1$, we have screening; if $\varepsilon<1$ --- antiscreening (asymptotic freedom). In the case of an ordinary matter, its dielectric and magnetic properties are independent. But vacuum should be Lorentz-invariant. Signals should propagate with velocity 1: \begin{equation} \varepsilon \mu = 1\,. \label{Dielmag:Lorentz} \end{equation} Therefore, diamagnetic vacuum ($\mu<1$) means screening, and paramagnetic one ($\mu>1$) --- asymptotic freedom. When we switch magnetic field $B$ on, the vacuum energy changes by \begin{equation} \Delta E_{\text{vac}} = \left(\mu^{-1} - 1\right) \frac{B^2}{2} V\,. \end{equation} We shall show that \begin{equation} \Delta E_{\text{vac}} = - \beta_0 \frac{g^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}} \cdot \frac{B^2}{2} V\,. \label{Dielmag:DEvac} \end{equation} This means \begin{equation} \mu = \varepsilon^{-1} = 1 + \beta_0 \frac{g^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}}\,, \label{Dielmag:mu} \end{equation} and therefore \begin{equation} e^2(\Lambda') = \left[1 + \beta_0 \frac{e^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}} \right]\,e^2(\Lambda)\,, \end{equation} i.e., $\beta_0$ is indeed the 1-loop $\beta$-function coefficient. The vacuum energy of a charged scalar field (describing particles and antiparticles) is \begin{equation} E_{\text{vac}} = 2 \sum_i \frac{\omega_i}{2} = \sum_i \omega_i\,. \label{Dielmag:Escal} \end{equation} For a charged fermion field, it is the energy of the Dirac sea \begin{equation} E_{\text{vac}} = - \sum_i \omega_i\,. \label{Dielmag:Efermi} \end{equation} Therefore, in general we can use~(\ref{Dielmag:Escal}) multiplied by $(-1)^{2s}$. \subsection{Pauli paramagnetism} \label{S:Pauli} How the vacuum energy changes when we switch the magnetic field $B$ on? There are two effects: spin and orbital, and they can be considered separately. First we discuss interaction of the spin magnetic moment with the magnetic field. Without the field, a massless particle has the energy $\omega=k$. When the magnetic field is switched on, the energy becomes \begin{equation} \omega = \sqrt{k^2 - g_s s_z e B}\,, \label{Pauli:omega} \end{equation} where $g_s$ is the gyromagnetic ratio for our spin-$s$ particle (we shall discuss this in Sect.~\ref{S:QED} in more detail). Suppose the magnetic field is along the $z$ axis. Massless particles only have 2 spin projections $s_z = \pm s$. The vacuum energy change at the order $B^2$ is \begin{equation} \begin{split} \Delta E_{\text{Pauli}} &{}= (-1)^{2s} \int \frac{V\,d^3 k}{(2\pi)^3} \left[\sqrt{k^2 + g_s s e B} + \sqrt{k^2 - g_s s e B} - 2 k\right]\\ &{} = - (-1)^{2s} V \frac{(g_s s e B)^2}{4} \int \frac{d^3 k}{(2\pi)^3} \frac{1}{k^3} = - (-1)^{2s} V \frac{(g_s s e B)^2}{8 \pi^2} \int \frac{dk}{k}\\ &{} = - 2 (-1)^{2s} (g_s s)^2 \frac{e^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}} \cdot \frac{B^2}{2} V\,, \end{split} \label{Pauli:DE} \end{equation} where only modes with momenta between $\Lambda'$ and $\Lambda$ are included. Let's stress once more that what we are calculating is the \emph{vacuum} energy: there are no particles, only