# What is the gauge field in Bose-Einstein condensation?

The Hamiltonian for bosons has $\phi^{\dagger}\phi$ terms in it which makes it U(1) invariant.

Bose-Einstein Condensation apparently breaks such symmetry by choosing a definite phase, even though I can't really see when exactly this happens.

Would it make sense to try and make the U(1) symmetry local? If so, in analogy with the Standard Model, a gauge field would need to be added in order to maintain the local phase symmetry.

What would this gauge field be?

• What do you mean "would it make sense to try"? What effects do you hope to explain with that? – ACuriousMind Nov 4 '15 at 20:18
• What about the other way round, what kind of effects would emerge if there were a local phase invariance? – SuperCiocia Nov 4 '15 at 20:21

I will try to address the first point raised by the OP, i.e. the occurrence of spontaneous symmetry breaking in Bose-Einstein condensation. The free boson gas is described by the hamiltonian: $$H_V=\int_V\frac{d^sx}{2m}\big|\nabla\phi(x)\big|^2.$$ The ground state satisfies $H_V\Psi_0 = 0,\ \forall V$ and hence $\nabla \phi(x)\Psi_0=0,\ \forall x.$ Exploiting the invariance of $\Psi_0$ under translations, $$0=-i\nabla \phi(x)\Psi_0=[P,\phi(x)]\Psi_0=P\phi(x)\Psi_0.$$ By uniqueness of the translation invariant state $$\phi(x)\Psi_0=c\Psi_0,$$ with a constant $c$ fixed by $c=\left(\Psi_0,\phi(x)\Psi_0\right)\equiv \langle\phi\rangle.$ The correlation function: $$\langle\phi(x)^\ast\phi(y)\rangle=\big(\Psi_0,\phi(x)^\ast\phi(y)\Psi_0\big) = \big(\phi(x)\Psi_0,\phi(y)\Psi_0\big)= |c|^2= |\langle\phi\rangle|^2,$$ gives the average density $n\equiv\langle\phi(x)^\ast\phi(x)\rangle=|\langle\phi\rangle|^2$, and we have: $$\langle\phi\rangle=\sqrt{n}e^{i\theta},\qquad \theta\in[0,2\pi).$$ The ground state is characterised by the (experimentally settable) average density $n$ and by the phase $\theta$, $\Psi_0=\Psi_{n,\theta}\equiv\Psi_\theta$, since is in fact labeled by $\langle\phi\rangle$, which is the symmetry breaking order parameter. Indeed, under the action of the $U(1)$ symmetry group, $\Psi_\theta$ is not gauge-invariant: $$\langle \beta^{\lambda} (\phi)\rangle_{\theta} = \langle e^{i\lambda}\phi(x)\rangle_\theta = \big(\Psi_\theta, e^{i\lambda}\phi(x)\Psi_{\theta} \big)=e^{i\lambda}\langle\phi\rangle_\theta = \sqrt{n}e^{i(\lambda+\theta)}=\langle\phi\rangle_{\theta+\lambda}.$$

As comment on the second issue: the local version'' of this global $U(1)$ invariance is indeed meaningful and plays a role, for example, in the Landau-Ginzburg theory of superconductivity or as the massless counterpart of scalar electrodynamics.

Local $U(1)$ invariance of the lagrangian is obtained by minimal coupling: replace the global parameter $\lambda$ with a function $\lambda(x)$, replace the ordinary derivative with a covariant derivative $D=\nabla-iA$, where $A$ is the gauge connection, transforming in the $U(1)$ adjoint $A\mapsto A+\nabla \lambda$.

Since the gauge connection has the meaning of force carrier, one also adds a gauge invariant kinetic term for $A$, $B^2$, $B=\nabla \times A$.

• Thanks for the answer. In the case of a BEC, would local U(1) invariance result in a gauge field? What is the gauge field here? – SuperCiocia Nov 5 '15 at 9:17
• BEC occurs (from a purely theoretical point of view) in a gas of noninteracting bosons described by the above hamiltonian, which is only global $U(1)$ invariant. Forcing local $U(1)$ invariance has the meaning of introducing a coupling with the gauge field $A$ described above. For example this is meaningful in the Landau-Ginzburg theory of superconductivity, where it describes the interaction of Cooper pairs with the e.m. field. Adding minimal coupling to the BEC hamiltonian would introduce an interaction with no apparent meaning. – Brightsun Nov 5 '15 at 11:37
• OK, I'm just trying to get a feel for why there is meaning. What would boson couple to? W+ and W- bosons are charged so they could couple to the EM potential right? – SuperCiocia Nov 5 '15 at 12:25
• Yes, $W^\pm$ couple to e.m. interaction vector bosons, i.e. photons. Adding minimal coupling to the BEC bosons would couple them, if you want, to the e.m. vector potential $A$, effectively giving them a e.m. charge. This is the same thing happening in quantum mechanics, when you add a magnetic field by letting $\Pi = p +\alpha A$ be the kinematic momentum, in contrast with canonical momentum. – Brightsun Nov 5 '15 at 12:45
• OK so we'd make it local only if we wanted to take into account the EM interaction of the BEC? But it is always in an magnetic trap so does that not introduce $A$ somehow? – SuperCiocia Nov 5 '15 at 13:30