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My question is about the BRST quantization of a point particle in Polchinski, Vol.1, Section 4.2.

The BRST quantization starts from the effective action for the gauge-fixed path-integral. But for the world-line approach of quantizing relativistic point particles, there is a problem.

There are no relativistic position operators, and therefore there are no position eigenstates $|x\rangle$ for relativistic point particles. This is one of the reasons why people abandoned the "relativistic quantum mechanics" and drifted towards quantizing field theory instead.

Considering this problem, the path-integral of a relativistic point particle does not make sense because $\langle t,x|t',x'\rangle$ is not defined. Does it make any physical sense to talk about the BRST quantization of a relativistic point particle? Is the section of the BRST quantization of a relativistic point particle in Polchinski merely a heuristic example for pedagogical purposes?

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  • $\begingroup$ Where in the corresponding section of Polchinski do you think the existence of position eigenstates is needed? Please include the relevant parts into the question so that one may answer the question without opening Polchinski. Note also that the $X^\mu(\tau)$ coordinates of the relativistic point particle are not position operators in the sense that is prohibited by relativity. $\endgroup$ – ACuriousMind Feb 5 '17 at 3:29
  • $\begingroup$ Can you elaborate the last sentence of your comment? $\endgroup$ – Xiaoyi Jing Feb 5 '17 at 3:38
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    $\begingroup$ The theory "of the relativistic point particle" doesn't know that it is a relativistic point particle. Purely formally, it's a theory of 4 scalar fields $X^\mu$ living on a 1-dimensional spacetime parameterized by $\tau$. There are certain assumptions that go into the no-go theorems for relativistic position operators that these scalar fields simply don't fulfill (and they don't lend themselves to being interpreted as actual position operators, consequently). As I said, please edit your question to pinpoint the exact step in the BRST procedure that you think fails. $\endgroup$ – ACuriousMind Feb 5 '17 at 3:45
  • $\begingroup$ I already said it in my post. I thought that there is no way to define $|X^{\mu}>$ and therefore I thought that it does not make sense to talk about the BRST quantization of a relativistic point particle. $\endgroup$ – Xiaoyi Jing Feb 5 '17 at 3:49
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You are not right. There is no orthogonal basis $|x,t\rangle$ for fixed $t$ in relativistic quantum mechanics, but there is a state $|x,t\rangle=|x^{\mu}\rangle$. In quantum field theory this state is expressed by $\phi(x)|0\rangle$, but this is a one-particle state, so can be expressed in the one-particle space, spanned by the momenta eigenstates $|\vec{p}\rangle$.

$$ |x^{\mu}\rangle=(2\pi)^{-3/2}\int \frac{d^{3}\vec{p}}{\sqrt{2E_p}}e^{ip_{\mu}x^{\mu}}|\vec{p}\rangle $$

The main point is that the state $|x\rangle$ does not imply the existence of a hermitian operator $x$, so this state can exist without the existence of $x$-operator. In fact, this operator can't exist since the Hamiltonian is bounded from bellow. You can see more about it here.

If the state $|x^{\mu}\rangle$ exist, makes perfect sense to have path integrals over trajectories $x_{\mu}(\tau)$, if we gauge away the parametrization at the end.

$$ \langle x_2^{\mu}|x_1^{\mu}\rangle=\int_{x(0)=x_1}^{x(1)=x_2}\frac{\left[dx(\tau)\right]}{\mathbb{diff}}\,e^{-S[x]} $$

What you will discover from that path integral is that $\langle x_2|x_1\rangle \neq 0$ even for space-like intervals, and by Lorentz symmetry you may deduce that the same happens for hyper-surface of simultaneity. This means that we cannot take $|x\rangle \rightarrow |\vec{x},t\rangle$ as an orthogonal basis for $t$ fixed.

This is why there is no wave function $\psi(x)$ description for the states. For the momentum wave-function there will be no problem, and a wave-function $\phi(p)$ is totally acceptable. For instance, $P$ is a hermitian operator for the one-particle Hilbert space in the free theory (free means no mixing with multi-particles states).

Actually the BRST-approach is far from be just pedagogical. You can obtain the field theory description by defining a field operator $\Psi(x^{\mu})$ and impose the equation of motion:

$$ Q\Psi=0 $$

This is useful to obtain non-linear generalization (i.e. how to put interactions) for particles in the presence of supersymmetry and/or gauge symmetry. The gauge transformations of the field will be described by the generalization of the BRST-exact fields $\delta\Psi=Q\Lambda+g\left[\Psi,\Lambda\right]$ and the equation of motion will be:

$$ Q\Psi + g\Psi^2=0\,. $$

You can see more about it here.

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  • $\begingroup$ I've seen some books saying that the local states cannot be defined in relativistic quantum mechanics. Theory of Group Representations and Applications by A.O. Barut As far as I could remember, it's from induced representation theory of Poincare algebra. Perhaps you are not right. $\endgroup$ – Libertarian Monarchist Bot Feb 28 '18 at 3:26
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    $\begingroup$ You cannot make state localized in the sense of $|x,t\rangle$ and $|y,t\rangle$ being orthogonal each other, i.e. localize a particle in a single point in space. But there is nothing prohibiting the existence of state $|x^{\mu}\rangle$ that have the right Poincare transformation. Actually this state is unique! $\endgroup$ – Nogueira Feb 28 '18 at 3:37
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    $\begingroup$ You can check that this is the state $\phi(x)|0\rangle$, but you can also work out the $p$ basis and construct the state $$ |x^{\mu}\rangle=\int \frac{d^{3}\vec{p}}{\sqrt{2E_p}}e^{ip_{\mu}x^{\mu}}|\vec{p}\rangle $$ $\endgroup$ – Nogueira Feb 28 '18 at 3:39
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    $\begingroup$ I think that you are not understanding what they are saying,. Localized states are not $|x^{\mu}\rangle$, are states that are eigenstates of $\vec{x}$, and this two things are completely different. $\endgroup$ – Nogueira Feb 28 '18 at 3:43
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    $\begingroup$ Just try to understand the state that I wrote. Do boost, translation on this state and check that this state behaves under this transformations as $x^{\mu}$. Again, this is not a localized state. $\endgroup$ – Nogueira Feb 28 '18 at 3:45

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