An example is applying magnetic flux through the axis of a cylinder (2D system with periodic boundary condition).

When changing flux from 0 to 1 flux quanta adiabatically, it seems that we can change one quantum state (eg. ground state) to another state (eg. excited state), but in some case it just goes back to itself.

What properties of the state determine these two different behaviors?

  • $\begingroup$ @Qmechanic: The tag "adiabatic" is for thermodynamics.Other tags like "quantum adiabatic" or "quasi static" are more appropriate. $\endgroup$
    – Tim
    Commented Aug 12, 2014 at 19:09
  • $\begingroup$ I am not sure I understand the question. Quantum mechanical systems are changing their state in discrete quanta, by definition. What can be changed adiabatically, is the wave function of the system (by going from pure to mixed states), which is a mathematical quantity describing the system. The system itself, however, can never be changed that way. If you want to change an actual physical spin, you have to do it in terms of a suitable multiple of 1/2. I understand that this may not be your question, though? $\endgroup$
    – CuriousOne
    Commented Aug 12, 2014 at 19:15
  • $\begingroup$ @CuriousOne Take your spin 1/2 as an example, what I mean by adiabatic process would be: at t=0, system at state |s_z=1/2>, which is the ground state of a spin under magnetic field $\vec{B}$ in $z$ direction. Then I slowly rotate $\vec{B}$ in x-z plane. The ground state would change accordingly. Then when $\vec{B}$ rotate 360 degree to original $z$ direction. The ground state is still |S_z=1/2>. But in some case, it is not. $\endgroup$
    – Tim
    Commented Aug 12, 2014 at 19:33
  • $\begingroup$ But the only quantum transitions that you will ever measure will be the absorption or release of a whole photon. The individual spin will flip from -1/2 to 1/2 or the other way, no matter what the expectation value of the wave function does. If you are building an NMR or ESR experiment, of course, you are never looking at a single spin, but only at the superposition of many spins. In that case you are not even measuring the expectation value of the wavefunction, but only the diagonal elements and the strongest off-diagonal elements of the density matrix for T>0. $\endgroup$
    – CuriousOne
    Commented Aug 12, 2014 at 23:34


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