# Hamiltonian open string

Homework Statement
Trying to derive the open string Hamiltonian given as ##H=\frac{1}{2}\sum_{n\in\mathbb{Z}}\alpha_{-n}\cdot\alpha_{n}(2.72)## (Becker Becker Schwartz; string theory) using the solution for the open string
##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)##
Relevant Equations
My Hamiltonian
##H=\frac{T}{2}\int_{0}^{\pi}(\dot{X}^{2}+X^{'2}). \tag{2.69} ##

And my open string solution as
##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
##\dot{X}=l_{s}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}e^{-im\tau}\cos(m\sigma)## Is derivative with respect to ##\tau## and
##{X}^{'}=-il_{s}\sum_{m\neq0}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)## with respect to ##\sigma##
On ***page 38*** of Becker Becker Schwarz, we're given ***equation 2.69*** which is the Hamiltonian for a string given as $$H=\frac{T}{2}\int_{0}^{\pi}(\dot{X}^{2}+X^{'2})$$

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 I am 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
$$H=\frac{1}{2}\sum_{n\in\mathbb{Z}}\alpha_{-n}\cdot\alpha_{n}(2.62)$$
The issue that I am 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 I am 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 I am not sure what went wrong here whether it was my mistake in doing m=-p or evaluating my integrals incorrect which I don't think is it the case.

Also the preview section is not working so I’m not sure how my equations looked or not unfortunately. I’ve tried different browsers and my phone to try and use the preview function but it didn’t work

Last edited:

You need to use appropriate delimiters for LaTeX code to render. See https://www.physicsforums.com/help/latexhelp/
thank you ! :) I’ve updated my post and it looks readable now !

• berkeman
If I am 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
$$H=\frac{T}{2}l^{2}_{s}\sum_{m,n\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}-\frac{T}{2}l^{s}_{2}\sum_{m,p\neq 0}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2}$$
While waiting for someone who actually knows this stuff to chime in, I will just note that there appears to be a mistake in the above where it looks like you let ##\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma = \int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma = \frac{\pi}{2}##

Note that $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)$$ $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} - \delta_{m,-p} \right)$$

--------------------------------------------

In your first summation symbol in ##H##, you have a typographical error where the index ##n## should be ##p##.

$${X}^{'}=-il_{s}\sum_{m\neq0}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$
OK. But note that you can include ##m = 0## in the summation since the summand is zero for ##m = 0##. So you can write
$${X}^{'}=-il_{s}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$

While waiting for someone who actually knows this stuff to chime in, I will just note that there appears to be a mistake in the above where it looks like you let ##\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma = \int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma = \frac{\pi}{2}##

Note that $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)$$ $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} - \delta_{m,-p} \right)$$

--------------------------------------------

In your first summation symbol in ##H##, you have a typographical error where the index ##n## should be ##p##.
This is really insightful as I was not aware you can express these orthogonality relationships in such way. I have normally been taught that $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ but considering that cosine is even then it’s also fair to to include the last delta function $delta_{m,-p}$ if I am thinking about it correctly right ? Taking into account these relations then I think this is the correct way to include my negative frequency modes as opposed to doing the incorrect redefining with $m=-p$.
OK. But note that you can include ##m = 0## in the summation since the summand is zero for ##m = 0##. So you can write
$${X}^{'}=-il_{s}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$ Yea this makes sense !Allowing my sum to include all integers is valid as running m=0 will kill off the term not changing my definition for $X^{‘}$.

I believe these tips will give me the correct expression now ! they’ve been really helpful, I’ll post my updated expression later on today as I believe I’m on a much better path now !

One quick question though and I’m sorry if this is extremely silly but why are we normally taught(at least I was) and emphasized in some sources the following statement $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ as opposed to $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} +\delta_{m,-p} \right)$$ ? The latter feels more of a general case, why isn’t this relation emphasized more then consider it also takes into account that cosine is an even function.

I believe these tips will give me the correct expression now ! they’ve been really helpful, I’ll post my updated expression later on today as I believe I’m on a much better path now !
OK, good.

