Casimir effect

The sum of all natural numbers

One may see the following mysterious identity that the sum of all natural numbers equals to $-\frac{1}{12}$:   \begin{equation} 1 + 2 + 3 + \cdots = -\frac{1}{12}\,.\tag{1}\end{equation} The rigorous mathematical interpretation relies on the so-called Riemann zeta function $\zeta(s)$ and its analytic continuation. This is not the focus of this post, so we just highlight the main conclusions: \begin{equation}\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}\tag{2}\end{equation} is only valid when $\mathfrak{Re} (s) > 1$. After analytic continuation, we can compute $\zeta(-1)=-\frac{1}{12}$. Therefore, mathematically, Eq.(1) should be interpreted in the language of analytic continuation.

In this post, we explain Eq. (1) by applying a scheme that is commonly used in physics when treating a divergent series $\sum_{n=1}^{\infty}a_n$:

  1.  Multiply each term $a_n$ by some regulator such as $e^{-n\epsilon}$ to make the series converge. 
  2. Compute the result of $\sum_{n=1}^{\infty}a_ne^{-n\epsilon}$. 
  3. Take the limit of $\epsilon\rightarrow 0$ at the end. 
Therefore,  to solve $\sum_{n=1}^{\infty}n$, we compute \begin{equation}\sum_{n=1}^{\infty}n e^{-n\epsilon} = -\frac{d}{d\epsilon}\sum_{n=1}^{\infty} e^{-n\epsilon}= -\frac{d}{d\epsilon} \frac{1}{1-e^{-\epsilon}} = -\frac{d}{d\epsilon}\left[\frac{1}{\epsilon}+\frac{1}{2}+\frac{\epsilon}{12}+O(\epsilon^2)\right]=\frac{1}{\epsilon^2}-\frac{1}{12} + O(\epsilon^2)\,.\tag{3}\end{equation} Now let $\epsilon\rightarrow 0$, we have \begin{equation} 1 + 2 + 3 + \cdots = \infty -\frac{1}{12}\,,\tag{4} \end{equation} which leads to a much more intuitive interpretation of Eq.(1): of course, $1 + 2 + 3 + \cdots$ is infinite; Eq. (4) simply says the sum of all natural numbers equals to an infinity plus a constant $-\frac{1}{12}$, which is still infinity. 

Remarks:

  • Those who learned calculus are familiar with the division of two infinities, through which one can prove that $\ln n < n^a < a^n < n! < n^n$ as $n\rightarrow\infty$. In contrast, Eq. (4) is useful in physics when computing the minus between two infinities, in which $\frac{1}{\epsilon^2}$ often cancels with another infinity and only the constant $-\frac{1}{12}$ matters.
  • Multiplying any other regulator besides  $e^{-n\epsilon}$ gives the same constant $-\frac{1}{12}$. Interested readers can find the proof in Terry Tao's post or Chapter 15. 3 of M. Schwartz's QFT book
  • Note that the scheme we applied above is not mathematically rigorous: when taking the limit $\epsilon\rightarrow 0$ to get Eq. (4) from Eq. (3), we assume $\lim_{\epsilon\rightarrow 0}\sum_{n=1}^{\infty} n e^{-n\epsilon} = \sum_{n=1}^{\infty}\lim_{\epsilon\rightarrow 0} n e^{-n\epsilon}$, which is NOT true. However, in physics, the computations with such scheme often lead to results that perfectly agree with experiments. We will provide a detailed explanation in the next section. Here, the short answer is: the divergent series $\sum_{n=1}^{\infty}a_n$ itself is also "problematic" when used to model our real world. It looks like we make two wrongs in such a way that our final result is correct : )
As we see below, the constant $-\frac{1}{12}$ is the key to explain the Casimir effect in quantum physics.

Casimir effect in one dimension

Casimir effect refers to a small attractive force between two parallel uncharged conducting plates in the vacuum. A simple phenomenological explanation is from the vacuum energy: According to quantum physics, our vacuum is not empty but full of virtual electromagnetic waves with all possible frequencies.

Fig. 1  Casimir effect in one dimension

Consider the one dimensional example for simplicity. As shown in Fig. 1, the two plates are located at $x=0$ and $x=a$, which divides the space into three regions. On one hand, in region II between two conducting plates, there are only standing waves of discrete frequencies ($\hbar=c=1$) \begin{equation}\omega_n = n\frac{\pi}{a}\quad \text{for } n \in \mathbb{N}\,,\tag{5}\end{equation} and thus the vacuum energy in region II is \begin{equation} E(a) = \sum_{n=1}^{\infty}\omega_n \,.\tag{6}\end{equation} On the other hand, the electromagnetic fields in region III are of all continuous frequencies. Since the region III ($x > a$) is infinitely large, we imagine that there exists some boundary at $x = L$ where  $L\rightarrow \infty$, making the region of length $L-a$. As a result, the density of states is $\frac{L-a}{2\pi}$ and the vacuum energy in region III is \begin{equation} E(L-a) = \frac{L-a}{2\pi} \int_{-\infty}^{\infty} dk\,|k|= \frac{L-a}{\pi} \int_0^{\infty} d\omega \,\omega\,.\tag{7}\end{equation} The difference of vacuum energy between regions II and III causes a force on the plate at $x=a$. Such force can be computed by (Remember that $dE=TdS-pdV$) \begin{equation}F = -\frac{\partial }{\partial a}\left[ E(a) + E(L-a)\right] =  -\frac{\partial }{\partial a}E_{r}(a)\,,\tag{8}\end{equation} where we have \begin{equation} \boxed{E_r(a)\equiv \sum_{n=1}^{\infty}\omega_n -\frac{a}{\pi} \int_0^{\infty} d\omega \,\omega}\,. \tag{10} \end{equation} by dropping out the term in $E(a)+E(L-a)$ that contain $L$ but not $a$ under the derivative to $a$. 

