Related post Causality and Quantum Field Theory

In Peskin and Schroeder's QFT p28, the authors tried to show causality is preserved in scalar field theory.

Consider commutator $$ [ \phi(x), \phi(y) ] = D(x-y) - D(y-x) \tag{2.53} $$ where $D(x-y)$ is the two-point correlation function, $$D(x-y):= \langle 0 | \phi(x) \phi(y) | 0 \rangle = \int \frac{d^3 p}{ (2\pi)^3} \frac{1}{ 2E_{\mathbf{p}}} e^{-ip(x-y)}\tag{2.50}$$

P&S argued that each term in the right-hand-side of (2.53) is Lorentz invariant, since $$\int \frac{d^3p }{ (2\pi)^3} \frac{1}{2E_{\mathbf{p}}} = \int \frac{ d^4 p }{ (2\pi)^4} (2\pi) \delta(p^2-m^2)|_{p^0>0} \tag{2.40}$$ is Lorentz invariant.

Since there exists a continuous Lorentz transformation in the spacelike interval $(x-y)^2<0 $ such that $(x-y) \rightarrow - (x-y) $ and $D(y-x)=D(x-y)$, (2.53) equals zero in the spacelike interval. In timelike interval, since such continuous Lorentz transformation does not exist, (2.53) is non-zero in general.

My question is, consider a non-continuous Lorentz transmation in the timelike interval, $PT$, namely time reversal times parity transformation. I can also let $(x-y) \rightarrow - (x-y) $. Why (2.53) in the timelike interval is non-zero?

I guess $PT$ will let (2.40) go to $p^0<0$ branch. But I am not sure if it breaks the Lorentz invariant of (2.40) and (2.50).

  • $\begingroup$ $p\cdot(x-y)$ is invariant under parity, but isn't invariant under time reversal. Under time reversal, for $p$ the spatial components are reversed, for $x$ and $y$ the temporal component is reversed. $\endgroup$ – Jia Yiyang May 11 '14 at 15:24
  • $\begingroup$ I thought about it. However, a Lorentz vector transforms as $$p^0 = \Lambda^{0}_{\nu} p^{\nu} $$. For time reversal, $\Lambda=diag(-1,1,1,1)$. Will $p^0$ change a sign, but not $p^i$? $\endgroup$ – user26143 May 11 '14 at 15:39
  • $\begingroup$ it should go the other way: we have the physical definition of $p$, and knows how it ought to transform under time reversal, and then we write down the time reversal matrix for p, i.e. $\Lambda=diag(1,-1,-1,-1)$. There's nothing wrong to have a transformation matrix for $p$ different from that of $x$. $\endgroup$ – Jia Yiyang May 11 '14 at 15:55
  • $\begingroup$ Does it mean, follow the definition of tensor, $V^{\mu} \rightarrow \frac{ \partial x'^{\mu} }{\partial x^{\nu}} V^{\nu} = \Lambda^{\mu}_{\nu} V^{\nu}$ for any vector $V^{\mu}$, the universal expression of $\Lambda$ is only for continuous transformation, where we can define partial derivatives. For non continuous transformation, the transformation has to be worked out in case-specific ways. $\endgroup$ – user26143 May 11 '14 at 16:29
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    $\begingroup$ I think so. Another example is vector versus pseudovector, they transform in the same way under rotation, but different under parity. $\endgroup$ – Jia Yiyang May 12 '14 at 1:52

I'm digging this thread out just to clarify some things for those who might have a similar question.


We cannot use $\mathcal T$. Space-like four-vectors are essentially like $(0,x,y,z)$, so we can ignore the time and do three-dimensional rotations to get $(0,-x,-y,-z)=-(0,x,y,z)$.

A la Valter Moretti

As Valter Moretti already pointed out, you cannot just apply $\mathcal P\mathcal T$ to get $(x-y)\to-(x-y)$, because $D(x-y)$ is not invariant under $\mathcal T$.

So the challenge is really to do $(x-y)\to-(x-y)$ using only proper orthochronous Lorentz transformations $SO(1,3)_+$ and $\mathcal P$. This is only possible for space-like four-vectors.

The point about space-like four-vectors is that there is a Lorentz-frame where $t=0$ (boost with $\beta=\frac{t}{|\vec x|^2}$), and in such a frame the parity transformation $$\mathcal P:(0,x',y',z')\to(0,-x',-y',-z')=-(0,x',y',z')$$ looks just like an inversion. So what you can do for space-like four-vectors is $$ (t,x,y,z) \overset{\Lambda}{\to}(0,x',y',z') \overset{\mathcal P}{\to}-(0,x',y',z') \overset{\Lambda^{-1}}{\to}-(t,x,y,z) $$

The difference between this transformation and $\mathcal P\mathcal T$ is that the latter takes all four-vectors to their inverses, whereas the former only a (three dimensional) subspace of the four-dimensional Minkowski space.

