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In an earlier blog post we asked if one can smoothen out the infinite momenta, or infinitesimal wavelengths in Feynman diagrams.
We found now an intuitive procedure where the sums of the path integrals are calculated on-the-fly and summed, so that short wavelengths never show up.
Let us model the annihilation with one space dimension x and time t. The electron is described as a wave of momentum p which moves to the right.
e- ------------->
<------------- e+
------------------------------> x
Similarly, the positron is a wave which moves to the left with momentum -p.
The Huygens principle states that we can calculate the phase of the wave simply by following a crest of the wave and only using the paths of the "rays" of the wave. The ray at time t is the pair of momentum vectors (p, -p). We have one spatial dimension for each of the particles, as well as time. In a path, the momenta stay opposite numbers.
Note 1. Feynman's principle of using conservation of momenta in treelike Feynman diagrams to prune the integrals is actually an application of the Huygens principle. The input to the collision consists of waves of momenta p. It is enough to calculate paths where the momentum stays constant until the next vertex. This is like the principle: just calculate the movement of the crests of the waves along rays, by Huygens.
Note 2. Has anyone proved the Huygens principle for the Dirac equation, under an electric potential?
The electron will meet the positron at a spacetime point A_0 and its wave function will do an infinite number of cycles before that. We have there a singularity.
Let us assume that the electric potential between the particles is nonzero only when their distance is less than R.
The following diagram shows the strategy which we will use in the calculation of the sum of the path phases.
^ A_0
| | / \ |
| | / \ |
| | / \ |
| | |
| /| |\
| e- R-area e+
|
------------------------------------------> x
Let us keep A_0 as fixed spacetime point for a while, and let the actual annihilation point A vary. If A is to the right of A_0, there is still a path that goes through the spacetime point A_0.
The total amplitude at A_0 is an integral over all such paths, weighed on the distance that the paths have from each other outside the R-area.
Suppose that the total amplitude at A_0 is ψ. What is the amplitude if we move A_0 some time Δt later? From the geometry in the diagram above, we see that it just means shifting the whole picture Δt upward. The amplitude is
exp(i E Δt) ψ,
where E is the total energy of the electron. That is, there is just a simple phase shift!
Similarly, if we move A_0 a distance Δx to the right, the phase will be exp(i p Δx) bigger, where p is the original momentum of the electron. That is all.
The summation made the wave function of the electron to remain the same as it would be if there were no positron flying around. There are no high momenta. Just the momentum p with which the electron enters the experiment.
This sounds too good to be true. We assumed that the "input" to the R-area, the electron wave function, is a simple plane wave. But since the electron will eventually get annihilated, its wave function does decay at some rate, when we move to later times.
The real wave function is something like
ψ = exp(i (E t - p x)) * f(t),
where f is a decreasing function on t. The Fourier decomposition of ψ is probably quite well-behaved.
What did we just show? Even though the annihilation of the electron and the positron is a violent act, the violent act happens at a random location of spacetime. Summing over all the paths yields a very smooth wave function.
Proposition 3. Let the wave function at a spacetime point x be a (very complex) function of its "inputs", the fluxes that enter its calculation area (R-area in the above example). If the inputs vary just by a phase shift factor, then the wave function value will vary just by that phase shift factor. QED.
In Feynman diagrams, this means that we can restrict ourselves to the momenta p which are given in the input particles.
Does Proposition 3 help us to get rid of the singularities in vacuum polarization loops?
The complex function in this case is the amplitude that a high-momentum virtual pair is born in some smallish zone inside the collision area. What are the input fluxes? They are the fluxes to the collision area.
In quantum mechanics, there is a princple that we can rotate the phase of any input flux by a constant angle, and we will get the same physical results. Does this help us?
Suppose that we are colliding two fluxes of electrons, with momenta p and q.
e- ------------> <----- e-
p q
The spacetime diagram looks like this:
*
| / \ |
| / \ |
| / \ |
|/ \|
/| |\
/ | | \
e- p e- q
The star marks a collision. What is the amplitude of the left electron wave function at the star? It is a complicated integral of pairs of left electron and right electron paths, where the left electron goes through the spacetime point marked with the star.
