Superfluid rotating frame of reference I'm currently studying a text about Bose-Einstein condensates (BECs) and vortices. When they want to study whether a vortex will be formed, they look at the fact wether it's enegetically favorable. 
To do that they go to a rotating frame of reference (since you rotate your BEC to generate these), where the energy is given by:
$$\widetilde{E}[\Psi]=E[\Psi]-L[\Psi]\cdot\Omega,$$
Where $\Omega$ is the rotational velocity which you apply to the BEC.
Now the argument that the term $L[\Psi]\cdot\Omega$ should be substracted comes from the fact that in a rotating frame your system loses that fraction of rotational energy. Now I was wondering if anyone knew where the form of $L[\Psi]\cdot\Omega$ came from?
I know that in a rotating frame of reference you have that $ \vec{v}=\vec{v}_r+\vec{\Omega}\times\vec{r}$. If you fill this in into the kinetic energy and use some basic definitions, you get that the extra effect of rotation is given by:
$$\frac{1}{2}I\Omega^2=\frac{1}{2}J\Omega.$$
This yields half of the value that is used in the book, is there something that I'm missing or not seeing right ?
 A: The $L\cdot\Omega$ term comes directly from the change of frame of reference, especially from the transformation from the static frame of reference to the rotating frame of reference. 
Let $\mathcal{R}\equiv(x,y,z)$ the initial static frame, and $\widetilde{\mathcal{R}}\equiv(x',y',z')$ the rotating frame at a constant velocity $\mathbf{\Omega}=\Omega\,\hat{z}$. We study the mouvement of a free particle which hamiltonian reads :
$$\hat{H}=\frac{\mathbf{\hat{p}}^2}{2m}\quad\text{expressed in the}\;\mathcal{R}\;\text{frame.}$$
The particule state $\vert\psi(t)\rangle$ follows the Schrodinger equation :
$$\hat{H}|\psi(t)\rangle=\mathrm{i}\hbar\frac{\partial}{\partial t}|\psi(t)\rangle\quad\text{in the}\;\mathcal{R}\;\text{frame.}$$
If we change from $\mathcal{R}$ to $\widetilde{\mathcal{R}}$ frame, there exists an unitary transformation $\hat{U}(t)$ so that the particule state is now $|\tilde{\psi}(t)\rangle=\hat{U}(t)|\psi(t)\rangle$. To derive the corresponding Schrodinger equation on $|\tilde{\psi}(t)\rangle$, one can start to calculate :
$$\mathrm{i}\hbar\frac{\partial}{\partial t}|\tilde{\psi}(t)\rangle=\mathrm{i}\hbar\left[\frac{d\hat{U}}{dt}|\psi(t)\rangle+\hat{U}\frac{\partial}{\partial t}|\psi(t)\rangle\right]$$
Then, by using the Schrodinger equation in the $\mathcal{R}$ frame and the fact that $|\psi(t)\rangle=\hat{U}^\dagger|\tilde{\psi}(t)\rangle$, we have :
$$\hat{\tilde{H}}|\tilde{\psi}(t)\rangle=\mathrm{i}\hbar\frac{\partial}{\partial t}|\tilde{\psi}(t)\rangle\quad\text{in the}\;\widetilde{\mathcal{R}}\;\text{frame}$$
$$\text{with}\quad\hat{\tilde{H}}=\hat{U}\hat{H}\hat{U}^\dagger+\mathrm{i}\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger$$
By simple assumptions on momentum conservation, one can take :
$$\hat{U}(t)=\exp\left(\frac{\mathrm{i}\Omega t}{\hbar}\hat{L}_z\right)$$
where $\hat{L}_z=\hat{x}\hat{p}_y-\hat{y}\hat{p}_x=-\mathrm{i}\hbar\frac{\partial}{\partial\varphi}$ momemtum operator along $z$. $\hat{L}_z$ commutes with $\hat{H}$ given the fact the kinetic energy is invariant by rotation transformation.
After some calculations, one can show that : 
$$\mathrm{i}\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger=-\Omega\,\hat{L}_z=-\mathbf{\Omega}\cdot\mathbf{\hat{L}}$$
