Why the perturbation Hamiltonian of induced dipole in field still $ d \cdot E$? [closed]

Question

We used to calculate the potential energy of permanent dipole in the uniform field is like $-p \cdot E$ ... and when we put a perturbation Hamiltonian of an external field on the neutral atom, that perturbation Hamiltonian is also the same expression. But the neutral atom doesn't have a permanent dipole but instead has an induced dipole..right??

so.. back in classical mechanics... we know the amount of energy shift(or work done) of induced dipole when we increase the E field is ${1 \over 2} \alpha E^2 = {1 \over 2} p \cdot E$. so... why does perturbation Hamiltonian drop ${1 \over 2}$ factor??

• although I am aware that when we diagonalize that perturbation Hamiltonian, the energy shift has ${1 \over 2}$ factor but... still I couldn't get that why induced dipole's potential is written as ${- d \cdot E}$

Answer 1

You are confusing two different kinds of energies.

Both in classical mechanics and quantum mechanics, the term $-\mathbf d \cdot \mathbf E$ is interaction energy between the charges making up the dipole $\mathbf d$ and the external electric field $\mathbf E$; we can think of this term as an approximate way to express the Coulomb interaction energy between charges in the dipole and charges outside the system. It does not matter how this dipole came about, whether it is permanent or induced; the interaction energy between the two systems is given by this formula. This energy does not take into account increase/decrease in internal energy of the system that has this dipole moment due to internal forces between parts of the dipole.

The formula for energy of induced dipole $\frac{1}{2}\alpha E^2$ is from a different concept of dipole energy; it takes into account also internal energy stored in the polarized system due to internal forces between parts of the dipole. It is work that has to be expended in order to depolarize the dipole in the external field; or energy that the external electric field has done to polarize the dipole.

The difference between these two concepts is similar to the difference between potential energy of a particle in an external field, e.g. $$ V = q_1\phi(\mathbf r_1) $$ for charge $q_1$ in given potential $\phi$ of some other charges, and potential energy of the whole system particle + the other particles making up the external field :

$$ E_p = \frac{1}{2}\sum_{i}\sum_k{'} q_i\phi_k(\mathbf r_i) $$ where $\phi_k(\mathbf r_i)$ is potential due to particle $k$ at position of particle $i$, the factor $1/2$ is to prevent counting the same interaction term twice and ' means the cases $i = k$ are skipped in the sum.