Apparently this result is used in many areas physics including the extra dimensions of string theory, which is not the scope of the question. The result is apparently also used to understand the Casimir force and there are many different types of experimental validation for this force. But how does this result of adding all the natural numbers and getting -1/12 accomplish this?

  • 4
    $\begingroup$ This particular sum is also discussed here, here and here, and on Math.SE here. See also this Phys.SE post. Also related Phys.SE post here, here and links therein. $\endgroup$ – Qmechanic Aug 6 '17 at 6:31

Actually the true equation is:

$$\int_0^\infty n dn - \sum_{n=1}^\infty n = \frac{1}{12}$$

Since the Casimir energy is the difference between the allowed wavelengths had the walls not been there and the allowed wavelengths with walls being there.

Actually the above sum is $\infty-\infty$ so you must regularise it using:

$$\lim_{\alpha\rightarrow 0}\left(\int_0^\infty ne^{-\alpha n} dn - \sum_{n=1}^\infty ne^{-\alpha n}\right) = \frac{1}{12}$$


In QED the energy of electromagnetic waves are quantized $E=(n+1/2)\hbar\omega$. When calculating the total energy of vacuum you need to add up them altogether, and this leads to the infinite integer summation in 1-d QED, if consider 3-d case the regulation results in $\zeta(-3)$. For detail description, please refer to: https://en.wikipedia.org/wiki/Casimir_effect; https://en.wikiversity.org/wiki/Quantum_mechanics/Casimir_effect_in_one_dimension.  

  • $\begingroup$ so were do we think that -1/12 is used ? -1/12 being the sum of all the natural numbers. Is it substituted into some otherwise not well behaved series ? or is it something else? $\endgroup$ – Sedumjoy Aug 6 '17 at 3:58
  • 2
    $\begingroup$ Sorry I didn't fully understand your comment? I mean, the -1/12 is the sum of natural numbers, which is calculated by Riemann zeta function $\zeta(-1)$. This is the 1-d box case. Actually the wiki discussed the 3-d case which shows as $\zeta(-3)=\sum_n n^3$ $\endgroup$ – Kangle Aug 6 '17 at 16:13
  • $\begingroup$ Yes...I see it now......the 2nd article explains it in detail....did not know this article even existed or I would not have asked the question....oh well next time I better look harder. sorry $\endgroup$ – Sedumjoy Aug 6 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.