# Electric-field/ length gauge : momentum displacement property

I am learning about the Theory of intense laser-induced molecular dissociation. To set the notations, the radiation field Hamiltonian is for a single charged particule : $$H = \frac{1}{2m}(p-q\hat{A}(r))^2 + V(r) + \hbar \omega (\hat{a}^{\dagger}\hat{a}+1/2)$$ with $$\hat{A}$$ the vector potential. The length gauge (Electric-field) has been introduced along side the property that, with the long-wavelength approximation :$$\hat{O} p \hat{O}^\dagger = p + q\hat{A}$$ $$\hat{A} = \beta\{\hat{a} + \hat{a}^{\dagger}\}\epsilon$$ with the gauge transformation $$\hat{O} = e^{-\frac{i}{\hbar}qr\cdot \hat{A}} = e^{-\frac{i}{\hbar}\beta q r\cdot \epsilon[\hat{a} + \hat{a}^\dagger]}$$, $$\beta = (\frac{\hbar}{2\epsilon_0 \omega V})^{1/2}$$ where $$\omega = c|k|$$ is the frequency of the mode and $$V$$ the finite volume leading to a discretization of the laser mode.

But I get, with $$p = -i\hbar\nabla_r$$ : $$\hat{O} p \hat{O}^\dagger = qr\cdot \nabla_r \hat{A} + q\hat{A} = q\hat{A}$$

Edit, derivation : \begin{align} \hat{O} p \hat{O}^\dagger &= \hat{O} (-i\hbar)\nabla_r \hat{O}^\dagger = \hat{O} (-i\hbar)\nabla_r e^{\frac{i}{\hbar}qr\cdot \hat{A}}\\ &= \hat{O} (-i\hbar) \frac{i}{\hbar}q(\nabla_r r\cdot \hat{A}(r))\hat{O}^\dagger \\ &= q(\hat{A}(r) + r\cdot \nabla_r\hat{A}(r)) \\ &= q\hat{A} \end{align}

since the dot product is bilinear

• Could you show the steps of your derivation? Jun 26 at 22:38
• Hi, would you mind sharing a reference regarding this derivation? Jun 29 at 19:46
• Sorry, the only resource I have read only give the property
– mle
Jun 29 at 20:31
• Thanks, sorry for the bother Jun 30 at 12:45

Thanks for adding your derivation. You made a common mistake, which is to treat $$\hat{O}$$ as an ordinary exponential function. However, here you are exponentiating an operator, which calls for more care. Formally, this can be treated as a matrix exponential, which is defined in terms of its Taylor expansion: $$e^\hat{x} = 1 + \hat{x} + \frac{\hat{x}^2}{2!} + \dots$$ Then you get \begin{align} \hat{O} \vec{p} \hat{O}^\dagger &= \left( 1 - \frac{i}{\hbar}q\vec{r} \cdot \vec{A} + \dots \right) \vec{p} \left( 1 + \frac{i}{\hbar}q\vec{r} \cdot \vec{A} + \dots \right)\\ &= \vec{p} -\frac{iq}{\hbar} \left[ \vec{r} \cdot \vec{A}, \vec{p} \right] + \dots\\ &= \vec{p} -\frac{iq}{\hbar} \left[ r_j , \vec{p} \right]A_j -\frac{iq}{\hbar} r_j \left[ A_j, \vec{p} \right] + \dots \end{align} where I have adopted the Einstein summation convention. Immediately you see the standalone $$\hat{p}$$ showing up. The canonical commutation relation, $$\left[ r_j, p_k \right] = i\hbar\delta_{jk}$$, can be substituted into the first commutator. Acting with the second commutator on a test (wave) function $$\psi(\vec{r},t)$$ lets you simplify to $$\hat{O} \vec{p}\hat{O}^\dagger = \vec{p} + q\vec{A} + q r_j \cdot \nabla_\vec{r} A_j + \dots$$ You already concluded that the third term vanishes in your case. What remains is to convince yourself that the higher-order terms (represented by $$\dots$$) do not modify the result.
• Ty, the answer is quite clear. Be careful with you Einstein notation index $i$, since it is already taken by the complex numbers.