empty modes. \subsection{Landau levels} \label{S:LL} Now let's discuss the effect of magnetic field $B$ on the orbital motion. In order not to have complications related to spin, we consider a massless charged scalar field $\varphi$. Its Lagrangian is \begin{equation} L=(D_\mu\varphi)^+ D^\mu\varphi\,. \label{LL:L} \end{equation} For magnetic field $B$ along the $z$ axis, we can choose the vector potential as $A_y = B x$, $A_x=A_z=0$. Then the equation of motion is \begin{equation} \left[ \nabla^2 - e^2 B^2 x^2 - 2 i e B x \frac{\partial}{\partial y} + E^2 \right] \varphi = 0\,. \label{LL:EOM} \end{equation} Its solutions have the form \begin{equation} \varphi = e^{i (k_y y + k_z z)} \varphi(x)\,. \label{LL:sol} \end{equation} The equation for $\varphi(x)$ has the same form as the Schr\"odinger equation for harmonic oscillator: \begin{equation} \left[ - \frac{1}{2} \frac{\partial^2}{\partial x^2} + \frac{\omega^2}{2} x^2 - E_n \right] \psi_n = 0\,. \label{LL:osc} \end{equation} The oscillator energies are \begin{equation} E_n = \omega \left(n + \tfrac{1}{2}\right)\,. \label{LL:En} \end{equation} Comparing~(\ref{LL:EOM}) with~(\ref{LL:osc}), we see that the energies of our massless particle in magnetic field $B$ are \begin{equation} E^2 = k_z^2 + 2 e B \left(n + \tfrac{1}{2}\right)\,, \label{LL:LL} \end{equation} and the corresponding wave functions are \begin{equation} \varphi = e^{i(k_y y + k_z z)} \psi_n\left(x - \frac{k_y}{eB}\right)\,. \label{LL:phi} \end{equation} In other words, $E^2$ consists of discrete Landau levels of transverse motion plus free motion along the magnetic field. Each Landau level has a high degree of degeneracy. In order to find it, let's put our particle into a large box $V=L_x\times L_y\times L_z$. Then the allowed longitudinal momenta are \begin{equation*} k_z = \frac{2\pi}{L_z} n_z\,; \end{equation*} therefore, the number of allowed modes in the interval $d k_z$ is \begin{equation*} d n_z = \frac{L_z\,d k_z}{2\pi}\,. \end{equation*} Similarly, the allowed values of $k_y$ are \begin{equation*} k_y = \frac{2\pi}{L_y} n_y\,. \end{equation*} As we see from~(\ref{LL:phi}), $k_y$ is related to the $x$ coordinate of the center of the Larmor orbit. It must be inside our box: \begin{equation*} \frac{k_y}{eB} \in [0, L_x]\,. \end{equation*} Therefore, \begin{equation*} n_y \in \left[0, \frac{eB\,L_x L_y}{2\pi}\right]\,. \end{equation*} Energy does not depend on $k_y$. Hence the degeneracy of each Landau level is \begin{equation} \frac{eB\,L_x L_y}{2\pi}\,. \label{LL:degen} \end{equation} It is equal to the magnetic flux through our box ($B\,L_x L_y$) measured in flux quanta $2\pi/e$. The spectrum of $E_\bot^2=E^2-k_z^2$ at $B=0$ is continuous. The number of states in the interval $d E_\bot^2$ is \begin{equation*} \frac{L_x L_y}{4\pi} d E_\bot^2\,. \end{equation*} When the magnetic field $B$ is switched on, each interval $\Delta E_\bot^2 = 2eB$ is contracted into a single Landau level (Fig.