In Classical Field Theory fields are sections of bundles over spacetime. In particular we almost always consider vector bundles. Some examples are:

  1. Scalar fields: these are sections of the trivial bundle $M\times \mathbb{R}$, that is, mappings $\psi(x) = (x, \overline{\psi}(x))$. We often in this case just speak of the field $\psi : M\to \mathbb{R}$, since it is virtualy the same. These can also be complex valued, by switching $\mathbb{R}$ with $\mathbb{C}$.

  2. Vector fields: these are sections of the tangent bundle $TM$. They can also be characterized by their coordinate representations $A^\mu$ in charts together with transformation rules to ensure the resulting object is indeed a section of $TM$.

  3. Tensor fields: these are sections of the tensor bundle $T^r_s M$, being $(r,s)$ the tensor type. Again we can represent these fields in coordinates $T^{\mu_1\dots\mu_r}_{\phantom{\mu_1\dots\mu_r}\nu_{1}\dots\nu_{s}}$ by specifying compatibility conditions - the so-called tensor transformation rules.

Now let's turn to QFT. How does one determine the type of the field? I mean, a quantum field is an operator valued function or also one operator valued distribution.

How does one classify the fields as scalar, real, complex, vector, tensor and so on? The fact is that while in Classical Field Theory each field type has a different target space, that is, a different type of value it takes on, in QFT all fields are operator valued.

Just to give one simple example where things get blurry. A scalar field $\phi(x)$ in Classical Field Theory is real if $\varphi(x)^\ast = \varphi(x)$. In QFT it is not clear what is a real scalar field, since both the real scalar field and the complex scalar field are operator valued (there isn't one $\mathbb{R}$-valued and another $\mathbb{C}$-valued).

I believe the same would happen to a vector field $A$. It also is operator valued, so both the scalar field $\varphi(x)$ and the vector field $A(x)$ takes value on $\mathcal{L}(\mathcal{H})$ the space of linear operators on the Hilbert space $\mathcal{H}$.

So considering all field types have same target space in QFT, how does one distinguish among the different field types?

  • $\begingroup$ QFT fields are classified according to the Wigner classification, which classify fields according to their little groups and spin representation. $\endgroup$
    – Slereah
    Commented Mar 27, 2017 at 5:01
  • 2
    $\begingroup$ @Slereah No, Wigner's classification classifies the particles/unitary representations of the Poincaré group, not the fields/finite-dimensional representations of the Poincaré group. $\endgroup$
    – ACuriousMind
    Commented Apr 1, 2017 at 10:00

1 Answer 1


Ignoring issues with the fields being operator-valued distributions rather than functions, a $X$-valued tensor field is a section of $T^r_s M \otimes X$, where you simply tensor every fiber with the space the field takes values in. So there's no issue - the $T^r_s M$ part still determines what the transformation of the field under Poincaré transformations is, just that every component of that tensor is no longer a real or complex number, but a value in $X$.

For a quantum field, $X$ should be thought of as the $C^\ast$-algebra of observables, and since these algebras carry by definition a notion of "complex conjugation", you can write down the equation $\phi = \phi^\ast$ without any problems.

A basic assumption of QFT - one of the Wightman axioms - is that given a representation $\rho$ of the $C^\ast$-algebra on the space of states, the finite-dimensional representation $\sigma_\text{fin}$ of the Poincaré group on $T^r_s$ and the unitary representation $U$ of the Poincaré group on the space of states, we have that $$ U(\Lambda,a)\rho(\phi(x))U(\Lambda,a)^{-1} = \rho(\sigma_\text{fin}(\Lambda,0)^{-1}\phi(\Lambda x + a)),$$ which is a compatibility condition between the Poincaré representation on the space of fields and on the space of states.


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