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

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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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
asked Jul 24, 2014
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)

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user ACuriousMind
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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
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!

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
You misunderstood what I (clumsily) meant by "does not change". I'll try and write an answer.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user ACuriousMind
Yes a precise statement of this invariance in terms of a target vector space of the Lorentz rep is precisely my question.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira

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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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
answered Jul 24, 2014 by (835 points)
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"?

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
@bechira. Yes. It's simply the codomain of the function $\phi$. Namely $\phi:\mathbb R^{3,1}\to V$. So, for example, for scalar field, $V = \mathbb R$. For a vector field $V = \mathbb R^{3,1}$. For a tensor field of rank $(k,\ell)$, $V = T^k_\ell(\mathbb R^{3,1})$ where this last notation just means the vector space of tensors of rank $(k,\ell)$ on $\mathbb R^{3,1}$.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
Please see the update to my question. I think in classical field theory this is obvious, my main concern is whether the same is valid in a quantum theory.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
I think in this answer you just said what I said in my first comment in the comment section.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
@bechira I think I understand your question now. You're right that my last comment holds only in the classical case. In the quantum case, the target space is more complicated because the fields are operator-valued. The actual target space, then, will be a space of e.g. tensor operators in the case of a tensor field etc. The formalization of that was the content of a question I myself asked a while ago which you will probably find illuminating. physics.stackexchange.com/questions/73532/tensor-operators

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
@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.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
Yes I can see the utility, I'm concerned whether such $R$ is still a valid representation in the usual sense.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
@bechira Yes it is. It is precisely the same representation as in the classical case. The fact that the fields are operator-valued doesn't change the fact that "the indices transform" in the specified way, namely according to the representation that acts on the target space of the classical fields.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
But does the dimension N representation still act on a vector space of dimension N if we don't have well-defined "scalar multiplication by operator valued fields"?

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user bechira
Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
@joshphysics I always love your posts since you contribute great content. But I think you might be a little sloppy here: what you call "$\mathrm{Lin}(V)$" is not exactly the codomain for a group representation (it should be $\mathrm{GL}(V)$ -- the set of all invertible linear operators on $V$). Just a minor detail...

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Alex Nelson
@AlexNelson I very much appreciate the praise and close readings of my responses to make sure I don't do sloppy things. Fortunately, in this case I don't think I was being sloppy. Recall that the codomain need only be a superset of the range. en.wikipedia.org/wiki/Codomain Nonetheless, I'll change it because $\mathrm{GL}(V)$ is certainly more informative since it's a smaller codomain.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics
@AlexNelson Incidentally, now that we've started this praise train, I feel the same way about your posts as well. Glad you're on SE.

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user joshphysics

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