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The quantum adiabatic theorem states that: A parametric system remains in its instantaneous eigenstate (with a phase difference) if one of the parameters of the Hamiltonian changes slow enough. This is very counter-intuitively to me giving the following example: Considering a material in a magnetic field, even we turn on the magnetic field as slow as possible, the eigenstates and their energies will change non-trivially, they do not remain the same. I don't know how to understand this point.

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3 Answers 3

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Imagine a Hamiltonian of the form $\hat H = \hat H_0 + \lambda \hat V$ where $\hat V$ is not assumed to be small. If $\lambda = 0$, then we can find a set of energy eigenstates $|\psi_{n}\rangle$ such that $\hat H_0|\psi_n\rangle = E_n|\psi_n\rangle$. Now we let $\lambda = 1$, at which point we get a completely different set of eigenstates $|\phi_n\rangle$, where $(\hat H_0 + \hat V)|\phi_n\rangle = \xi_n |\phi_n\rangle$. To be clear, these are two different Hamiltonians with two different sets of eigenvectors corresponding to two different sets of eigenvalues.

Now imagine that $\lambda$ is a function of $t$, such that $\lambda(0)=0$ and $\lambda(T)=1$ for some large time $T$. Physically, we are imagining that the $\hat V$ part of the Hamiltonian is being turned on very, very slowly. The adiabatic theorem states that if $\lambda'(t)$ is sufficiently small, then if the system is initially in some eigenstate $|\psi_n\rangle$ at $t=0$, it will evolve into the corresponding eigenstate $|\phi_n\rangle$ at $t=T$ (up to a phase factor).

In particular, if a system is in the ground state of $\hat H_0$ at $t=0$, then it will be found in the ground state of $\hat H_0 + \hat V$ at $t=T$.

What is meant by "sufficiently slowly" depends on the gap separating $|\psi_n\rangle$ from its nearest neighbors. In particular, the theorem fails if the spectrum of the Hamiltonian is continuous$^\dagger$ (so there is no gap between $|\psi_n\rangle$ and its neighbors) or if $|\psi_n\rangle$ is degenerate (or becomes degenerate at any point).


$^\dagger$Though this result can be generalized.

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    $\begingroup$ Thank you so much for your answer! Is it because, in my example, the operation of turning on the magnetic field, not an adiabatic process? Because the electron wave-function is a plane wave before turning on the field and then become an oscillator after. I can't see how the corresponding eigenstate could evolve into each other. $\endgroup$
    – Albert
    Oct 6, 2020 at 16:40
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    $\begingroup$ @Albert Are you imagining a free electron in a uniform magnetic field? $\endgroup$
    – J. Murray
    Oct 6, 2020 at 17:37
  • $\begingroup$ Yes, in the simplest case. $\endgroup$
    – Albert
    Oct 6, 2020 at 19:03
  • $\begingroup$ @Albert You have to be careful - the free particle Hamiltonian has a continuous spectrum and the Landau levels corresponding to a constant magnetic field are extremely degenerate. However, note that an electron in the $n^{th}$ Landau level with $B(t=0)=B_0$ will remain in the $n^{th}$ Landau level (even though the wavefunction and energy will change) if $B$ is changed sufficiently slowly. $\endgroup$
    – J. Murray
    Oct 6, 2020 at 19:55
  • $\begingroup$ @J.Murray, changing of $B$ will change the degeneracy of Landau levels (LLs). Won't it result in shuffling of electrons to/from different LLs? $\endgroup$ Feb 9, 2021 at 7:05
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J. Murray already gave you a detailed (very nice) answer.

Intuitively: adiabatic processes are (by definition) something that live in the limbo between dynamics and statics.

In quantum mechanics the "dynamics" is given by the Schrodinger equation, the "statics" by the eigenvalue equation for the Hamiltonian (i.e. the energy eigenstates of the Hamiltonian are the "stationary states").

Now, the adiabatic theorem is telling you that if the Hamiltonian is "modulated" slowly enough (adiabatically), then the evolution is such that if you are in an energy eigenstate of the initial Hamiltonian at $t=0$, you will remain in the corresponding energy eigenstate of the "modulated Hamiltonian" at $t>0$.

This does not mean that "the eigenstates and their energies will change non-trivially": they can change, typically in a non-trivial way, but continuously (i.e. are continuously "modulated" by the adiabatic parameter $\lambda$ of J. Murray's answer).

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  • $\begingroup$ Thank you for your answer. If I understood correctly, the point is that the states can change but they have the same instantaneous eigenstate. $\endgroup$
    – Albert
    Oct 6, 2020 at 19:02
  • $\begingroup$ The state is (if you start from an eigenstate) an istantaneous eigenstate of the Hamiltonian at that time (remember that the Hamiltonian slowly changes in time, and so its eigenstate). $\endgroup$
    – Quillo
    Oct 7, 2020 at 8:20
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Conversely; application;two magnetic emitters then, within each others range (gap) will pull at each others field. hence using these references above, a ‘quick switch and build up ‘ of the wave front attachment densor field attractors vs. the slow build up of the e.states will provide an unattached force wave of disturbed e. state mag vector that a slowly building force state can capture/attract/ride/pull

So in effect it captures and reacts to densor field that has no connection and gets a pull pulse ‘free’ as if climbing through an ether. An electric charge will vibrate the gas that throws and pulls the resultant mag field giving even more free pulse pull to the containment device. you could also use a ceramic with ferro/mag emitters that respond in sequence to timed directional electron flow. At the correct clipping rate, it will pull on its own container in one direction, due to the non attachment of the first emission/wave. The key is how to change the eignestates, i don't think you can measure them unless you use some kind of third source standing wave emmision that it interferes with and observe the resulting interference patterns as pulse timing changes.

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