3
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Could you show the steps of your derivation? $\endgroup$
    – Anyon
    Commented Jun 26, 2023 at 22:38
  • $\begingroup$ Hi, would you mind sharing a reference regarding this derivation? $\endgroup$
    – capitn96
    Commented Jun 29, 2023 at 19:46
  • 1
    $\begingroup$ Sorry, the only resource I have read only give the property $\endgroup$
    – mle
    Commented Jun 29, 2023 at 20:31
  • $\begingroup$ Thanks, sorry for the bother $\endgroup$
    – capitn96
    Commented Jun 30, 2023 at 12:45

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Ty, the answer is quite clear. Be careful with you Einstein notation index $i$, since it is already taken by the complex numbers. $\endgroup$
    – mle
    Commented Jul 1, 2023 at 9:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .