Propagator at time-like separations

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On page 68, equation (8.13) of Srednicki's QFT book is the equation for the scalar propagator:

[tex]\Delta(x-x')=i\theta(t-t') \int \frac{d^3k}{2(2\pi)^3E_k}e^{ik(x-x')}

+i\theta(t'-t) \int \frac{d^3k}{2(2\pi)^3E_k}e^{-ik(x-x')}
[/tex]

where the exponential is the product of 4-vectors and k is on-shell.

My question is to evaluate the integrals, can you set t'=t in the exponential by choosing a frame where the two events happen simultaneously?

Because Srednicki says that these integrals are the same ones as in equation (4.12) on page 46, which turn out to be modified Bessel functions. However, equation (4.12) assumed space-like separations by evaluating the integral in a frame where t=t', thereby simplifying the expression. However, when you do this, isn't the expression only correct for space-like separations, and not time-like separations?
 

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  • #2
tom.stoer
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Because Srednicki says that these integrals are the same ones as in equation (4.12) on page 46, which turn out to be modified Bessel functions. However, equation (4.12) assumed space-like separations by evaluating the integral in a frame where t=t', thereby simplifying the expression. However, when you do this, isn't the expression only correct for space-like separations, and not time-like separations?
As soon as you have a function which is analytic in some region of the complex plane an analytic continuation is possible. For a complex function f(z) evaluated for f(x), i.e. Im z=y=0 it should be possible to continue it y=Im z and Re z=0. Look at sin(z) with sin(x) and sinh(y) as an example
 
  • #3
A. Neumaier
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[tex]\Delta(x-x')=i\theta(t-t') \int \frac{d^3k}{2(2\pi)^3E_k}e^{ik(x-x')}

+i\theta(t'-t) \int \frac{d^3k}{2(2\pi)^3E_k}e^{-ik(x-x')}
[/tex]

My question is to evaluate the integrals, can you set t'=t in the exponential by choosing a frame where the two events happen simultaneously?
No. The equal-time case is ill-defined since \theta is undefined there.
You need to distinguish the cases t<t' and t>t', where the \theta evaluates differently.
 
  • #4
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My main concern is that the integral:

[tex]
\int \frac{d^3k}{E_k}e^{ik(x-x')} [/tex]

doesn't converge if you set the distance between two events to be zero (which would correspond to a time-like separation).

You can sort of see it immediately because the measure d^3k behaves as k^2dk, and E_k is of the order k, so the integral will oscillate like:

[tex]ke^{-iE_k(t-t')}dk [/tex]

However, if you allow the distance x-x' to be nonzero but t-t' to be zero, then you get:

[tex]ke^{ik(x-x')}=ke^{ik|x-x'|cos \theta} [/tex]

and integrating that over [tex]cos \theta [/tex] actually gets rid of the k term, since:

[tex]ke^{ik|x-x'|cos \theta}=\frac{k}{ik|x-x'|} (ik|x-x'|e^{ik|x-x'|cos \theta}) [/tex]

and the term in the second parenthesis is easily integrated over theta to give a sine function (even so, I still fail to see how this expression converges, but it is not as bad as having a linear term k that grows to infinity).

All of the integrals in the expression:

[tex]
\Delta(x-x')=i\theta(t-t') \int \frac{d^3k}{2(2\pi)^3E_k}e^{ik(x-x')}

+i\theta(t'-t) \int \frac{d^3k}{2(2\pi)^3E_k}e^{-ik(x-x')}

[/tex]

are Lorentz-invariant, so you should be able to evaluate them in any class of frames (time-like, space-like, and light-like) and the result should hold in any frame of the same class.

I'm also a little confused about the [tex]\theta(t-t') [/tex] functions. These aren't Lorentz invariant are they? They're Lorentz invariant for events that are time-like in separation where causality is conserved. But shouldn't the propagator as a whole be Lorentz invariant?
 
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  • #5
A. Neumaier
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My main concern is that the integral:

[tex]
\int \frac{d^3k}{E_k}e^{ik(x-x')} [/tex]

doesn't converge
This integral exists not as Lebesgue integral but in the sense of distributions.

