- #1
jack476
- 328
- 125
- TL;DR Summary
- What does it mean when a Lagrangian has more Goldstone bosons than there are spontaneously broken generators?
Although strictly speaking this is motivated by a classical field theory problem, I felt that the quantum physics subforum would be more appropriate because it pertains to topics which to my understanding are almost entirely dealt with in the context of QFT.
The most commonly stated form of Goldstone's theorem is that there is one massless scalar boson corresponding to each spontaneously broken generator of the global symmetry group. In most situations there is exactly one Goldstone boson for each broken generator, but in the book I'm working through, The Classical Theory of Gauge Fields, the authors Rubakov and Wilson point out that the theory actually says that there are at least as many Goldstone bosons as there are broken generators. It is possible for there to be more Goldstone bosons than there are broken generators.
As an example, consider the following Lagrangian for a pair of complex scalar fields ##\phi_1## and ##\phi_2##:
$$\mathcal{L} = \partial_{\mu}\phi_1^*\partial_{\mu}\phi_1 + \partial_{\mu}\phi_2^*\partial_{\mu}\phi_2 - \lambda\big(\phi_1^*\phi_1-\phi_2^*\phi_2 - v^2\big)^2$$ where ##v## is a real constant. The global symmetry group is the group of matrices with the form $$\omega = \begin{bmatrix} e^{i\alpha} & 0 \\0 & e^{i\beta} \end{bmatrix}$$
Note that the generators of small transformations are ##T^1 = \begin{bmatrix} i & 0 \\ 0 & 0 \end{bmatrix} \text{ and } T^2 = \begin{bmatrix} 0 & 0 \\ 0 & i \end{bmatrix}##.
The equations for the ground state are:
$$\begin{align*}
\frac{\partial V}{\partial \phi_1} = 4{\lambda}\phi_1^*\big( \phi_1^*\phi_1-\phi_2^*\phi_2 - v^2 \big) &= 0 \\
\frac{\partial V}{\partial \phi_2} = -4{\lambda}\phi_2^*\big( \phi_1^*\phi_1-\phi_2^*\phi_2 - v^2 \big) &= 0
\end{align*}$$
Other than the trivial vacuum ##\phi_1^{(v)}= \phi_2^{(v)} = 0##, the ground states must satisfy ##\phi_1^*\phi_1-\phi_2^*\phi_2 - v^2## so they take the form ##\phi^{(v)} = \begin{bmatrix}ae^{i\alpha_0} \\ be^{i\beta_0}\end{bmatrix}## where ##a_1, a_2 \in \mathbb{R}## and ##a^2-b^2 = v^2##. The phase angles ##\alpha_0## and ##\beta_0## are arbitrary and we will assume that they are both zero. First consider small excitations about the ground state with ##\phi_1^{(v)} = a## and ##\phi_2^{(v)} = b##, which spontaneously breaks the entire symmetry group. Let ##\chi_1, \chi_2, \theta_1, \theta_2## be small real scalar fields, then these vacuum excitations take the form ##\phi_1 = a+\chi_1+i\theta_1## and ##\phi_2 = b+\chi_2+i\theta_2##. Substitute these into ##\mathcal{L}## and expand to quadratic order to get:
$$\mathcal{L}^{(2)} = \big(\partial_{\mu}\chi_1\big)^2 + \big(\partial_{\mu}\chi_2\big)^2 + \big(\partial_{\mu}\theta_1\big)^2 + \big(\partial_{\mu}\theta_2\big)^2- 4\lambda\big(a^2\chi_1^2+b^2\chi_2^2\big)$$
We get two massive scalars and two massless scalars, so that we have one massless scalar for each broken generator. Now consider the ground state ##\phi_1^{(v)} = v, \phi_2^{(v)} = 0##, so that small vacuum excitations have the form ##\phi_1 = v+\chi_1+i\theta_1## and ##\phi_2 = \chi_2+i\theta_2##. Now the symmetry is only partially spontaneously broken, with the unbroken subgroup being the group of matrices with the form ##\omega = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\beta} \end{bmatrix}##. The quadratic Lagrangian is:
$$\mathcal{L}^{(2)} = \big(\partial_{\mu}\chi_1\big)^2 + \big(\partial_{\mu}\chi_2\big)^2 + \big(\partial_{\mu}\theta_1\big)^2 + \big(\partial_{\mu}\theta_2\big)^2- 4\lambda v^2\chi_1^2$$
Only one of the two generators is broken by the ground state but now there are three massless scalar bosons. This doesn't violate Goldstone's theorem, but by my understanding it is still considered unusual. How should I interpret the fact that there are more Goldstone bosons than there are broken generators? The only two things that really stand out to me are 1.) that the condition ##a^2-b^2 = v^2## means that there are two disconnected families of vacuum states and 2.) that although ##\big(\phi_1^{(v)}, \phi_2^{(v)}\big) = (a, 0)## is a valid ground state, ##\big(\phi_1^{(v)}, \phi_2^{(v)}\big) = (0, b)## is not because then ##-b^2 = v^2## which is not possible for ##b, v \in \mathbb{R}##.
