Abstract

We consider the response of the QCD ground state at finite baryon density to a strong magnetic field $B$. We point out the dominant role played by the coupling of neutral Goldstone bosons, such as ${\ensuremath{\pi}}^{0}$, to the magnetic field via the axial triangle anomaly. We show that, in vacuum, above a value of $B\ensuremath{\sim}{m}_{\ensuremath{\pi}}^{2}/e$, a metastable object appears---the ${\ensuremath{\pi}}^{0}$ domain wall. Because of the axial anomaly, the wall carries a baryon number surface density proportional to $B$. As a result, for $B\ensuremath{\gtrsim}{10}^{19}\text{ }\text{ }\mathrm{G}$ a stack of parallel ${\ensuremath{\pi}}^{0}$ domain walls is energetically more favorable than nuclear matter at the same density. Similarly, at higher densities, somewhat weaker magnetic fields of order $B\ensuremath{\gtrsim}{10}^{17}--{10}^{18}\text{ }\text{ }\mathrm{G}$ transform the color-superconducting ground state of QCD into new phases containing stacks of axial isoscalar ($\ensuremath{\eta}$ or ${\ensuremath{\eta}}^{\ensuremath{'}}$) domain walls. We also show that a quark-matter state known as ``Goldstone current state,'' in which a gradient of a Goldstone field is spontaneously generated, is ferromagnetic due to the axial anomaly. We estimate the size of the fields created by such a state in a typical neutron star to be of order ${10}^{14}--{10}^{15}\text{ }\text{ }\mathrm{G}$.

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