I just have a trouble making a full analogy between Lorentz Algebra Representation in Quantum Field Theory (QFT) and SU(2) representation in Quantum Mechanics (QM).

To make my point, I will write few things that I think is true for the case of QM. We first start by looking at the rotation matrices in Classical Mechanics, represented by matrices $R \in SO(3)$.

Then, we associate unitary matrices with $R$, $D(R)$, and these matrices form $SU(2)$ group. Now, we look at the algebra of $SU(2)$ to find fundamental commutation relationships among the generators of $D(R)$, namely, $$[J_i,J_j] = i\epsilon_{ijk}J_k$$

Then we look for different representations of these generators characterized by different angular momentums (which defines the dimension of the vector space that generators act).

The representation that we use, then also gives an explicit expression for our unitary matrices $D(R)$ by $$D(R) = \exp(\frac{i\vec{J}\cdot\hat{n}}{\hbar}).$$

Also, I can define vectors and tensors by this unitary matrix, $D(R)$. For instance, vector $V_i$ transforms by $$D(R)^{-1} V^i D(R) = R_{\:j}^i V^j.$$

Now, I want to similarly understand QFT's case with the Lorentz group. (I am currently following QFT text by Srednicki).

I start with Lorentz matrices $\Lambda$, and associate it with unitary matrices, $U(\Lambda)$. I have a similar definition of 4-vector in QFT as in QM: $$U(\Lambda)^{-1} V^i U(\Lambda) = \Lambda_{\:j}^i V^j.$$

I can also define the generators of $U(\Lambda)$, $M^{\mu\nu}$, and derive its fundamental commutation relations, $$[M^{\mu\nu},M^{\rho\sigma}]=\cdots.$$

Now, making complete analogy with QM, I expect to find representation of $M^{\mu\nu}$ and the representation of $U(\Lambda)$ by exponentiating $M^{\mu\nu}$.

But instead, we proceed by looking for the representation of $\Lambda$, instead of $U(\Lambda)$ like in QM. For instance, as for left Weyl-spinor representation, I find representation $L(\Lambda)$: $$U(\Lambda)^{-1} \psi_a(x) U(\Lambda) = L_a^{\:b}(\Lambda) \psi_b(\Lambda^{-1} x).$$

Now, I have a generator $S_L$ (which is now not necessary to be Hermitian (unlike QM)), which gives $L(\Lambda)$ when exponentiated (rather than $U(\Lambda)$ (unlike QM)).

I do not get explicit expression (unlike QM) for $U(\Lambda)$, so I do not know what to think of them or its generators $M^{\mu\nu}$. For instance, I get expressions that involve both $M^{\mu\nu}$ and $S_L^{\mu\nu}$ ((whereas in QM, since I looked for a representation of $D(R)$ (rather than $R$), quantity analogous to $M^{\mu\nu}$ and $S_L^{\mu\nu}$ were the same thing)).

I do know that there is no finite unitary representation of the Lorentz algebra, so I think that must be the missing piece in my understanding. I would like to make a complete analogy with QM, could anyone please be of help?

Thank you.

  • $\begingroup$ If you know that there is no finite-dimensional unitary rep of the Lorentz group, what missing piece are you searching for? What is your specific question? (You seem confused by the fact that there are two "simultaneous" representations of the Lorentz group in QFT. Those that act directly on the (classical) fields as finite-dimensional representations (your $L(\lambda)$), and the unitary representations upon the Hilbert spaces of state under which the fields transform as operators (by your $U\Lambda$). You don't get a complete analogy to QM because this doesn't happen in QM. $\endgroup$
    – ACuriousMind
    Commented Apr 8, 2015 at 14:19
  • $\begingroup$ Your definition of a "vector" is also off, already in the QM case - is $V^i$ is a vector under rotation, it transforms as $V\mapsto D(R) V$, or, in components, $V^i \mapsto D(R)^i_j V^j$. That in QFT the "operator transformation" $U(\lambda)\phi U(\lambda)^\dagger$ and the "vector transformation" $\phi\mapsto L(\Lambda)\phi$ coincide is one of the Wightman axioms. $\endgroup$
    – ACuriousMind
    Commented Apr 8, 2015 at 14:22
  • $\begingroup$ @ACuriousMind: Yes, that is what I am confused about. Are you saying the right hand side (finite dimensional representation) is not a finite dimensional "Hilbert" space? So $\psi_a(x)$ has a infinite dimensional representation in the Hilbert space and finite dimensional representation in classical vector space? Also, you are saying Wightman axiom tells us, if I find an explicit infinite dimensional representation of $\psi_a(x)$ and $U(\Lambda)$, left hand side will give exactly same result? Thank you ACuriousMind! $\endgroup$ Commented Apr 8, 2015 at 14:39
  • $\begingroup$ Closely related physics.stackexchange.com/q/799610/226902 $\endgroup$
    – Quillo
    Commented Jan 31 at 13:08

