinate in spacetime there wouldn't need to be delta functions in the commutation relations! – bianchira Jul 24 '14 at 23:44
You misunderstood what I (clumsily) meant by "does not change". I'll try and write an answer. – ACuriousMind Jul 24 '14 at 23:48
Yes a precise statement of this invariance in terms of a target vector space of the Lorentz rep is precisely my question. – bianchira Jul 24 '14 at 23:50

If $\mathcal H$ is the Hilbert space of the QFT, then \begin{align} U:\mathrm{SO}(3,1)\to \mathscr U(\mathcal H) \end{align} where $\mathscr U(\mathcal H)$ is the set of unitary operators on $\mathcal H$. In other words, $U$ is a unitary representation on the Hilbert space of the theory. If $V$ is the target space of the fields begin considered, then \begin{align} R:\mathrm{SO}(3,1)\to \mathrm{GL}(V), \end{align} where $\mathrm{Lin}(V)$ is the vector space of linear operators on $V$. In other words, $R$ is a representation on the target space of the fields in the theory. The mapping \begin{align} x\to \Lambda x \end{align} is the defining representation of $\mathrm{SO}(3,1)$ on $\mathbb R^{3,1}$. The statement that the field transforms in a particular way, namely the equation you wrote down, simply says that the action by conjugation of the Lorentz group on field operators by $U$ agrees with the composite of the representation on the target space and the inverse of the defining representaton.
Can you be more explicit about what you're calling $V$? By "The target space of the fields" do you mean "the target space consisting of the fields"? – bianchira Jul 25 '14 at 0:15
@bechira I should add, however, that even thought the values of the fields are rather more complicated, the same representation $R$ can still be used to describe the transformation of the target space. I think that the easiest way to see this is to notice that in a given basis, the indices on tensor operators transform in the same way as the indices on tensors. In fact, this is kind of the whole point of tensor operators, namely that their indices transform in the same way as their "classical" counterparts. – joshphysics Jul 25 '14 at 1:12
Yes I can see the utility, I'm concerned whether such $R$ is still a valid representation in the usual sense. – bianchira Jul 25 '14 at 1:14