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Particle/Pole correspondence in QFT Green's functions

+ 7 like - 0 dislike
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The standard lore in relativistic QFT is that poles appearing on the real-axis in momentum-space Green's functions correspond to particles, with the position of the pole yielding the invariant mass of that particle. (Here, I disregard complications tied to gauge-fixing and unphysical ghosts)

Schematically, I take this to mean:

$$\text{one particle state} \Leftrightarrow \text{pole on real axis}$$

It is easy to show the correspondence going $\text{particle} \Rightarrow \text{pole} $. This is done in many textbooks, e.g. Peskin and Schroeder's text, where a complete set of states diagonalizing the field theoretic Hamiltonian is inserted into a $n$-point correlator, and a pole is shown to emerge.

However, how does one make the argument going the other way $\text{particle} \Leftarrow \text{pole} $? That is, the appearance of a pole in a Green's function indicates an eigenstate of the Hamiltonian, with the eigenvalue corresponding to the particle energy?


This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot

asked Jul 2, 2014 in Theoretical Physics by QuantumDot (195 points) [ revision history ]
edited Jan 9 by Dilaton
Just to add extra information, it is possible to show within non-relativistic potential scattering theory (for well-behaved potentials) the particle/pole correspondence both ways: essentially, existence of zeros of the Jost function establishes that wavefunctions will give a normalizable bound state, and also would give a pole on the real axis of the corresponding partial wave $s$-matrix eigenvalue. What is the generalization of this to relativistic field theory?

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot

2 Answers

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For simplicity, take $\hbar=1$ and consider a Hermitian scalar, renormalized field $\phi(x)$; other fields are treated analogously. Then $$G(E)=i\int_0^\infty dt e^{itE}\langle \phi(0)\phi(t,0)\rangle =i\int_0^\infty dt e^{itE}\langle \phi(0)e^{-itH}\phi(0)\rangle\\ =\Big\langle\phi(0)\int_0^\infty dt ie^{it(E-H)}\phi(0)\Big\rangle =\langle \phi(0)(E-H)^{-1}\phi(0)\rangle =\psi^*(E-H)^{-1}\psi,$$ where $\psi=\phi(0)|vac\rangle$, since the vacuum absorbs the other exponential factors. The spectral theorem for the self-adjoint operator $H$ guarantees a spectral decomposition $\psi=\int d\mu(E')\psi(E')$ into proper or improper eigenvectors $\psi(E')$ of $H$ with eigenvalus $E'$, where $d\mu$ is the spectral measure of $H$. Orthogonality of the eigenvectors gives $$G(E)=\psi^*(E-H)^{-1}\psi =\int d\mu(E')\psi(E')^*(E-E')^{-1}\psi(E') =\int d\mu(E')\frac{|\psi(E')|^2}{E-E'}.$$ Therefore $$(1)~~~~~~~~~~~~~~~~~~~~~~~~~~G(E)=\int\frac{d\rho(E')}{E-E'},$$ with the measure $d\rho(E')=d\mu(E')|\psi(E')|^2$. For negative $E$, the Greens function is finite; since the measure $\rho$ is positive, this implies that the integral is well-behaved and has no other singularities apart from those explicitly visible in the right hand side of (1). Note that (1) holds for every self-adjoint Hamiltonian, no matter what its spectrum is. It thus describes the most general singularity structure an arbitrary Greens function can have.

In general, the spectral measure consists of a discrete part corresponding to the bound states and a continuous part corresponding to scattering states. (There might also be a singular spectrum, which is usually absent and does not affect the main conclusion.) Formula (1) therefore says that (in the absence of a singular spectrum) the only possible singularities of the Greens function are poles and branch cuts. Moreover, (1) implies that $G(E)$ has a pole precisely at those discrete eigenvalues $E'$ of $H$ for which $\psi(E')$ is nonzero, and branch cuts precisely at the part of the continuous spectrum in the support of the measure $d\rho$.

In particular, any pole of $G(E)$ must be a discrete eigenvalue of $H$, whereas the converse only holds under the qualification that $\psi=\phi(0)|vac\rangle$ has a nonzero projection to the corresponding eigenspace of $H$.

The above argument applies to quantum field theories after elimination of the center of mass degrees of freedom. Thus we consider the subspace of states in which the spatial momentum vanishes. On this subspace, the bound states have a finite multiplicity only, as each mass shell is intersected with the line in momentum space defined by vanishing spatial momentum.

