- #1
chingel
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Homework Statement
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?
Homework 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
$$\langle i | \mathbf{A \cdot p} | j\rangle \propto \langle i | \mathbf{A \cdot x} | j\rangle .$$
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.