The most commonly stated form of Goldstone's theorem is that there is one massless scalar boson corresponding to each spontaneously broken generator of the global symmetry group. In most situations there is exactly one Goldstone boson for each broken generator, but in the book I'm working through, The Classical Theory of Gauge Fields, the authors Rubakov and Wilson point out that the theory actually says that there are at least as many Goldstone bosons as there are broken generators. It is possible for there to be more Goldstone bosons than there are broken generators.
As an example, consider the following Lagrangian for a pair of complex scalar fields ##\phi_1## and ##\phi_2##:
$$\mathcal{L} = \partial_{\mu}\phi_1^*\partial_{\mu}\phi_1 + \partial_{\mu}\phi_2^*\partial_{\mu}\phi_2 - \lambda\big(\phi_1^*\phi_1-\phi_2^*\phi_2 - v^2\big)^2$$ where ##v## is a real constant. The global symmetry group is the group of matrices with the form $$\omega = \begin{bmatrix} e^{i\alpha} & 0 \\0 & e^{i\beta} \end{bmatrix}$$
We can write the pair of complex scalars as the two elements of a column ##\phi## in the fundamental representation space of ##U(2)##, and then we can write the kinetic energy and the term ##\phi_1^*\phi_1-\phi_2^*\phi_2## as:
$$ \partial_{\mu}\phi_1^*\partial_{\mu}\phi_1 + \partial_{\mu}\phi_2^*\partial_{\mu}\phi_2 = \begin{bmatrix} \partial_{\mu}\phi_1^* & \partial_{\mu}\phi_2^* \end{bmatrix}\begin{bmatrix}\partial_{\mu}\phi_1 \\ \partial_{\mu}\phi_2\end{bmatrix} \quad \text{,} \quad \phi_1^*\phi_1-\phi_2^*\phi_2 = \begin{bmatrix} \phi_1^* & \phi_2^* \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix}\phi_1 \\\phi_2\end{bmatrix}$$
Let ##\omega = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in U(2)##, then the kinetic energy is clearly invariant under ##\phi \to \omega \phi## since under this transformation ##\phi^\dagger \to \phi^{\dagger} \omega^{\dagger}## and ##\omega \in U(2) \implies \omega^{\dagger} = \omega^{-1}.## However, ##\phi_1^*\phi_1-\phi_2^*\phi_2## is not invariant under ##\phi \to \omega \phi## unless ##\omega## also has the property that ##\omega^{\dagger} J \omega = J## where ##J = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}##, meaning that the ##U(2)## symmetry is explicitly broken down to ##U(2) \cap SU(1,1)##. The conditions ##\omega^{\dagger}\omega = I## and ##\omega^{\dagger}J\omega = J## set up a system of six equations in the four variables ##a, b, c, d##, though we only need the first four:
$$\begin{align*}
a^*a+c^*c &= 1 \\
a^*b+c^*d &= 0 \\
b^*b+d^*d &= 1 \\
a^*a-c^*c &= 1
\end{align*}$$
Add the first equation to the fourth to get ##2a^*a = 2## which means that ##a = e^{i\alpha}##, then the first equation becomes ##1+c^*c =1## so ##c = 0##, which in turn means that the second equation becomes ##a^*b = 0## meaning that ##b=0##, and then finally the third equation becomes ##d^*d = 1## so ##d=e^{i\beta}##. Therefore the global symmetry group of ##\mathcal{L}## is the subgroup of ##U(2)## which contains matrices of the form ##\omega = \begin{bmatrix} e^{i\alpha} & 0 \\ 0 & e^{i\beta} \end{bmatrix}##.
