# Scalar operators In Quantum Field Theory

I am trying to learn Quantum Field Theory and I am stuck in a basic point.

What is the definition of a scalar operator in QFT? That is, how does it transform under a Poincare transformation? Why do we want them?

You can assume there is no spin etc. to simplify things. Please be explicit.

• I don't agree with the close-vote that this has attracted. – Kyle Kanos Nov 11 '14 at 2:25

Under an active Lorentz transformation, a scalar field transforms as,

$$\phi(x) \to \phi'(x)=\phi(\Lambda^{-1}x)$$

Suppose at $x_0$, the field has the value $\phi_0$. Now I rotate the field by some angle - say $\pi/2$. If I were to evaluate the old field at that point, it would give me value that's probably not $\phi_0$. So to express the new field in terms of the old field, I need to take the point, and do the inverse transformation so the field 'thinks' it's at that old point. On the other hand, under a passive transformation which constitutes relabling our coordinates, then it would be $\Lambda$ rather than $\Lambda^{-1}$.

• Under a passive transformation, is it true that: $U^{-1}(\Lambda) \varphi(\Lambda x) U(\Lambda) = \varphi(x)$ – user1 Nov 10 '14 at 22:00
• I don't accept the answer, because it doesn't answer my question completely. Assuming the formula I have written above is correct, I can't see why it is required for theory to be lorentz invariant. I think we want a condition on S opeator, but I can,t immediately see the connection. – user1 Nov 11 '14 at 18:22

A scalar field/operator is a mathematical construct. We "want" it because it may be used to model real particles. See e.g. Wikipedia, where composite particles such as pi meson or the fundamental particle Higgs boson can be modeled by a scalar field.

Taking a step back from QFT, an example of a scalar field could be something like the temperature of a room, which assigns a (single component) number (temperature) to each point in that room. This is in contrast with a multi-component vector field, e.g. air velocity at each point in the same room.

How should such a scalar field transform under rotation (as a special form of a Lorentz transform)? Well, if you take the same room and rotate it around an axis by, say theta degrees, you are not really changing the temperature at any point in that room. In fact, you would get the same effect if you instead rotated the coordinates by the negative of theta degrees. Therefore $$T(x) \rightarrow T(\Lambda^{-1}x)$$ which means if you apply the transformation on T, it is as if you apply the inverse of the same transformation on the coordinate x.

Now why should the scalar field transform as such for Lorentz transformations, and not any arbitrary transformation? Because we want to merge Quantum mechanics (which we know is true in its regime) with special relativity (which we know is true in its own regime) to come up with a more comprehensive theory (which is true in a broader regime, also encompassing the other two regimes). Now special relativity requires Lorentz invariance for scalar fields, because (using the temperature example again) two observers using different coordinates x and x' should agree on the temperature of the same point in the room, although expressed using two different coordinate systems x and $$x'=\Lambda^{-1}x$$