In QM, we solve for the eigen kets of the Hamiltonian operator $\hat{H}$ and say that the state of my system lies in a linear superposition of these eigenstates $\{|n\rangle\}$ as the relation implies $|\psi\rangle=\sum_{n=0}^{\infty}\left |n\rangle\langle n|\psi \right \rangle$ and then everything about the system, its momentum, position etc. can be inferred by the action of corresponding operators on $|\psi\rangle$. Similar is done for the case of a simple harmonic oscillator.

  1. But, what then is the meaning of coherent states in this context and what relation do coherent states hold with the state of the system $|\psi\rangle$? Do coherent states just provide us with basis to express out state in?

  2. Why is the Hamiltonian operator so special in the case of generating basis or solving for the state of a quantum system? why can't we write the state of the system in the momentum basis for example without solving for the eigenbasis of $\hat{H}$? Because if we can do so, for any system, $\hat{p}|\phi\rangle=\phi|\phi\rangle$ would give us the state of the system as $|\psi\rangle=\int |\phi\rangle\left \langle \phi|\psi \right \rangle d\phi$ would give us the same state for every system.


1 Answer 1


To your first question, yes. They are just a basis for Hilbert space. They can be used to express any state as a superposition of basis elements and the same is true for the states of any Hermitian operator (any observable). So you could express an arbitrary state in terms of momentum eigenstates, etc.

To you second question the states provided by the Hamiltonian are special because they are related to the time evolution of the wave function. Just writing a function as a superposition of other functions may not seem very meaningful at first glance, but here the Hamiltonian is the operator that is responsible for time evolution. Since the energy eigenstates are eigenstates of the Hamiltonian there is no mixing of energy states in time after an energy measurement. This is the nature behind their "coherence". Measure E placed the system in an H eigenstate, time evolution of the wavefunction is trivial (a scalar phase times the energy state of the previous measurement). All future E measurements will be the same. Measure position and the wavefunction immediately after measurement is a Dirac delta function. To get the correct time evolution you need to write the delta function as a superposition if energy eigenstates. Time evolution will not be trivial, the state mixture will change in time and future measurements of position will differ from the last measurement. This is not coherent.

Hamiltonian eigenstates do in fact have a special status in QM. Some of my statements may change if there is a time dependent potential, but hopefully the gist is clear.

  • $\begingroup$ Thanks for the answer @ggcg. What about this "why can't we write the state of the system in the momentum basis for example without solving for the eigenbasis of $\hat{H}$? Because if we can do so, for any system, $\hat{p}|ϕ⟩=ϕ|ϕ⟩$would give us the state of the system as |ψ⟩=∫|ϕ⟩⟨ϕ|ψ⟩dϕ would give us the same state for every system. " $\endgroup$ Jun 5, 2018 at 12:14
  • $\begingroup$ You absolutely can write the state of a system (at an instant) in terms of the momentum states (that is implicitly addressed in my response). But you will not be able to evolve the state in time for a general system. $\endgroup$
    – user196418
    Jun 5, 2018 at 12:22
  • $\begingroup$ Wont that give me the same state for all the systems? $\endgroup$ Jun 5, 2018 at 12:25
  • $\begingroup$ It darn well better! The point of the superposition is to give a "representation" of a given state in terms of other states. The question I think you want to ask is what have I gained by doing a superposition. In QM the coefficients of the expansion are probability amplitudes for measuring a given value associated with the observable for which the basis is an eigenstate. So, it you expand an arbitrary function in terms of a momentum basis the coefficient cn will give the likelihood of measuring a value of pn, etc. Again, the energy states provide a natural basis for time evolution. $\endgroup$
    – user196418
    Jun 5, 2018 at 12:57
  • $\begingroup$ @NamanAgarwal, do you mean "same state for all the systems" or same state for all the basis choices? This is different. $\endgroup$
    – user196418
    Jun 5, 2018 at 23:02

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