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Linear Chain of oscillators (Quantum Treatment)

  1. Feb 17, 2014 #1
    I am reading Field Quantization by Greiner and Rienhardt and I am trying to prove that
    [itex] i\hbar \frac{d\hat{b_k}}{dt} = [\hat{b_k},H]=\hbar \omega_k \hat{b_k}[/itex] where [tex]
    \hat{b_k} = \frac{1}{2} \sum_n u^{k*}_n(q_n(t)+ \frac{i}{\omega_km}\hat{p}_n(t))[/tex] and
    [tex] H= \sum_{n=1}^{N}\frac{1}{2m}\hat{p^2_n} + \sum_{n=1}^{N} \frac{\kappa}{2}(\hat{q_{n+1}}-\hat{q_n})^2[/tex] lastly [tex] u_n^k= \frac{1}{\sqrt{n}} e^{ikan}[/tex] I am interested in doing it the brute force way namely using the Hamilton as specifically stated above and the definition of [itex] \hat{b_k}[/itex]. Without showing all my work, just by directly plugging the equations into the Heisenberg equation of motion(commutator). I arrived at the following equation [tex] \frac{1}{2}\sum_n \frac{u^k_n}{2m}[q_n,p_n^2] + \frac{1}{2} \sum_n \frac{i\kappa u^{k*}_n}{2 \omega_k m}[p_n,(q_{n+1}-q_n)^2] [/tex] which according to my calculation reduces to [tex] \frac{1}{2}\left(\sum_n u^{k*}_n \hbar(\frac{i\hat{p_n}}{m}-\omega_k(q_{n+1}-q_n))\right)[/tex]
    The problem I am facing is I need to get rid of the [itex] q_{n+1} [/itex] term but I do not know how to.
  2. jcsd
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