When do we consider states under a $U(1)$ transformations to be physically different? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-20T05:16:51Z https://physics.stackexchange.com/feeds/question/327013 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/327013 1 When do we consider states under a $U(1)$ transformations to be physically different? jinawee https://physics.stackexchange.com/users/21817 2017-04-16T18:54:29Z 2017-04-16T22:50:37Z <p>Consider the Goldstone model of a complex scalar field $\Phi$. It has $U(1)$ global symmetry, so if we apply the transformation $\Phi \to e^{i\alpha} \Phi$ the Lagrangian is left invariant. </p> <p>Furthermore, we have an infinite set of possible vacua all with the same non-zero vacuum expectation value. But the vacuum changes under a $U(1)$ transformation, so $U(1)$ symmetry is spontaneously broken.</p> <ul> <li>In this case we assume that for each value $\alpha$, $e^{i\alpha} |0 \rangle$ corresponds to a different state, right?</li> <li>But since we can't measure a phase, wouldn't it be more natural to consider them the same states? On the other hand, if I think of the real and imaginary parts as two independent fields, I would say that they shouldn't be the same states.</li> </ul> <p>Let's now couple $\Phi$ to a gauge field, such that the Lagrangian is invariant under local $U(1)$ transformations. Then we do consider $e^{i\alpha} | \Psi \rangle$ to be all the same states, right? But then, doesn't this mean that the Higgs vacuum is unique?</p> https://physics.stackexchange.com/questions/327013/when-do-we-consider-states-under-a-u1-transformations-to-be-physically-diffe/327019#327019 2 Answer by Everett You for When do we consider states under a $U(1)$ transformations to be physically different? Everett You https://physics.stackexchange.com/users/7616 2017-04-16T19:25:19Z 2017-04-16T19:25:19Z <p>You are wrong from this assumption:</p> <blockquote> <p>We assume that for each value $\alpha$, $e^{i\alpha}\Phi | 0 \rangle$ corresponds to a different state.</p> </blockquote> <p>The wave function of the symmetry broken vacuum (the Higgs vacuum) is not $e^{i\alpha}\Phi | 0 \rangle$ as you assumed, but a <strong>coherent state</strong> $|v\rangle$ (a boson condensate), obtained by applying the shift operator $D(v)$ to the symmetric vacuum state $|0\rangle$:</p> <p>$$|v\rangle=D(v)|0\rangle=e^{v\Phi^\dagger-v^*\Phi}|0\rangle,$$</p> <p>where $v=|v| e^{\text{i}\alpha}$ is the vacuum expectation value (VEV) of the boson operator $\Phi=\int\text{d}^dx\Phi(x)$. Note that in the Higgs vacuum state $|v\rangle$, the operators $\Phi$ and $\Phi^\dagger$ are raised to the exponent and the phase factor $e^{\text{i}\alpha}$ is not exposed as an overall factor of the wave function. One can indeed show that $\forall x:\langle v|\Phi(x)|v\rangle=v$ using the boson commutation relation $[\Phi(x),\Phi(x')^\dagger]=\delta(x-x')$ and the definition of the symmetric vacuum state $\Phi(x)|0\rangle=0$ (see this <a href="https://en.wikipedia.org/wiki/Coherent_states" rel="nofollow noreferrer">Wikipedia page</a> for more about coherent states). The Higgs vacuum state transforms nontrivially under the U(1) symmetry action: $|v\rangle\to|ve^{\text{i}\theta}\rangle$. The new vacuum is orthogonal to the old one $\langle v|v e^{\text{i}\theta}\rangle=\delta(\theta)$ (in the thermodynamic limit). So the two states are indeed physically different under the U(1) transformation.</p>