Understanding Microcausality in QFT

[JD_PM]

TL;DR

I want to understand why the invariance of $\Delta(x-y)$ implies that



$$[\phi(x), \phi(y)]=i \hbar c \Delta(x-y)=0, \ \ \ \ \forall \ \ (x-y)^2<0$$



Which means that any two points x and y, with space-like separation, commute

I stumbled upon the concept of microcausality while studying how to covert the (equal time) commutation relations of the (given as an example) Real Klein Gordon Field (QFT Mandl & Shaw second edition, page 40) $[\phi(\vec x, t), \phi(\vec x', t)]=0$ into its manifestly covariant form (QFT Mandl & Shaw second edition, page 47):

$$[\phi(\vec x, t), \phi(\vec y, t)]=i \hbar c \Delta(\vec x - \vec y,0)=0$$

Then Mandl & Shaw asserted that 'The invariance of $\Delta(x-y)$ implies that

$$[\phi(x), \phi(y)]=i \hbar c \Delta(x-y)=0, \ \ \ \ \forall \ \ (x-y)^2<0$$

Which means that any two points x and y, with space-like separation, commute' (i.e. , as vanhees71 wrote in (4.1), $[\mathscr{O}_1 (x), \mathscr{O}_2 (y)]=0$). Then microcausality means that 'measurements of the fields at two points with space-like separation must not interfere with each other'.

I didn't really understand Mandl & Shaw's point so I checked Tong's notes (2.6.1) together with his lecture (at 1:06:46 he starts explaining what I do not get). There he explains microcausality based on the light cone but I still do not get what he means... Could you please explain it to me?

Thank you

 

[PeterDonis]

The Mandl & Shaw argument is that, if we know that field operators at the spacetime points $(\vec{x}, t)$ and $(\vec{y}, t)$ commute for any $t$, then the operators must also commute for any pair of spacetime points that are spacelike separated. The reason is simple: for any pair of spacetime points that are spacelike separated, there will be some frame in which those points will have coordinates $(\vec{x}, t)$ and $(\vec{y}, t)$ for some $\vec{x}$ and $\vec{y}$ and some $t$. In that frame, we already know the operators at those two points commute, from the formula already given; but since the commutator is frame invariant, the operators must commute in any frame if they commute in one. Thus, field operators must commute at any pair of spacelike separated points.

 

[JD_PM]

Thank you PeterDonis

The reason is simple: for any pair of spacetime points that are spacelike separated, there will be some frame in which those points will have coordinates $(\vec{x}, t)$ and $(\vec{y}, t)$ for some $\vec{x}$ and $\vec{y}$ and some $t$. In that frame, we already know the operators at those two points commute, from the formula already given; but since the commutator is frame invariant, the operators must commute in any frame if they commute in one. Thus, field operators must commute at any pair of spacelike separated points.

Is that what Tong is trying to explain by means of the light cone?

 

[PeterDonis]

Is that what Tong is trying to explain by means of the light cone?

It seems to be, yes.

 

[samalkhaiat]

Summary:: I want to understand why the invariance of $\Delta(x-y)$ implies that
[\phi(x), \phi(y)]=i \hbar c \Delta(x-y)=0, \ \ \ \ \forall \ \ (x-y)^2&lt;0

No, Lorentz invariance alone implies that \Delta (x) is a function of x^{2} only for x^{2} &lt; 0, and of x^{2} and \epsilon (x^{0}) (the sign of the time) for x^{2} \geq 0. So, for space-like x, Lorentz invariance allows you to write \Delta (x) = F (x^{2}). However, since \Delta (x) is an odd function (\Delta (-x) = - \Delta (x)) and a function of x^{2} cannot be odd, we conclude that F(x^{2}) = 0.

 

[JD_PM]

Hi samalkhaiat

Mmm I am afraid I do not understand your point, but your #5 made me think and read more so thank you.

I've read the following reasoning (in natural units)

Let us compute $[\phi(x), \phi(y)]$ explicitly

$$[\phi(x), \phi(y)] = \int \frac{d^3 p}{(2 \pi)^3}\frac{1}{2 \omega_{\vec k}}\Big(e^{-ik \cdot (x-y)}-e^{ik \cdot (x-y)}\Big)=$$ $$=\frac{-1}{(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}} \sin(kx)=$$ $$=\Delta^+(x-y)-\Delta^+(y-x)$$ $$=\Delta (x-y)$$

Where:$$\Delta^+(x-y):=\frac{-i}{2(2\pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}} e^{-ik\cdot(x-y)}, \ \ \ k_0=\omega_{\vec k}$$Then the idea is that the two $\Delta_s^+$ must cancel each other out. Peskin & Schroeder (page 28) explain why based on Lorentz Transformations: when $(x-y)^2<0$, we can perform the following (continuous) LT on $\Delta^+(y-x)$

$$(x-y) \rightarrow -(x-y)$$

(I understand that we can do this because each term by itself is Lorentz Invariant). Thus the two terms cancel out and we get that causality is preserved.

I now have two questions:

1) Does $(x-y) \rightarrow -(x-y)$ simply mean a Lorentz boost from right to left? What I have in mind is a space-like distance $(x-y)$ between two events, having two frames in the ending points of such a distance. We can perform a Lorentz boost either from frame 1 to frame 2 (from left to right) or from frame 2 to frame 1 (from right to left). The only difference between these two boosts is the negative sign. Do you all agree with this interpretation?

2)

No, Lorentz invariance alone implies that \Delta (x) is a function of x^{2} only for x^{2} &lt; 0, and of x^{2} and \epsilon (x^{0}) (the sign of the time) for x^{2} \geq 0.

Could you please explain it further? I understand that $\Delta (x)$ is a real odd function but nothing else.

 

[PeterDonis]

Let us compute $[\phi(x), \phi(y)]$ explicitly

The argument that @samalkhaiat is making does not require computing the commutator explicitly; it relies solely on what Lorentz invariance means for spacelike and timelike intervals, respectively.

Peskin & Schroeder (page 28) explain why based on Lorentz Transformations: when $(x-y)^2 < 0$, we can perform the following (continuous) LT on $\Delta^+(y - x)$

This property does not just apply to $\Delta^+(y - x)$; it applies to any function of $(x - y)$ that is Lorentz invariant, for spacelike $(x - y)$.

(I understand that we can do this because each term by itself is Lorentz Invariant).

I don't think that's the reason. The reason is that spacelike vectors form one continuous family, while timelike vectors are split into two continuous families (future and past) that are not continuously connected to each other; you have to do a discrete transformation (called "time reversal") to go from one to the other.

The general facts I have pointed out are what underlie the argument @samalkhaiat is making.

 

[PeroK]

Could you please explain it further? I understand that $\Delta (x)$ is a real odd function but nothing else.

I asked the same question here:

https://www.physicsforums.com/threads/qft-commutator-for-spacelike-separation.983741/

 

[samalkhaiat]

Mmm I am afraid I do not understand your point,

My point is very simple: Lorentz invariant function is a function of Lorentz invariant argument
Outside the light-cone, x^{2} is the only Lorentz invariant which can be formed from the space-like x. So, if you are told that \Delta (x) is a Lorentz invariant function, i.e., \Delta (x) = \Delta (\Lambda x), then for space-like x, \Delta can only be a function of x^{2}. Thus, you can write \Delta (x) = F(x^{2}) for all x^{2} &lt; 0. If you also know that \Delta (x) = - \Delta (-x), then you conclude that \Delta (x) = 0.

Within the light-cone (or on the light-cone), x^{2} and the sign function \epsilon (x^{0}) = \frac{x^{0}}{|x^{0}|} are the only two invariants which can be constructed from the time-like (or the light-like) x. Therefore, for x^{2} \geq 0, the most general Lorentz invariant function has the form \Delta (x) = F(x^{2}) + \epsilon (x^{0}) G (x^{2}) . Again, if you know that \Delta (x) is an odd function, then \Delta (x) = \epsilon (x^{0}) G(x^{2}). In fact, near the light-cone (i.e., for small x^{2}), explicit calculations show that \Delta (x) \approx - \frac{1}{2\pi} \epsilon (x^{0}) \left( \delta (x^{2}) - \frac{m^{2}}{2} \theta (x^{2}) \right) . This shows that \Delta (x) has a delta-function singularity as well as finite discontinuity on the light-cone.

