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Apr 29, 2020 at 12:46 answer added PPIP timeline score: 1
Jan 16, 2020 at 20:03 comment added Artem Alexandrov I am not sure, I just suggest
Jan 16, 2020 at 19:58 comment added dfgoe55 Why you said that in term of creation-annihilation operators the equivalence is obvious? How can I prove that?
Jan 16, 2020 at 17:17 comment added dfgoe55 I have used the canonical commutation relations between $q$ and $p$ to rewrite the term $(pq^2 + q^2p)$ but the problem is that the constant I have written is a complex number and therefore the energy of the system is shifted by a complex quantity, that is obviously wrong.
Jan 16, 2020 at 17:04 comment added Artem Alexandrov What is $\hbar$? How does it appear?
Jan 16, 2020 at 16:31 comment added dfgoe55 I put the Hamiltonian in this form $H = (\frac{p}{\sqrt{2m}} + \sqrt{\frac{m}{2}} q^2)^2 + (\frac{m}{2} \omega q + \frac{\alpha + i \hbar \beta}{\omega \sqrt{2m}})^2 - constant^2$. Under an proper change of variables It seems that is the Hamiltonian of a quantum harmonic oscillator without any strange term.
Jan 15, 2020 at 17:32 comment added Artem Alexandrov It is useful to start from the rewriting hamiltonian in terms of creation-annihilation operators. This makes Hamiltonian more clear but consumes lot of time. It seems that you should use perturbation theory but if you provide the text of the problem may be it ll be useful. Also, if you rewrite Hamiltonian in terms of creation-annihilation operators, it seems that equivalence is obvious
Jan 15, 2020 at 17:26 history edited lurscher
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Jan 15, 2020 at 17:20 review First posts
Jan 15, 2020 at 20:34
Jan 15, 2020 at 17:16 history asked dfgoe55 CC BY-SA 4.0