1 Answer 1


The confusion here arises because we are not fully analogous to non-relativistic QM here.

Given a (quantum or classical) field $\phi$, we usually specify whether it is a "scalar", "spinor", "tensor", whatever field. This refers to a finite-dimensional representation $\rho_\text{fin}$ of the Lorentz group the field transforms in as an element: $$ \phi \overset{\Lambda}{\mapsto} \rho_\text{fin}(\Lambda)\phi$$ But, simultaneously, the quantum field is an operator on the Hilbert space of the theory, and on the Hilbert space there must exist a unitary representation $U$. More precisely, every component $\phi^\mu$ of the quantum field is an operator, and hence transforms as operators do: $$ \phi^\mu \overset{\Lambda}{\mapsto} U(\Lambda)\phi^\mu U(\Lambda)^\dagger$$ It is now one of the Wightman axioms that $$ U(\Lambda)\phi U(\Lambda)^\dagger = \rho_\text{fin}(\Lambda)\phi$$ or, in components $$ U(\Lambda)\phi^\mu U(\Lambda)^\dagger=\rho_\text{fin}(\Lambda)^\mu_\nu\phi^\nu$$ It is by this assumption that it suffices to give the finite-dimensional representation of the quantum field to also fix the accompanying unitary representation on the infinite-dimensional Hilbert space it is an operator on. The infinite-dimensional representations are characterized by Wigner's classification through their mass and spin/helicity. Since the finite-dimensional representations on the fields are also characterized by spins, the mass (from the kinetic term of the field) and the spin of the field (from its finite-dimensional representation) fix the unitary representation the particles it creates transform in.

All of this is often brushed under the rug because for the Lorentz invariant vacuum $\lvert\Omega\rangle$, we have $$ \phi \lvert \Omega \rangle \overset{\Lambda}{\mapsto} \rho_\text{fin}(\Lambda) \phi \lvert\Omega\rangle$$ so knowing the finite-dimensional representation suffices to know how all states the field creates from the vacuum transform, and since the Fock spaces are entirely build out of such states, this is all the practical knowledge about the unitary representation that is usually needed.

  • $\begingroup$ Thanks for your answer! Could you possibly expand upon the following point? > It is by this assumption that it suffices to give the finite-dimensional representation of the quantum field to also fix the accompanying unitary representation on the infinite-dimensional Hilbert space it is an operator on. I.e. how exactly does this "fix" the accompanying unitary representation? $\endgroup$
    – balu
    Commented Feb 20, 2018 at 20:36
  • 1
    $\begingroup$ @balu Actually, this does not fully fix the corresponding unitary representation - a unitary representation is determined uniquely by mass and spin as per Wigner's classification. The spin is determined by the finite-dimensional Lorentz representation, but the mass is an additional parameter the field (or rather the Lagrangian) carries with it. $\endgroup$
    – ACuriousMind
    Commented Feb 20, 2018 at 20:44
  • $\begingroup$ Wow...this is amazing! So basically we are talking about two different representations at the same time? One in which the field itself is the vector space to be operated on and one in which the field operates on the associated Hilbert space? $\endgroup$ Commented Oct 9, 2021 at 13:09
  • 1
    $\begingroup$ Schwartz's book on QFT said we need infinite dimensional representations and I was wondering forever how eg. a (1/2, 0) representation is infinite dimensional when it's 4-dimensional. So this is the key to it? $\endgroup$ Commented Oct 9, 2021 at 13:11

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