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Expectation values of interacting fields

+ 6 like - 0 dislike
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I was motivated to ask this question by the equality claimed in equation 10.3.3 of Weinberg's volume 1 of QFT books.

My interpretation of that,

If $O_s$ is a quantum field of spin $s$, $\psi_s$ is the free field of spin $s$, $|p,\sigma\rangle$ is a one-particle state of some interacting theory and $|0\rangle$ is the vacuum state, then there should exist a constant $N$ such that,

$\langle 0|O_s|p,\sigma\rangle = \frac{N}{(2\pi)^3} \langle 0|\psi_s|p,\sigma\rangle$.

If I understand his equation 10.3.6 then that seems to say that the field $O_s$ will be said to have been "renormalized" if $N$ can be set to $1$.

So is one saying that all effect of interactions are absorbed into an overall factor at the level of matrix elements?

That sounds very surprising to me - and it seems that Weinberg claims that it follows merely from the fact that $O_s$ and $\psi_s$ have to transform under the same irreducible representation of the Poincare group.

I would be glad if someone can elaborate this point.

Also what happens if one replaces $|p,\sigma\rangle$ by multi-particle states (it's not clear to me as to what the complete set of labels is that will be required to index the continuum of multiparticle states, and clearly total momentum or/and the invariant mass is not enough).

This post has been migrated from (A51.SE)
asked Nov 19, 2011 in Theoretical Physics by Anirbit (585 points) [ no revision ]
@Squark But the claim here seems to be stronger - as in the matrix elements are proportional and not just that the interacting matrix element is a sum over certain free field matrix elements - thats the point which I don't understand. How does the symmetry constrain to proportionality rather than each being a linear sum of terms like the other? ... somehow seems like that the irreducibility of the representation should allow for the Schur's Lemma to do something...

This post has been migrated from (A51.SE)
@Squark I don't really get it. Since $O_s$ and $\psi_s$ both transform under the same representation of the Poincare group it only means that both are linear transforms of each other via some $U(\lambda,a)$ - a spin-s representation of the element of the Poincare group corresponding to a Lorent tranformation by $\Lambda$ and a space-time translation of $a$ - so i would have thought that matrix element for the interacting field is a sum of the corresponding matrix elements of the free field between the vacuum and to whatever $U(\lambda,a)$ takes the given 1-particle state to.

This post has been migrated from (A51.SE)
There is no direct relation between the free field and the interacting field. The relation is that Poincare symmetry applies to both and as result this matrix element has the same form for both (since symmetry alone is sufficient to determine it modulo an overall factor).

This post has been migrated from (A51.SE)
@Squark Can you elaborate on how the Poincare invariance is fixing the matrix element to this simple form? Its not obvious that these matrix elements for an arbitrary complicatedly interacting field should have had anything to do with that of the free field.

This post has been migrated from (A51.SE)
It's not really surprising. Poincare invariance imposes strong constraints on the matrix elements. In case of this simple matrix element its form is completely determined by the symmetry. This is because all of the 1-particle states are related by Poincare transformations. More complicated matrix elements are also constrained by symmetry but less strongly and hence will differ from the free field case

This post has been migrated from (A51.SE)
Not all effects of interactions are absorbed into an overall factor at the matrix elements. First of all, interactions lead to a non trivial dynamics/evolutions of the field content (occupation numbers). It is renormalization that is fulfilled as renormalization of mass, charge, and the field coefficients.

This post has been migrated from (A51.SE)

1 Answer

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The answer about this equation in Weinberg's book is rather simple. You have to evaluate $\langle 0|{\cal O}_l(0)|q_1,s\rangle$ being $l$ the index due to the corresponding Lorentz transformation. This means that this matrix element should transform as an element of the Lorentz group. Weinberg's assumption is that the corresponding representation is irreducible. If this is true, being this a matrix element between the vacuum and a single particle state, the only factor that has such a property of transformation under Lorentz group is the one of the free particle state and that can be always singled out. What remains is the contribution of the interaction that amounts to a constant. This is just a requirenment of Lorentz invariance. Let us make some examples:

Scalar field:

$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}\sqrt{2E}}N_s$$

Spin-1/2 field:

$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}}u_l(q_1,s)N_f$$

and so on.

But this is a well-known theorem in quantun filed theory: It is named Lehmann-Symanzik-Zimmerman (LSZ) theorem that Weinberg puts forward in a rather discorsive way in his book at that stage. LSZ theorem permits one to pass from N-point functions to S-matrix elements and to obtain cross sections and decay rates. So, you can write down the partition function of your favourite theory, compute N-point functions for a given reaction and, from it, using LSZ theorem obtain the corresponding observable. Another good presentation of this reduction formula is given on Peskin and Schroeder staring from pag.222.

This post has been migrated from (A51.SE)
answered Nov 22, 2011 by JonLester (376 points) [ no revision ]

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