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  • For a LC harmonic oscillator, the energy spectrum is evenly spaced by

$$ \Delta E = \hbar \omega \quad \omega = {1\over \sqrt{LC} } $$

  • For two inductively coupled LC harmonic oscillators with mutual inductance $M$, starting from:

$$\begin{cases} \quad (i\omega L_1 + {1\over i\omega C_1})I_1 + i\omega M I_2 &= U \\ \quad (i\omega L_2 + {1\over i\omega C_2})I_2 + i\omega M I_1 &= 0 \end{cases} $$

  • The eigen-energies are (calculated by Mathematica):

$$ \left[ \left[W = \sqrt{\frac{1}{2}} \sqrt{-\frac{L_{1} C_{1} + L_{2} C_{2} - \sqrt{L_{1}^{2} C_{1}^{2} + 4 \, C_{1} M^{2} C_{2} - 2 \, L_{1} C_{1} L_{2} C_{2} + L_{2}^{2} C_{2}^{2}}}{C_{1} M^{2} C_{2} - L_{1} C_{1} L_{2} C_{2}}} \right], \left[W = \sqrt{\frac{1}{2}} \sqrt{-\frac{L_{1} C_{1} + L_{2} C_{2} + \sqrt{L_{1}^{2} C_{1}^{2} + 4 \, C_{1} M^{2} C_{2} - 2 \, L_{1} C_{1} L_{2} C_{2} + L_{2}^{2} C_{2}^{2}}}{C_{1} M^{2} C_{2} - L_{1} C_{1} L_{2} C_{2}}} \right] \right] $$

  • In the case where $L_1 = L_2, C_1 = C_2$, the above becomes:

$$ \left[ \left[W = \frac{1}{\sqrt{L_{1} C_{1} + C_{1} M}}\right], \left[W = \frac{1}{\sqrt{L_{1} C_{1} - C_{1} M}}\right] \right] $$

  • The question is: Now that there are two different frequencies (close to each other when $M$ is small), how does the energy spectrum look like?

  • I have found this: https://en.wikipedia.org/wiki/Quantum_LC_circuit, but the spectrum is not discussed.

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When the two (identical) oscillators are decoupled, the energy spectrum is identical to that of a harmonic oscillator albeit each energy level is doubly degenerate. In the case of the coupled oscillator, it can be viewed as this degeneracy being lifted. As the coupling $M$ increases, the splitting is higher. enter image description here

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  • $\begingroup$ this is also what I would guess through physical reasoning or educated guess, but did you get the above though actual derivation ? $\endgroup$ – Rick Sep 25 '20 at 7:06
  • $\begingroup$ You found the eigenenergies, so I am confused as to what other derivation you’re seeking $\endgroup$ – Superfast Jellyfish Sep 25 '20 at 7:09
  • $\begingroup$ same way as you get the spectrum from quantum mechanics textbook, for example as done in Griffith. $\endgroup$ – Rick Sep 25 '20 at 7:21
  • $\begingroup$ Coupled oscillators can generally be solved by a coordinate transform analogous to going to centre of mass frame of reference. That’s a separate question however. $\endgroup$ – Superfast Jellyfish Sep 25 '20 at 8:08

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