# What does $U^{-1}(\Lambda)\phi(\Lambda y) U(\Lambda) = \phi(y)$ physically mean?

In QFT, let $$U(\Lambda)$$ denote a unitary representation of the Lorentz group. Let $$\phi(x^{\mu})$$ be scalar field operators in the Hilbert space; in other words:

$$U^{-1}(\Lambda)\phi(x^{\mu}) U(\Lambda) = \phi(\Lambda^{-1}x).$$

What does this equation physically implies? Does this mean if we have a scalar field $$\phi(x^{\mu})$$ in one coordinate $$x^{\mu}$$, how it looks like in coordinate $$\Lambda x$$? If so, this confuses me.

Because if we denote $$x=\Lambda y$$ Then the equation above can be written as $$U^{-1}(\Lambda)\phi(\Lambda y) U(\Lambda) = \phi(y).$$ Which one is the old field and which one is the one transformed? How should we understand it intuitively? Is this active or passive transformation? I've read some about active and passive transformation, but they don't help me understand this question.

The "new field" is always $$\phi'(x) = U(\Lambda)^{-1} \phi(x) U(\Lambda)$$ Using the properties above, for a scalar field, we find $$\phi'(x) = \phi(\Lambda^{-1} x)$$

A passive transformation is a change of coordinates and doesn't involve unitary operators, so lets deal with an active Lorentz transformation. This means that we have a point $$x$$ and we have a Lorentz transformation $$\Lambda$$. For example one could choose $$\Lambda$$ to be a rotation about some axis; then $$\Lambda x$$ would be the vector $$x$$ rotated about the axis of rotation.

Now in relativistic quantum theory, there is, for every Lorentz transformation $$\Lambda$$, an unitary operator $$U(\Lambda)$$ on Hilbert space such if $$\psi_x$$ is the wave function of a particle centered at position $$\vec{x}$$ at time $$x^0$$, then $$(U(\Lambda) \psi_x)$$ is a the wave function of a particle at position $$\vec{\Lambda^{-1} x}$$ at time $$(\Lambda^{-1} x)^0$$. In particular, if $$\psi_x$$ describes a particle of spin zero, then

$$U(\Lambda) \psi_x = \psi_{\Lambda^{-1}x}$$

Let $$\phi(x)$$ by a field operator annihilating particles of spin zero at position $$x$$, i.e.

$$\phi(x) \psi_x = \Omega$$

where $$\Omega$$ is the vacuum vector. Acting on both sides with $$U(\Lambda)$$ we get

$$U(\Lambda) \phi(x) U(\Lambda)^{-1} U(\Lambda) \psi_x = U(\Lambda) \phi(x) U(\Lambda)^{-1} \psi_{\Lambda^{-1} x} = \Omega \ .$$

That is, the operator $$U(\Lambda)\phi(x) U(\Lambda)^{-1}$$ annihilates a particle at position $$\Lambda^{-1}x$$. Hence

$$U(\Lambda) \phi(x) U(\Lambda)^{-1} = \phi(\Lambda^{-1}x) \ .$$

This equation hence just means that if we rotate an operator, we obtain a rotated operator.