In terms of the original question, the direction ''pole $\to$ particle'' holds without qualification, whereas the direction ''particle $\to$ pole'' (claimed by the OP to be the obvious part) is not universally correct but only holds under a nondegeneracy condition.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Arnold Neumaier
answered Jan 3 by Arnold Neumaier (12,355 points) [ no revision ]
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If I am not mistaken, this is still $\text{evalue}\Rightarrow\text{pole}$. To go the other way, you'd have to start with something that already has a pole $$G(E) \sim \frac{-g}{E-\lambda} + \text{regular}$$ and then show that this implies $\hat{H}|\psi\rangle = \lambda |\psi\rangle$.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot
Should not the $e^{-itE}$ inside the bracket instead read $e^{-it\hat{H}}$?

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot
@QuantumDot: I corrected the typo. The argument goes both ways. From your assumption and ny equation one can see that my right hand side, expanded in a spectral basis, must have a pole at $\lambda$, which is not the case if there is no discrete eigenvalue in $H$, since the rhs has poles only there.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Arnold Neumaier
Are you sure? To rephrase the question in a couple different ways, (1) how do we know that $\mathcal{FT} \langle\hat\phi(x)\hat\phi(y)\rangle$ wouldn't "accidentally" run to infinity? (2) Why couldn't there be more poles in $\mathcal{FT} \langle\hat\phi(x)\hat\phi(y)\rangle$ than as warranted by the spectral resolution of $(E-\hat{H})$?

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot
@QuantumDot: yes, I am sure. The rhs of my expression, when expanded into a Stieltjes integral over the spectrum of $H$, has poles exactly at the discrete eigenvalues and a branch cut at the continuous spectrum. This is a standard result from complex analysis. You can also see it by integrating over a tiny circle containing your $\lambda$ and using Cauchy's integral theorem.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Arnold Neumaier
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@QuantumDot: (2) is guaranteed by the fact that for negative $E$, the Greens function is finite, and the measure $\rho$ is positive. I added the response to your (1) to my answer.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Arnold Neumaier
Oh, I see; the point is that the 2-point Green function has the property that it can be given so explicitly by the spectral decomposition, that together with constraints on $\rho$, there is no room for any additional singularities to develop. I am now happy with the answer. I will now, on my own, try to understand how this would work for higher-point functions. Thanks for your patience and time.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot
+ 2 like - 0 dislike

Particles are very rarely eigenstates of the Hamiltonian.

In particle physics, only (dressed) electrons (and protons?) might be eigenstates. All the other particles have a finite life-time, implying that they decay, which means that they are obviously not eigenstates.

What we want to call a particle is something which has a clear signature in a correlation function and a long life-time compare to its energy. If the theory is perturbative (e.g. QED), the free theory already gives an intuition of what we are looking for: a pole of a Green's function very close to the real axis. Once again, these appear as poles.

In strongly coupled theory (e.g. low-energy QCD), it is much less clear what particles will be if we start from the bare action (quarks+gluons). Nevertheless, there are still particles (stable or unstable, protons and neutrons for instance). There are also resonances (when the life-time is too short).

Finally, there are also strongly-coupled theories where no (quasi)-particles exist. They appear frequently in condensed-matter, and are very difficult to study since one does not have a nice picture in term of scattering, propagation, etc. to figure out what's going on.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user Adam
answered Jul 2, 2014 by Adam (95 points) [ no revision ]
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I am perfectly aware of this technicality, and for that reason I refer, in the question, only to poles on the real axis, which correspond to strictly stable particles. Also, I am looking for a non-perturbative argument (possibly involving axiomatic field theory) for poles corresponding to particles.

This post imported from StackExchange Physics at 2017-01-09 20:54 (UTC), posted by SE-user QuantumDot

'' free (quasi)-particles in such theories are those without interaction (coupling). They make sense since it is from them the interacting theory is built, isn't it? '' No. Quasiparticles are collective modes constructed from the interacting theory in a way that they are effectively noninteracting at low accuracy, thus weakly interacting at higher accuracy, so that perturbation theory can be applied. 

@ArnoldNeumaier: I completely agree with you and I don't understand your "No", neither somebody's downvote to my comment since in my papers I speak precisely about these possibilities.

Well, your comment is misleading since you claimed that it is from the quasiparticles that the interaction is built, which is just the opposite of what I asserted.

You can see how what you write is read by others. Maybe you should learn to express yourself more clearly. Note that you can edit all your comments at any time.

Most recent comments show all comments

If it is obsolete, would you agree if I hide it completely?

No, I disagree. I my comment I was speaking of the second answer where I found a strange statement. This statement has not been clarified yet.

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