The potential between the left electron and the right electron is agnostic to the phase of the right electron.
What is the amplitude if we move the star to the right some distance x? The phases of the right electron do not matter - the phase of the left electron changes by exp(i p x).
Similarly, the phase of the right electron changes by exp(i q x).
What if the collision produces some other particles? Let us have a path where the collision produces a real pair e- and e+. What is the amplitude of the e+ at a certain spacetime point within the calculation area?
A e+ k
/
*
/ \
/ \
e- p e- q
Since e+ can have any momentum k, we have various paths where the slope of the line between * and A varies. The amplitude of a path at A is probably something like a product of the amplitudes of the electrons and something that depends on k. This is in the spirit of Feynman diagrams.
What if we move A some distance x right? The paths that lead to the new position are like the old paths, but shifted by x. The amplitude at the new position will be the old value multiplied by exp(i p x + i q x).
We showed that the amplitude of the virtual positron behaves very well. Its Fourier decomposition does not contain much weight on high momenta.
Theorem 4. A natural ultraviolet cutoff scale for any momentum in a treelike Feynman diagram is the total energy of the collision, as long as the running fine structure constant remains < 1.
Proof. Let us consider a single Feynman diagram of the process. That is, we consider a set of paths with the same topology.
Suppose that the amplitude of a path is calculated from a function of type
input1 * input2 * f (input1, input2),
where f is a function whose value does not depend on the phase of the inputs, only on the absolute value. The inputs are plane waves.
If we move the path by a translation in spacetime, the phase of the inputs changes but not their absolute value.
It is a general principle in quantum mechanics that physical processes are agnostic of the phase of inputs. Therefore, all f satisfy the condition above.
The wave function at a spacetime point depends in a complex fashion on a sum of different paths. But the sum only varies in spacetime according to
input1 * input2
= exp(i (Et - kx)) * exp(i (E't - k'x))
= exp(i ((E + E') t - (k + k') x)).
The sum only contains one Fourier mode!
Momentum is conserved at the vertexes of a Feynman diagram. In a treelike diagram, all inputs to all vertexes have their momentum constrained by the momentum which entered in the original real particles into the experiment. That is why we can claim that the "natural cutoff scale" is the total energy. We can forget about the potentially infinite energy that is transiently available in pair annihilation.
If the diagram contains loops, then an arbitrary momentum can circle the loop and we cannot constrain momentums this easily. We will look at loops in the next section.
Where does the complexity then creep into processes? When we are summing over diagrams of different topologies, they have different Fourier modes and their sum becomes more complicated.
Also, the inputs are not perfect plane waves. Their Fourier decomposition contains all kinds of modes.
The proof of this theorem leaves open one possibility of divergence. Suppose that f can hugely amplify infinitesimal high frequency components in the inputs, so that their contribution becomes relevant.
If the fine structure constant is < 1, functions f in Feynman diagrams cannot hugely inflate amplitudes. QED.
What happens if the inputs have the energy of a Landau pole? Then the contribution from complex diagrams may grow without bounds. Our cutoff scheme removes infinities from a single diagram. But if complex diagrams contribute more than simple ones, then there is an explosion.
A vacuum collapse may happen if complex diagrams contribute more than simple ones, that is, if the coupling constant is > 1.
The Feynman integral on a virtual loop
Above we argued that in treelike Feynman diagrams, the wave functions are smooth for all vertexes. Our argument was symmetry: the inputs to the collision are plane waves. If we move any path in timespace, the amplitude from it changes only by a phase factor of the inputs.
If there is a loop, we cannot constrain the momenta of inputs. Let us study a simple virtual pair loop.
If there is a loop, we cannot constrain the momenta of inputs. Let us study a simple virtual pair loop.
z
/ photon
/
||
e- || e+
||
/
/ photon plane wave
The diagram above shows a loop in a more realistic fashion than usual. The electron and the positron are born at the same point and annihilate at the same point. They fly the exact same line. We want to calculate the photon amplitude at points close to z.
The vacuum atom model says that there must be an integer number of electron wavelengths in the loop, but we ignore that for now and allow any momentum and length in the loop.
If the wave function of the input photon is a plane wave, then we can move any individual path and the amplitude of the path only changes by the phase factor of the input.
Energy and momentum conservation dictate that the output has to be the same plane wave as the input, except that a phase shift is possible.
But what about the diverging Feynman integral on the loop? Where did it go? The Feynman integral is ill-defined, and therefore one cannot really calculate with it. We may use any regularized version of it. We showed that any reasonable path integral just amounts at most to a phase shift on the photon plane wave. It may also be that the Feynman integral converges, after all, if the integration on the loop momentum k is performed last.
Feynman tried to calculate the amplitude at the annihilation point of a virtual loop. He noticed that the integral diverges. A regularization industry was born and has flourished for 70 years. But one can argue that since the amplitude at the annihilation point is not physically observable, the sum of the amplitudes at the annihilation point is not definable at all.
Conjecture 5. One should not try to sum the intermediate point amplitudes in a path integral. It does not make sense to talk about the amplitude of a wave function at a spacetime point if the wave function is not of a real observable particle.
If Conjecture 5 is true, then the regularization industry has been chasing a wild goose.
Conjecture 5 is relevant for the famous Schrödinger cat problem. When we put a cat and the radioactive atom in the box, it is, in a sense, an abstract scattering experiment. The system may take different paths until the box is opened, which corresponds to measuring the real particles that exit the scattering experiment. The path integral method suggests the following: the paths cannot be meaningfully summed during the experiment. It does not make sense to say that the cat is dead and alive "at the same time". The only meaningful wave function is the sum of the paths when we measure the cat afterwards.
Energy and momentum conservation dictate that the output has to be the same plane wave as the input, except that a phase shift is possible.
But what about the diverging Feynman integral on the loop? Where did it go? The Feynman integral is ill-defined, and therefore one cannot really calculate with it. We may use any regularized version of it. We showed that any reasonable path integral just amounts at most to a phase shift on the photon plane wave. It may also be that the Feynman integral converges, after all, if the integration on the loop momentum k is performed last.
Feynman tried to calculate the amplitude at the annihilation point of a virtual loop. He noticed that the integral diverges. A regularization industry was born and has flourished for 70 years. But one can argue that since the amplitude at the annihilation point is not physically observable, the sum of the amplitudes at the annihilation point is not definable at all.
Conjecture 5. One should not try to sum the intermediate point amplitudes in a path integral. It does not make sense to talk about the amplitude of a wave function at a spacetime point if the wave function is not of a real observable particle.
If Conjecture 5 is true, then the regularization industry has been chasing a wild goose.
Conjecture 5 is relevant for the famous Schrödinger cat problem. When we put a cat and the radioactive atom in the box, it is, in a sense, an abstract scattering experiment. The system may take different paths until the box is opened, which corresponds to measuring the real particles that exit the scattering experiment. The path integral method suggests the following: the paths cannot be meaningfully summed during the experiment. It does not make sense to say that the cat is dead and alive "at the same time". The only meaningful wave function is the sum of the paths when we measure the cat afterwards.
What if the input is not a plane wave, but changes smoothly in timespace?
__________
________| |______
__________
________| |______
0 z 1
Let us calculate a 1-dimensional toy example, where the input is a constant 1 from 0 to 1 on the x axis, and the end of the loop is at x = z. The momentum |p| > 1. The point z may be anywhere in the x axis.
We assume that in the interval there is a photon flux of a negligible momentum, starting from 0. The photon is transformed into a virtual pair at some point of the interval, and the pair flies to z. The electron in the pair has an arbitrary momentum |p| > 1 and the positron the exact opposite momentum.
One may ask how the particles can meet at all, if they have opposite momenta. But here we imagine that a momentum corresponds to a plane wave which fills the whole spacetime.
Let us calculate the amplitude of the virtual electron at z. We cannot use the translation argument of Theorem 4, because the input varies in space.
The input to the loop comes from any point in the interval (0, 1) and the loop ends at z. We want to sum the amplitudes
We assume that in the interval there is a photon flux of a negligible momentum, starting from 0. The photon is transformed into a virtual pair at some point of the interval, and the pair flies to z. The electron in the pair has an arbitrary momentum |p| > 1 and the positron the exact opposite momentum.
