# Hamiltonian Open String

On page 38 of Becker Becker Schwarz, we're given equation 2.69 which is the Hamiltonian for a string given as $$H=\frac{T}{2}\int_{0}^{\pi}(\dot{X}^{2}+X^{'2}).\tag{2.69}$$ Considering the open string we have $$X^{\mu}(\tau,\sigma)=x^{\mu}+l^{2}_{s}p^{\mu}\tau+il_{s}\sum_{m\neq0}\frac{1}{m}\alpha^{\mu}_{m}e^{-im\tau}\cos(m\sigma)$$ where we can calculate our terms $$\dot{X}=l_{s}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}e^{-im\tau}\cos(m\sigma)$$ and $${X}^{'}=-il_{s}\sum_{m\neq0}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$ remembering that $$\alpha^{\mu}_{0}=l_{s}p^{\mu}$$. If im correct, plugging our expressions into our Hamiltonian gives us $$H=\frac{T}{2}l^{2}_{s}\sum_{m,n\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma-\frac{T}{2}l^{s}_{2}\sum_{m,p\neq 0}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma$$ Evaluating our integrals gives us $$H=\frac{T}{2}l^{2}_{s}\sum_{m,n\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}-\frac{T}{2}l^{s}_{2}\sum_{m,p\neq 0}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}$$ By equation 2.72 I know that I should get $$H=\frac{1}{2}\sum_{n\in\mathbb{Z}}\alpha_{-n}\cdot\alpha_{n}..\tag{2.72}$$ The issue that Im stuck on is based on my equation that I found $$H=\frac{T}{2}l^{2}_{s}\sum_{m,n\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}-\frac{T}{2}l^{s}_{2}\sum_{m,p\neq 0}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}$$ I can use $$m=-p$$ I think to get $$H=\frac{T}{2}l^{2}_{s}\sum_{p\in\mathbb{Z}}\alpha_{-p}\cdot\alpha_{p}\frac{\pi}{2}-\frac{T}{2}l^{s}_{2}\sum_{p\neq 0}\alpha_{-p}\cdot\alpha_{p}\frac{\pi}{2}$$ but im not sure how to get equation 2.72 from here. In addition if I write out my sums the only term that survives is the m=0 terms im not sure what went wrong here whether it was my mistake in doing $$m=-p$$ or evaluating my integrals incorrect which I dont think is it the case.

• Hi and welcome to PSE. What is the variable(s) that we are integrating with respect to in (2.69)? I haven't done the calculation, but it seems to me that an integration wrt $\tau$ would solve most of your problems... Jul 23, 2022 at 12:17
• @schris38 hello !! I’m integrating over \sigma so I only have my cos(m\sigma)cos(n\sigma) and sin(m\sigma)sin(n\sigma) being integrated. I’m not sure how I would integrate over \tau though in this case Jul 23, 2022 at 13:43

The orthogonality relations for sine and cosine functions should be proportional to the Kronecker $$\delta$$ symbol. Namely $$\int d\sigma \cos(m\sigma)\cos(n\sigma)\sim \delta_{mn}$$ $$\int d\sigma \sin(m\sigma)\sin(n\sigma)\sim \delta_{mn}$$ with the appropriate proportionality constants. This is where your mistake is located I think. This would kill one of the sums and identify the indices of the Fourier coefficients.
• Hello !! Thank you, I appreciate your comment and advice. So I’ve plugged in the orthogonality relations and as you said, it kills off one of the sums resulting in let’s say the first term looking like $$\sum_{m\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{m}e^{-2im\tau}$$ but now from here I’m unsure how to proceed to get my first alpha as a negative mode and combine with the second sum $$\sum_{m\neq 0}\alpha_{m}\cdot\alpha_{m}e^{-2im\tau}$$ Jul 23, 2022 at 15:09
• Normally you should multiply $\dot{X}$ with its complex conjugate and the same goes for $X'$. However, I think both $\dot{X}$ and $X'$ are thought to be real and hence the respective creation operator (or c.c. of the Fourier coefficient) one would found in the mode expansion of the complex conjugates of $X$, can be replaced by $\alpha_{-n}$. The complex exponentials accompanying the respective creation operator contain the opposite sign in their exponents. This is another hint for deriving the expression Jul 23, 2022 at 16:00
• I see how this would work, but it brings up an issue in determining when I should use complex conjugate or not. I do agree with your integral relation between the sines and cosines but I feel like this should be derivable using your integral relation on its own. I’m not sure how to incorporate the negative modes without using a substitution in my sum like $m=-p$ but I don’t think that’s a valid substitution considering it gives me the wrong answer. Jul 24, 2022 at 16:11
• It is a valid substitution. Any substitution is valid. But, I think, due to the fact that the Lagrangian must be real in any case and on top of that you require the field also be real in the case of open strings, it is okay to use the $a_m^{\dagger}=a_{-m}$ suggestion. It is valid Jul 25, 2022 at 6:49
• Anything you do is legit, as long as you do not violate any mathematical rule (i.e. substituting the integral of the cosines with an expression that does not involve the Kronecker $\delta$), but in the end, you want to bring the expression in the form of the book's equation (without meaning that anything else is not correct)... But this is the form the authors want to exhibit Jul 25, 2022 at 6:57