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Goldstones from a wavefunctional perspective?

+ 2 like - 0 dislike
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I'm familiar with the two standard proofs of Goldstone theorem, namely the 1PI effective action approach and the current-algebra-ish operator proof. Then I was trying to view Goldstone modes from a wavefunctional perspective but got myself quite confused.

Consider the Mexican hat potential with a complex scalar $\phi$. If one writes in polar coordinates $\phi = \rho e^{i\theta}$, then the quadratic term of the Lagrangian has the form $\sim (\partial_\mu \theta)^2$, and people usually at this point conclude there's a massless mode corresponding to angular excitation. However, if we go one step further and make explicitly what these excitations are, there seems to be trouble.

In the wavefunctional picture, a state is a functional $\Psi [\theta(p)]$ if we work in momentum space, and for free fields an energy eigenstate a solution to the collection of infinite number (labeled by $p$) of harmonic oscillator Schroedinger equations

$$\Big(-\frac{d^2}{d\theta^2_p}+\omega^2(p) \theta(p)^2\Big)\Psi [\theta(p)] =E_p\Psi [\theta(p)] \cdots (1)$$

A occupation number state $|n_{p_1}, n_{p_2},\ldots\rangle$ ($n_{p_i}$ particles with momentum $p_i$) is represented by the functional 

$$\Psi_{\{n_{p_i}\}} [\theta(p)] = \prod_{p\neq p_i}\psi_0(\theta(p))\prod_{p_i}\psi_{n_{p_i}}(\theta(p_i)) \cdots(2),$$

where $\psi_n$ is the nth excited state of a harmonic oscillator.

The above is true for all quadratic theories, and if there were no restrictions on $\theta$, $(\partial_\mu \theta)^2$ type of action indeed gives gapless excitations for $p\approx0$. However there is a restriction on $\theta$, that is $\theta\sim\theta+2\pi$, so we are really dealing with oscillators on a circle, which means there is a gap due to periodicity no matter how small $p$ is. Why do we still have Goldstones then? 

asked Dec 26, 2016 in Theoretical Physics by Jia Yiyang (2,465 points) [ no revision ]

I do not understand why you think that there is a restriction on $\theta$? "Periodicity" of the factor $\exp{(i\theta)}$ in $\phi(x,y,z)$ does not limit the values of $\theta$: $\theta$ belongs to $(-\infty,+\infty)$, no?

@VK, then one would still impose periodic boundary condition $\psi(\theta(p))=\psi(\theta(p)+2\pi)$, which is no difference.

@JiaYiyang: Periodic exponential is different from "periodic" $\psi(x,y,z,t)$ since the latter contains a generally non periodic $\rho(x,y,z,t)$.

OK, $\sin(x)$ is a periodic function of $x$, but it does not impose any conditions on $x$ per ce.

@VK, in my case, 'x' itself lives on a circle. If you want to be precise, $\psi(\theta)=f(e^{i\theta})$ for some function f.

@JiaYiyang: So it is your own definition, not an objective constraint to the variable.

As to me, your $\theta$ is a function to be found from the equations, not from your "definition".

Is the function $f=\left(e^{i\theta}\right)^a$ periodic?

1 Answer

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I think I understand this now. The same question can be asked about the Goldstone modes on Heisenberg ferromagnet, at each lattice point the wavefunction is also defined on a compact space (say spin one half, then a two-component spinor is a wavefunction defined on two isolated points). On a Heisenberg ferromagnet everthing about Goldstones can be analysed exactly, and we know how the gapless dispersion relation comes about:

Denote the translation-invariant vacuum state by $|0\rangle$, which is $|0\rangle = |\cdots\downarrow\downarrow\downarrow\cdots\rangle$. There is the 0-momentum one-Goldstone state $$|G_0\rangle = S^{\text{total}}_x|0\rangle=\cdots + |\cdots\uparrow\downarrow\downarrow\cdots\rangle+|\cdots\downarrow\uparrow\downarrow\cdots\rangle+|\cdots\downarrow\downarrow\uparrow\cdots\rangle+\cdots$$

And the finte-momentum Goldstone states are 

$$|G_k\rangle = \cdots + |\cdots\uparrow\downarrow\downarrow\cdots\rangle+e^{ika}|\cdots\downarrow\uparrow\downarrow\cdots\rangle+e^{2ika}|\cdots\downarrow\downarrow\uparrow\cdots\rangle+\cdots, $$

$k$ being lattice momentum and $a$ being lattice spacing. So we see the compactness of the base space of the wavefunction does not matter, the wavefunction on each site merely serves as a "carrier" on which a position-dependent phase can be implemented, and the important prerequisite is that there exists a 0-momentum Goldstone degenerate with $|0\rangle$ (that is, the broken symmetry charge does not annihilate the true vacuum $|0\rangle$). 

The complex $\phi^4$ theory is entirely analogous, the state I wrote down in the original post (eqn (2) with $n_{p_i}=0$) is the true vacuum instead of the 0-momentum Goldstone, and doesn't correspond to any point on the gapless branch of the dispersion relations. The usual textbook argument using $(\partial_\mu \theta)^2$ is only a sloppy (or slick, depending on your perspective) way of deriving the dispersion relation for Goldstones.

answered Jul 10 by Jia Yiyang (2,465 points) [ revision history ]
edited Jul 12 by Jia Yiyang

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