# A question about causality and Quantum Field Theory from improper Lorentz transformation

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).

• $p\cdot(x-y)$ 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. May 11, 2014 at 15:24
• 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 $p^i$? May 11, 2014 at 15:39
• 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 $x$. May 11, 2014 at 15:55
• 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. May 11, 2014 at 16:29
• I think so. Another example is vector versus pseudovector, they transform in the same way under rotation, but different under parity. May 12, 2014 at 1:52

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

# Summary

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 $(0,x,0,0)$.

# Conclusion

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.

• Hi, does this inversion transformation should be thought of as a inversion of the axis? Apr 21, 2018 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 $$D(x-y)= D(y-x)\:.$$