2
$\begingroup$

I have a few doubts about my understanding of spontaneous symmetry breaking. To keep it simple, I will take a global U(1) transformation on a complex scalar field as an example. The questions are marked in bold and numbered, in order to encourage structured, easy-to-read answers that will hopefully be useful to other students as well.

The kinetic Lagrangian for complex scalar fields $\mathcal{L} = \partial_\mu \phi\ \partial^\mu \phi^*$ is invariant with respect to a global U(1) transformation $\phi(x) \rightarrow \phi'(x) = e^{i \alpha} \phi(x)$. Let's add a potential $V(|\phi|^2)$ that respects this symmetry, but makes the vacuum expectation value of $\phi$ non-vanishing:

$$V(|\phi|^2)=\frac{\lambda}{2}\left(|\phi|^2 - \frac{v^2}{2} \right)\tag{1}$$

However (1) our usual machinery for the treatment of the Lagrangian is designed for fields with potentials taking their minimum value at $\phi = 0$. This bit comes from Griffiths. Why is that true?. Anyway this can be solved easily by shifting the field ($\tilde{\phi} = \phi - v/\sqrt{2}$) and rewriting the Lagrangian:

$$\mathcal{L} = \partial_\mu \tilde{\phi}\ \partial^\mu \tilde{\phi}^* - \frac{\lambda v^2}{4} \left( \tilde{\phi} + \tilde{\phi}^* \right)^2 + \mathcal{O} \left( \text{cubic terms in the fields} \right)\tag{2}$$

(2) In the following, I will ignore the cubic terms in the field, as done in this source for example. Do we ignore these terms, because they correspond to interactions and are not relevant to our present study? A complex field can be written as two real scalar fields ($\tilde{\phi} = \tilde{\phi}_1 + i \tilde{\phi}_2$), and in this form the Lagrangian becomes:

$$\mathcal{L} = \partial_\mu \tilde{\phi}_1 \partial^\mu \tilde{\phi}_1 + \partial_\mu \tilde{\phi}_2 \partial^\mu \tilde{\phi}_2 - \frac{\lambda v^2}{2} \tilde{\phi}_1^2\tag{3}$$

So in this picture we now have a massive, real scalar field $\tilde{\phi}_1$ with mass $m_{\tilde{\phi}_1}=\lambda v^2$, and a massless, real scalar field $\tilde{\phi}_2$ (corresponding to a Goldstone mode).

Let's go back to eq. 2 and perform the global U(1) transformation on $\tilde{\phi}$:

$$\mathcal{L}'= \partial_\mu \tilde{\phi}\ \partial^\mu \tilde{\phi}^* - \frac{\lambda v^2}{4} \left(e^{2 i \alpha} \tilde{\phi} \tilde{\phi}+ e^{-2 i\alpha} \tilde{\phi}^* \tilde{\phi}^* + 2\tilde{\phi} \tilde{\phi}^* \right) \tag{4}$$

(3) Is this strange-looking expression correct? Is this why we say that the symmetry "has been broken", since the Lagrangian is now not invariant under this transformation?

(4) If that is correct, how is that a symmetry "breaking" at all? The original U(1) transformation was with respect to $\phi$, and is still valid. It seems to me more that you could say something like: the symmetry observed in the Lagrangian when expressed in terms of $\phi$ is a deceiving one, if the statement in question (1) is correct. Or said differently, it is a hidden one when the Lagrangian is expressed in terms of $\tilde{\phi}$.

Any other relevant comment is of course welcome. Thank you very much in advance for your answers!

Julien.

$\endgroup$
  • $\begingroup$ Which of these four is the actual question? $\endgroup$ – Gabriel Golfetti Mar 29 at 10:20
  • $\begingroup$ A state spontaneously breaks symmetry. The underlying Lagrangian is always symmetric. So you can't just play with the Lagrangian, you have to think about the state of the system (the quantum ground state, in this example). $\endgroup$ – Mark Mitchison Mar 29 at 10:37
1
$\begingroup$

(1) our usual machinery for the treatment of the Lagrangian is designed for fields with potentials taking their minimum value at $\phi = 0$. This bit comes from Griffiths. Why is that true?

Because we expand the quantum field with a Fourier transform, with the Fourier components roughly corresponding to creation and annihilation operators. If the field's expectation value is nonzero, that corresponds to an extremely large occupancy in the zero-momentum mode. Instead of computing two-to-two scattering, you'd be computing (a-billion-plus-two)-to-(a-billion-plus-two) scattering.

(2) In the following, I will ignore the cubic terms in the field, as done in this [source][1] for example. Do we ignore these terms, because they correspond to interactions and are not relevant to our present study?

Yes.

(3) Is this strange-looking expression correct? Is this why we say that the symmetry "has been broken", since the Lagrangian is now not invariant under this transformation?

Yes.

(4) If that is correct, how is that a symmetry "breaking" at all? The original U(1) transformation was with respect to $\phi$, and is still valid. It seems to me more that you could say something like: the symmetry observed in the Lagrangian when expressed in terms of $\phi$ is a deceiving one, if the statement in question (1) is correct. Or said differently, it is a hidden one when the Lagrangian is expressed in terms of $\tilde{\phi}$.

Indeed, the original symmetry is still there, but it is now hidden. In terms of $\tilde{\phi}$, it's no longer a linear transformation. As a result, you lose lots of the properties we would expect from symmetries. For example, at the most basic level, if you had a lot of real scalar fields with a nontrivial $U(1)$ symmetry, then you would expect at least two particles to have the same mass. But here you don't get that. Or, with a $U(1)$ symmetry you could couple some particles to a massless gauge field, but here you can't.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.