How can the Cauchy integral and Fourier integral produce the same result?

[Micha]

Why is Avodyne's remark confusing you? What you say is in agreement with what he said, isn't it?

 

[Micha]

Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?

Edit: I'd say, it is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.

 

[OOO]

Why is Avodyne's remark confusing you? What you say is in agreement with what he said, isn't it?

Thanks for pointing that out, Micha. My brain has become a knot. I'm not even able to read properly. :rofl:

 

[jostpuur]

Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?

Edit: I'd say, it is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.

I tried to ask about this here https://www.physicsforums.com/showthread.php?t=176563

I don't feel like I got satisfying answer. The answer seems to be, that the propagator doesn't necessarily mean anything. It just works when used correctly to compute scattering amplitudes.

 

[Hans de Vries]

Thanks for pointing that out, Micha. My brain has become a knot.

At the end all the confusion is unlikely to have any influence on real calculations since these are all done in momentum space, not in position space, and non physical results are massaged away with other prescriptions.

Leaking outside the light cone with exp-m at t=0 would mean instantaneous propagation at infinite speed over micron size distances in the case of neutrinos. The size of living species.Regards, Hans

 

[Avodyne]

Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?

The Feynman propagator has no direct physical meaning. It simply appears as a component in the calculation of infinite-time scattering amplitudes.

I'd say, [the Feynman propagator] is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.

I don't think this is correct; at least, I've never seen a calculation that shows it to be correct. (One has to be careful about the meaning of position in quantum field theory, so there are some subtleties. But it doesn't even have the right dimensions.)

 

[meopemuk]

Hi Avodyne,

The Feynman propagator has no direct physical meaning. It simply appears as a component in the calculation of infinite-time scattering amplitudes.

I agree with you completely. Propagators are formal quantities used in calculations of the S-matrix amplitudes. Position-space propagators cannot be interpreted as propagation amplitudes (from point to point) or time-dependent wave functions. Such interpretation can be found in some QFT textbooks, but it 1) has zero experimental support; 2) leads to numerous theoretical contradictions.

Eugene.

 

[Hans de Vries]

For those interested: from Pauli's famous 1940 paper, Spin and Statistics:Pauli's Spin and Statistics

To be compared with Feynman's:

Feynman's propagator in position space.

Although Pauli's propagators are worse. (zero'th order Bessels rather than first order). Pauli, quote, "expressively postulates" commutation outside the light cone to overrule the Green's function.

Peshkin & Schroeder's remarks about anti-particles canceling the non-causality stem from the second link. Chapter 18 of "Fundamental processes": Taking only one pole violates relativity, any physical process has diagrams with the other pole as well (anti-particle) to restore Lorentz invariance.Regards, Hans

 

[Haelfix]

Hi Hans, what paper or book is that Feynman link too? Hellishly hard to find some of his old papers nowdays.

Anyway that should settle the confusion as expected.

 

[Hans de Vries]

Hi Hans, what paper or book is that Feynman link too? Hellishly hard to find some of his old papers nowdays.

Anyway that should settle the confusion as expected.

It's in this nice book from his 1959-60 Caltech lectures:

The Theory of Fundamental Processes


Regards, Hans

PS: more copies here: amazon.com

 

[Micha]

Anyway that should settle the confusion as expected.

Notice, that in the link Feyman does take serious the leaking out of the lightcone of the propagator named after him as a physical effect.

If the modern view is apperently different, ok.

EDIT: What I ask myself, is, how to we design an experiment to check this?

 

[meopemuk]

EDIT: What I ask myself, is, how to we design an experiment to check this?

That's exactly the point. How can you measure propagators?

Eugene.

 

[Avodyne]

Feynman clearly thought of the propagator as representing the amplitude for a particle to start at one spacetime point and end at another. When faced with the difficulty that this amplitude does not vanish outside the lightcone, he made up something about precise measurement of position leading to pair production, and this making it OK that his amplitude was nonzero outside the lightcone.

None of this holds up to close scrutiny. Feynman was making up the formalism (for what we now call Feynman diagrams) as he went along; he had no deep justification. Only later did Dyson show how you could get Feynman's formalism from quantum field theory. But one of the things you lose when you do this is the notion that the propagator is an actual amplitude.

We can calculate that amplitude in quantum field theory for a free massive particle. (There are extra issues in the massless case.) For simplicity of notation, I will work in one space dimension, and set hbar=c=1. The generalization to more dimensions is obvious.

We know what the one-particle momentum eigenstates are:
$$|k\rangle = a^\dagger(k)|0\rangle.$$
The only issue is normalization. Let us use the commutation relations
$$[a(k),a^\dagger(k')]=f(k)\delta(k-k'),$$
where f(k) is a positive-definite function that is otherwise arbitrary. Common choices in the literature include f(k)=1 and f(k)=(2pi)³2E(k), where E(k)=(k²+m²)^1/2. But any positive-definite function is acceptable; this is simply a matter of convention. I will leave f(k) unspecified. The one-particle momentum eigenstates then have the normalization
$$\langle k'|k\rangle = f(k)\delta(k'-k).$$
Correspondingly, the completeness relation is
$$\int {dk\over f(k)}|k\rangle\langle k| = 1.$$

Now we need to decide what a position eigenstate is. Certainly two eigenstates at different positions should be orthogonal, so we have
$$\langle x'|x\rangle = h(x)\delta(x'-x),$$
where h(x) is a postivie-definite normalization function analogous to f(k). The choice h(x)=constant is the only one consistent with translation invariance, so we will take h(x)=1:
$$\langle x'|x\rangle = \delta(x'-x).$$

Next, we need to know the inner product of a position eigenstate and a momentum eigenstate. We will take
$$\langle x|k\rangle =g(k)e^{ikx}.$$
The x dependence is again the only one consistent with translation invariance. Given a choice of f(k), we can determine g(k) as follows:
$$\delta(x'-x)=\langle x'|x\rangle =\int {dk\over f(k)}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{ik(x'-x)}.$$
This only holds if |g(k)|²/f(k)=1/2pi, so we will make that choice. Note that a one-particle position eigenstate can be expressed as a linear combination of one-particle momentum eigenstates; there is no "pair production", because there are no interactions to produce any pairs.

Now let's compute the propagation amplitude. This is given by
$$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$
This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk.

Also, this does not vanish outside the lightcone. This has nothing whatsoever to do with pair production, because we have done the calculation entirely within the one-particle subspace.

So, is it a problem? Only if you can measure it. Can you? It depends on what you mean by "measurement" in quantum field theory.

 

[meopemuk]

Now let's compute the propagation amplitude. This is given by
$$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$
This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk.

Also, this does not vanish outside the lightcone. This has nothing whatsoever to do with pair production, because we have done the calculation entirely within the one-particle subspace.

So, is it a problem? Only if you can measure it. Can you? It depends on what you mean by "measurement" in quantum field theory.

I think your derivation is correct and what you got is exactly the "amplitude of finding the particle at point x' at time t if it was released from point x at time 0". Indeed, this amplitude does not vanish outside the lightcone. And I am ready to accept that this fact can be measured, in principle. But there is no contradiction with the principle of causality yet. You still need to prove that superluminal propagation of wavefunctions can be used for sending signals back to the past. Can you do that?

Eugene.

 

[OOO]

Now let's compute the propagation amplitude. This is given by
$$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$
This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk.

Also, this does not vanish outside the lightcone.

I think this is no surprise, since the Hamiltonian corresponding to the energy E(k)=(k²+m²)^1/2 is known to be non-local. The acausal transition amplitude you calculate is just an expression of this fact. Wasn't the traditional escape from this nightmare to question, whether this is the right one-particle Hamiltonian ?

 

[meopemuk]

I think this is no surprise, since the Hamiltonian corresponding to the energy E(k)=(k²+m²)^1/2 is known to be non-local. The acausal transition amplitude you calculate is just an expression of this fact. Wasn't the traditional escape from this nightmare to question, whether this is the right one-particle Hamiltonian ?

There are many good reasons to believe that E(k)=(k²+m²)^1/2 is the correct 1-particle Hamiltonian:

1. This form of the Hamiltonian follows from Wigner's theory of irreducible representations of the Poincare group;

2. This form is used throughout QFT with great success in calculations of scattering cross-sections, etc.

I think that the "escape from this nightmare" should be sought in another direction. Most importantly, there is no nightmare yet. The superluminal propagation is not a paradox by itself. The only real paradox is violation of causality, e.g., if one can build a machine that influences the past. I haven't seen a convincing proof that one can build such a machine by using superluminally propagating wave functions.

Eugene.

 

[OOO]

There are many good reasons to believe that E(k)=(k²+m²)^1/2 is the correct 1-particle Hamiltonian:

1. This form of the Hamiltonian follows from Wigner's theory of irreducible representations of the Poincare group;

2. This form is used throughout QFT with great success in calculations of scattering cross-sections, etc.

I think that the "escape from this nightmare" should be sought in another direction. Most importantly, there is no nightmare yet. The superluminal propagation is not a paradox by itself. The only real paradox is violation of causality, e.g., if one can build a machine that influences the past. I haven't seen a convincing proof that one can build such a machine by using superluminally propagating wave functions.

Eugene.

I for one find the prospect of describing the propagation of a single particle by
an equation like

$$i\partial_t \psi = \sqrt{-\partial_x^2+m^2} \psi$$

somewhat "itchy". Of course, my objection doesn't mean much. And maybe I'm just too narrow-minded.

 

[meopemuk]

I for one find the prospect of describing the propagation of a single particle by
an equation like

$$i\partial_t \psi = \sqrt{-\partial_x^2+m^2} \psi$$

somewhat "itchy".

There is no way around it. The principle of relativity (the Poincare group) + quantum mechanics lead directly to this equation. All details of the proof can be found in first five chapters of http://www.arxiv.org/abs/physics/0504062

Eugene.

 

[OOO]

There is no way around it. The principle of relativity (the Poincare group) + quantum mechanics lead directly to this equation. All details of the proof can be found in first five chapters of http://www.arxiv.org/abs/physics/0504062

Eugene.

Have you ever considered Google ads

 

[Avodyne]

I think your derivation is correct and what you got is exactly the "amplitude of finding the particle at point x' at time t if it was released from point x at time 0". Indeed, this amplitude does not vanish outside the lightcone. And I am ready to accept that this fact can be measured, in principle.

Well, I'm not even sure that's possible. We really need a better model of what it means to measure something. The obvious thing to do is model particle detectors as external sources coupled to the field. I strongly suspect that this will render the effect unobservable.

But there is no contradiction with the principle of causality yet. You still need to prove that superluminal propagation of wavefunctions can be used for sending signals back to the past. Can you do that?

Certainly not! And I don't think it's possible, but this can only be answered in the context of a specific model of measurement.

 

[Micha]

Avodyne, I think your result breaks Lorentz invariance.
The measure dk for the integral is only the spatial momentum component, and this will we different for different Lorentz frames. I suppose, this has to do with the fact, that you excluded pair production.

Do you agree and if so, do you have any good reasons, why we should trust the formula anyway?

 

[Avodyne]

Avodyne, I think your result breaks Lorentz invariance.

That depends on what you mean by "Lorentz invariance". The real problem is defining what we mean by a particle at a definite position. My states of definite position are eigenstates of the Newton-Wigner position operator, as explained by Eugene in http://www.arxiv.org/abs/physics/0504062, but there is no comparable "time" operator. So I am treating space and time differently at the very beginning. Lorentz transformations do not connect these "space" and "time" labels.

On the other hand, the calculation is entirely within the quantum field theory of a free scalar field, which is manifestly Lorentz invariant.

 

[Avodyne]

Here is another calculation one could do. Suppose we have a free scalar field $$\varphi(x,t).$$ We could couple it to a time-dependent source $$J(x,t)$$ by adding a term $$\textstyle\int dx\,\varphi(x,t)J(x,t)$$ to the hamiltonian. Suppose J(x,t) is zero for t<0 at all x, and for |x|>L for all t, where L is some fixed finite length. (I'm still in one dimension.) Suppose also that the theory is in its ground state for t<0. At t=0, the source turns on, and the state changes. Now compute the time-dependent expectation value of the field. It will of course be zero for t<0. I conjecture that it will remain exactly zero for |x|>L+t; that is, outside the lightcone of the disturbance by the source.

This calculation can be done exactly. I will do it when I get a chance.

 

[meopemuk]

We really need a better model of what it means to measure something. The obvious thing to do is model particle detectors as external sources coupled to the field. I strongly suspect that this will render the effect unobservable.

I don't think it is necessary to model particle detectors and measuring apparatuses. Once we have the position-space wave function (e.g., the one that you calculated) we already have by definition the probability amplitude for measuring particle position at a given point in space. It doesn't matter what kind of physical device will be actually used for this measurement. It is only important that this is a position-measuring device, and this fact is already incorporated in our choice of the position-space representation for the wave function.

It is true that in reality physical observables cannot be measured with unlimited precision. So, the quantum-mechanical assumption about precise measurements of observables is, of course, an idealization. But I guess that without such an idealization the theory would be a complete mess.

Eugene.

 

[meopemuk]

Avodyne, I think your result breaks Lorentz invariance.

It is possible to prove that the inner product of Avodyne's wave functions

$$\langle \psi | \phi \rangle = \int d^3r \psi^*(\mathbf{r}) \phi(\mathbf{r})$$

is the same in all moving reference frames. Boost transformations of position-space wave functions are rather tricky, but this can be done. So, the Lorentz invariance is not violated.

