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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 $?

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Related: physics.stackexchange.com/q/4994 –  twistor59 May 29 '13 at 14:46
@twistor59 I have seen that, but it doesn't seem to answer my question –  smiley06 May 29 '13 at 14:49
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. –  Qmechanic May 29 '13 at 15:09
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. –  Prahar May 29 '13 at 16:12
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) ? –  smiley06 May 30 '13 at 7:48
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1 Answer

up vote 3 down vote accepted

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 $B_2 = \frac{1}{6}$

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Thank you, this is perfect as far as this question is concerned. –  smiley06 May 30 '13 at 19:11
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 ? –  smiley06 May 30 '13 at 19:17
Am I mistaken somewhere above in the derivations ? –  smiley06 May 30 '13 at 19:25
First, I made an error on the last line, it is $f(0) =0$ and not $B_1= 0$ !!! See the edit. –  Trimok May 31 '13 at 9:20
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. –  Trimok May 31 '13 at 9:29
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