I want to understand what adiabaticity in Quantum Mechanics means. I have attained the next information:

Adiabatic process: gradually changing conditions allow the system to adapt its configuration, so the probability density is modified by the process. If the system begins in an eigenstate of the initial Hamiltonian, will end in the eigenstate that corresponds to the final Hamiltonian.

The adiabatic theorem states that quantum jumps are preferably avoided and that the system tries to retain your state and quantum numbers.

An adiabatic change is one that occurs at a rate much slower than the difference in frequency between the eigen states of energy. In this case, the energy states of the system do not make transitions, so the quantum number is an invariant.

I don't understand completely what these sentences mean. I want to state it in the most simple terms possible.

  • $\begingroup$ What about those sentences do you not understand? Have you seen any examples? $\endgroup$
    – jacob1729
    Commented Mar 11, 2021 at 17:31
  • $\begingroup$ I am particularly interested in adiabaticity for quantum rings. Regarding the second sentence, I do not understand what quantum jumps mean, and why are they preferably avoided. In the third sentence, what does it mean for the energy states to not make transitions? $\endgroup$
    – alefisto
    Commented Mar 11, 2021 at 17:36

1 Answer 1


I am not sure what specific applications of the adiabatic theorem you are looking at for quantum rings, but I can give you a general overview of the quantum adiabatic theorem and break down some of what those words mean.

As review, remember that energy states and levels in quantum mechanics are represented by eigenstates and eigenvalues, respectively, of your time independent Hamiltonian $\hat{H}$. Thus if we our state $| \psi_E \rangle$ is in an energy eigenstate, then it satisfies the following: $$ \hat{H}\ | \psi_E \rangle = E\ | \psi_E \rangle $$

where $E$ would be the energy corresponding to our energy eigenstate.

Now let's say our Hamiltonian now carries some explicit time dependence $\hat{H}(t)$; for example maybe the mass of our particle is now changing in time. In simplest terms, what the quantum adiabatic theorem states is that if your Hamiltonian is changing slowly enough (we'll define this in a second), then if you start in an energy eigenstate $| \psi_E(t = 0)\rangle$, then you will remain in an energy eigenstate $|\psi_E(t)\rangle$ for all time $t$. Thus, you will always have a well defined instantaneous energy for all time (which is what they meant when they said you retain your state and quantum numbers).

If your Hamiltonian is not changing slowly, then in general you will you will have: $$ |\psi(t)\rangle = \sum c_i |\psi_E(t)\rangle $$ which means your state is now in a superposition of the instantaneous energy states of your system. I guess in the language of your original question, your state can now "jump" to other simultaneous energy levels, since it will no longer remain in just one.

So how slow is "slow enough"? Sakurai goes through a full derivation to find that the adiabatic approximation holds if the time scale for changes of your Hamiltonian is much larger than the inverse energy of your eigenstate, $\tau \gg \hbar /E$. This is what is meant by "An adiabatic change is one that occurs at a rate much slower than the difference in frequency between the eigen states of energy." This may have been a bit formal, but it gets to the meat behind the quantum adiabatic theorem. Hope this helped.

  • $\begingroup$ great, thanks a lot $\endgroup$
    – alefisto
    Commented Mar 12, 2021 at 2:39

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