Eugene.

 

[Avodyne]

It doesn't matter what kind of physical device will be actually used for this measurement. It is only important that this is a position-measuring device ...

The issue (as I see it) is whether we can actually build such a device.

In the "source model" of measurement, described above, I can show that, as I conjectured, there is not a disturbance outside the lightcone. If our ability to manipulate particles corresponds to having sources that we control, then I think this result shows that the fact that the wave function does not vanish outside the lightcone cannot be measured.

Here's a sketch of the proof. In the interaction picture, the time evolution operator in the source model is
$$U(t) = \exp\!\!\left[-{i\over\hbar}\int_0^t dt'\int_{-L}^{+L}dx' \,J(x',t')\varphi(x',t')\right].$$
Usually, the right-hand side must be time-ordered; however, because the commutator of two fields is a c-number, we can drop the time-ordering, up to an overall c-number phase factor. (See, e.g., Merzbacher, 3rd edition, p.339, for a similar analysis of a harmonic oscillator.) So the state at time t is $$|\psi(t)\rangle =U(t)|0\rangle,$$ and the expectation value of the field at time t is
$$\langle\psi(t)|\varphi(x,t)|\psi(t)\rangle = \langle 0|U^\dagger(t)\varphi(x,t)U(t)|0\rangle,$$
where $$\varphi(x,t)$$ is the free field. Now we use
$$e^A B e^{-A}=B + [A,B] + {\textstyle{1\over2}}[A,[A,B]]+\ldots\;.$$
Because the commutator of two fields is a c-number, this expansion terminates with the second term, and we get
$$\langle\psi(t)|\varphi(x,t)|\psi(t)\rangle = \langle 0|\varphi(x,t)|0\rangle +{i\over\hbar}\int_0^t dt'\int_{-L}^{+L}dx'\,J(x',t')[\varphi(x',t'),\varphi(x,t)].$$
Of course $$\langle 0|\varphi(x,t)|0\rangle$$ is zero; the interesting part is the second term. But we know the commutator vanishes outside the lightcone, so the expectation value of the field at a particular point in spacetime will not register the disturbance by the source until a speed-of-light signal can get there. The same will be true of any operator built out of fields at a particular spacetime point, because we will always end up with a commutator of fields somewhere in every term.

 

[meopemuk]

The issue (as I see it) is whether we can actually build such a device.

What about a ruler?

In the "source model" of measurement, described above, I can show that, as I conjectured, there is not a disturbance outside the lightcone. If our ability to manipulate particles corresponds to having sources that we control, then I think this result shows that the fact that the wave function does not vanish outside the lightcone cannot be measured.

What is the relevance of the scalar field $\phi(x,t)$ to measurements of position? In my opinion, it is not relevant at all.

The position observable of a single particle is represented by the corresponding Newton-Wigner position operator $\hat{\mathbf{r}}$. Eigenvalues of this operator are found in the arguments of the particle wave function $\psi(\mathbf{r},t)$. The square of the modulus of this wave function is the probability density for measuring position. So, it contains all information that one would need to interpret position measurements.

Eugene.

 

[Avodyne]

What about a ruler?

A ruler is made out of local excitations of quantum fields. The source, a classical field, is a simple model of these. (It may, of course, be too simple.)

What is the relevance of the scalar field $\phi(x,t)$ to measurements of position?

In QFT, position is defined as the argument of the quantum fields, just like time.

The position observable of a single particle is represented by the corresponding Newton-Wigner position operator

The Newton-Wigner operator is a global object that cannot be written as the integral of a local density (in this respect, it is unlike momentum or energy). So it's not at all obvious to me how you would build a device that measures it.

 

[Haelfix]

Here we go with the Newton-Wigner position operator schtick again.

The literature on that particular is vast, and needless to say controversial and murky to the nth degree. It is completely undefined for interacting field theories, and likely the operator itself is a nogo by various powerful theorems.

Likewise, going back a few posts the physicist will simply object to the propagator argument by Avodyne b/c its a one particle free field approximation. The real physical situation requires a hard cut off on E(k) to make sense at all, and to turn on interactions.

So even though I completely agree the path integral shouldn't be thought off as saying something like Feynman originally thought (eg amplitudes between space like separated paths), he is perfectly justified in pointing out that upon a measurement, you will have pair creation ambiguities. No amount of free field handwaving gets around this.