~\ref{F:Landau}). The number of states in each interval is the Landau level degeneracy~(\ref{LL:degen}). \begin{figure}[ht] \begin{center} \begin{picture}(52,52) \put(26,28.5){\makebox(0,0){\includegraphics{ll.eps}}} \put(6,2){\makebox(0,0){$B=0$}} \put(36,2){\makebox(0,0){$B$}} \put(8,50){\makebox(0,0)[l]{$E_\bot^2/(2eB)$}} \put(5,6){\makebox(0,0)[r]{0}} \put(5,16){\makebox(0,0)[r]{1}} \put(5,26){\makebox(0,0)[r]{2}} \put(5,36){\makebox(0,0)[r]{3}} \put(5,46){\makebox(0,0)[r]{4}} \put(47,11){\makebox(0,0)[l]{$1/2$}} \put(47,21){\makebox(0,0)[l]{$3/2$}} \put(47,31){\makebox(0,0)[l]{$5/2$}} \put(47,41){\makebox(0,0)[l]{$7/2$}} \end{picture} \end{center} \caption{Continuous spectrum at $B=0$ and Landau levels} \label{F:Landau} \end{figure} The vacuum energy of our massless charged scalar field in the magnetic field $B$ is \begin{equation} E_{\text{vac}} = \sum_{n=0}^\infty f\left(n + \tfrac{1}{2}\right)\,, \label{LL:Evac} \end{equation} where \begin{equation} f(x) = \frac{eBV}{(2\pi)^2} \int\limits_{-\infty}^{+\infty} \sqrt{k_z^2 + 2 e B x}\,d k_z\,. \end{equation} \subsection{Euler summation formula} \label{S:Euler} Integrals are more convenient than sums. If we have a smooth function $f(x)$ (i.e., its characteristic length is $L\gg1$), then, obviously, \begin{equation} \sum_{n=0}^N f\left(n + \tfrac{1}{2}\right) \approx \int\limits_0^{N+1} f(x)\,dx\,. \label{Euler:approx} \end{equation} But how to find a correction to this formula? Let's re-write this integral as a sum of integrals over unit intervals: \begin{equation*} I = \int\limits_0^{N+1} f(x)\,dx = \sum_{n=0}^N \int\limits_{-1/2}^{1/2} f\left(n + \tfrac{1}{2} + x\right)\,dx\,. \end{equation*} The smooth function $f(x)$ can be expanded in Taylor series in each interval: \begin{equation*} I = \sum_{n=0}^N \int\limits_{-1/2}^{1/2} \left[f\left(n + \tfrac{1}{2}\right) + \frac{1}{2} f''\left(n + \tfrac{1}{2}\right) x^2 + \cdots \right]\,dx \end{equation*} (terms with odd powers of $x$ don't contribute). Calculating the integrals, we get \begin{equation*} I = \sum_{n=0}^N f\left(n + \tfrac{1}{2}\right) + \frac{1}{24} \sum_{n=0}^N f''\left(n + \tfrac{1}{2}\right) + \cdots \end{equation*} The second sum in the right-hand side is a small correction (because $f''\sim f/L^2$); therefore, we can replace it by an integral (see~(\ref{Euler:approx})): \begin{equation*} I = \sum_{n=0}^N f\left(n + \tfrac{1}{2}\right) + \frac{1}{24} \int\limits_0^{N+1} f''(x)\,dx + \cdots = \sum_{n=0}^N f\left(n + \tfrac{1}{2}\right) + \left. \frac{1}{24} f'(x)\right|_0^{N+1} + \cdots \end{equation*} Finally, we arrive at the Euler summation formula \begin{equation} \sum_{n=0}^N f\left(n + \tfrac{1}{2}\right) = \int\limits_0^{N+1} f(x)\,dx - \left. \frac{1}{24} f'(x)\right|_0^{N+1} + \cdots \label{Euler:Sum} \end{equation} The correction here is of order $1/L^2$. It is easy to find a few more corrections, if desired. \subsection{Landau diamagnetism} \label{S:Landau} The vacuum energy in magnetic field $B$~(\ref{LL:Evac}) can be re-written using the Euler formula~(\ref{Euler:Sum}) as \begin{equation} E_{\text{vac}} = \int\limits_0^\infty f(x) - \left. \frac{1}{24} f'(x) \right|_0^\infty\,. \label{Landau:Evac} \end{equation} The integral here is the vacuum energy at $B=0$ (when the spectrum of $E_\bot^2$ is continuous, Fig.