## Why the i \epsilon?

Finally, I felt I had something about Physics that I wanted to write about. The i \epsilon terms sitting in the propagator of a QFT, in the Lippmann-Schwinger equation and in Chapter 4 of Peskin and Schroeder have been bothering a couple students including me at the department for a while now. I am not qualified enough in Quantum Field Theory to make any serious comments on this, but I just had some thoughts regarding the i \epsilon. They may be wrong, and I request readers to correct me if there are mistakes, or if they have something to add to this.

At first look, the i \epsilon looked like some bizzare mathematical trick, put in by hand, to give meaning to integrals. “Oh, this integral diverges, but we want it to converge, so we just throw in an i \epsilon”. A lot of us were pretty dissatisfied with this. Also, there was this question too — there are these i \epsilon terms in (a) the propagator, (b) in the Lippmann-Schwinger equation, (c) in Peskin-Schroeder’s derivation relating the interacting ground state with the free-theory ground state, and (d) in the derivation of the path integral formalism from canonical QM — are they all stuck there for the same purpose?

The first time i \epsilon bothered us was in Peskin-Schroeder’s derivation of a relation between the free-field ground state and the interacting-field ground state, where he says “let us take time to infinity in a slightly imaginary direction”. Now, the question was, why should time become imaginary? A long argument on VoIP with Naveen Sharma was adjourned with this: “The T \to \infty( 1 + i \epsilon) is a mathematical trick to supress the contribution of all other states and solve for the interacting ground state in terms of the free-field ground state.”

Then came Prof. Weinberg’s notes on the Lippmann-Schwinger equation. As he explained in class, and as was explained in his notes, the right choice of ± i \epsilon in the Lippmann-Schwinger Green’s function fixes whether we are choosing in-states or out-states. i.e. states with the +i \epsilon in the Green’s function’s denominator satisfy the condition that they look like free particles in the asymptotic past, while states with the -i \epsilon look like free particles in the asymptotic future. A similar argument, with a bit more detail, is presented in his book “The Quantum Theory of Fields” volume 1, in Chapter 3. He also has made a reference to B. A. Lippmann and J. Schwinger, Phys. Rev. Vol 79, No. 3 (1950)..

So I briefly looked at the Lippmann-Schwinger paper, where they actually derive the equation. Then they make a comment: “simulating the cessation of interaction, arising from the separation of component parts of the system, by an adiabatic decrease in the interaction strength as t → ± \infty. The latter can be represented by a factor exp(-\epsilon |t|/ħ) where \epsilon is arbitrarily small.” Aha! So that epsilon came from an adiabatic (slow) decrease of interaction strength! But why are we forced to kill that interaction “by hand”? [PS: Loophole — I still don’t know the adiabatic theorem] I don’t know enough, but I’d ordinarily expect a “factor killing the interaction” to sit in the interaction Hamiltonian rather than outside it (see eqn 1.51 of the Lippmann-Schwinger paper).

At least now, the \epsilon factor had some physical meaning — it came from the adiabatic killing of the interaction, rather than being just some “pole-pushing technology”.

More came today. There is the same epsilon in the Fourier transform of a \theta function (Heaviside step function). One may write:

This is something that I was supposed to know from Electrical Engineering, but we used to “throw away” the epsilon — it didn’t matter much there I guess. Really, it’s just the Fourier transform of a decaying exponential (which every electrical engineer, from IIT Madras at least, would know) with the characteristic length taken to infinity. And then, today we worked out the Feynman propagator for the scalar field. I should’ve known this long back, but I learned it today, that really, the epsilon in the propagator comes from the \theta function’s Fourier transform.

So it seems like the epsilons — at least in (a), (b) and (c) — are present to impose causality.

And then, I learned something more today: An i\epsilon is going to make the Hamiltonian non-Hermitian (eg: See the Green’s function in the Lippmann-Schwinger: it’s effectively adding a small non-Hermitian component to the Hamiltonian). And we see that Hermiticity of the Hamiltonian is required for time-reversal symmetry:

Thus, if my logic is right, the i \epsilon is necessary to break the time-reversal-invariance in the system so that we can talk about an “in” state and distinguish that from an “out” state. Of course, this is unphysical in most situations as far as we know, so we do away with the \epsilon at the end.

Now, this brings me to a couple of questions:

- Does that mean the weak interaction has a non-Hermitian Lagrangian density? [Need to check; sounds like a No]
- We’re always time ordering in quantum mechanics. A naïve look gives me the impression that time ordering breaks time-reversal invariance. Then why are our theories time-invariant?

Anyway, so much for an epsilon!

## Janakiraman 12:56 am

onMarch 18, 2011 Permalink | Log in to ReplyCouple of suggestions. I think of the i\epsilon in the propogator as something that results due to the laplace transform of the propogator. You cannot take a fourier transform it is not absolutely integrable no matter what. Another the hamiltonian needs to be hermitian no matter what, since your energy eigen values can’t be complex. The time reversal symmetry is true if your Hamiltonian is real. This is why this property is lost when we have the vector potential corresponding to the B field (though the Hamiltonian is hermitian).