One may ask how the particles can meet at all, if they have opposite momenta. But here we imagine that a momentum corresponds to a plane wave which fills the whole spacetime.
Let us calculate the amplitude of the virtual electron at z. We cannot use the translation argument of Theorem 4, because the input varies in space.
The input to the loop comes from any point in the interval (0, 1) and the loop ends at z. We want to sum the amplitudes
exp(i p * (z - x))
for all 0 < x < 1 and |p| > 1.
The integral on x is
ψ(z) = i / p * (exp(i p (z - 1)) - exp(i p z)) = i / p * exp(i p z) * (1 / exp(i p) - 1).
If z is 0 or 1, then the integral on p diverges logarithmically. Otherwise, the integral is essentially of the form
exp(i p) / p,
where |p| > 1. It is a Dirichlet integral and converges. We have shown that a simple path integral converges, unless the end point is at the border of the input area.
k k + p ->
~~~~~~>========= z
<- p
virtual loop
What if we the input is a plane wave of momentum k, in one dimension? In the diagram, the double line === depicts the virtual loop. We need to integrate
exp(i k x) * exp(i (k + p) (z - x))
= exp(i (k z + p (z - x)))
over all x and |p| > 1. Let us integrate over a limited range x_0 < x < x_1. We require x_0 < z < x_1.
----------------------------------------> x
x_0 z x_1
The integral over the range x0, x1 is
i / p * exp(i (k z + p (z - x_0)))
- i / p * exp(i (k z + p (z - x_1))).
The result rotates with the end points x_0 and x_1. The result is not very sensible. But let us integrate on 1 < p then. Let us look at the first term.
Let q = p (z - x_0). We need to calculate
∞
ψ_1 (z) = exp(i k z) ∫ i (z - x_0) * 1 / q * exp(i q) dq.
z - x_0
Let g(y) be the integral of the Dirichlet function
∞
g(y) = ∫ sin(x) / x dx.
y
The real part of the integral above is
∞
-y ∫ sin(q) / q dq,
y
where y = z - x_0.
What happens when y tends to infinity? If y = 2n π, then g(y) is roughly
2 / (2n π) - 2 / (2(n + 1) π) + …
= 1 / π * (1 / n - 1 / (n + 1) + …)
= 1 / π * 1 / 2n.
That is, at such points, g(y) is roughly 1 / y. Thus, g(y) generally is cos(y) / y with large y. The real part of the integral is thus -cos(y). We may guess that the imaginary part is -i sin(y). The integral would then be -exp(i y).
ψ_1 (z) = - exp(i k z) exp(i (z - x_0)).
Unfortunately, ψ_1 (z) does not converge. We could have guessed that. A virtual loop can start at a remote position, and affect the amplitude at z without any damping. There is no obvious reason why remote areas should contribute to the amplitude at z less than close areas.
It is not physically reasonable that a high-energy virtual particle flies over great distances. Can we restrict the contribution of long loops?
The idea of assuming a infinite plane wave for the photon is bad. Let us look at a more realistic collision example.
e- ----------------->
^
|
z
|| virtual pair
||
|
<------------------- e+
We are not interested in the amplitude of just any virtual electron at z. The virtual electron has to be from the collision with the nearby positron. We need to develop a way of handling the conditional probabilities that are involved in collisions.
Infrared divergences
Electron self-energy involves an infrared divergence. Is there some principle which allows us to establish a cutoff to low-frequency modes in the Fourier decomposition of any particle in a Feynman diagram?
The inputs themselves cannot contain an infrared divergence in them, because then the inputs would contain infinite energy. Can we prove that no process can create an infrared divergence if the inputs do not have an infrared divergence?
Can we use the proof of Theorem 4 to remove infrared divergences?
An infrared divergence would happen if the function f would hugely amplify the infinitesimal infrared components in the inputs. But we saw that such f do not exist if the fine structure constant is < 1.
Theorem 6. A natural cutoff scale for infrared divergences in a treelike Feynman diagram is the total energy of the collision. We assume that the running fine structure constant is < 1. QED.
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