1
$\begingroup$

In the text Gerry Knight "Introduction to Quantum Optics" they start with electric and magnetic fields $$E_x(z,t) = \bigg(\frac{2 \omega^2}{V \epsilon_0}\bigg)q(t) \sin(kz)$$ and $$B_y(z,t) = \bigg(\frac{\mu_0 \epsilon_0}{k} \bigg) \bigg(\frac{2 \omega^2}{V \epsilon_0} \bigg)^{\frac{1}{2}}\dot{q}(t)\cos(kz).$$

What I don't understand is how they simply identify $q$ and $\dot{q}$ with the position and momentum operators simply because the Hamiltonain is the same form as that of the Harmonic oscillator: $$H = \frac{1}{2} \int dV [\epsilon_0 E_x^{2}(z,t) + \frac{1}{\mu_0}B_y^{2}(z,t)] = \frac{1}{2}(\dot{q}^2 + \omega^2 q^2).$$ Whereas in the equations of electric and magnetic field, $q$ and $\dot{q}$ are not even functions of position or momentum. Also, do we then simply assume that they obey the commutation relation $[\hat{q}, \hat{p}] = i \hbar$ since we have identified them with the position and momentum operators which obey this commutation relation?

Thanks for any assistance.

$\endgroup$
  • $\begingroup$ It's not clear from your question whether you think of $q$ as physical position (what you measure with a ruler) or as a generalized coordinate. It is the latter. Perhaps that helps? $\endgroup$ – garyp Jul 7 '17 at 12:14
  • $\begingroup$ This is the key observation of classical field theory: it is formally identical to a chain of notional coupled oscillators, an infinity of them. The q s, at each point in spacetime represent displacement of each such oscillator from equilibrium. In this language, they can be easily quantized, through second quantization methods. $\endgroup$ – Cosmas Zachos Jul 7 '17 at 14:16
  • $\begingroup$ Related 293088. $\endgroup$ – Cosmas Zachos Jul 7 '17 at 14:19
  • 1
    $\begingroup$ Yes, on the right track, but there is a wrinkle... the oscillator at each z is coupled to its neighbors (the gradient terms in Maxwell's eqns...) only in k space they are decoupled, and then the classical expression is what you are visualizing. It is easier to see this with phonons, whose index i corresponds to your z, and their k to your k, etc... But, yes, this is the idea of field quantization.... $\endgroup$ – Cosmas Zachos Jul 8 '17 at 14:23
  • 1
    $\begingroup$ I should think so... They are considering a Single phonon (oscillator normal mode) for fixed k and ω... But the Hamiltonian should superpose an infinity of them, for all such k and ω... I'm not familiar with the text and its methods... maybe alternate texts could help? $\endgroup$ – Cosmas Zachos Jul 10 '17 at 18:58

Your Answer

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