7
$\begingroup$

Do amplitudes and correlation functions in string theory satisfy the cluster decomposition principle?

Note added: Even without local observables such as correlation functions, one can define the cluster decomposition principle on the S-matrix elements that are observable in principle even in string theory (as far as I know). It requires that scattering amplitudes factorize for clusters of particles that are very far from each other. Technically that should correspond to the fact that any connected S matrix element $S_C$ contain one $\delta^4(p^{cluster}_{total})$ for the total momentum conservation of the cluster, but not any other $\delta^4(p^{cluster}_{subset})$. More info on this point can be found in S-matrix theory books, or e.g. in the QFT book by Weinberg in chapter 4, volume I.

$\endgroup$
2
  • $\begingroup$ How do you want to formulate the cluster decomposition principle? The version discussed on wikipedia doesn't work well for string theory, which lacks gauge invariant local operators. $\endgroup$ – user1504 Apr 2 '15 at 20:51
  • $\begingroup$ @user1504 One can define the cluster decomposition principle on the S-matrix elements that are observable even in string theory. It requires that scattering amplitudes factorize for clusters of particles that are very far from each other. Technically that should correspond to the fact that any connected S matrix element $S_C$ contain one $\delta^4(p^{cluster}_{total})$ for the total momentum conservation of the cluster, but not any other $\delta^4(p^{cluster}_{subset})$. More info on this point can be found in S-matrix theory books, or e.g. in the QFT book by Weinberg in chapter 4, volume I. $\endgroup$ – TwoBs Apr 3 '15 at 10:22
3
+150
$\begingroup$

When the Hamiltonian of theory is constructed out of creation and annihilation operators the S-matrix automatically satisfies the cluster decomposition principle given that in momentum space the coefficient of the interaction contains only one delta function. However this does not apply to the (first quantized) string theory because there even though the world sheet action is constructed out of fields which in turn can be expanded in terms of creation and annihilation operators of string excitation modes, the spatial coordinated distance is of the order of the Plank length, so the question itself is not well posed for a single string state. In other words the excitation modes of a single string state cannot be asked whether they satisfy the cluster decomposition principle simply because they describe physics over a confined scale of order of the plank length (lets focus on closed strings for the moment), so $\lim_{\sigma\rightarrow\infty}$ does not even make sense and the cluster decomposition principle is not a physical requirement.

What is required is that string theory when considered at long distances, and as a matter of fact infinite distances, only then should satisfy the cluster decomposition principle. And in this respect it does. This can be most easily seen in two remarkable different but equivalent ways.

$\bullet$ The requirement of the vanishing of the weyl anomaly on the world sheet, leads to an effective low energy action through the requirement of the vanishing of the beta functions in the presence of background fields. This action for the bosonic string is $$ S\propto\int d^{26}X \sqrt{-G}e^{-2\Phi}\left( R - \tfrac{1}{12}H_{\mu\nu\rho}H^{\mu\nu\rho} + 4 \partial \Phi\cdot\partial \Phi\right) $$ Once the fields in this action are written in terms of free field interaction picture annihilation and creation operators, the S-matrix satisfies the cluster decomposition principle (when we consider $\lim_{X\rightarrow\infty}$) due to the presence of a single delta function arising from translational invariance.

$\bullet$ By considering string field theory instead. In flat space time and for a free string field theory, elements of the fock space take the form $$ \left| \Psi\right\rangle = \int d^{26}p e^{ip\cdot X}\left( c_1 T(p)+ c_1 A_\mu (p)\partial^\mu X + c_0 \chi(p)\right)\left|0\right\rangle + \ldots $$ This again should satisfy the cluster decomposition principle for correlations between string different string states (not same string state with different excitations)

$\endgroup$
4
  • $\begingroup$ Thank you for the your answer. Concerning your first statement, I must say that, actually, it isn't enough for a Hamiltonian, in order to satisfy the cluster decomposition principle, to be expressed in terms of creation and annihilation operators. The cluster decomposition principle put a further constraint on the coefficients in this expansion, namely that they contain a single delta function of the momenta. See Weinberg. sec.4.4 for a discussion. As for the rest of the answer I will add an independent comment below. $\endgroup$ – TwoBs Apr 7 '15 at 8:36
  • $\begingroup$ Moreover, I think it is a somewhat misleading to take the infinite distance limit. Only a subset of the asymptotic particles, each cluster as a whole in fact, must be set to infinite distance, whereas the distance among particles within the cluster should be kept finite. I don't see why e.g. the massive states should be simply decoupled when a cluster of particles becomes very close to each other and yet form a cluster infinitely separated from the rest. $\endgroup$ – TwoBs Apr 7 '15 at 8:40
  • $\begingroup$ Regarding your second comment I should have emphasized that being able to take the infinite distance limit is a necessary but no sufficient condition. Regarding the first, I believe the effective action / string field action above does satisfy the delta function requirement. I'll edit the answer as to emphasize these two important facts $\endgroup$ – Ali Moh Apr 7 '15 at 8:48
  • $\begingroup$ I think you have not addressed my second criticism. We know that at low-energy you recover some QFT limit where cluster decomposition trivially applies. The interesting configuration to check is instead when one has Planckian or subPlanckian distances between a set of of particles (no low-energy limit can be taken) whereas some other particles are very separated from the previous ones (but perhaps they are close to each other again by subPlanckian distances) $\endgroup$ – TwoBs Apr 8 '15 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.