1
$\begingroup$

The problem of the particle in the external magnetic field is one of the most famous and oldest problems in quantum mechanics. Hamiltonian of the particle in the constant magnetic field is: $$ \mathbf{H} = \frac{1}{2 m} (\mathbf{p} - e \mathbf{A})^2 $$ And the solution for a particular choice of gauge $A = (0 \ \ Bx \ \ 0)$, are the $$ \psi(x, t) = e^{- i \omega t + i p_y y + i p_z z} f_n(x-x_0) $$ Where $$ x_0 = -p_y / eB \qquad f_{n}(x) = \sqrt{\frac{1}{2^{n} n !}}\left(\frac{e B}{\pi}\right)^{1 / 4} e^{-\frac{1}{2} e B x^{2}} H_{n}(\sqrt{e B} x) $$ And this quantization leads to plethora of remarkable effects, like the Magnetic susceptibility of Fermi gas, Quantum hall effect, some properties of graphene, e.t.c.

However, I haven't seen investigation of the particle, for example, heavy quark in external chromomagnetic field. In general, the problem seems to be much more difficult but in case there are only components, corresponding to the Cartan subalgebra the solution should reduce to the electromagnetic picture.

As far as I understand, there would be quantization of each color factor of the wavefunction separately. And the total wavefunction is a product of all separate factors.

Is there any notion of gyromagnetic phenomena for the color-charged particles. Barnet effect https://en.wikipedia.org/wiki/Barnett_effect, Einstein–de Haas effect https://en.wikipedia.org/wiki/Einstein%E2%80%93de_Haas_effect?

I apologise for probably having ill-formulated the questions. I would be grateful for any response or corrections!

$\endgroup$
3
$\begingroup$

The classical equation for motion in a chromomagnetic gauge field are called the Wong equations (S.K.Wong, Il Nuovo Cimento 65A (1970) 689) describe the gauge theory generalization of charge. There is an article here that has an account and more references.

$\endgroup$

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.