I'm also a little confused about the [tex]\theta(t-t') [/tex] functions. These aren't Lorentz invariant are they? They're Lorentz invariant for events that are time-like in separation where causality is conserved. But shouldn't the propagator as a whole be Lorentz invariant?
Look at the discussion of the propagator in
http://en.wikipedia.org/wiki/Propagator_(Quantum_Theory)#Scalar_propagator
One can deform the integral into the complex domain to make it convergent, and then take the undeformed limit at the end. Since there are two poles causing trouble, this can be done in four topologically different ways, whose limits give four Lorentz invariant distributions, two of which involve the Heaviside function. The invariant way of viewing the Heaviside function is by the concept of time-ordering of operators, which is well-defined because of the causal commutation relations. See Weinberg's QFT book, Vol.1, p.145.
 
  • #6
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This integral exists not as Lebesgue integral but in the sense of distributions.
Does this mean if I numerically feed the integral into a computer for some value of the spacetime separation, it can't solve it?

That seems odd, because when you go into momentum space, which is usually what you do to solve perturbative QFT problems, there aren't any problems. But when you try to solve these things in coordinate space, you have to redefine what an integral is to get the right answer? Is there any advantage to doing QFT calculations in coordinate space as opposed to momentum space?

Look at the discussion of the propagator in
http://en.wikipedia.org/wiki/Propagator_(Quantum_Theory)#Scalar_propagator
One can deform the integral into the complex domain to make it convergent, and then take the undeformed limit at the end. Since there are two poles causing trouble, this can be done in four topologically different ways, whose limits give four Lorentz invariant distributions, two of which involve the Heaviside function. The invariant way of viewing the Heaviside function is by the concept of time-ordering of operators, which is well-defined because of the causal commutation relations. See Weinberg's QFT book, Vol.1, p.145.
That's a good link. I think on both the retarded and advanced propagator, they have it wrong though: the Green's function vanishes in the opposite region than what's stated.
 
  • #7
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Does this mean if I numerically feed the integral into a computer for some value of the spacetime separation, it can't solve it?
It depends on the program, I guess. If you feed it into a Gaussian quadrature rule, I don't think you'll get meaningful results. But those who need to calculate these kinds of integrals surely have methods adapted to this purpose that work well.

That seems odd, because when you go into momentum space, which is usually what you do to solve perturbative QFT problems, there aren't any problems.
You still have only distributional integrals. This is covered up by the +i*epsilon prescription, where you take the limit at the end.


Is there any advantage to doing QFT calculations in coordinate space as opposed to momentum space?.
Only if you have space-dependent external fields, since then the momentum representation is awkward.
 
  • #8
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You still have only distributional integrals. This is covered up by the +i*epsilon prescription, where you take the limit at the end.
Well usually what happens is you Wick rotate, which moves the pole off the real line.

What's weird is that the propagator in position space:

[tex]\int \frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} [/tex]
is not Wick rotatable. The poles are at [tex]k_0=E_k-i\epsilon [/tex] and [tex]k_0=-E_k+i\epsilon [/tex], so that to Wick rotate, you have to complete the contour via infinite arcs in the 1st quadrant and the 3rd quadrant, to connect the real axis with the imaginary axis and miss all the poles. But one of those arcs will be divergent for any x and y, since the arcs are at opposite extremes in the imaginary axis.

It's only when all the [tex]e^{ik(x-y) [/tex] terms disappear via integration over position coordinates that the momentum integral over any circular arc at infinity will give a result of zero.

However a problem could arise when you have expressions like:

[tex]\int d^4k \frac{1}{(k-p)^2-m^2+i\epsilon}\frac{1}{k^2-m^2+i\epsilon} [/tex]


Since ko is shifted to the right by po in the (k-p) term, then you could have a pole above the real-axis and to the right of the origin so that you can't complete the contour in the 1st quadrant. I don't understand how one can Wick rotate that expression.
 
  • #9
A. Neumaier
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Well usually what happens is you Wick rotate, which moves the pole off the real line.
This doesn't invalidate my statement, which is in Minkowski space.

