How can negative integers be used in deriving the Hamiltonian for open strings?

In summary, on page 38 of Becker Becker Schwarz, we are 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 can calculate our terms for $$\dot{X}$$ and $${X}^{'}$$ using the given expression for $$X^{\mu}(\tau,\sigma)$$ and the fact that $$\alpha^{\mu}_{0}=l_{s}p^{\mu}$$. Plugging these into the Hamiltonian gives an expression involving sums over integers. By equation 2.72,
  • #1
Overtoad123
6
3
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
 
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  • #3
Orodruin said:
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 !
 
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  • #4
Overtoad123 said:
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##.
 
  • #5
Overtoad123 said:
$${X}^{'}=-il_{s}\sum_{m\neq0}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$
OK. But note that you can include ##m = 0## in the summation since the summand is zero for ##m = 0##. So you can write
$${X}^{'}=-il_{s}\sum_{m\in\mathbb{Z}}\alpha^{\mu}_{m}e^{-im\tau}\sin(m\sigma)$$
 
  • #6
TSny said:
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 $m=-p$.
TSny said:
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.
 
  • #7
Overtoad123 said:
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.

Overtoad123 said:
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.
 
  • #8
TSny said:
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.
 
  • #9
TSny said:
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.
vanhees71 said:
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 :smile:! 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🤔
 
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  • #10
No, you simply don't need negative integers, and the integral gives with usual orthogonality conditions for the cosine.
 
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  • #11
vanhees71 said:
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
 
  • #12
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.
 
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  • #13
vanhees71 said:
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.
 
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Related to How can negative integers be used in deriving the Hamiltonian for open strings?

1. What is a Hamiltonian open string?

A Hamiltonian open string is a type of string theory that describes the behavior of open strings, which have two endpoints that can move freely in space. It uses the Hamiltonian formalism, a mathematical framework for describing the dynamics of a system, to study the motion and interactions of these strings.

2. How does a Hamiltonian open string differ from other types of string theories?

Unlike other string theories, which use the Lagrangian formalism, a Hamiltonian open string takes into account the energy of the string's endpoints and their interactions with other strings. This allows for a more precise description of the string's behavior and its interactions with other particles.

3. What is the significance of the Hamiltonian in open string theory?

The Hamiltonian plays a crucial role in open string theory as it represents the total energy of the string and governs its dynamics. It is derived from the string's Lagrangian and is used to calculate the string's equations of motion and interactions with other strings and particles.

4. Can Hamiltonian open string theory be applied to other areas of physics?

Yes, Hamiltonian open string theory has applications in various areas of physics, including quantum gravity, particle physics, and cosmology. It provides a framework for studying the behavior of strings and their interactions with other particles, which can help us better understand the fundamental laws of the universe.

5. What are some current developments in Hamiltonian open string theory?

There are ongoing developments in Hamiltonian open string theory, including efforts to incorporate it into a more comprehensive theory of quantum gravity. There is also research being done on the implications of this theory for black hole physics and the holographic principle. Additionally, there is ongoing work on the mathematical foundations of the theory and its applications to other areas of physics.

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