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  Is the time derivative of the WKB phase globally defined?

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Let $Q$ and $\mathcal{L}$ be smooth n-dimensional manifolds and $\iota_t:\mathcal{L}\rightarrow T^*Q$ a time-dependent Lagrangian embedding that is smooth in $t$ and satisfies the Bohr-Sommerfeld quantization condition for a fixed value of $\hbar$. Thus, if $\vartheta$ is the canonical 1-form on $T^*Q$ and $m_{\iota_t}$ is the Maslov class of the embedding $\iota_t$,the real cohomology class $$ [\iota_t^*\vartheta]-\frac{\pi\hbar}{2}m_{\iota_t}$$ takes values in $\hbar\mathbb{Z}$.

Fix a $t\in\mathbb{R}$. Because $\iota_t^*\vartheta$ is closed, we can choose an open cover $\{\Lambda_\alpha\}$ for $\mathcal{L}$ such that $\iota_t^*\vartheta=d S_t^\alpha$ on $\Lambda_\alpha$. Because the Bohr-Sommerfeld condition holds, the $S^\alpha_t$ can be chosen such that on $\Lambda_{\alpha\beta}=\Lambda_\alpha\cap\Lambda_\beta$, $$S^\alpha_t-S^\beta_t=\frac{\pi\hbar}{2}m_{\alpha\beta} ~\text{mod}~2\pi\hbar,$$ where the $m_{\alpha\beta}$ are the transition functions for the time-$t$ Maslov principal $\mathbb{Z}$-bundle over $\mathcal{L}$.

Now my question. I believe that, at least in a small open neighborhood of $t$, $m_{\alpha\beta}$ can be regarded as smooth integer-valued functions of time and that the $S^\alpha_t$ can be chosen to be smooth in $t$. I am therefore lead to the conclusion that $\frac{d}{dt}S^\alpha_t$ is a globally defined function on $\mathcal{L}$. Are my beliefs incorrect? Is it true that $\dot{S}_t=\frac{d}{dt} S_t$ is a well-defined function on $\mathcal{L}$?

I have a partial answer already. It turns out that $\dot{S}_t$ is globally defined in the special case where $\iota_t=\phi_t\circ\iota_0$, with $\phi_t$ the flow map of a globally Hamiltonian vector field on $T^*Q$; there is an explicit expression for $\dot{S}_t$ in terms of $\iota_t$ and the Hamiltonian function associated with $\phi_t$.

This post imported from StackExchange MathOverflow at 2015-05-27 22:07 (UTC), posted by SE-user Josh Burby
asked Jan 16, 2014 in Theoretical Physics by Josh Burby (120 points) [ no revision ]
retagged May 27, 2015

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