Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,549 answers , 19,357 comments
1,470 users with positive rep
410 active unimported users
More ...

Critical Dimension of Bosonic Strings and Regularization of $\sum_{n=1}^\infty n$

+ 3 like - 0 dislike
21 views

If $D$ is critical dimension of Bosonic strings, a particular derivation goes like the following, where we arrive finally at $$ \frac{D-2}{2}\sum_{n=1}^\infty n + 1 = 0. $$ Now mathematically this is clearly a divergent series, but using zeta function regularization here we are taking $$ \sum_{n=1}^\infty n = \zeta(-1) = -\frac{1}{12}. $$ And obtain $ D = 26 $ where $\zeta $ is the analytic continuation of the zeta function we know. But it makes no sense in putting $ s = -1 $ in the formulae $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. $$ As the above is only valid for $ Re(s) > 1 $. So what is going on in here? Can anyone give me a reasonable explanation about obtaining $ -1/12 $?

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06
asked May 29, 2013 in Theoretical Physics by smiley06 (40 points) [ no revision ]
retagged Apr 19, 2014 by dimension10
Related: physics.stackexchange.com/q/4994

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user twistor59
@twistor59 I have seen that, but it doesn't seem to answer my question

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06
This particular sum is also discussed here and here, and on Math.SE here. See also this Phys.SE post. Also related Phys.SE post here.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Qmechanic
There is a mention of this sum and its regularization in page 39 here - damtp.cam.ac.uk/user/tong/string/two.pdf. The statement is: Quantization of the string action requires going to $d=2+\epsilon$, to get rid of the ultraviolet divergences. This introduces a cosmological constant term in the action. However, this breaks Weyl invariance. In order to restore this Weyl invariance, one must introduce a counterterm. It is this counterterm that precisely cancels the divergence in the sum above. I have never worked this out myself, so if my understanding is wrong, please correct me.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Prahar
I understand the mathematical part saying $\zeta(-1) = -1/12 $ but still I am pretty much in the dark about getting rid of such infinite terms in context of physics, can anyone provide some details perhaps in an answer so that a student of mathematical background with a knowledge of basic QM and QFT can understand (if it is possible of course to give such an explanation) ?

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06

1 Answer

+ 3 like - 0 dislike

A way to do this is using regularization by substracting a continuous integral, ,with the help of the Euler-MacLaurin formula:

You can write :

$$ \sum_{Regularized} =(\sum_{n=0}^{+\infty}f(n) - \int_0^{+\infty} f(t) \,dt) = \frac{1}{2}(f(\infty) + f(0)) + \sum_{k=1}^{+\infty} \frac{B_k}{k!} (f^{(k - 1)} (\infty) - f^{(k - 1)} (0))$$ where $B_k$ are the Bernoulli numbers.

With the function $f(t) = te^{-\epsilon t}$, with $\epsilon > 0$, you have $f^{(k)}(\infty) = 0$ and $f(0) = 0$, so with the limit $\epsilon \rightarrow 0$, you will find : $$\sum_{Regularized} = - \frac{B_1}{1!} f (0) - \frac{B_2}{2!} f' (0) = - \frac{1}{12}$$

because $f(0) = 0$ and $B_2 = \frac{1}{6}$

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok
answered May 30, 2013 by Trimok (950 points) [ no revision ]
Most voted comments show all comments
Thank you, this is perfect as far as this question is concerned.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06
But we have the Hamiltonian as $$ H = \frac{1}{2}\sum_{n=1}^\infty \alpha_{-n}\alpha_n $$ And then replacing $\alpha_n$ with terms having $\sqrt{n} $ we arrive at $$ \frac{D-2}{2}\sum_{n=1}^\infty n $$ Now it seems to be the normal $\sum_n $ and not $\sum_{Regularised} $ in the expression of Hamiltonian because that is further obtained by using normal Fourier series $$ X^\mu(\tau,\sigma) = X^\mu_{COM} + \sum_{m\neq 0} \frac{1}{m}\alpha^\mu_m e^{-im\tau}\cos(m\sigma) $$ in the expression $$ H = \frac{T}{2}\int_0^\pi (\dot{X}^2+ X'^2)d\sigma $$ So how do we have $ \sum_{Regularised} $ above ?

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06
Am I mistaken somewhere above in the derivations ?

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user smiley06
First, I made an error on the last line, it is $f(0) =0$ and not $B_1= 0$ !!! See the edit.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok
You cannot escape some kind of regularization. The best you can do, is to find a acceptable physical explanation, for this regularization. The same kind of calculus is used, for instance, in the 1-dimensional Casimir effect. The modes between the plates are quantized, and they are not quantized outside the plates, so, to have the net result, you must make the difference of the continous modes and the discrete modes. So you can have a kind of analogy here, sometimes people speak of Casimir Energy even in the string domain.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok
Maybe a way to see that, is to consider a kind of infinite string, where only a finite part (a segment) is concerned by discrete modes. If you accept this, it will be an analogy with the 1-dimensional Casimir effect. Of course, we have to be cautious about the limit of these analogies, but I have nothing more clever. Just an additional remark, fermionic and bosonic calculus are different, in original Casimir effect, as in string theory.

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok
I have made an other edit, because this was not rigourous enough. Sorry

This post imported from StackExchange Physics at 2014-03-26 12:48 (UCT), posted by SE-user Trimok

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...