Consider $N$ coupled oscillators whose Hamiltonian is $$\hat{H}=\sum_j^N\frac{\hat{p}_j^2}{2m} +\frac{1}{2}K(\hat{x}_{j+1}-\hat{x}_j)^2,$$ which represents a system like ($K$ is the spring constant):

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Are the following commutation relations $$[\hat{x}_j,\hat{p}_{j'}]=i\hbar\delta_{jj'},$$ $$[\hat{x}_j,\hat{x}_{j'}]=0,$$$$[\hat{p}_j,\hat{p}_{j'}]=0.$$ true?

If yes, why? I'd think that these commutation relations are true for uncoupled oscillators that don't affect each other, but not true for coupled oscillators.

  • $\begingroup$ Are you those operators in the Schrödinger or Heisenberg picture? $\endgroup$ Feb 4 at 14:18
  • 1
    $\begingroup$ Yes, they are true. How did you quantize the corresponding classical system? Run to your classical FT review. They are related to the decoupled normal modes by precisely the same linear algebraic maps as their classical origins! $\endgroup$ Feb 4 at 14:39

1 Answer 1


The Hilbert space of the system is the direct product of the Hilbert spaces corresponding to separate oscillators. Operators for every specific oscillator are transformations in its own Hilbert space, which do not affect other Hilbert spaces.

However, the above is true for the operators taken at the same time instant. If we now work in Heisenberg picture, write Heisenberg equations of motion for operators and solve them, the operators at time $t$ will be linear combinations of the operators at some previous time $t'$ and vice versa. These operators taken at different time moments will not commute, generally speaking, even for the same oscilator.


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