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I studied the Higgs mechanism a couple of times now and one question that always comes to my mind is the imaginary part of the mass in the Higgs potential.

The Higgs potential can be written as $$V = -\mu^2 \lvert\phi\rvert^2 + \lambda \lvert \phi \rvert^4$$ where the $\mu$ term is identified as mass term. Plugged in in the Lagrangian ${\cal L}=\ldots -V$, one can obtain spontaneous symmetry breaking for $\mu^2<0$. My question now is, how should I interpret a imaginary mass term in a physical way?

Chiral Anomaly
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    Related: https://physics.stackexchange.com/q/17944/2451 , https://physics.stackexchange.com/q/302138/2451 and links therein. – Qmechanic Sep 10 '21 at 08:35

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People often use the name "mass term" for a quadratic term in the lagrangian. That careless habit comes from free field theories, where it really is a mass term. But in general, the theory's physical predictions are determined by the whole theory, not just by one term in the lagrangian.

The mass of the Higgs particle is real and positive. It's not equal to (the square root of) the coefficient of any individual term in the lagrangian. Loosely, you can think of the mass as the energy-cost of small oscillations about the ground state. The ground state corresponds to the mininum of the potential, and small oscillations about that minimum cost some energy. That energy-cost is the particle's mass. It's affected by the coefficients of both terms, $-\mu^2$ and $\lambda$.

You can get some even better intuition about this by considering a simple lattice QFT model of spontaneous symmetry breaking (SSB). I described the intuition in another answer.

Chiral Anomaly
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Just to augment @ChiralAnomaly's answer with some algebra, we can compute a squared mass$$\frac{\partial^2V}{\partial\phi\partial\phi^\ast}=4\lambda\phi^\ast\phi-\mu^2$$for each solution of $\frac{\partial V}{\partial\phi}=0$. If $\mu^2\ge0$, $|\phi|^2=\frac{\mu^2}{2\lambda}$; if $\mu^2<0$, $\phi=0$. In both cases$$\frac{\partial^2V}{\partial\phi\partial\phi^\ast}=|\mu^2|\ge0,$$with equality iff $\mu^2=0$. See Sec. 4.3.3 here for a multiple-fields generalization to mass matrices (despite the name, their eigenvalues are squared masses).

J.G.
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