~\ref{F:Landau}). The shift of the vacuum energy due to the magnetic field is \begin{equation} \Delta E_{\text{Landau}} = \frac{1}{24} f'(0) = \frac{e^2 B^2 V}{48 \pi^2} \int \frac{dk}{k} = \frac{1}{3} \frac{e^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}} \cdot \frac{B^2}{2} V\,, \label{Landau:DE} \end{equation} where only modes with momenta between $\Lambda'$ and $\Lambda$ are included% \footnote{Many subtleties have been swept under the carpet in this derivation sketch. A somewhat more accurate derivation~\cite{N:81} yields the same result.}. \subsection{The QED result} \label{S:QED} The full $\Delta E_{\text{vac}}$ in QED is the sum of the spin contribution~(\ref{Pauli:DE}) and the orbital one~(\ref{Landau:DE}): \begin{equation} \Delta E_{\text{vac}} = - \beta_0 \frac{e^2}{(4\pi)^2} \log\frac{\Lambda^2}{\Lambda^{\prime2}} \cdot \frac{B^2}{2} V\,, \label{QED:DE} \end{equation} where \begin{equation} \beta_0 = 2 \sum_s (-1)^{2s} \left[ (g_s s)^2 - \frac{n_s}{6} \right] \label{QED:beta0} \end{equation} is the sum over all charged fields, and $n_s$ is the number of polarization states: $n_0=1$, $n_{s\neq0}=2$. Let's demonstrate that a Dirac particle (e.g., electron) has gyromagnetic ratio $g_{1/2}=2$. For a massless particle, the Dirac equation is \begin{equation} \D\psi = 0\,,\qquad D_\mu = \partial_\mu - i e A_\mu\,. \label{QED:Dirac} \end{equation} Let's multiply it by $\D$: \begin{equation*} \D^2 \psi = 0\,,\qquad \D^2 = \partial^2 - e^2 A^2 - 2 i e A^\mu \partial_\mu - i e \gamma^\mu \gamma^\nu \partial_\mu A_\nu \end{equation*} (in the last term, $\partial_\mu$ acts only on $A_\nu$). If we suppose that $\partial\cdot A=0$, \begin{equation*} \D^2 = D^2 - \frac{ie}{4} F_{\mu\nu} [\gamma^\mu,\gamma^\nu]\,. \end{equation*} We choose $A^\mu=(0,0,B x^1,0)$, then $F_{12}=-F_{21}=-B$ and \begin{equation*} \D^2 = D^2 + i e B \gamma^1 \gamma^2 = D^2 + 2 e B s_z\,. \end{equation*} The equation of motion in magnetic field $B$ (directed along $z$) becomes \begin{equation} \left[ \nabla^2 - e^2 B^2 x^2 - 2 i e B x \frac{\partial}{\partial y} + 2 e B s_z + E^2 \right] \psi = 0\,. \label{QED:EOM} \end{equation} Its energy spectrum is \begin{equation} E^2 = k_z^2 + 2 e B \left(n + \tfrac{1}{2}\right) - 2 e B s_z \label{QED:E} \end{equation} (see Sect.~\ref{S:LL})% \footnote{Strictly speaking, we should consider spin and orbital effects together, using~(\ref{QED:E}). It is easy to see that in the order $B^2$ they can be treated separately.}. Comparing this with~(\ref{Pauli:omega}), we see that \begin{equation} g_{1/2} = 2\,. \label{QED:g} \end{equation} So, the electron contribution to $\beta_0$~(\ref{QED:beta0}) is $-4/3$; if a charged scalar particle exists, it contributes $-1/3$. Note that for $s=1/2$ the spin effect overweights the orbital one. However, due to the factor $(-1)^{2s}$, the spin effect leads to diamagnetism, and the orbital one to paramagnetism. This is because here we are interested in the Dirac see. In physics of metals, we are interested in positive-energy electrons (below the Fermi surface), and