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In A. Zee's Quantum field theory in a nutshell p. 24, it says the Klein-Gordon propagator depends on the sign of $x^0$. Here $x=(x^0,x^1,x^2,x^3)$ with Minkowski sign convention $(+,-,-,-)$.

$$D(x)=-i\int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_k x^0-\vec{k}\cdot\vec{x})}\theta(x^0)+e^{i(\omega_k x^0-\vec{k}\cdot\vec{x})}\theta(-x^0)].\tag{I.3.23}$$

I don't understand why the sign of $x^0$ is Lorentz invariant as Zee claims. A typical Lorentz transformation is like $(x{^0})'=\gamma(x^0-v x^1)$. So there must be transformations that change the sign of $x^0$ by carefully choosing the form of $x^1$.

Qmechanic
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ruima86
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    $x^0$ is just time. The proper Lorenz transformation does not change the time sign. It is obvious that your $vx^1$ is in fact $V\cdot 0$, so the time gets dilated. – Vladimir Kalitvianski Jul 19 '16 at 20:19
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    I don't see why $x^1$ is obviously zero. If event A happens earlier than event B in one frame, it is totally possible that A happens later than B in another frame, provided they happen at different places. – ruima86 Jul 19 '16 at 20:57
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    The proper Lorenz transformation involves different RF with different velocities $V$, but no spacial shift. In your $D(x)$ the coordinates $x^i$ are independent variables. They do not form an invariant interval. – Vladimir Kalitvianski Jul 19 '16 at 21:06
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    Let me address the question in this way. If $D(x)$ is the propagator in S frame, and is written in the form shown in my question. What is the corresponding $D'(x')$ in S' frame which moves relative with S at a speed $V$ – ruima86 Jul 19 '16 at 21:18
  • You should differ two types of interval but propagator is the same analytic function of interval and you can obtain both answers – Artem Alexandrov Jan 11 '20 at 14:03

2 Answers2

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OP got a point: Zee presumably only meant orthochronous Lorentz transformations when he claimed that the sign of $x^0$ [...] do not change.

Nevertheless, eq. (I.3.23) in Zee's book (and mentioned by OP) is directly derived from the previous equation $$ D(x-y)~=~\int_{\mathbb{R}^4}\! \frac{d^4k}{(2\pi)^4} \frac{e^{ik\cdot(x-y)}}{k^2-m^2+i\epsilon}, \tag{I.3.22} $$ which is manifestly Lorentz invariant (which can be seen by performing the same Lorentz transformation on the $k^{\mu}$-integration variables as the $x^{\mu}$-variables), so eq. (I.3.23) better also be invariant under the full Lorentz group. In fact, eq. (I.3.23) can be rewritten as $$ D(x)~=~-i\int_{\mathbb{R}^3}\! \frac{d^3\vec{k}}{(2\pi)^3} e^{-i(\omega_{\vec{k}}|x^0|\pm\vec{k}\cdot\vec{x})},\tag{I.3.23'} $$ thereby avoiding the Heaviside step function. Eq. (I.3.23') is clearly invariant under time reversal $T$ and parity $P$. One can check that eq. (I.3.23') is also invariant under restricted (and thereby all) Lorentz transformations.

See also this & this related Phys.SE posts.

Qmechanic
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If you consider the propagator to be the following(check it for sign of metric):

$$D(x-y)=-i\int \frac{d^3k}{(2\pi)^32\omega_k}[e^{-i(\omega_k (x^0-y^0)-\vec{k}\cdot(\vec{x}-\vec{y}))}\theta(x^0-y^0)+e^{i(\omega_k (x^0-y^0)-\vec{k}\cdot(\vec{x}-\vec{y})}\theta(-(x^0-y^0)]$$

Then you are obviously right. The time-ordering of $x^0$ and $y^0$ is obviously not Lorentz invariant when the separation is space-like. It is clearly stated in many textbooks but let us provide just few references for those who are curios: see here-page 211 ,here-page 32, here-chapter4 or here-chapter10. I definitely recommend chapter 4 of Sakurai's old text.

But now let's get back to your question. We know that sign of time coordinate doesn't change under Lorentz transformation. So if we have $x^0>0$ and $y^0>0$ in one frame it will remain so. But the sign of their difference is not invariant, indicating that it can change when separation is space-like(as we said in above paragraphs). Now suppose that you take $y$ in the above formula to be zero. What happens after a Lorentz transformation? It is simple: zero remains zero, and if $x^0$ was positive or negative it will remain so. The result is that in this special case $sign{(x^0)}$ is Lorentz invariant.

Indeed the statement that $sign(x^0-y^0)$ is Lorentz invariant is a silly statement that you will never find in any book.

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    This answer makes no sense from the second paragraph onward. There can be an event $(x^0,x^1,x^2,x^3)$ which goes to $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$ under a Lorentz transformation with $x^0\geq 0$ and $x^{\prime 0}\leq 0$ or vice versa as indicated in the question. If the four vector $x^\mu$ is timelike, only then a proper "orthochronus" Lorentz transformation will preserve the sign of $x^0$. There is no such restriction on $x^\mu$ inside the propagator. Propagator exists and is non zero even for a spacelike $x^\mu$. – BoundaryGraviton Jul 20 '16 at 07:49
  • Lorentz transformation doesn't change the sign of time coordinate, why should it be dependent on time-like or space-like property?. The $\Lambda^0_{0}$ is positive! This is independent of properties of the $x^\mu$ it's acting on. See comments of Vladimir above. –  Jul 20 '16 at 07:59
  • In the references that I have provided it is clearly stated that the sign of difference between time coordinates is not Lorentz invariant!. This is not at odds with the fact that individual time coordinate's doesn't change sign. –  Jul 20 '16 at 08:04
  • "...as indicated in the question". Well, This is wrong! , That's what is causing the problem. I tried to show it. The sign of $x^0$ can't change. That is general property of Lorentz transformation. –  Jul 20 '16 at 08:07
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    If $(x^0,0,0,0)$ goes to $(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})$ then $x^0\geq 0$ goes to $x^{\prime 0} \geq 0$. Because then the vector is a time like vector. However the argument inside the propagator can even be space like. In such a case, $\Lambda^0_i$'s can be negative to lead to a negative $x^{\prime 0}$ even if we have a positive $x^0$. The proof of proper orthochronus LTs preserving the sign of the time coordinate applies only to time like vectors. See the link given by QMechanic – BoundaryGraviton Jul 20 '16 at 08:10
  • I agree with Hegde. $sign(x^0)$ is Lorentz invariant only if $x$ is time-like vector. But the argument of the propagator can be space-like. – ruima86 Jul 20 '16 at 14:19
  • My bad!. I agree with you. I suggest contacting the author. Here is the Link to errata page. –  Jul 20 '16 at 15:24