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Positivity of residues and unitarity in scattering amplitudes

+ 6 like - 0 dislike
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I am reading "Superstring Theory" by Green, Schwarz, Witten. In the introduction, about the Veneziano amplitude (below eq. 1.1.16/17), they say that

The residues of poles must be positive in a relativistic QFT, for unitarity and absence of ghosts.

Now, I have the following questions:

1) By "unitarity", are they referring to the optical theorem (which follows from unitarity of S matrix) or to something more general?

2) If they are referring to the Optical Theorem, then one can use it to get positivity constraints for processes with the same initial and final states only $M(A \rightarrow A)$. But the Veneziano amplitude holds in general (right?), so I cannot see how they can say that unitarity implies positivity also for residues coming from processes where initial and final states are different, like $M(A \rightarrow B)$.

[less important 3) Presence of ghosts prevents unitarity. Is absence of ghosts sufficient to ensure unitarity? (references)]


This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user BLS

asked Nov 14, 2016 in Theoretical Physics by BLS (60 points) [ revision history ]
edited Jan 9 by Dilaton
comment to $3)$: no, more like the other way around. In the early stages of gauge theories, before the introduction of the Fadeev-Popov fields, calculations were performed without ghosts. Feynman and others found that the theory was not unitary. In this sense, you need ghosts for unitarity, just only you don't want them on the external lines. Or put it another way, presence of ghosts in the external lines prevents unitarity; absence of ghosts in the internal lines might prevent unitarity as well!

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user AccidentalFourierTransform
By "ghosts" I mean negative norm states, that's surely what you don't wanna have, don't you?

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user BLS
well, you do want to have them, but on internal lines (to cancel unphysical polarisations). You do not want to have them on external lines, but negative norm states circulating on internal loops is the only way to make nonabelian gauge theories work.

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user AccidentalFourierTransform
useful answers to the first questions?

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user BLS
@BLS what you are actually looking for comes under the category of dual resonance models. Unitarity actually forces you to consider only positive residues which is based on the requirement of positive decay width and basically avoids negative lifetime. A negative residue implies the presence of ghost state and violate unitarity. For reference consider ' the birth of string theory' by Cappelli, Castellani.

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user ved

1 Answer

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See Weiberg's QFT, Vol. I, section 10.3, the equation below 10.3.6. Cleaning up the notation a bit, and using the modern normalisation of spinors together with the mostly minus metric convention, the the pole of propagator for a general field is $$ \Delta_{\ell\ell'}(p)=\frac{|Z|^2}{p^2-m^2+i\epsilon} M_{\ell\ell'} \tag{1} $$ where $M$ is the usual projection matrix, $$ M_{\ell\ell'}=\sum_\sigma u_\ell(p,\sigma)u^*_{\ell'}(p,\sigma) \tag{2} $$

Using the fact that polarisation vectors are always orthogonal to each other (i.e., $u^\dagger v=0$), we see that $$ Mv=0 \tag{3} $$

On the other hand, using the fact that $u^\dagger u=|u|^2>0$, we see that $$ Mu=|u|^2u \tag{4} $$

Therefore, as the $u,v$ polarisation vectors are a basis, we see that $M\ge 0$ is a non-negative matrix (its eigenvalues are either zero or positive).

But there is a subtlety hiding in $|u|^2>0$: the polarisation vectors $$ u(p)\equiv \langle 0|\psi(0)|p\rangle \tag{5} $$ have positive norm if and only if $|p\rangle$ has positive norm. The same thing can be said about the polarisation vectors of antiparticles, defined as $$ v(p)\equiv \langle \bar p|\psi(0)|0\rangle \tag{6} $$ where the bar indicates antiparticle (opposite charge). The objects $u,v$ are the usual polarisation vectors one includes in the external lines of scattering amplitudes (in the case of spin $j=1$ particles, the usual notation is $\varepsilon^\mu$, but it is the same concept).

Therefore, the residue of the propagator $M$ is non-negative only if the physical sector has positive norm, that is, if the asymptotic states $|p\rangle$ have positive norm. If you have an external line with $|u|^2<0$, then the matrix $M$ becomes indefinite (it's no longer non-negative).

Note that in gauge theories, there are unphysical states as well, but these should never appear on external lines. These states are allowed to have negative norm. For example, in the case of spin $j=1$ particles, the polarisation vectors are space-like, but the longitudinal states $\varepsilon^\mu=p^\mu$ are time-like. But if we use $\varepsilon^\mu=p^\mu$ on an external line, the amplitude is zero (by the Ward identity).


Remark: the polarisation vectors $u,v$ are only a basis for the physical Hilbert space. Ghosts are orthogonal to the physical $u,v$, and they have negative norm, $u^\ell u_\ell^*<0$. As an example, consider once again the longitudinal polarisations for spin $j=1$ particles: the physical polarisation vectors are space-like, while the longitudinal polarisation is time-like. Together, these four vectors span $\mathbb R^4$, while the physical polarisations only generate space-like vectors.

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user AccidentalFourierTransform
answered Nov 22, 2016 by AccidentalFourierTransform (375 points) [ no revision ]
If you need me to add more details to any specific step or to clarify some statement, feel free to ask me to (and I'll do my best).

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user AccidentalFourierTransform
OK, thank you, I feel like this is going to be a satisfactory answer, once more details are included. Can you clarify the steps after (5)? and can you define $\nu$ precisely (I am a little bit confused) ?

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user BLS
@BLS please let me know if it's more clear now. BTW, the $v$ is a vee "V", not the Greek letter "nu".

This post imported from StackExchange Physics at 2017-01-09 20:50 (UTC), posted by SE-user AccidentalFourierTransform

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