Suppose we have two scalar fields $\varphi, \kappa$. Next, suppose there is a region in space where they are mix with each other, i.e., we have a lagrangian $$ \tag 1 L_{\text{int}} = A \varphi \kappa $$ By taking into account their kinetic term, we have following EOMS: $$ \left(\omega^{2} + \partial_{\mathbf{r}}^2 - \begin{pmatrix}0 & A \\ A & 0\end{pmatrix}\right)\begin{pmatrix}\varphi\\ \kappa\end{pmatrix} = 0 $$ It gives rise to particle oscillations.

Next, suppose we have a beam of $\varphi$ particles propagating along $z$ axis. After entering the domain (say, at $z=0$) in which there is the interaction $(1)$ it begins to oscillate into $\kappa$ particle; this is because it is constructed from two "mass" eigenstates $\psi_{\pm} = \frac{1}{\sqrt{2}}(\varphi \pm \kappa)$ with the dispersion relations $$ k_{\pm} = \sqrt{\omega^2 \mp A} $$

I want to calculate the probability of oscillation at $z>0$. It turns out that it is proportional to $$ P_{\varphi\to\kappa}\sim |e^{-ik_{+}z}-e^{-ik_{-}z}|^2 $$ It turns out that for $|A|>\omega$ one of the momenta $k_{+}$, $k_{-}$ becomes imaginary, and the probability doesn't behave as oscillating function. By the other words, the refractive index for one of "mass" eigenstates becomes imaginary.

What is the physical reason for this? Is the oscillation interpretation valid in this case?

  • $\begingroup$ You might already know this, but this reminds me very much of Rabi oscillations in Quantum Mechanics. However while I've forgotten the derivation, the result seems very nearly the same, except that this problem doesn't occur. I'd need to go over it again. At any rate, a probability (and not a probability density, I assume) that goes to infinity is obviously problematic. Are your calculations too involved to put up here? $\endgroup$ – Philip Cherian May 26 '17 at 23:04
  • $\begingroup$ @PhilipCherian: the basis for which EOMs are diagonal is $$ \begin{pmatrix} \varphi ' \\ \kappa' \end{pmatrix} = \begin{pmatrix} \cos(\pi/4) & \sin(\pi/4) \\ -\sin(\pi/4) & \cos(\pi/4) \end{pmatrix}\begin{pmatrix}\varphi \\ \kappa \end{pmatrix} = T \begin{pmatrix}\varphi \\ \kappa \end{pmatrix} $$ The probability to find the state $\kappa$ generated by initial state $\varphi$ at the distance $z$ reads $$ P_{\varphi \to \kappa} = \left|(0, \ 1)T^{T}\text{diag}(e^{-ik_{+}z}, e^{-ik_{-}z})T\begin{pmatrix}1 \\ 0\end{pmatrix}\right|^{2}, $$ giving the result. $\endgroup$ – Name YYY May 27 '17 at 22:02
  • $\begingroup$ @PhilipCherian : physically such problem appears when considering axion-photon conversion in presence of external magnetic field. $\endgroup$ – Name YYY May 27 '17 at 22:05
  • $\begingroup$ Related 290475. $\endgroup$ – Cosmas Zachos Jul 30 '17 at 0:42

Consider your potential as a quadratic form, $$ V= (\varphi,\kappa ) \begin{pmatrix}\omega^{2} & -A \\ -A & \omega^{2}\end{pmatrix} \begin{pmatrix}\varphi\\ \kappa\end{pmatrix} . $$

For $|A|< \omega^2 $ , it is concave, so stable, with the minimum at the origin of the fields. Oscillations occur with the two eigenfrequencies you found in the directions (1,1) and (1,-1).

For $|A| = \omega^2 $, it amounts to an infinite trough, with the flat direction along (1,1). (Such trough potentials afflict SUGRA).

For $|A| > \omega^2 $, it is an unstable saddle; since the energy is unbounded below, the system will transition down the saddle slope, again along the (1,1) direction, "forever", just like unstable 2-state molecules. So your state is readily $\varphi+\kappa$.

No oscillations, and soon your hamiltonian is irrelevant/inapplicable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.