Why can a Lorentz transformation not take $\Delta x^\mu$ to $-\Delta x^\mu$ when $\Delta x$ is time-like? [duplicate]

Question

In Peskin and Schroder's Quantum field theory text (page 28, the passage below Eq. 2.53), it asserts that for a spacelike interval $(x-y)^2<0$, one can perform a Lorentz transformation taking $(x-y)\rightarrow -(x-y).$ This is argued by noting that both the points $(x-y)$ and $-(x-y)$ lies on the spacelike hyperboloid.

The specific Lorentz transformation matrix $\Lambda^{\mu}{}_{{\nu}}$ that acting on $(x-y)^\mu\equiv \Delta x^\mu$ takes it to $-\Delta x^\mu$ is given by $\Lambda=diag(-1,-1,-1,-1)$. Is that correct?

For $(\Delta x)^2\equiv \Delta x_\mu\Delta x^\mu>0$, why the Lorentz transformation $\Lambda=diag(-1,-1,-1,-1)$ will not take $\Delta x^\mu$ to $-\Delta x^\mu$?

Answer 1

The argument given in Peskin and Schröder is at the very least confusing. The "Lorentz-invariant" measure $\frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{2 E_p}$ is only invariant under proper orthochronous Lorentz transformations, i.e. those continuously connected to the identity (P&S seem aware of this when they talk about a "continuous Lorentz transformation"). In particular, $E_p$ would flip the sign under your transformation and so you would not get the desired zero for the commutator from that.

What P&S presumably have in mind is this: Take a space-like vector $s$ with components $(s^0,s^1,s^2,s^3)$, where we can w.l.o.g. take $s^2 = s^3 =0$, $s^0,s^1$ positive and where "space-like" then means $s^1 > s^0$. The boost with velocity $v = 2\frac{s^0 s^1}{s_0^2 - s_1^2}$ is given by $$ \begin{pmatrix}\sqrt{1+v^2} & v \\ v & \sqrt{1+v^2}\end{pmatrix}$$ and will send $(s^0,s^1)$ to $(-s^0,-s^1)$ if and only if $(s^0)^2 - (s^1)^2 < 0$ (this is a tedious, but straightforward computation). That there is no proper orthochronous transformation doing this for a time-like vector follows directly from e.g. this earlier question of yours.

An easier argument for why $[\phi(x),\phi(y)]$ vanishes is as follows:

If $x$ and $y$ are at space-like separation, there is a proper orthochronous Lorentz transformation that sets $x^0 = y^0$, i.e. makes both events simultaneous. Then, we have that \begin{align} [\phi(x),\phi(y)] & = \int \left(\mathrm{e}^{-\mathrm{i}p(x-y)} - \mathrm{e}^{-\mathrm{i}(y-x)}\right)\frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{2 E_p} \\ & = \int\left( \mathrm{e}^{-\mathrm{i}\vec p (\vec x-\vec y)} - \mathrm{e}^{-\mathrm{i}\vec p(\vec y-\vec x)}\right)\frac{\mathrm{d}^3 p}{(2\pi)^3} \frac{1}{2 E_p}\end{align} and the simple integral substitution $\vec p\mapsto -\vec p$ in the second term then yields zero for the commutator. Note that we do not use "Lorentz invariance" for this substituiton, you just observe the transformation flips the integral boundaries $\int^\infty_{-\infty}$ to $\int^{-\infty}_\infty$ and that $\mathrm{d}^3 p$ also flips its sign so we get no overall sign change.