Quantization of relativistic point particle, string style

1. Aug 6, 2008

jostpuur

I don't have the Zwiebach's string theory book myself, but I paid a visit to a library, and took a glance on it. The chapter 5 was about relativistic point particle. Now, did I understand correctly, that the string people actually have a technique to quantize a relativistic point particle? I thought that the basic QM and QFT texts are forbidding that procedure, declaring it impossible.

The way how Zwiebach was using the proper time $$\tau$$ seemed dangerous. It's ratio to the time $$t=x^0$$ depends on the particle's velocity. So if you in quantum mechanics parametrize the wave function with such proper time, $$\psi(x,\tau)$$, aren't you having different fourier amplitudes propagating with "different speeds in time" then? It looks very messy. I have difficulty seeing what's happening with that kind of approach.

Zwiebach seemed to be mainly interested in the operators. There was no discussion about spatial probability densities. Am I wrong to guess that despite the fact that string theorists have a technique to quantize a relativistic point particle, they have nothing to say about the probability densities of relativistic particles?

2. Aug 6, 2008

Demystifier

You are right. Essentially, their quantization of the relativistic particle reduces to a derivation of the Klein-Gordon equation. They do not provide a probabilistic interpretation in the configuration space. (However, the probability in the momentum space is well defined, which is sufficient for most of the practical purposes.) At least this is so in the mainstream research in string theory. For my non-mainstream approach see however
http://xxx.lanl.gov/abs/0806.1431
and references therein.

3. Aug 6, 2008

humanino

You run into trouble only if you insist that you have a single particle. That's another way of seeing

Chapter 5 is pretty short really, and half of it is non-relativistic. It only aims at showing where the idea for the simplest free lagrangian comes from. But all this is at a level even before the Nambu-Goto string. It's important in string theory since "string generate interactions", or "once you have defined the free sector, the theory is done" : interactions results form topological re-arrangements of the worldsheet.

4. Aug 7, 2008

jostpuur

I don't see how the problem about proper time, which I attempted to describe in the original post, could be dealt with by moving into multi-particle formulation. The multi-particle formulation, and the proper time stuff seem to be separate subjects.

5. Aug 7, 2008

Demystifier

The point is that the action does not depend on the choice of the parameter tau. Consequently, the Hamiltonian defined with respect to tau must vanish. In the particular case of a relativistic particle, this Hamiltonian is
H=p^2-m^2
where p is the 4-momentum. (Note that H is NOT the energy of the particle.) Therefore, we have equation
p^2-m^2=0
which is a well-known classical equation.
In the quantum case, this becomes
[p^2-m^2]|psi>=0
In the Schrodinger picture p becomes a derivative operator (times i\hbar), so the last equation becomes the Klein-Gordon equation. It describes the wavefunction psi(x), where x is the spacetime coordinate. Thus, psi(x) is NOT a function of the proper time tau.

If the Hamiltonian was not constrained to vanish, we would have a Schrodinger-like equation
H psi(x,tau) = i\hbar \partial_\tau psi(x,tau)
so psi would depend on both x^0=t and tau as two independent variables. But this is not the case. The constraint H=0 implies the constraint that psi does not depend on tau.

Last edited: Aug 7, 2008
6. Aug 11, 2008

jostpuur

I made mistake when talking about the chapter 5 (the relativistic point particle). It was the chapter 11 (the relativistic quantum point particle) that I meant. I looked at the chapter 11 in the library, but I didn't remember the number in home and tried to check it from amazon.com and then confused the two chapters with similar names.

There are equations like

$$H(\tau) = \frac{1}{2m^2}\big(p^I(\tau)p^I(\tau) + m^2\big)\quad\quad\quad (11.1)$$

for the Heisenberg picture, and

$$i\frac{\partial}{\partial\tau}\psi(\tau,p^+,\vec{p}_T) = \frac{1}{2m^2}(p^Ip^I + m^2)\psi(\tau, p^+, \vec{p}_T)\quad\quad\quad (11.34)$$

for the Schrödinger picture. The wave function seems very much to be a function of the proper time, although I'm not understanding this notation very well yet...

7. Aug 13, 2008

Demystifier

Jostpuur, (11.34) can be written as
$$H \psi = i \partial_{\tau} \psi$$
However, as I explained in the previous post, psi does not depend on tau because this equation should be accompanied by another equation, the Hamiltonian constraint
$$H \psi = 0$$