RotatingFrame

\def\bra#1{\left\langle{#1}\right|} \def\ket#1{\left|{#1}\right\rangle} \def\braket#1#2{\left\langle{#1}\mid{#2}\right\rangle}

Schrodinger equation in a rotating frame of reference

The time-dependant Schrodinger equation is given by:

i\hbar\psi_t = \hat{H}\psi

$\hat{H}$ may vary. In an inertial frame we have:

\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V}

A rotating frame of reference is non-inertial, so there may be extra terms appearing in the hamiltonian.

$F \equiv (x, y, z)$ $F'\equiv (x',y',z')$ $\vec{\Omega} = \Omega \hat{z}$ $F$ $\ket{\psi(t)}$ in a potential is governed by the hamiltonian:

\hat{H} = \frac{\hat{p}^2}{2m} + \hat{V} \hspace{1cm}\rightarrow\hspace{1cm} \hat{H}\ket{\psi(t)} = i\hbar \frac{\partial}{\partial t}\ket{\psi(t)}

$F$ $F'$ $\hat{U(t)}$ $\ket{\tilde{\psi}(t)} = \hat{U}(t)\ket{\psi(t)}$ . We can now calculate the new evolution rule in the following way:

i\hbar\frac{\partial}{\partial t}\ket{\tilde{\psi}(t)} = i\hbar\frac{\partial}{\partial t} \left [ \hat{U}(t)\ket{\psi(t)} \right ] = i\hbar \hat{U}(t) \frac{\partial \ket{\psi(t)}}{\partial t} + i\hbar\frac{\partial \hat{U}(t)}{\partial t} \ket{\psi(t)}

\implies i\hbar\frac{\partial}{\partial t}\ket{\tilde{\psi}(t)} = \hat{U}(t)\hat{H}\ket{\psi(t)} + i\hbar \frac{\partial \hat{U}(t)}{\partial t}\ket{\psi(t)}

\implies i\hbar\frac{\partial}{\partial t}\ket{\tilde{\psi}(t)} = \left [\hat{U}(t)\hat{H}\hat{U}^{\dagger}(t) + i\hbar \frac{\partial \hat{U}(t)}{\partial t}\hat{U}^{\dagger}(t) \right ]\ket{\tilde{\psi}(t)}

$F'$ is given by:

i\hbar\frac{\partial}{\partial t}\ket{\tilde{\psi}(t)} = \hat{H'}\ket{\tilde{\psi}(t)} \hspace{1cm}\leftarrow\hspace{1cm} \hat{H'} = \hat{U}\hat{H}\hat{U}^{\dagger} + i\hbar\frac{\partial \hat{U}}{\partial t}\hat{U}^{\dagger}

Since angular momentum is the generator of rotations, we can write the unitary operator for the transformation like so:

\hat{U}(t) = \exp\left(\frac{i\Omega t}{\hbar} \hat{L}_z \right)

$\hat{L}_z$ $z$ $\hat{L}_z$ $\hat{H}$ , this gives us the new hamiltonian to be:

\hat{H'} = \hat{U}\hat{H}\hat{U}^{\dagger} + i\hbar\frac{\partial \hat{U}}{\partial t}\hat{U}^{\dagger}

\implies \hat{H'} = \hat{H}\hat{U}\hat{U}^{\dagger} + i\hbar\left(\frac{i\Omega \hat{L}_z}{\hbar}\right)\hat{U}\hat{U}^{\dagger}

\implies \hat{H'} = \hat{H} - \Omega\hat{L}_z

More generally, we can write it as:

\begin{align*} \hat{H'} &= \hat{H} - \vec{\Omega} \cdot \hat{L} \\ &= \frac{\hat{p}^2}{2m} + \hat{V}(r, \ \theta - \Omega t, \ z) - \vec{\Omega} \cdot \hat{L} \end{align*}