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Harmonic oscillator

  1. Nov 22, 2013 #1
    Let a+,a- be the ladder operators of the harmonic oscillators. In my book I encountered the hamiltonian:

    H = hbarω(a+a-+½) + hbarω0(a++a-)
    Now the first term is just the regular harmonic oscillator and the second term can be rewritten with the transformation equations for x and p to the ladder operators as:
    hbarω0(a++a-) = x/(√(2hbar/mω))
    My question is: Does this last term just represent a translation in the origin of the harmonic oscillator i.e. the potential is mω2(x-x0)^2 where x0 is determined by ω0? If so how do I see that algebraically?
     
  2. jcsd
  3. Nov 22, 2013 #2

    hilbert2

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    If the potential energy function in the usual harmonic oscillator Hamiltonian is ##V(x)=kx^{2}##, and in the case of this problem it is ##V(x)=kx^{2}+k'x##, you can complete the square to write the potential in form ##V(x)=a(x-x_{0})^{2}+b##. To find ##a##,##x_{0}## and ##b##, you just require that like powers of ##x## in both sides of equation ##a(x-x_{0})^{2}+b=kx^{2}+k'x## have the same coefficients.
     
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