Wick rotation is (like most of the manipulations in standard QFT treatments) a formal trickery whose rigorous justification is nontrivial.
 
  • #10
tom.stoer
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Which Wick rotation are you discussing here?
- Wick rotation of n-point functions in position space (which is defined via analytic continuation etc.)
- Wick rotation in the PI which is a formal trick in QFT
 
  • #11
A. Neumaier
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Which Wick rotation are you discussing here?
- Wick rotation of n-point functions in position space (which is defined via analytic continuation etc.)
- Wick rotation in the PI which is a formal trick in QFT
Let me be more explicit. Wick rotation is always analytic continuation, but justifying the permissibility of the analytic continuation is usually not a one-liner.

Only for Lebesgue integrals it is easy (and independent of quantum mechanics). For distributional integrals it is a bit more difficult. Note that even the Wick-rotated Euclidean integral
[tex]\int d^4k}\frac{e^{ik(x-y)}}{-k^2-m^2} [/tex]
is still ill-defined as a Lebesgue integral, and makes sense only in distribution.

Wick rotation of more complex fractions is done by first using a partial fraction decomposition and then rotating each term separately. Probably this also has a rigorous mathematical justification, but I haven't seen one.

Wick rotation in the context of QFT is not hard to justify to be valid for n-point functions in the free case. That an interacting n-point function can be continued to the Euclidean case is a highly nontrivial result (Osterwalder-Schrader theorem).

That this carries over to path integrals is even more difficult (discussed in the Quantum physics book by Glimm and Jaffe for free theories and for 1+1D scalar field theory), and unknown in higher dimension (I believe). Similarly, that one is allowed to do lattice gauge theory in Euclidean space and then derive implications for Minkowski space is currently a purely formal procedure without any justification.
 
  • #12
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My knowledge of Wick rotations and imaginary time is really limited, so I'd appreciate it if you could tell me if my interpretation is correct.

It is my understanding that you replace time t by [tex]-i\tau[/tex], so that

[tex]<q,t|q',t'>=<q|e^{-iH(t-t')}| q'> [/tex]

gets replaced by:

[tex]<q,-i\tau|q',-i\tau'>=<q|e^{-H(\tau-\tau')}| q'> [/tex]

You can now do everything you do in real time to evaluate this expression, such as insert a complete set of momentum and position space states to get the path integral. The only difference then is that instead of [tex]e^{iS} [/tex] in the integrand you get [tex]e^{-S_E} [/tex], where for the free field case, [tex]S_E [/tex] is actually the integral of the Hamiltonian with respect to time instead of the Lagrangian with respect to time.

If you now include a source with a Wick rotation via:

[tex]<q,-i\tau|q',-i\tau'>=<q|e^{-H(\tau-\tau'')+\int j(x'')\phi(-i\tau'')d^4x''}| q'> [/tex]

and solve for your free-field propagator, you get the Wick rotated expression:

[tex]\int d^4k}\frac{e^{ik(x-y)}}{-k^2-m^2} [/tex]

with the propagator pole off the axis.

Therefore you mention a partial fraction decomposition and Wick rotating each fraction individually. But it seems if you start with imaginary time, then everything is Wick rotated at once, since your 2-point free propagator has no singularities.

Also, ultimately you are interested in real times. Does this mean that for the end result you have to subsitute back i*t for [tex]\tau [/tex]?
 
  • #13
A. Neumaier
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It is my understanding that you replace time t by [tex]-i\tau[/tex], so that
[tex]<q,t|q',t'>=<q|e^{-iH(t-t')}| q'> [/tex]
gets replaced by:
[tex]<q,-i\tau|q',-i\tau'>=<q|e^{-H(\tau-\tau')}| q'> [/tex]
[etc.]
On a formal level, what you do is correct. But on a rigorous level, you partially replace here well-defined quantities with ill-defined quantities; e.g., e^{-Hs} with real s is well-defined only for s>=0 (unless H is bounded, which is never the case if scattering is possible).

If you are concerned with whether the Wick rotation is well-defined, you must account for all such things. So each application of Wick rotation needs to be checked for whether the transformation is applicable.
 

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