(x-y) \rightarrow -(x-y)

That is neither continuous nor connected Lorentz transformation. It is a discrete space-time reversal (x^{0} , x^{i}) \to (-x^{0} , -x^{i}). Space reflection (x^{0} , x^{i}) \to (x^{0} , -x^{i}); time reversal (x^{0} , x^{i}) \to (-x^{0} , x^{i}) and space-time reversal form disjoint subsets and are not continuously connected to the identity. In English, x \to – x does not belong to the proper orthochronous Lorentz group SO^{\uparrow}(1,3). By the “Lorentz group”, we always mean the real semi-simple Lie group SO^{\uparrow}(1,3).

I understand that $\Delta (x)$ is a real odd function but nothing else.

Then you may want to consider doing the following exercises in the order presented:

Given the defining property of Lorentz transformation \eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}, prove the following identities

(i) \ (\Lambda k) \cdot (\Lambda x) = k \cdot x

(ii) \ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x

(iii) \ \left(\Lambda^{-1}k \right)^{2} = k^{2}

The above identities are very useful and repeatedly used to solve the following exercises

1) Given that \Delta (x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.

Show that \Delta^{*}(x) = \Delta (x) and \Delta(-x) = - \Delta (x).

2) Let \varphi (x) be a real scalar field (operator) satisfying the Klein-Gordon equation (\partial^{2} + m^{2}) \varphi (x) = 0. Assuming that the commutator \big[ \varphi (x) , \varphi (y) \big] is a Lorentz invariant c-number (i.e., function), show that \big[\varphi (x) , \varphi (0) \big] = C \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x} , with C being an arbitrary real constant.

 

[vanhees71]

The Mandl & Shaw argument is that, if we know that field operators at the spacetime points $(\vec{x}, t)$ and $(\vec{y}, t)$ commute for any $t$, then the operators must also commute for any pair of spacetime points that are spacelike separated. The reason is simple: for any pair of spacetime points that are spacelike separated, there will be some frame in which those points will have coordinates $(\vec{x}, t)$ and $(\vec{y}, t)$ for some $\vec{x}$ and $\vec{y}$ and some $t$. In that frame, we already know the operators at those two points commute, from the formula already given; but since the commutator is frame invariant, the operators must commute in any frame if they commute in one. Thus, field operators must commute at any pair of spacelike separated points.

To complete this argument you also have to mention that the Lorentz transformations by construction act locally, i.e., the field operators transform as the corresponding classical fields, i.e.,
$$\hat{U}(\Lambda) \hat{\phi}(x) \hat{U}^{\dagger}(\Lambda)=\hat{\phi}(\Lambda^{-1} x).$$

 

[JD_PM]

Thank you all, really really helpful replies.

Outside the light-cone, x^{2} is the only Lorentz invariant which can be formed from the space-like x.

Why? Please let me try to answer myself and see if you all agree.

This is what I've found related to it:

In an inertial frame $S$, the whole set of light rays (which are represented by the world-line $x(\tau)$ with slope of $1$) emanating from a given point forms a light-cone and if, for instance, we place such a point at the origin of our coordinate system at time zero (i.e. $x^0=0$), the light rays will form a cone of equation $x^2=0$ (i.e. $x^0 = \sqrt{(x^1)^2+(x^2)^2+(x^3)^2}$). If we perform a LT to the frame $S'$ we actually get $x'^2=0$, due to the invariance of the inner product under LTs.

Source: Kungliga Tekniska högskolan (KTH Physics).

Let us now have a time-like (space-like) event x^{2}; i.e. $x^2>0$ (i.e. $x^2<0$). Then performing a LT I get $x'^2>0$ ($x'^2<0$).

At least I think I understand now why x^{2} is a Lorentz invariant. However, I have to say the argument I presented doesn't explain why x^{2} is a unique Lorentz invariant (when $x$ is space-like of course)... .

Within the light-cone (or on the light-cone), x^{2} and the sign function \epsilon (x^{0}) = \frac{x^{0}}{|x^{0}|} are the only two invariants which can be constructed from the time-like (or the light-like) x.

