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Problems with Relativisic Quantum Mechanics of single-objects

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As done in several text on string theory, you can quantize the worldsheet action to get the quantum string. But first, they start with the quantum theory of a single point particle by quantization of a worldline action theory.

So, I want to draw a parallel between them. For the point particle we promote position and momentum to operators fulfilling ($\hbar=1$):

$$[x^\mu(\tau),p_\mu(\tau)]=i\delta^\mu_\nu$$

Also we have the Klein-Gordon equation which arise as the first-class constraint $p^2+m^2=0$ associated with reparametrization invariance of the worldline, yielding free particle momentum states $|p \rangle$.

As we learned in QFT courses this theory has many problems, such as negative energies, negative probabilities and causality violation (which are the main motivations they state to develop another theory, field theory). Note that with the usual BRST procedure we get a hamiltonian $H=\frac{1}{2}(p^2+m^2)$ and acording to Polchinki's book page 130:

The structure here is analogous to what we will find for the string. The constraint (the missing equation of motion) is $H = 0$, and the BRST operator is $c$ times this.

So $p_0<0$ states are in the state space too.

Now, we do the same for the string. We now have the operators satisfying $$[x^\mu(\tau,\sigma),p_\mu(\tau,\sigma')]=i\delta(\sigma-\sigma')\delta^\mu_\nu$$ and we can costruct a state space and so on. But the books never talk about the potential problems that could arise such as:

  1. Negative energies

  2. Negative Probabilities

  3. Causality violation

So, do these problems arise in the theory of a single string in the same way as arised in the relativistic single particle?

This post imported from StackExchange Physics at 2017-08-04 22:00 (UTC), posted by SE-user MoYavar
asked Jul 29 in Theoretical Physics by MoYavar (15 points) [ no revision ]
Related question by OP: physics.stackexchange.com/q/348773/2451

This post imported from StackExchange Physics at 2017-08-04 22:00 (UTC), posted by SE-user Qmechanic

1 Answer

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The quantization of a point particle analogous to the quantization of a single string does not have the issues you claim. The problem with negative energies etc. arises when you believe that the Klein-Gordon equation is an equation for a wavefunction in position space. But if we do the process analogous to string theory, we quantize the 1d Polyakov action $$ S_\text{P}[x,e] = \int \left(e^{-1}\dot{x}^\mu\dot{x}_\mu - m^2 e\right)$$ with the einbein $e$ as a dynamical variable analogous to the metric on the string worldsheet. This theory is a gauge theory and can be consistently quantized with the usual BRST procedure (cf. e.g. these notes), yielding free particle momentum states $\lvert p\rangle$. Notably, $x^\mu$ is a BRST-variant operator and therefore not an observable, so it is difficult/impossible to speak about "causality" or "position space wavefunctions" in this theory, since position is not a meaningful concept for these states.

The quantum theory of a single free string is a two-dimensional (super-)conformal field theory, the precise SCFT depending on which variant of the string (heterotic, type I, type II) we choose, likewise obtained through BRST quantization of the worldsheet action and some additional choices like the GSO projection. It has no issues with "negative probabilities" or similar things.

Furthermore, one can phrase ordinary QFT completely analogous to string theory in the so-called worldline formalism: The interactions of string theory were first inspired by the Feynman diagrams of QFT, leading to the definition of perturbative string theory that interactions are to be thought of as summing the worldsheet theory of the string over all possible "2d Feynman diagrams", essentially the Riemann surfaces. After that, it was realized that the sum over Feynman diagrams in a QFT can actually also be phrased as summing the worldline theories over the set of diagrams the QFT admits.

This post imported from StackExchange Physics at 2017-08-04 22:00 (UTC), posted by SE-user ACuriousMind
answered Jul 30 by ACuriousMind (700 points) [ no revision ]
States with $p_0<0$ are BRST-invariant. They are still there.

This post imported from StackExchange Physics at 2017-08-04 22:00 (UTC), posted by SE-user MoYavar
Sorry, what I should say is that $p_0$ is BRST-invariant. The BRST-charge $\Omega$ encodes the constraint $p^2+m^2$, so I cannot see the point of using BRST method, my question is not about any particuar quantisation method (but I had used the old covariant method in the question, ok), is about the quantum theory of a single particle. It does not matter what method you use, read equation (4.4.21) on page 141 in Polchinski's book. And why did you give a reference that is not sure how to ban the states with $k_0<0$ and the states $|\uparrow , k \rangle$, see the footnote at page 3.

This post imported from StackExchange Physics at 2017-08-04 22:00 (UTC), posted by SE-user MoYavar

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