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In the Heisenberg picture, operators evolve according to

$$ \partial_t A = \frac{1}{i\hbar} [A,H]. $$

My question is, does the following relation hold?

$$ (X_H)^2 = (X^2)_H $$

The system (Hamiltonian) in mind is the 1D harmonic oscillator,

$$ H = a^{\dagger}a + \frac 12 $$

I would like to say that it does, but the Heisenberg equations of motion for $x^2$ and $p^2$ are very messy (and I believe nonlinear).

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    $\begingroup$ So to be very clear, is your question if $X_H^2=(X^2)_H$? The subscript $H$ denotes here operator in Heisenberg picture. If so, you should write down the definition of e.g. $X_H=\ldots$ in terms of $X$ (the position operator in the Schrödinger picture). $\endgroup$ Oct 9, 2023 at 22:01
  • $\begingroup$ Yes, that is my question. Are you saying something like the relation $X_H = U^{\dagger} X_S U$. In which case it is clear that $(X_H)^2 =(X^2)_H$ ? $\endgroup$
    – Hurricane
    Oct 9, 2023 at 22:09
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    $\begingroup$ Indeed. Consider to write an answer yourself. $\endgroup$ Oct 9, 2023 at 22:14

1 Answer 1

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The answer can be seen from going back to the definition of the Heisenberg operators in terms of the Schrodinger operators.

$$ X_H = U^{\dagger}X_S U $$

Then, we have

$$ (X_H)^2 = U^{\dagger}X_S U U^{\dagger}X_S U = U^{\dagger}(X_S)^2 U = (X^2)_H $$

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