At least now I understand why x^{2} is an invariant within (i.e. $x^2>0$) or in the light-cone (i.e. $x^2=0$). I do not see how to prove that the sign function \epsilon (x^{0}) = \frac{x^{0}}{|x^{0}|} is also an invariant though (I didn't come up with the trick, maybe I just need more time and reading ; please feel free to provide a hint if you wish).

That is neither continuous nor connected Lorentz transformation. It is a discrete space-time reversal (x^{0} , x^{i}) \to (-x^{0} , -x^{i}). Space reflection (x^{0} , x^{i}) \to (x^{0} , -x^{i}); time reversal (x^{0} , x^{i}) \to (-x^{0} , x^{i}) and space-time reversal form disjoint subsets and are not continuously connected to the identity. In English, x \to – x does not belong to the proper orthochronous Lorentz group SO^{\uparrow}(1,3). By the “Lorentz group”, we always mean the real semi-simple Lie group SO^{\uparrow}(1,3).

Thanks to this comment I noticed that my knowledge on basic groups in Physics is quite poor... let's boost it.

Then you may want to consider doing the following exercises in the order presented:

Given the defining property of Lorentz transformation \eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}, prove the following identities

(i) \ (\Lambda k) \cdot (\Lambda x) = k \cdot x

(ii) \ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x

(iii) \ \left(\Lambda^{-1}k \right)^{2} = k^{2}

The above identities are very useful and repeatedly used to solve the following exercises

1) Given that \Delta (x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.

Show that \Delta^{*}(x) = \Delta (x) and \Delta(-x) = - \Delta (x).

2) Let \varphi (x) be a real scalar field (operator) satisfying the Klein-Gordon equation (\partial^{2} + m^{2}) \varphi (x) = 0. Assuming that the commutator \big[ \varphi (x) , \varphi (y) \big] is a Lorentz invariant c-number (i.e., function), show that \big[\varphi (x) , \varphi (0) \big] = C \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x} , with C being an arbitrary real constant.

Thank you very much for this, they are going to be really useful! And I can guarantee I am going to post them on PF!

 

[PeterDonis]

the argument I presented doesn't explain why $x^{2}$ is a unique Lorentz invariant

If all you have is a vector $x$, there aren't many things that could possibly be invariant, since there aren't many things period. All you have are the components of the vector and its squared norm.

The squared norm is always invariant.

The components themselves are obviously not invariant. But we can split them up further into signs and magnitudes, and note that the magnitudes are never invariant, the signs of the space components are never invariant, and the sign of the time component is only invariant for timelike and null vectors (because Lorentz transformations--more precisely proper orthochronous ones--cannot take a timelike or null vector from the future to the past light cone or vice versa since those two sets are disjoint, but they can take any spacelike vector to any other one since the spacelike vectors are a single continuous set).

And that's everything that could possibly be invariant.

 

[strangerep]

Then you [JD_PM]may want to consider doing the following exercises in the order presented: [...]

Heh, heh, you should write a "Theoretical Physics Challenge" thread, similar to the "Math Challenges".

 

[vanhees71]

I think the solution is in Pauli's lectures on QFT (old and outdated somwhat, but if it comes to the math of these free-particle invariant two-point functions it's very good) or also in Bogoliubov&Shirkov.

That the said function, which is explicitly invariant under proper orthochronous Lorentz transformations vanishes for space-like $x$ is not too difficult to see by calculating it in the reference frame where $x^0=0$. For time-like or light-like $x$ it's more complicated, leading to Bessel functions if I remember right ;-).

 

[vanhees71]

Are these in English?

I learned QED from that book. In my opinion, it is the best book ever written on QED. I still use it whenever I get stuck on something.

The Pauli Lectures are in English. It's a 6-volume set and just master pieces. Reading these lectures, it's clear, where Pauli comes from, namely from Sommerfeld's Munich school of theoretical physics :-)).

 

[JD_PM]

The Pauli Lectures are in English. It's a 6-volume set and just master pieces. Reading these lectures, it's clear, where Pauli comes from, namely from Sommerfeld's Munich school of theoretical physics :-)).

I'd love to check them. I guess they're not online...

 

[vanhees71]

They are not expensive:

https://www.amazon.com/s?k=pauli+le...aps,236&ref=nb_sb_ss_i_1_14&tag=pfamazon01-20