In the classical mechanics the conjugate momenta was typically a derivative of the Lagrangian, i.e. \begin{equation} p_i=\frac{\partial L}{\partial \dot q_i}.\tag{1} \end{equation}

However, in the QFT and the string theory, the conjugate momentum was often associated to the derivative of the action.

Related post:

  1. Derivation of total momentum operator QFT
  2. What is the physical meaning of the canonical momentum field in quantum field theory?

where as in the string theory, \begin{equation} P_0^\mu =\frac{\delta S}{\delta \dot X_\mu}\tag{2} \end{equation} which integrate through the world sheet, and one supposed the similar construction (integrate the Lagrangian density over the $p+1$ hyperplane) happen to the $p$-branes?

This kind of make sense, since the integration acted like a "path integral expectation". In the classical theory, the action presented and was responsible for the derivation of the equation of the motion.

But this is the conjugate momenta, not the equation of the motion, shouldn't $P_0^\mu$ be derived from the Lagrangien density as a local operator, just like a specific momenta $p_i$ could be associated to a specific point?

In the dimensional analysis, classically $[p_i]\sim[kg\cdot m/s]$ where the lagrangian carried the unit of energy$[kg \cdot m^2/s^2]$. But the action was unitless, with $\int d\sigma \int d\tau$ each presented a unit of length, $\partial^2$ presented an inverse of the length, $X$ was unit less and $[\dot X ] \sim [1/s]$ . Clearly the dimension of those two conjugate momenta didn't match. (In the nature unit $[p_i]\sim[E]$ where $[1/s]\sim [E]$. Since $\delta \dot X_\mu$ was a derivative at the denominator, $P_0^\mu\sim[E^{-1}]$ )

What's the difference between the conjugate momenta in the classical and in the string qft?

  • $\begingroup$ Who writes eq. (2)? Which author? Which reference? Which page? $\endgroup$
    – Qmechanic
    Commented Sep 7, 2022 at 6:16
  • $\begingroup$ @Qmechanic String Theory and M-Theory: A Modern Introduction Eq. 2.70. It might be an abbreviation. $P_0^\mu =\frac{\delta S}{\delta \dot X_\mu} =T\dot X^\mu$ $\endgroup$ Commented Sep 7, 2022 at 8:00
  • 1
    $\begingroup$ Yes, the formally wrong eq. (2) occasionally appears in the literature, presumably because the authors' end results happen to be correct. $\endgroup$
    – Qmechanic
    Commented Sep 8, 2022 at 7:01

1 Answer 1

  1. In field theory with Lagrangian $$\begin{align}L&[q(\cdot,t),v(\cdot,t);t]\cr ~=~&\int \!d^nx~{\cal L}\left(q(x,t),\partial q(x,t),\partial^2q(x,t), \ldots;\right. \cr &\left. v(x,t),\partial v(x,t),\partial^2 v(x,t), \ldots; x,t\right),\end{align}\tag{A} $$ the canonical momentum is defined as $$ p(x,t) ~:=~\frac{\delta L[q(\cdot,t),v(\cdot,t);t]}{\delta v(x,t)}, \tag{B}$$ cf. my related Phys.SE answer here.

  2. If the Lagrangian density contains at most first order derivatives, eq. (B) simplifies to a partial derivative of the Lagrangian density $$ p(x,t) ~=~\frac{\partial{\cal L}\left(q(x,t),\partial q(x,t); v(x,t);x,t\right)}{\partial v(x,t)} .\tag{C}$$

  3. The action is given as $$S[q]~:=~\int_{t_i}^{t_f} \! dt~ \left. L[q(\cdot,t),v(\cdot,t);t]\right|_{v=\dot{q}}.\tag{D}$$

  4. It does not make sense mathematically to write$^1$ $$ p(x,t)~=~\frac{\delta S[q]}{\delta \dot{q}(x,t)}\qquad\qquad (\leftarrow \text{Wrong!}) \tag{E}$$ since $q$ and $\dot{q}$ are not independent variables in the action, cf. e.g. this Phys.SE post. So OP's eq. (2) is formally wrong.


$^1$ For what it's worth, eq. (E) is nevertheless dimensional consistent, cf. e.g. this Phys.SE post.


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