$$ \partial_{\mu}\phi_1^*\partial_{\mu}\phi_1 + \partial_{\mu}\phi_2^*\partial_{\mu}\phi_2 = \begin{bmatrix} \partial_{\mu}\phi_1^* & \partial_{\mu}\phi_2^* \end{bmatrix}\begin{bmatrix}\partial_{\mu}\phi_1 \\ \partial_{\mu}\phi_2\end{bmatrix} \quad \text{,} \quad \phi_1^*\phi_1-\phi_2^*\phi_2 = \begin{bmatrix} \phi_1^* & \phi_2^* \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix}\phi_1 \\\phi_2\end{bmatrix}$$
Let ##\omega = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in U(2)##, then the kinetic energy is clearly invariant under ##\phi \to \omega \phi## since under this transformation ##\phi^\dagger \to \phi^{\dagger} \omega^{\dagger}## and ##\omega \in U(2) \implies \omega^{\dagger} = \omega^{-1}.## However, ##\phi_1^*\phi_1-\phi_2^*\phi_2## is not invariant under ##\phi \to \omega \phi## unless ##\omega## also has the property that ##\omega^{\dagger} J \omega = J## where ##J = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}##, meaning that the ##U(2)## symmetry is explicitly broken down to ##U(2) \cap SU(1,1)##. The conditions ##\omega^{\dagger}\omega = I## and ##\omega^{\dagger}J\omega = J## set up a system of six equations in the four variables ##a, b, c, d##, though we only need the first four:
$$\begin{align*}
a^*a+c^*c &= 1 \\
a^*b+c^*d &= 0 \\
b^*b+d^*d &= 1 \\
a^*a-c^*c &= 1
\end{align*}$$
Add the first equation to the fourth to get ##2a^*a = 2## which means that ##a = e^{i\alpha}##, then the first equation becomes ##1+c^*c =1## so ##c = 0##, which in turn means that the second equation becomes ##a^*b = 0## meaning that ##b=0##, and then finally the third equation becomes ##d^*d = 1## so ##d=e^{i\beta}##. Therefore the global symmetry group of ##\mathcal{L}## is the subgroup of ##U(2)## which contains matrices of the form ##\omega = \begin{bmatrix} e^{i\alpha} & 0 \\ 0 & e^{i\beta} \end{bmatrix}##.
The equations for the ground state are:
$$\begin{align*}
\frac{\partial V}{\partial \phi_1} = 4{\lambda}\phi_1^*\big( \phi_1^*\phi_1-\phi_2^*\phi_2 - v^2 \big) &= 0 \\
\frac{\partial V}{\partial \phi_2} = -4{\lambda}\phi_2^*\big( \phi_1^*\phi_1-\phi_2^*\phi_2 - v^2 \big) &= 0
\end{align*}$$
Other than the trivial vacuum ##\phi_1^{(v)}= \phi_2^{(v)} = 0##, the ground states must satisfy ##\phi_1^*\phi_1-\phi_2^*\phi_2 - v^2## so they take the form ##\phi^{(v)} = \begin{bmatrix}ae^{i\alpha_0} \\ be^{i\beta_0}\end{bmatrix}## where ##a_1, a_2 \in \mathbb{R}## and ##a^2-b^2 = v^2##. The phase angles ##\alpha_0## and ##\beta_0## are arbitrary and we will assume that they are both zero. First consider small excitations about the ground state with ##\phi_1^{(v)} = a## and ##\phi_2^{(v)} = b##, which spontaneously breaks the entire symmetry group. Let ##\chi_1, \chi_2, \theta_1, \theta_2## be small real scalar fields, then these vacuum excitations take the form ##\phi_1 = a+\chi_1+i\theta_1## and ##\phi_2 = b+\chi_2+i\theta_2##. Substitute these into ##\mathcal{L}## and expand to quadratic order to get:
$$\mathcal{L}^{(2)} = \big(\partial_{\mu}\chi_1\big)^2 + \big(\partial_{\mu}\chi_2\big)^2 + \big(\partial_{\mu}\theta_1\big)^2 + \big(\partial_{\mu}\theta_2\big)^2- 4\lambda\big(a^2\chi_1^2+b^2\chi_2^2\big)$$
We get two massive scalars and two massless scalars, so that we have one massless scalar for each broken generator. Now consider the ground state ##\phi_1^{(v)} = v, \phi_2^{(v)} = 0##, so that small vacuum excitations have the form ##\phi_1 = v+\chi_1+i\theta_1## and ##\phi_2 = \chi_2+i\theta_2##. Now the symmetry is only partially spontaneously broken, with the unbroken subgroup being the group of matrices with the form ##\omega = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\beta} \end{bmatrix}##. The quadratic Lagrangian is:
$$\mathcal{L}^{(2)} = \big(\partial_{\mu}\chi_1\big)^2 + \big(\partial_{\mu}\chi_2\big)^2 + \big(\partial_{\mu}\theta_1\big)^2 + \big(\partial_{\mu}\theta_2\big)^2- 4\lambda v^2\chi_1^2$$
Only one of the two generators is broken by the ground state but now there are three massless scalar bosons. This doesn't violate Goldstone's theorem, but by my understanding it is still considered unusual. How should I interpret the fact that there are more Goldstone bosons than there are broken generators? The only two things that really stand out to me are 1.) that the condition ##a^2-b^2 = v^2## means that there are two disconnected families of vacuum states and 2.) that although ##\big(\phi_1^{(v)}, \phi_2^{(v)}\big) = (a, 0)## is a valid ground state, ##\big(\phi_1^{(v)}, \phi_2^{(v)}\big) = (0, b)## is not because then ##-b^2 = v^2## which is not possible for ##b, v \in \mathbb{R}##.
Last edited: