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In superconductivity, the BCS condensate can be described in terms of 2 creation operators (the 2 electrons of the pair) acting on the vacuum. I'm wondering whether a similar description can be given for the Higgs condensate. If yes, what state is created by $a^{\dagger} |0 \rangle$? which 4-momentum, which mass?

Edit: in this question, Luboš Motl mentioned that the condensate can be seen as:

$$ |0\rangle = \exp(C\cdot a^\dagger_{p=0}) |h=0\rangle $$

But are these creator operators the ones of the scalar Higgs field? If yes, they are supposed to create object of mass $m_H$ which seems contradictory with the requirement $p^\mu = 0$. So what would be these guys?

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  • $\begingroup$ as an experimentalist I am only familiar with outlines of theories. I do remember that in Technicolor theories the Higgs is considered to be a mechanism of combined exchanges from techni particles. I do not know whether your hypothesis would fit these. $\endgroup$ – anna v Nov 28 '15 at 4:26
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I might be totally wrong, but my understanding is that the condensate by definition cannot be written as a state created by a creation operator. Afaik, the condensate isn't even a state at all but an expectation value of the field operator in the ground/vacuum state (vacuum expectation value or "vev"). Given a field operator $\Phi$ you get:

$vev=\langle\Phi\rangle(x) =\langle0 |\Phi(x) |0\rangle=\langle0 |\Phi_0(x)+\delta\Phi(x) |0\rangle = \langle0 |\Phi_0(x)|0\rangle+\langle0 |\delta\Phi(x) |0\rangle$

where the splitting of $\Phi$ in $\Phi_0$ and $\delta\Phi$ is such that

$\langle0 |\Phi(x) |0\rangle=\langle0 |\Phi_0(x)|0\rangle$ and $\langle0 |\delta\Phi(x) |0\rangle=0$

Physically, $\Phi_0(x)$ describes the condensate and $\delta\Phi(x)$ the fluctuations around the condensate. Now, if we want to quantize the field, we write the commutator relations not for the whole field $\Phi(x)$ but for the fluctuating part $\delta\Phi(x)$ only (although many QFT books assume $\Phi_0=0$ such that the field is just its fluctuation). Also the creation and annihilation operators enter in the mode expansion of $\delta\Phi(x)$, not $\Phi(x)$. But since $\langle0 |\delta\Phi(x) |0\rangle=0$ the creation operators inside $\delta\Phi(x)$ will never generate a condensate.

This is a somewhat floppy answer and I'm not even completely sure whether it is correct, so I'd be glad to get corrections and comments!

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  • $\begingroup$ Thanks "Photon" for your answer. Although I'm familiar with the maths of the spontaneous symmetry breaking of the electroweak theory, what triggered my question is the superconductivity theory (BCS) where, the condensate of cooper pairs is described as a coherent state (yes a macroscopic quantum state) given by smoothing like $$|\psi \rangle = \sum_k a^{\dagger}_{k\uparrow} a^{\dagger}_{-k\downarrow} |0 \rangle.$$ Since both phenomena are similar, a Bose-Einstein condensate, I imagine that Higgs condensate can be of a similar form (but with spin-0 creator of course). $\endgroup$ – Paganini Dec 25 '14 at 10:07

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