Quantum mechanics transitions in an electromagnetic field

1. Sep 3, 2017

chingel

1. The problem statement, all variables and given/known data
This is problem (7.1) from John A. Peacock "Cosmological Physics".

Show that the first-order perturbation term for quantum mechanics with an electromagnetic field, $(e/m) \mathbf{A \cdot p}$ is proportional to the electric dipole moment. What is the interpretation of the $A^2$ term?

2. Relevant equations
We have the hamiltonian
$$H = \frac{1}{2m} (\mathbf{p}-e\mathbf{A})^2 + e\phi + V.$$

In the book they adopt the Coulomb gauge, where $\nabla \cdot \mathbf{A}=0$ and they say that to first order
the perturbation is $H' = \mathbf{A \cdot p}$. This means that $\phi=0$ also I assume?

The text also says (where it refers to this problem), that the transition rate is calculated with the element (using the Fermi's golden rule I assume) $\langle i | \mathbf{A \cdot p} | j\rangle$, and we must show that it is proportional to the dipole moment
$$\langle i | \mathbf{A \cdot p} | j\rangle \propto \langle i | \mathbf{A \cdot x} | j\rangle .$$

3. The attempt at a solution
If for simplicity I take $V=0$, so that $|i\rangle = \int d^3x\ e^{ip_j x}\,|x\rangle$ and it is an eigenvector of the momentum operator, I get
$$\langle i | \mathbf{A \cdot p} | j\rangle = \mathbf{A \cdot p_j}\ \delta^3 (\mathbf{p_j}-\mathbf{p_i}).$$
It seems that because of the delta function there are no transitions, what am I missing?

Also in this case I don't think that $\langle i | \mathbf{A \cdot x} | j\rangle$ will be proportional to the above result:
$$\langle i | \mathbf{A \cdot x} | j\rangle = \int d^3x\ \int d^3y\ e^{i(p_jx-p_iy)} A \cdot x \;\delta^3(\mathbf{x}-\mathbf{y})$$
$$\langle i | \mathbf{A \cdot x} | j\rangle = \int d^3x\ e^{i(p_j-p_i)x} A\cdot x.$$
I don't think this expression is equal to the delta function one above.

2. Sep 8, 2017

PF_Help_Bot

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