# Usage of Lagrangian not Lagrangian density for quantum string in Polchinski

In section 1.3 p. 18 while working out the energy spectrum of quantum string Polchinski uses Lagrangian instead of Lagrangian density while in QFT the usual convention is to use Lagrangian density (to keep the covariance intact). With the convention used there, calculations are done using expressions that are integral of spatial coordinate $$\int d \sigma$$. Is there any reason to choose this odd way of solving the field-action-variation problem?

This same problem is done using the usual density route in Tong's note and Becker book.

• He's quantizing the string using canonical quantization. When we canonically quantize a theory we work with the Lagrangian/Hamiltonian NOT the densities (even in QFT). Commented May 2, 2021 at 10:17
• @PraharMitra maybe because of limited exposure to QFT I haven't seen Lagrangian/Hamiltonian play a central role in canonical quantization. Everything is done using $\phi(x,t)$ and its conjugate momentum. Commented May 3, 2021 at 17:29
• I'm not sure how limited your exposure to QFT is, but its the very first thing we learn in QFT. See for instance chapter 2 of peskin, schroeder (chap 1 is a historical introduction so chap 2 is the first real discussion of QFT). Already in that chapter, you will see how he works with the Hamiltonian to canonically quantize the theory. Commented May 3, 2021 at 17:33
• You said you have studied the conjugate momentum, but even that is defined as the derivative of the Lagrangian (NOT Lagrangian density) with respect to ${\dot \phi}$ so even here one has to work with a Lagrangian, not the density. Commented May 3, 2021 at 17:34
• Just to be clear, what I'm saying is that while its true that generally in QFT one works with the Lagrangian density, when doing canonical quantization or deriving the conjugate momentum then one has to work with the Lagrangian and Hamiltonian, NOT the densities. Commented May 3, 2021 at 17:38

• The Lagrangian density $${\cal L}$$,

• the Lagrangian $$L=\int_0^{\ell}\!d\sigma~ {\cal L}$$,

• and the action $$S=\int\!d\tau~ L$$

contain the same information, and one is in principle free to use any of them. The reason why Polchinski chooses the Lagrangian $$L$$ in eq. (1.3.11) (which is integrated over $$\sigma$$) is because it is more convenient when he (in the next step) wants to separate the string $$X^-(\tau,\sigma)$$ into a mean value

$$x^-(\tau)~:=~\frac{1}{\ell}\int_{0}^{\ell}\! d\sigma~ X^-(\tau,\sigma)\tag{1.3.12a}$$

(which is integrated over $$\sigma$$), and the rest

$$Y^-(\tau,\sigma)~:=~X^-(\tau,\sigma)-x^-(\tau). \tag{1.3.12b}$$