To solve the Casimir force, we need to compute Eq. (10) which is a a real example from physics that involves the minus of two infinities. So we apply the scheme introduced in the previous section and compute the following quantity for Eq. (10): \begin{equation} E_r(a) = \lim_{\Lambda\rightarrow \infty} \left[\sum_{n=1}^{\infty}\omega_n\,e^{-\omega_n/\Lambda} -\frac{a}{\pi} \int_0^{\infty} d\omega \,\omega\,e^{-\omega/\Lambda}\right]\,.\tag{11}\end{equation} The first term in Eq. (10) is \begin{equation} \sum_{n=1}^{\infty}\omega_n\,e^{-\omega_n/\Lambda} =\frac{\pi}{a}\left[\sum_{n=1}^{\infty} n\, e^{-\frac{\pi}{a\Lambda}n}\right]=\frac{\pi}{a}\left[\left(\frac{a\Lambda}{\pi}\right)^2-\frac{1}{12}+O(1/\Lambda^2)\right] =\frac{a}{\pi}\Lambda^2 - \frac{\pi}{12a},\end{equation} where we use the result of Eq. (3) derived in the previous section. The second term is simply \begin{equation} \frac{a}{\pi} \int_0^{\infty} d\omega \,\omega\,e^{-\omega/\Lambda} =  \frac{a\Lambda^2}{\pi} \int_0^{\infty} dt \,t\,e^{-t}=\frac{a\Lambda^2}{\pi}\Gamma(2)=\frac{a\Lambda^2}{\pi}\,.\end{equation} Now we see explicitly that the infinity$\frac{a\Lambda^2}{\pi}$ cancels between two terms in Eq. (11), making $E_r(a)$ finite. $E_r(a)$ is mainly from the constant $-1/12$ when summing all natural numbers as discussed in the previous section. So we have \begin{equation} \boxed{E_r(a) = -\frac{\pi}{12a}}\,,\tag{12}\end{equation} and the force \begin{equation} F = -\frac{\pi}{12a^2}\,,\end{equation} where the negative sign indicates that it is an attractive force.

Why does the scheme work? Note that both the summation over $\omega_n$ in Eq. (6) and the integral over $\omega$ in Eq. (7) are from zero to infinity. However, in reality, each physics law is only valid within some energy scale $\Lambda$ and cannot describe the physics above the energy scale $\Lambda$. In Prof. A. Zee's word, "Ignorance is no shame" (in Chapter III.1 of his QFT book). So $\Lambda$ is a physics concept rather than a mathematical trick and the scheme in Eq. (11) is to suppress the contributions of high frequencies above which our theory is not applicable. Alternatively, we can also directly exclude the high frequencies above $\Lambda$ by a hard cutoff: \begin{equation} E_r(a) =\lim_{\Lambda\rightarrow\infty}\left[\sum_{n=1}^{\lfloor a\Lambda/\pi \rfloor} \omega_n -\frac{a}{\pi} \int_0^{\Lambda} d\omega \,\omega\right]\,.\end{equation} It yields the same result as Eq. (12), as shown in Chapter 15.2 of M. Schwartz's QFT book.

With the physics concept of $\Lambda$, we can expand the total vacuum energy as $E(a)=\gamma_2\Lambda^2 + \gamma_1\Lambda + \gamma_0 + O(1/\Lambda)$ since $E\sim\int d\omega \omega\sim\omega^2$. By dimension analysis, we have $\gamma_2\sim a^1$, $\gamma_1\sim a^0$ and $\gamma_0\sim a^{-1}$. The first term does not contribute to the Casimir force: when the region II shifts by $dx$, region III always shifts by $-dx$. The second term vanishes when taking derivative to $a$. Only the third term survives, leading to $F\sim \frac{1}{a^2}$. This can be viewed as a physics proof that the Casimir effect does not reply on $\Lambda$: using any other regulator or hard cutoff should give the same result. 

Miscellaneous:
  • Eq. (10) is also the difference of vacuum energy only in region II before and after placing the plates. Casimir's original paper in 1948 started directly from this difference in real three dimensional case. It is still the total vacuum energy of the system but shifted by a constant of the normal vacuum energy in free space. In quantum field theory jargon, the constant for the shift is counterterm and the vacuum energy before and after shift is bare and renormalized.
  • All the above discussions are the hidden context behind the native substitution of Eq. (1) to Eq. (6) in some physics literature.

Casimir effect in real world

Vacuum energy is a phenomenological explanation to Casimir effect. On the contrary, Casimir effect only proves the existence of fluctuations of virtual electromagnetic fields in the vacuum rather than the reality of vacuum energy.

Note that the standing waves come from the boundary condition that the electric fields vanish at the conducting plates. This is achieved by the response of electrons in the conducting plates to the virtual electromagnetic fields. Therefore, the electron-photon interaction strength, which is the fine structure constant $\alpha$, must appear in the result of Casimir effect. If $\alpha$ is zero, there should be no Casimir force. Indeed, Prof. Jaffe's paper shows that the standard result of Casimir force computed from vacuum energy is the $\alpha\rightarrow\infty$ limit. Casimir force is simply the relativistic retarded Vander Waals force.

Comments

  1. analytic continuation means "解析延拓“?
    ”there are only standing waves of discrete frequencies“ in region II:
    two plates set as electromagnetic boundary condition with zero magnitude, then it is purely electromagnetic feature to show static wave ,right? You mentioned, due to quantum physics, vacuum has such electromagnetic field in region II.

    Nice content and nice paper quality formatting!

    ReplyDelete

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