 

[strangerep]

In Avodyne's earlier calculation of the propagator,

$$[a(k),a^\dagger(k')]=f(k)\delta(k-k'),$$

Those k's are 3-vectors, right? If we were talking about 4-vectors,
we'd need in general an extra factor like $$\delta^{(4)}(k^2 - m^2)$$.
The latter is what normally causes a $$1/E_k$$ to appear when a
4-momentum integral is reduced to a 3-momentum integral (and the latter
depends in turn on a choice of contour).

Now we need to decide what a position eigenstate is. Certainly two
eigenstates at different positions should be orthogonal, so we have
$$\langle x'|x\rangle = h(x)\delta(x'-x),$$

I think there's an error in the above. If your x's are 4-vectors, then the
simplistic expression above is not justified. But if your x's are 3-vectors, then
the way you act on them with $$e^{-iHt}$$ below is incorrect. (You
can't move between equal-time points via a pure time-translation.)

Now let's compute the propagation amplitude. This is given by
$$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$
This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk.

It would all need to be re-done with 4-momentum integrals, and 4-positions, using
the $$\delta^{(4)}(k^2 - m^2)$$ in appropriate places. Note also that
introducing such a constraint in 4-momentum space would also imply some kind
of admissability constraint on the set of 4-position eigenstates.

Also, this does not vanish outside the lightcone. [...]

It's well-known that the Feynman propagator does not have causal support,
unlike the Pauli-Jordan function. But all that matters is that the matrix elements
of all states in our Hilbert space (in position basis) reflect only causal relationships.
I think you'd find that the matrix element between two arbitrary position states which
are mutually outside each other's light cones are either zero, or the two states are not
in the same Hilbert space.

 

[strangerep]

Here we go with the Newton-Wigner position operator schtick again.

The literature on that particular is vast, and needless to say controversial
and murky to the nth degree. It is completely undefined for
interacting field theories,

Is that just because 4D interacting theories are themselves undefined, or for some
other reason?

and likely the operator itself is a nogo by various powerful theorems.

Which theorems?

 

[Avodyne]

Those k's are 3-vectors, right?

Yes. This is standard; see any QFT book.

I think there's an error in the above. If your x's are 4-vectors, then the simplistic expression above is not justified.

They're 3-vectors.

But if your x's are 3-vectors, then the way you act on them with $$e^{-iHt}$$ below is incorrect. (You can't move between equal-time points via a pure time-translation.)

The time evolution operator can be applied to any state. I have defined a perfectly sensible linear combination of single-particle momentum eigenstates. I want to call it a position eigenstate; but doing so does not mean that I am suddenly not allowed to time-evolve it in the way that any state is time-evolved.

If we don't call $|x\rangle$ a position eigenstate, then everything I did is straight out of the textbooks. So the only possible controversy is whether $|x\rangle$ "really" represents a particle localized exactly at the point x.

 

[meopemuk]

In QFT, position is defined as the argument of the quantum fields, just like time.

Is there any evidence that parameter x of quantum fields is related to particle positions? I don't know of any experimentally verifiable prediction of QFT that would suggest that x can be measured as the position. All experimental predictions of QFT are contained in the S-matrix. When S-matrix is calculated the parameter x gets integrated out. So, in my opinion, x is simply an integration variable in QFT.

Moreover, in quantum theory (including QFT) each observable should have a corresponding Hermitian operator. If we assume that x are eigenvalues of some Hermitian operator X, then what is X?

The Newton-Wigner operator is a global object that cannot be written as the integral of a local density (in this respect, it is unlike momentum or energy). So it's not at all obvious to me how you would build a device that measures it.

Why do you think that operators of observables should be written as integrals of a local density?

Eugene.

 

[meopemuk]

It would all need to be re-done with 4-momentum integrals, and 4-positions, using
the $$\delta^{(4)}(k^2 - m^2)$$ in appropriate places.

Why? Obeying the Poincare invariance (which is the true mathematical expression of the principle of relativity) does not require you to perform all calculations in a 4-dimensional notation. A theory is relativistically invariant if and only if it has a representation of the Poincare group of inertial transformations. In most cases the 3D notation is entirely appropriate for relativistic theories and their comparison with experiment. The 4D notation is often (e.g., in this case) redundant and confusing.

Eugene.

 

[meopemuk]

So even though I completely agree the path integral shouldn't be thought off as saying something like Feynman originally thought (eg amplitudes between space like separated paths), he is perfectly justified in pointing out that upon a measurement, you will have pair creation ambiguities. No amount of free field handwaving gets around this.

I read this argument (particle localization = pair creation) very often, but I've never seen a detailed explanation or proof. Could you please clarify?

Eugene.