# Lippmann-Schwinger equations

1. Apr 30, 2010

### Yat-lo Lay

I am currently reading Weinberg's "The Quantum Theory of Fields", vol I.

On page 112, he tries to show why "+ie" term is the correct choice to add in the denominator of the Lippmann-Schwinger equations for the scattering "in" states. The proof is based on the residue theorem of a complex variable. The idea is to show the integral associated with the Lippmann-Schwinger equations approaches zero when the time tends to -infinity.

So for this purpose and with the energy as the variable of integration, Weinberg uses the residue theorem and extends its integration domain to a semi-circle contour on the upper complex half-plane. Then he argues that integral is indeed zero when the energy goes to +/- infinity and time goes to -infinity.

Here is the part that I don't understand. Physically, energy should have a lower bound, so the domain of integration of the integral associated with the Lippmann-Schwinger equations should be from the finite energy lower bound to +infinity. But the integral that Weinberg proves zero is an integral with energy running from -infinity to +infinity.

On page 114, Weinberg uses similar argument.

So my question is why Weinberg sets the energy integration range from -infinity to +infinity but not from a finite lower bound to +infinity?

2. Apr 30, 2010

### ansgar

but that IS how the residue theorem works..

3. Apr 30, 2010

### Yat-lo Lay

Yes, I know, mathematically, that's how the residue theorem works. But my question is why it can be applied to the physical situation where the energy lower bound is finite not negative infinity.

4. Apr 30, 2010

### ansgar

what you integrate over is d^3 p not energy

change to spherical coordinates

p^2 dp d\Omega

p from 0 to infty

this integration is equal to a circle

5. Apr 30, 2010

### Yat-lo Lay

But this is not what Weinberg uses. If you read his book (page 112 & 114), then you will understand my question.

6. May 1, 2010

### humanino

ansgar is right, and it is equation (3.1.4) in my volume
$$\int\text{d}\alpha\cdots=\sum_{n_1\sigma_1n_2\sigma_2\cdots}\int\text{d}^3p_1\text{d}^3p_2\cdots$$

Think about it physically : once you fix a particle type, its mass is given, and integration over asymptotic states must be over momenta. Energy is just conserved.

7. May 1, 2010

### Yat-lo Lay

Absolutely. If Weinberg uses momentum as the integration variable like what you and ansgar show, then I will have no question. The thing is Weinberg uses energy, and considers its integration range from -infinity to +infinity, that's what I don't understand.

If you read the paragraphs under equations (3.1.21) on page 112 and (3.2.6) on page 114, you can see what I am talking about.

8. May 3, 2010

### Count Iblis

The Hamiltonian should be bounded from below. The energy variable in your equations, let's call it omega (I don't have Weinberg's book) presumably appears as follows:

Integral d omega/(2 pi) f(omega) exp(i omega t)/(omega + i epsilon)

f(omega) consists of matrix element times some factors like

1/(omega - E_final +i epsilon) etc.

Now this you can obtain by writing the transition amplitude as a fourier integral over omega. So, it should be clear that the omega integral runs from minus to plus infinity. Then what happens is that the omega integral will implement delta functions and you'll end up with something proportional to delta(E_in - E_fin).

9. May 3, 2010

### Yat-lo Lay

I know the proof for the Lippmann-Schwinger equations in the non-relativistic case. Normally people would first consider its Green function G(x, y) in the position coordinates. Then expand G(x, y) as a Fourier integral in terms of the momentum variable. Using the fact that the integrand of the Fourier integral is an even function of momentum, one is allowed to change the momentum integration range from (0, +infinity) to (-infinity, +infinity).

The above proof can be found in many elementary quantum mechanics textbooks.

However, Weinberg's proof is different, there is no Fourier integration. He basically uses the residue theorem for the integral with energy as the integration variable running from -infinity to +infinity. Since energy is bounded below and the integrand does not appear as an even function of energy, I am puzzled of what justifies the energy going down to -infinity in his equation.

If you get a chance to see Weinberg's book, you will know exactly what I mean.

10. May 3, 2010

### Count Iblis

Ok, but then can you explain precisely where the integration over energy comes from in the first place? You can have expressions that involve integrations over all of the k space, like e.g. for the propagator, in some expressions you have an integration involving theta(k0), sometimes it will be sign(k0). There is no general rule that says that all integrals over d^4 k must have a theta(k0) in the integrand.

11. May 4, 2010

### Yat-lo Lay

I totally understand we could have different expressions for the same or similar integral. But it seems you still don't know what I want to point out. I can understand because you have not yet read Weinberg's book.

As I said, if you have Weinberg's book, I am sure you will know what I mean and will know how he ends up with the integration over energy. I can't tell you precisely in words only without writing down the equations from the beginning (it takes 3 pages in Weinberg's book p.110 to p.112).

12. May 4, 2010

### Haelfix

I just skimmed the pages in question, and I don't understand what you are worried about. The dalpha measure is a shorthand for the complete set of momentum measures, with a sum over species and helicity.

So look at eqn 3.1.21, and wherever you see a Ealpha, replace it with the sum of the one particle energies. So eg the zeroth component of each (p1 + p2 + ... ) = Ealpha.

13. May 4, 2010

### Yat-lo Lay

The sum of all free 1-particle energies must be greater than zero. So the integration lower limit should not be minus infinity.

14. May 4, 2010

### humanino

The result of the integral does not depend on E which is an internal variable to the integration. Obviously when E -> -E we have dE -> -dE, but as integration then runs backwards, so the rest of the integrand must be even.

15. May 5, 2010

### Yat-lo Lay

By definition, a function f(E) is an even function of E if f(E) = f(-E)

And for even function f(E):

[integrate f(E) from E= 0 to E=+infinity] is equal to (1/2) * [integrate f(E) from E=-infinity to E=+infinity].

So we can extend the integration range from (0, +infinity) to (-infinity, +infinity) when the integrand is an even function.

However, the integrand in eq (3.1.21) contains the exponential factor exp(-iEt). Obviously exp(-iEt) is not equal to exp(iEt), so not an even function.

The integrand in eq (3.1.21) also contains a function g(E). There is no constraint on g(E) that it has to be either an even or odd function.

The overall integrand in eq (3.1.21) also does not appear to be an even function of E.

So how can we extend this integration with respect to energy E from (0, +infinity) to (-inifinity, +infinity) ?

And if we can't extend it to (-inifinity, +infinity), how can we use the residue theorem with a large semi-circle integration contour on the upper complex E half-plane (so that the contour for E runs from -infinity to +infinity on the real-axis) to evaluate the integral in eq (3.1.21)?

There is one more thing to point out. For massive particles, the lower limit should be the sum of their rest masses M. So the correct integration range should actually run from E=M to E=+infinity.

Anyway, it seems I am unable to make people in this forum understand what I want to point out. This is probably due to my inability to present my question in a comprehensible way. I apologize for it and I think I should not proceed further with this thread.

Thank you.

Last edited: May 5, 2010
16. May 5, 2010

### Count Iblis

To be clear here, does eq (3.1.21) contain an integral from zero to infinity which Weinberg extends to an integral from minus infinity to plus infinity without further explanation?

Or is eq (3.1.21) an integral from minus infinity to plus infinity?

17. May 6, 2010

### Yat-lo Lay

The integral measure of eq (3.1.21) is supposed to be dp1dp2dp3 where p1, p2 and p3 are the 3 independent components of momentum. Essentially p1, p2, p3 can run from -infinity to +infinity.

However Weinberg does not continue to stay with the momentum variable to evaluate the integral. Rather he uses energy E. But he does not show how he carries out the change of variable from momentum to energy.

Finally he allows the energy E in the integral to run from -infinity to +infinity so that he can use the residue theorem to calculate it. But there is no explanation of why he allows the energy E to run from -infinity to +infinity.

If you can find the book maybe from a library or something, you will have a much more clear picture of what I am talking about here.

Last edited: May 6, 2010
18. May 6, 2010

### xepma

I think the function $g(\alpha)$ is what solves your problem. It represents the 'shape' of the wave-packet; it determines the amplitude of the states corresponding to an energy $E_\alpha$, as explained above eq 3.1.8. It's some smooth function, and non-zero over a finite-range of energies. So you may assume it is simply zero for energies which are not accessible.

With that in mind the integration over energy can be safely extended to the negative domain.