They claim that the usual wisdom about the Hagedorn phase transition is incorrect. The usual wisdom is that the exponential growth of the density of states implies that at temperatures higher than the Hagedorn temperature, the second quantized spacetime partition sum diverges because the Boltzmann factor is not sufficient to suppress the exponential growth of the number of states. Alternatively, the Hagedorn transition can be seen as the appearance of a tachyon in the channel related by modular transformations.

Dienes and Lennek argue that this widely believed argument is wrong.

- In the UV picture, the divergence is wrong in string theory, they say, because it effectively leads to you to treat string theory as field theory with many fields, which means that the one-loop integrals over "tau" are made over "Im(tau)" being positive and "Re(tau)" smaller than one-half. The correct stringy integral involves the fundamental domain of the modular group "SL(2,Z)", and if this domain is used, the divergence at the (old) Hagedorn point disappears, they claim.

- In the IR picture based on the existence of tachyon, they note that the tachyon is actually projected out by the GSO projections, and consequently, it has no physical consequences. The usual Hagedorn phase transition is therefore absent in heterotic string theories and other tachyon-free superstring theories. They affirm that the Hagedorn behavior in type 0 theories, for example, still holds.

The authors also say that one of the conventional Hagedorn transitions does occur but it is a 10th order phase transition where "10" is indeed the spacetime dimension "D" (and it would be replaced by "D-1" if the dimension were odd).

I am curious what others think about this paper.

I think there is a danger of confusing two very different concepts of "state", which unfortunately use the same English word. And to me, the authors clearly had made such a confusion.

ReplyDeleteTo clarify, the two different concept of states:

1.Number of possible quantum states under certain constraints, e.g., energy below a certain level. Each quantum state can certainly be either occupied or un-occupied. This concept of state is otherwise known as degree of freedom. When people talk about "density of state", they are likely talking about this number of degree of freedom.

2.Number of possible states a system can be in. You obtain this by considering all possible combinations of all quantum states, each could be in either an occupied or un-occupied state for Fermions. And for bosons it's more complicated: the occupation of each quantum state could be from zero to N. Clearly this number of states is always exponential, if you multiply all possible states of each degree of freedom up.

For example, a system can have two quantum states A and B, but it could be in 4 different states, depends on whether each of A and B is occupied or not.

I do not know what states the author is refering to when he said "exponential increase". Only the number of state of the system will increase exponentially. The number of state in the sense of first kind, the degree of freedom

never increase exponentially, it at most increase polynormially. And when you talk about density of state, you would be talking about density of degree of freedoms.The quantum degree of freedoms do not increase exponentially for two reasons:

one, it would have lead to the total possible states of the system to increase hyper-exponentially, in the form of exp(exp(alpha)), which lead to an entropy which is exponential. And temperature of the system then no longer has a good definition.

two, the quantum uncertainty principle determines that you really can't have quantum states too close together. When two quantum states are too similar to each other they become indistinguisable due to the uncertainty principle, and really count as just one state. You may have states with similar or even identical energy levels, but they must separate between them by other factors, like different spins. In that sense the density of state really can not go denser exponentially.

Quantoken

Dear Quantoken,

ReplyDeletewhy don't you pick up an introductory text on quantum mechanics and try to survive the first 10 pages or so? (Alternatively, pick up your medication.)

Your naive attempt of state counting betrays that you have no clue what a *density* of states is. You don't even seem to understand the word exponential. Here is a clue: Not everything that is large is exponential.

The distinction you make between "the two different concepts" is ridiculous. Ever heard of superposition of states? I didn't think so.

Your psycho:

ReplyDeleteAs always you drink too much of the stuff.

"Superposition of states" is just "superposition" of "states". It is not a different state itself, but merely a

statistical"superposition" of the "states" that the system can be in, "simutaneously". It does not change the total number of states a system can be in.In QM you have the concept of "mixed states" and "pure states". They are NOT different states for a system, but merely a description that possibility wise, how you would detect a system in a particular state.

In any one single measurement, you could only find the system in one of the states. You could never detect the system in two different states simutaneously in just

one measurement, just as you can not catch a photon simulatenously behaving both like a wave and like a particle. Only when you do multiple measurements, then by statistics the photon shows both a face of wave and a face of particle, and you could find the system sometimes in one state and some other times in another.Certainly none of these change the fact that there simply is NO exponentially increasing density of state. If you can cite one such example I can point out right away while it is wrong. Ever heard about entropy bound?

Quantoken