the spin effect gives Pauli paramagnetism, while the orbital one --- Landau diamagnetism. \subsection{The QCD result} \label{S:QCD} Now it is easy to obtain $\beta_0$ in QCD. We have chromomagnetic field instead of magnetic. Let's choose its colour orientation along an axis $a_0$ such that $t^{a_0}$ is diagonal (for the $SU(3)$ colour group with the standard choice of the generators $t^a$, $t^8$ is diagonal, and we choose $a_0=8$). The quark contribution follows from the electron one in QED. The contribution to $\beta_0$ is proportional to the charge squared. The sum of squares of colour ``charges'' of a quark is $\Tr t^{a_0} t^{a_0}$ (no summation). Recalling \begin{equation*} \Tr t^a t^b = T_F \delta^{ab}\,, \end{equation*} we arrive at the contribution \begin{equation} - \left(1 - \tfrac{1}{3}\right) 2 T_F n_f \label{QCD:quark} \end{equation} of $n_f$ quark flavours. If there were scalar quarks, each flavour would contribute \begin{equation*} - \frac{1}{3} 2 T_F\,. \end{equation*} What about gluons? First of all, we have to find their gyromagnetic ratio $g_1$. We shall do this for the $SU(2)$ colour group, because calculations are simpler in this case; the result will be valid for other colour groups, too. Let's consider the $SU(2)$ Yang--Mills equation \begin{equation} D^\nu G^a_{\mu\nu} = \left( \partial^\nu \delta^{ab} + g \varepsilon^{acb} A^{c\nu} \right) G^b_{\mu\nu} = 0\,. \label{QCD:YM} \end{equation} The external field is $A^3_\mu$. We linearize in the small components $A^{1,2}_\mu$. For $A^-_\mu=A^1_\mu-i A^2_\mu$ we get \begin{equation*} D^\nu G^-_{\mu\nu} + i g G^3_{\mu\nu} A^{-\nu} = 0\,, \end{equation*} where \begin{equation*} G^-_{\mu\nu} = D_\mu A^-_\nu - D_\nu A^-_\mu\,,\qquad D_\mu = \partial_\mu - i g A^3_\mu\,. \end{equation*} In the $D^\mu A^-_\mu=0$ gauge, the equation of motion becomes \begin{equation*} D^2 A^-_\mu - 2 i g G^3_{\mu\nu} A^{-\nu} = 0 \end{equation*} (we have used $[D_\mu,D_\nu]=-i g G^3_{\mu\nu}$). Our external field is oriented along $z$ in space and along 3 in colour: $G^3_{12} = -G^3_{21} = -B$. Using $A^{-2} = i s_z A^{-1}$ ($s_z=\pm1$), we finally obtain \begin{equation} \left[ D^2 + 2 i g B s_z \right] A^{-1} = 0\,. \label{QCD:EOM} \end{equation} This equation looks exactly the same as~(\ref{QED:EOM}), and hence $g_1=2$. Gluons with colour $a_1$ such that $t^{a_1}$ is diagonal don't interact with our chromomagnetic field (for the standard $SU(3)$ generators, $a_1=3$). All other gluons can be arranged into pairs with positive and negative ``colour charges'' (particles and antiparticles). The sum of their squares (both signs!) is $C_A$: in the adjoint representation \begin{equation*} \Tr t^a t^b = C_A \delta^{ab}\,, \end{equation*} i.e.\ we have to replace $2 e^2 \to C_A g^2$ in QED results. Finally, we arrive at \begin{equation} \beta_0 = \left(4 - \tfrac{1}{3}\right) C_A - \left(1 - \tfrac{1}{3}\right) 2 T_F n_f\,. \label{QCD:beta0} \end{equation} Pauli paramagnetism of the gluon vacuum $(g_1\cdot1)^2=4$ is stronger than its Landau diamagnetism $-1/3$. 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