# The string Poisson bracket

Where does the factor $\frac{1}{T}$ ($T$ is the string tension) in this Poisson bracket come from?

$$\{X^{\mu}(\tau,\sigma),\dot{X}^{\nu}(\tau,\sigma')\} ~=~ \frac{1}{T}\delta(\sigma-\sigma')\eta_{\mu\nu}.$$

I think I can see from remembering the definition of a Poisson bracket (for example in canonical coordinates) why in terms of momentum we have

$$\{P^{\mu}(\tau,\sigma),X^{\nu}(\tau,\sigma')\} ~=~ \delta(\sigma-\sigma')\eta_{\mu\nu}$$

but I don't see why this factor in the first equation has to be there.

In addition to deriving it by calculation, is there an intuitive physical way how one can see why the factor of inverse tension has to be there, similar to explaining the appearance of the tension in front of the integral in the action by the fact that it costs energy to stretch the world-sheet?

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I somehow feel very stupid that I dont see this :-/ – Dilaton Mar 15 '13 at 12:06
Dimensional analysis? – Michael Brown Mar 15 '13 at 12:17
@MichaelBrown Maybe ... but I am not too fond of purely dimensional arguments and nothing else ... ;-) – Dilaton Mar 15 '13 at 12:26
You might want to also mention what the context of this question is. – joshphysics Mar 15 '13 at 15:58
I put the string-theory tag on it as a start. – David Z Mar 15 '13 at 17:17

It comes from the normalization of the Polyakov action, \begin{align*} S=\frac{T}{2}\int d^2\sigma\, (\dot{X}^\mu\dot{X}_\mu-{X'}^\mu{X'}_\mu). \end{align*} The canonical momentum is \begin{align*} \frac{\partial \mathcal{L}}{\partial \dot{X}^\mu}=T\dot{X}_\mu, \end{align*} and this gives the equal time commutator (or Poisson bracket) that you wrote down, \begin{align*} [X^\mu(\tau,\sigma),\dot{X}^\nu(\tau,\sigma')]=\frac{i}{T}\eta^{\mu\nu}\delta(\sigma-\sigma'). \end{align*} The way I think about this is that as the tension of the string goes to zero, the string becomes less and less classical (alternatively, $T$ plays the role of $1/\hbar$ on the worldsheet).

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Matthew Dodelson has already given the main reason in his answer: The string action $S$ (either the Nambu-Goto or the Polyakov action) is proportional to the string tension $T_0$. So the canonical momentum density

$$\tag{\star} {\cal P}^{\tau}_I ~:=~\frac{\partial {\cal L}}{\partial \dot{X}^{I}}~=~\frac{T_0}{c^2} \eta_{IJ}\dot{X}^{J}$$

becomes proportional to the string tension $T_0$ as well. (The latter equality in ($\star$) is only true with further assumptions. See e.g. B. Zwiebach, A first course in string theory, for details.)

Finally, let us restore the correct factors of speed of light $c$ in the Poisson brackets:

$$\tag{\star\star}\{X^{I}(\tau,\sigma),\dot{X}^{J}(\tau,\sigma')\}_{PB} ~=~ \frac{c^2}{T_0}\delta(\sigma-\sigma')\eta^{IJ}.$$

Both sides of ($\star\star$) have now dimension of inverse mass $M^{-1}$. This is a consequence of the following dimensional analysis:

$$[\tau]~=~ \text{dim. of time} ~=:~T,$$

$$[\sigma] ~=~ \text{dim. of length} ~=:~L~=~ [X],$$

$$[c]~=~ \text{dim. of speed} ~=~\frac{L}{T},$$

$$[T_0] ~=~ \text{dim. of force} ~=~\frac{ML}{T^2},$$

$$[\text{Poisson bracket}] ~=~[\{\cdot, \cdot\}_{PB}] ~=~ \frac{1}{\text{dim. of angular momentum}}~=~\frac{T}{ML^2}.$$

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Thanks Qmechanic, I like these additional explanations. – Dilaton Mar 16 '13 at 13:57