Also, the number of space-time dimensions is not an assumption of string theory. This is very important. Unlike in GR, the number of dimensions in string theory is a derived quantity.
Bosonic string theory is a TOY theory, that is, it cannot describe nature because it has no fermions. So the fact that it is inconsistent (i.e. presence of a tachyon) is not a problem.
Ben, I thought the S matrix was by definition unitary regardless of string theory. (Disclaimer: Unfortunately, I am no mathematician) Possibly related general question: Does string theory need to be background independent? Has string FIELD theory been completely dropped?
Sorry to be dim, but I am no wiser. Do you think you could make this more comprehensible to ignorants like me? Like, a slightly fuller explanation?
Well, perhaps I should say that the evolution is goverened by a unitary S matrix. Well, there is a bosonic string field theory, which is fully background dependant, and has been solved recently by Martin Schnabl (I think I spelled his name right). So the objections that string theory is not background independant (cf Smolin's book) are not well-founded anymore, in my opinion. But I will freely admit to knowing very little about this subject. As for whether or not string theory NEEDS to be background independant, I don't know. If this is the ONLY objection to string theory, then I don't think it is a particularly well-founded one. I have never understood how to understand ``space-time'' in string theory. We are all comfortable saying that GR is a classical theory, an effective description of gravity at long distances, and that the geometry of space-time governs how we understand the gravitational dynamics. But this is not true on the quantum level, as I understand it. On the quantum level in string theory, we have gravitons, not geometry. So I don't really know what background independance MEANS for string theory. If AlphaNumeric or pbrane is following, I would like to hear your opinions on this, as mine are probably wrong Please Register or Log in to view the hidden image!
The S matrix is the operator which governs the evolution of initial states to final states in a quantum field theory. Ok, here goes. This is public, so you all will know how terrible a mathematician I am. Let there be a vector space \(I\) consisting of vectors \(\left| i \right\rangle\) which describe the initial states of some system. There is a dual vector space, \(F\), consisting of vectors \(\left\langle f \right|\) which describe the possible final states of a physical system. There is a good inner product and basis, etc. If we wish to find the probability that some initial state \(\left| i' \right\rangle\) evolves into some final state \(\left\langle f' \right|\) we take the inner product of these two states with the S matrix: \(\left|\left\langle f'\right| S \left| i'\right\rangle\right|^2 = P_{i' \rightarrow f'} \) I don't know if I've made this more clear or not...
I'm lost, but I don't blame Ben, for this, I blame Dirac! Ben: You state that the operator \(S:I \to F\) is a map from a vector space to its dual, fair enough - this is an isomorphism, right? But earlier you wrote the form \(\left\langle f'|S|i' \rangle \). Dirac gives us no information about the vector argument of S. By this form, it must be either \(|i' \rangle \) or \(|f' \rangle \). Or is \(\left\langle f'|S|i' \rangle \) a bilinear form e.g. an inner product? Even so, I do not understand the notation: if \( S: I \to F\), what is the justification for writing \(\left\langle f'|S|i' \rangle \), which implies that \(|i' \rangle \in I\), which is contrary to your opening assertion that \(|i \rangle \in I\). How can this be? What elements of \(I\) is \(S\) operating upon; is it \(|i \rangle\) or \( |i' \rangle\)? How are these guys related? Or have I wildly missed the point?
I think S maps I to I because then you can make the duality pairing <f,Si>. Anyway, it is not important in case of Hilbert spaces since F is isometric to I via the Riesz representation theorem.
Ahh I should have made my notation more clear. I just wanted to be clear that i' was some other vector (like i) living in the space I. I feel like a child wearing my fathers boots when talking about math like this. You are correct. I should write \(\left\langle f|S|i \rangle\). Apologies all around. Also, I think temur is correct in this point:
