# Why is the classical solution of an harmonic oscillator more like a coherent state than an eigenstate of the Hamiltonian?

In general the classical harmonic oscillator should be a superposition of eigenstates of Hamiltonian. Why does it always turn out to be a coherent state than any other kind of superposition?

Note: A coherent state is an eigenstate of the lowering operator $$\hat{a}$$, and it's expected value of position and momentum vary sinusoidally, while its product of uncertainties of position and momentum remains $$\hbar/4$$.

• There is nothing unusual about the classical limit not being a eigenstate. You encounter that for the first time with the free particle (you must sum over the plane-waves to get a wave-packet, right?). – dmckee --- ex-moderator kitten Sep 5 '19 at 18:47
• Can you share literature references on this subject? – my2cts Sep 5 '19 at 18:50
• @my2cts It has been mentioned on the site before: physics.stackexchange.com/questions/397474/… (and that post was linked in the comments of something recent). – dmckee --- ex-moderator kitten Sep 5 '19 at 18:54
• My comment is: please clarify what these statements are based on. – my2cts Sep 5 '19 at 20:45
• Schroedinger constructed these wavepackets expressly so that their centers replicate the phase-space motion of a classical oscillator. You are puzzled that they can achieve that? Can you formalize your technical question, if that is what it is? – Cosmas Zachos Sep 6 '19 at 14:58

The larger question is why would you expect the energy eigenstates to look like anything you have classical experience with? In some sense, saying a quantum state "has a definite energy $$E$$" is just language we use to describe the math. Useful language, but language nonetheless. It's really an assertion we place into the language when we say that eigenstates of the Hamiltionian "have" a definite energy. What does it even mean for a state to "have" an energy in quantum mechanics? We have to define the word "have an energy" for ourselves.