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I am having a hard time arguing that, after spontaneous breaking of a continuous symmetry of a field Lagrangian, local fluctuations around the vacuum can be interpreted as particles (without referring to analogies from condensed matter physics). I encountered a treatment which stated that, when the symmetry is unbroken, the corresponding current annihilates the vacuum

$$J^{\mu} | 0 \rangle = 0$$

while after spontaneous symmetry breaking, the current creates a state out of the vacuum with some momentum $k^{\mu}$

$$J^{\mu} | 0 \rangle = k^{\mu} | k \rangle $$

My question is why this second equality holds only after spontaneous symmetry breaking.

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    $\begingroup$ It has to do with the unbroken generators of the stabiliser subgroup. I don't have time to write it out, but you can find a proof for this, the quantum Nambu-Goldstone, in many good books, for example Hanzel & Martin, and I believe Peskin & Schroder and Weinberg too, though I don't have the page references to hand. $\endgroup$ – Flint72 May 23 '14 at 12:44
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It is proved by the so called Goldstone theorem which I believed in discussed in every QFT book. In any case, the intuitive picture is provided by the very definition of what a spontaneously broken (continuous) symmetry is about.

I am going basically to tell you what the book of Tom Banks say about (but, again, any QFT book contains it; the second volume by Weinberg has e.g. a very nice discussion about it). A continuous (exact) symmetry always leaves the action (that is the dynamical laws) invariant. This gives rise to a conserved current $J_\mu$ with $\partial_\mu J^\mu=0$ on the equations of motion. On the other hand, should the symmetry be broken spontaneously, the correlation functions such as $$\langle0|T \phi_{i_1}(x_1)\phi_{i_2}(x_2)\ldots|0\rangle $$ are not invariant under the action of the symmetry because not all the symmetry generators annihilate the vacuum. This is possible only iff $$ \int d^4x \partial_\mu \langle0|T J^\mu(x)\Phi(y)|0\rangle\neq 0 $$ for some local (possibly composite) operator $\Phi(x)$ charged under the symmetry for every broken currents. Taking the Fourier transform of this expression you get that $$ k_\mu \Gamma^\mu(k)=k^2 \Gamma(k)\neq 0 \qquad \mbox{for } k\rightarrow 0 $$ which means that the $J^\mu-\Phi$ two point function has a simple pole $1/k^2$ at $k=0$. In other words, the currents generate massless particles when acting on the vacuum, $J^\mu(x)|0\rangle\sim e^{ikx}k^\mu|k\rangle$

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