A la Peskin and Schroeder

You can actually achieve the same without using $\mathcal P$, that is only with $SO(1,3)_+$ transformations. This means we can continuously bring a fixed space-like vector $p$ to its inverse $-p$. Just do the following steps: \begin{align*} (t,x,y,z) &\overset{R_1}{\to}\left(t,\sqrt{x^2+y^2},0,z\right)\\ &\overset{R_2}{\to}\left(t,\sqrt{x^2+y^2+z^2},0,0\right)\\ &\overset{B\left(\beta=\frac{t}{|\vec x|^2}\right)}{\to}\left(0,\sqrt{x^2+y^2+z^2-t^2},0,0\right)\\ &\overset{R_\pi}{\to}-\left(0,\sqrt{x^2+y^2+z^2-t^2},0,0\right)\\ &\overset{\left(BR_2R_1\right)^{-1}}{\to}-\left(t,x,y,z\right) \end{align*} In view of this one should really say that space-like vectors are like $(0,x,0,0)$.


Space-like four-vectors should be thought of as $(0,x,0,0)$, and since there are three spacial dimensions, there is enough room to rotate this vector in any direction. This allows us to invert space-like vectors just by using proper ortochronous transformations $SO(1,3)_+$.

Time-like four-vectors are like $(t,0,0,0)$. There is only one time direction, and hence no rotations are possible. Hence the only way of getting $-t$ is to use time inversion $\mathcal T$.

Short, because there is only one time dimension, but more than one space dimensions, we can invert space-like four vectors by continuous Lorentz-rotations, but not time-like.

  • $\begingroup$ Hi, does this inversion transformation should be thought of as a inversion of the axis? $\endgroup$ – Slayer147 Apr 21 '18 at 22:28

The thesis is true, but I cannot understand well the claimed relation with the existence of "continuous" Lorentz transformations such that $x-y \mapsto y-x$. The argument essentially relies upon the invariance of the measure under the orthochronous Lorentz group.

Fix a four vector $x-y$ and consider $$D(x-y):= \int \frac{d \vec{p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{p}}} e^{-ip(x-y)}\:.$$ Since the measure $\frac{d \vec{p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{p}}}$ is $O(3,1)_+$ invariant, for any $\Lambda \in O(3,1)_+$ you have, $$D(x-y)= \int\frac{d \vec{p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{p}}} e^{-ip(x-y)} = \int \frac{d \vec{\Lambda p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{\Lambda p}}}e^{-ip(x-y)}= \int \frac{d \vec{p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{ p}}}e^{-i(\Lambda^{-1} p)(x-y)} = \int \frac{d \vec{p}}{ (2\pi)^3} \frac{1}{ 2E_{\vec{ p}}}e^{-ip (\Lambda (x-y))} = D(\Lambda (x-y))\:.$$ We conclude that, for every four vector $x-y$ and every $\Lambda \in O(3,1)_+$, it holds, $$D(x-y) = D(\Lambda(x-y))\:.$$

Remark. Since $O(3,1)= O(3,1)_+ \cup TO(3,1)_+$ and $ O(3,1)_+ \cap TO(3,1)_+= \emptyset$, and the considered measure is not invariant under $T$, just because of $$\int \frac{d\vec{p} }{ (2\pi)^3} \frac{1}{2E_{\vec{p}}} = \int \frac{ d^4 p }{ (2\pi)^4} (2\pi) \delta(p^2-m^2)|_{p^0>0}\:,$$ we conclude that

$\Lambda \in O(3,1)$ leaves invariant the measure if and only if $\Lambda \in O(3,1)_+$.

Notice that the measure is $P$-invariant since we are dealing with $O(3,1)_+$ and not $SO(3,1)_+$. However it is not $PT$ invariant.

Now there are two possibilities for $x-y \neq 0$:

(a) $x-y$ is spacelike. In this case, for that $x-y$ there is $\Lambda \in O(3,1)_+$ such that $\Lambda (x-y) = y-x$. Such $\Lambda$ is a spatial $\pi$ rotation around $x$ in the $3D$ rest frame defined by a timelike vector $u$ orthogonal to $x-y$. In this case we conclude that $$D(x-y)= D(y-x)\:.$$

(b) $x-y$ is not spacelike. In this case there is no $\Lambda \in O(3,1)_+$ such that $\Lambda(x-y)= y-x$, because $y-x$ is past directed if $x-y$ is future directed and vice versa and thus they cannot be connected by transformations of $O(3,1)_+$ by definition. In this case we cannot conclude that $$D(x-y)= D(y-x)\:.$$


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