One quick question though and I’m sorry if this is extremely silly but why are we normally taught(at least I was) and emphasized in some sources the following statement $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ as opposed to $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} +\delta_{m,-p} \right)$$ ? The latter feels more of a general case, why isn’t this relation emphasized more then consider it also takes into account that cosine is an even function.
I guess it's because we often see $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ in the context of Fourier series where ##m## and ##p## are non-negative integers.

While waiting for someone who actually knows this stuff to chime in, I will just note that there appears to be a mistake in the above where it looks like you let ##\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma = \int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma = \frac{\pi}{2}##

Note that $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)$$ $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} - \delta_{m,-p} \right)$$

--------------------------------------------

In your first summation symbol in ##H##, you have a typographical error where the index ##n## should be ##p##.
Note however that for real Fourier series in terms of ##\cos## and ##\sin## only positive integers ##n## and ##p## (and ##0## for the cos) already make a complete set.

OK, good.

I guess it's because we often see $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ in the context of Fourier series where ##m## and ##p## are non-negative integers.
So I get the following now $$H=\frac{T}{2}\int_{0}^{\pi}(\dot{X}^{2}+X^{'2})=H=\frac{T}{2}l^{2}_{s}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma-\frac{T}{2}l^{s}_{2}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma$$ Now using the relations that $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)$$ and $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} - \delta_{m,-p} \right)$$ we get $$\frac{T}{2}l^{2}_{s}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)-\frac{T}{2}l^{s}_{2}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\frac{\pi}{2} \left (\delta_{m,p} - \delta_{m,-p} \right)$$ The surviving terms are then $$\frac{T}{2}l^{2}_{s}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\pi\delta_{m-p}$$ which would give my expected Hamiltonian $$H=\frac{1}{2}\sum_{n\in\mathbb{Z}}\alpha_{-n}\cdot\alpha_{n}(2.62)$$ when I plug in what T is along with evaluating my krocker delta.
Note however that for real Fourier series in terms of ##\cos## and ##\sin## only positive integers ##n## and ##p## (and ##0## for the cos) already make a complete set.
Hello thanks for looking at my post I appreciate it ! so you're saying that in order to include negative integers for $n$ and $p$ then I must look at the complex representation of cosine and sine since $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ already forms a solution as opposed to $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} +\delta_{m,-p} \right)$$. The latter makes sense to me since we know that cosine is an even function allowing us to include the last term at least that's my understanding Last edited:
No, you simply don't need negative integers, and the integral gives with usual orthogonality conditions for the cosine.

• malawi_glenn
No, you simply don't need negative integers, and the integral gives with usual orthogonality conditions for the cosine.
So in other words I only need $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ since I have a real Fourier series ?? But if i don’t incorporate negative integers through integral then how would I go about introducing negative integers in my sum ?? Sorry if I’m a bit confused by the way at the moment I’m not sure why I wouldn’t need the negative

I don't know the context, but what do you need negative integers for? The functions ##\cos(m \sigma)## with ##m \in \mathbb{N}_0## already form a complete set.

• malawi_glenn
I don't know the context, but what do you need negative integers for? The functions ##\cos(m \sigma)## with ##m \in \mathbb{N}_0## already form a complete set.
I need negative integers in order to derive my Hamiltonian for the open string which is given as $$H=\frac{1}{2}\sum_{n\in\mathbb{Z}}\alpha_{-n}\cdot\alpha_{n}(2.62)$$

Altough TSny pointed out that since my sums in $$H=\frac{T}{2}\int_{0}^{\pi}(\dot{X}^{2}+X^{'2})=H=\frac{T}{2}l^{2}_{s}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma-\frac{T}{2}l^{s}_{2}\sum_{m,p\in\mathbb{Z}}\alpha_{m}\cdot\alpha_{p}e^{-i(m+p)\tau}\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma$$ run through negative integers as well then I should really consider $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} + \delta_{m,-p} \right)$$ $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \left( \delta_{m,p} - \delta_{m,-p} \right)$$ since these take care of the cases when I am summing over negative integers in my Hamiltonian above. I was originally only considering $$\int_{0}^{\pi}\cos(m\sigma)\cos(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ $$\int_{0}^{\pi}\sin(m\sigma)\sin(p\sigma)d\sigma =\frac{\pi}{2} \delta_{m,p}$$ which gave me the wrong answers as I was neglecting the case where one of the integers was negative and the other was positive.

• TSny and malawi_glenn