# 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 ?

• Apr 20, 2021 at 16:41

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 :

$$\mathrm{i}\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger=-\Omega\,\hat{L}_z=-\mathbf{\Omega}\cdot\mathbf{\hat{L}}$$

• thanks, but I don't quite see where the second term comes from in your equation for $\hat{\bar{H}}$. The one with the derivative to time of $\hat{U}$. The rest does indeed sound logic to me.
– Nick
Oct 28, 2014 at 7:04
• See the edit for a little bit more precise calculations. Oct 28, 2014 at 9:01
• Is the above derivation also valid for a varying $\boldsymbol{\Omega}$ or only for a constant value ?
– Nick
Oct 31, 2014 at 13:53
• A priori I don't see any reason why it shoudn't. You will have additional term in $\hat{\tilde{H}}$ coming from the time derivative of $\hat{U}(t)$ : $$\mathrm{i}\hbar\frac{d\hat{U}}{dt}\hat{U}^\dagger=-\Omega(t)\,\hat{L}_z-\frac{d\Omega}{dt}\,t\,\hat{L}_z$$ Oct 31, 2014 at 15:56
• That is true, but does the form of your unitary transformation $\hat{U}(t)$ hold for such cases ? It seems to me that there should come an integral to time with an time-ordered product. This might give problems when $\hat{\mathbf{L}}$ does not commute with itself at alle times. Or is this reasoning wrong ?
– Nick
Nov 2, 2014 at 15:33