# Using the Scalar Electrostatic Potential to Calculate Transition Probabilities

transition probabilites of atomic systems prone to some time-varying electromagnetic field are very often calculated using perturbation theory leading to expressions including the vector potential $\mathbf{A}$. This approach seems to be very complicated and I am wondering:

### Is using the electrostatic potential $U$ to calculate "electric transitions" (approximately) sufficient?

This was the short version of my question; below you will find some further notes that (hopefully) contain all information needed to clearify it.

In relativistic quantum mechanics, the action of an an electrodynamic field on an electron can be realized by minimal coupling using the replacement $\partial_\mu \rightarrow \partial_\mu - \mathrm{i}eA_\mu$. Thus, in a certain sense, (quantum) electrodynamics can be derived from the Dirac equation using the invariance of the free-electron action with respect to a phase change.

### Electromagnetic Fields and the Schrödinger Equation

Nevertheless, often one must ignore the beauty of gauge theory to be able to practically calculate something under certain simplifications using the non-relativistic Schrödinger-equation.

There, one can incorporate an external electromagnetic field applying the replacements $$H\left(\mathbf{p},\mathbf{x}\right) \rightarrow H\left(\mathbf{p}-e\mathbf{A}\left(t,\mathbf{x}\right),\mathbf{x}\right)+eU\left(t,\mathbf{x}\right)$$ to the Hamiltonian of the system where $\mathbf{A}$ is the known vector potential and $U$ is the potential corresponding to the electric field.

In the Coulomb gauge $\mathrm{div}\mathbf{A}=0$ one can always achieve $U\equiv 0$ if no sources are present and the overall Hamiltonian will look like $$H = H_0 -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$ where $H_0 = \frac{\mathbf{p}^2}{2m} + V(\mathbf{x})$ and the term in $\mathbf{A}^2$, the ponderomotive potential has been neglected.

### Time-Dependent Perturbation Theory

The last expression for $H$ has the form $$H = H_0(\mathbf{p},\mathbf{x}) + H_t(\mathbf{p},\mathbf{x},t)\ .$$ Using the Dyson series, one can formally write down the time-dependent solution in the interaction picture as $$|\psi_I(t)> = T\exp\left[\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}H_{t,I}\left(t^{\prime}\right)dt^{\prime}\right]|\psi_I(t_0)> .$$

Then, one takes only the first two terms into account to find that transition amplitudes are approximately given by $$A_{n\rightarrow m}\approx\frac{1}{\mathrm{i}\hbar}\int_{t_{0}}^{t}\left\langle m\right|H_{t,I}\left(t^{\prime}\right)\left|n\right\rangle dt^{\prime}\ \mathrm{with}$$ $$H_{t,I}\left(t\right) = e^{\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}H_t\left(t\right)e^{-\mathrm{i}\left(t-t_{0}\right)H_{0}/\hbar}\ .$$

### Relation to the Question

In the given case we have $$H_t = -\frac{e}{m}\mathbf{A}\left(t,\mathbf{x}\right)\mathbf{p}$$ and assuming that $$\mathbf{A}\left(\mathbf{x},t\right) = \mathbf{A}\left(\mathbf{x}\right) T(t)\ ,$$ and further using $$\mathbf{p}=\frac{\mathrm{i}m}{\hbar}\left[H_{0},\mathbf{x}\right]\ ,$$ one will have to solve $$A_{n\rightarrow m}\propto <m|\mathbf{A}\left(\mathbf{x}\right)\cdot \mathbf{x}|n>\int_{t_0}^t e^{-\mathrm{i}\omega_{mn}\tau}T(\tau) d\tau\ +\ \mathrm{h.c.}$$ with $\hbar \omega_{mn} = E_m - E_n$, where the $E_i$ are energy levels of the unperturbed $H_0$.

Using here the vector potential $\mathbf{A}$ is the correct way to incorporate the field, of course. For me, it seems extremely complicated to do so since to calculate the transition probabilities for an arbitrary field distribution one will have to expand $\mathbf{A}$ into vector spherical harmonics.

So, the use of a scalar potential like $U$ would greatly simplify computations and my question can now be re-asked as

### Why can't we just simply use $U(\mathbf{x})\cdot T(t)$ to calculate the transition amplitudes $A_{n\rightarrow m}$ for "electric excitations"?

Thank you very much for insights, comments and corrections.

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Also a minor quibble: the minimal substitution rule is still correct in the non-relativistic limit. In fact the formula you gave for $H$ is exactly that. –  BebopButUnsteady Jun 20 '11 at 23:02

Of course, one cannot find in general a gauge for which $\mathbf{A}=0$. So I was asking for some physically motivated approxomative approach using $U$ alone. However, I will accept your answer since the question seems too technical for this site. Greets –  Robert Filter Jun 23 '11 at 8:04
@Robert Filter: I'm sorry if that was not helpful. But I'm curious what you're looking for. No matter what your gauge fixing procedure is you can never specify the EM field with just $U$. So how can you have an approximation which converges to the right answer using $U$ alone? How would your approximation know which EM Field you were talking about? –  BebopButUnsteady Jun 23 '11 at 20:53