Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Causality preserved in Klein-Gordon equation

  1. Apr 30, 2017 #1
    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.
     
  2. jcsd
  3. May 1, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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).
     
  4. May 1, 2017 #3
    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?
     
  5. May 1, 2017 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Causality preserved in Klein-Gordon equation
  1. Klein-Gordon equation (Replies: 6)

  2. Klein-Gordon equation (Replies: 5)

Loading...