# Causality preserved in Klein-Gordon equation

• I
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.

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vanhees71
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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).

dextercioby
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
$$D'(z')=D(z).$$