Power-Zienau-Woolley Transformation

One can now show that the Coulomb gauge Hamiltonian ${\mathcal H}_{\rm Coul} (t)$ can be transformed into a Hamiltonian ${\mathcal H}(t)$ in any other gauge as specified by ${\bf g}_\perp$ and given by Eq. (VII.3.8). This is achieved by the so-called Power-Zienau-Woolley Transformation which is a unitary transformation of the Coulomb gauge Hamiltonian ${\mathcal H}_{\rm Coul} (t)$,

$${\mathcal H}(t) = \Lambda^{-1} {\mathcal H}_{\rm Coul} (t)\Lambda... ...da \equiv \exp \left[i \int d{\bf r} \mathbf{A}({\bf r},t) \P({\bf r}) \right].$$ (3.12)

A relation can be derived between $\mathbf{A}$ and $\mathbf{A}_{\rm Coul}$, cf. Woolley [8],

$$\mathbf{A}({\bf r},t) = \mathbf{A}_{\rm Coul}({\bf r},t) - \div\int d{\bf r}' \mathbf{A}_{\rm Coul}({\bf r}',t){\bf g}_\perp({\bf r}',{\bf r}).$$ (3.13)

If this is inserted into ${\mathcal H}(t)$, Eq. (VII.3.8), one obtains

$${\mathcal H}(t) = H(t) + H_{\rm rad}$$

$$H_{\rm rad} = \frac{1}{2}\int d{\bf r} \left[ \varepsilon_0 \mathbf{E}_\perp^2 + \mu_0^{-1} \mathbf{B}^2\right]$$ (3.14)

$$H(t) = \sum_n \frac{{\bf p}_n^2}{2m_n} + V_{\rm Coul}+ V_{\bf EP}+V_{\bf PP}+$$

$$V_{\rm Coul} \equiv \frac{1}{2}\sum_{n,m}\frac{q_nq_m}{4\pi\varepsilon_0\vert{\bf r}_n-{\bf r}_m\vert}$$

$$V_{\bf EP} \equiv - \int d {\bf r} \mathbf{E}_\perp \P_\perp$$

$$V_{\bf PP} \equiv \frac{1}{2\varepsilon_0}\int d {\bf r} {\bf P}_\perp({\bf r})^2.$$ (3.15)

Basically, apart from the magnetic terms the ${\bf pA}$ coupling is transformed away and one has instead a coupling not to the vector potential, but to the electric field $\mathbf{E}_\perp$. As a slight warning, here things can again get a little bit confusing: compare the discussion in Cohen-Tannoudji, Dupont-Roc and Grynberg [7], and the lecture notes by K.P. Marzlin, http://qis.ucalgary.ca/ pmarzlin/lectures/al0203/ who gives more detailed derivations. In fact, one has to interpret the field in the transformed Hamiltonian as a displacement field $D_\perp$ rather than the electric field $E_\perp$.