I Causality preserved in Klein-Gordon equation

[Silviu]

Hello! I am reading Peskin's book on QFT and in chapter one he shows that $[\phi(x), \phi(y)] = D(x-y) - D(y-x)$, with $D(x-y)$ being the propagator from $x$ to $y$. He says that if $(x-y)^2<0$ we can do a Lorentz transformation such that $(x-y) \to -(x-y)$ and hence the commutator vanishes and causality is conserved. Also he says that if $(x-y)^2>0$ we can't make such a Lorentz transformation. I am not sure how is he doing these Lorentz transformations and why you can do them in the first case but not in the second. Also, I am not sure I understand how can you do in the fist case this transformation just for the second propagator, while keeping the first one fixed.

 

[vanhees71]

First of all, it's always important to tell which propagator you are talking about. It's about Peskin+Schroeder Section 2.4. So here he defines $D$ to be the Wightman function (vev of a fixed-order field-operator product),
$$D(x-y)=\langle 0|\hat{\phi}(x) \hat{\phi}(y)|0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3 2 E_{\vec{p}}} \exp[-\mathrm{i} p\cdot(x-y)].$$
This is a Lorentz-scalar field, because $\mathrm{d}^3 \vec{p}/(2 E_{\vec{p}})$ is a Lorentz scalar and all other ingredients in the integral are manifestly Lorentz invariant (with Lorentz invariant here I mean invariance under the proper orthochronous Lorentz group). Since further there is no other four vector then $z=x-y$ this implies that
$$D'(z')=D(z')=D(z), \quad z'=\Lambda z, \quad \Lambda \in \mathrm{SO}(1,3)^{\uparrow}.$$
This implies that if the claim about space-like vectors is true you have
$$D(z)=D(-z) \quad \text{for} \quad z^2<0,$$
and this implies that the vev of the commutator which is $D(x-y)-D(y-x)$ by definition, indeed vanishes for space-like $x-y$.

Now let's prove the claim about space-like vectors. So let $z^2<0$. Then we can always use a coordinate system, where $z^0=0$. To see this just take $e_1=z/\sqrt{-z \cdot z}$ as one of the space-like basis vectors. The basis can be obviously constructed such that the change to this coordinate system is given by a proper orthochronous Lorentz transformation. So we have
$$z=\begin{pmatrix}0\\ z^1 \\ 0 \\ 0 \end{pmatrix}.$$
Now a rotation around the $z^3$-axis reads
$$z^{\prime 0}=z^0=0, \quad \vec{z}'=\begin{pmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}z^1 \\0 \\ 0 \end{pmatrix} = z^1 \begin{pmatrix} \cos \phi \\ -\sin \phi \\0 \end{pmatrix}.$$
For $\phi=\pi$ you get $\vec{z}'=-\vec{z}$, as claimed.

It's a good exercise to prove that this argument doesn't work for timelike vectors, because the sign of $z^0$ cannot be changed by a proper orthochronous Lorentz transformation (that's why it's called orthochronous).

 

[Silviu]

First of all, it's always important to tell which propagator you are talking about. It's about Peskin+Schroeder Section 2.4. So here he defines $D$ to be the Wightman function (vev of a fixed-order field-operator product),
$$D(x-y)=\langle 0|\hat{\phi}(x) \hat{\phi}(y)|0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3 2 E_{\vec{p}}} \exp[-\mathrm{i} p\cdot(x-y)].$$
This is a Lorentz-scalar field, because $\mathrm{d}^3 \vec{p}/(2 E_{\vec{p}})$ is a Lorentz scalar and all other ingredients in the integral are manifestly Lorentz invariant (with Lorentz invariant here I mean invariance under the proper orthochronous Lorentz group). Since further there is no other four vector then $z=x-y$ this implies that
$$D'(z')=D(z')=D(z), \quad z'=\Lambda z, \quad \Lambda \in \mathrm{SO}(1,3)^{\uparrow}.$$
This implies that if the claim about space-like vectors is true you have
$$D(z)=D(-z) \quad \text{for} \quad z^2<0,$$
and this implies that the vev of the commutator which is $D(x-y)-D(y-x)$ by definition, indeed vanishes for space-like $x-y$.

Now let's prove the claim about space-like vectors. So let $z^2<0$. Then we can always use a coordinate system, where $z^0=0$. To see this just take $e_1=z/\sqrt{-z \cdot z}$ as one of the space-like basis vectors. The basis can be obviously constructed such that the change to this coordinate system is given by a proper orthochronous Lorentz transformation. So we have
$$z=\begin{pmatrix}0\\ z^1 \\ 0 \\ 0 \end{pmatrix}.$$
Now a rotation around the $z^3$-axis reads
$$z^{\prime 0}=z^0=0, \quad \vec{z}'=\begin{pmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}z^1 \\0 \\ 0 \end{pmatrix} = z^1 \begin{pmatrix} \cos \phi \\ -\sin \phi \\0 \end{pmatrix}.$$
For $\phi=\pi$ you get $\vec{z}'=-\vec{z}$, as claimed.

It's a good exercise to prove that this argument doesn't work for timelike vectors, because the sign of $z^0$ cannot be changed by a proper orthochronous Lorentz transformation (that's why it's called orthochronous).

Thank you for you answer. I am just now sure why $D'(z')=D(z')=D(z)$. The first and the 3rd term are equal as D is a scalar field. But why does the equality in the middle holds?

 

[vanhees71]

By definition a scalar field obeys
$$D'(z')=D(z).$$
Now since there's no other vector involved, of which $D$ could depend, and thus you must have $D'=D$, i.e., also $D'(z')=D(z')$.

Another argument is that if you have no other vectors then $z$ a scalar function must be a function of $z^2$.