Wannier-stark state calculation (1 Viewer)

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In a tilt lattice, there are wannier-stark states

Is this state localized or extended? I think it should be extended because there is no lower bound of the potential

are they stable or quasi-stable?

How to calculate them?
 
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When the electric field is strong, the Wannier state is localized, see the article
by M Dignam in Phys. Rev. B in 1994.
They are quasi-stable, they are resonant state.
See also the Dignam's article
 
307
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When the electric field is strong, the Wannier state is localized, see the article
by M Dignam in Phys. Rev. B in 1994.
They are quasi-stable, they are resonant state.
See also the Dignam's article
Thanks a lot!

Why are they quasi-stable?

I take a tight-binding model, and numerically diagonalize the hamiltonian, and find out the eigenstates, i find that the eigenstates are all well-localized and of the same shape. Of course, except for those near the boundary.
 
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Thanks a lot!

Why are they quasi-stable?

I take a tight-binding model, and numerically diagonalize the hamiltonian, and find out the eigenstates, i find that the eigenstates are all well-localized and of the same shape. Of course, except for those near the boundary.
For one dimensional tight binding lattice, without interband tunneling, the Wannier-Stark state
is rigorous localized state. and the eigenenergy is equidistant.

If there exists interband tunneling, the WS state is a resonant state and quasi-stable.
 
307
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For one dimensional tight binding lattice, without interband tunneling, the Wannier-Stark state
is rigorous localized state. and the eigenenergy is equidistant.

If there exists interband tunneling, the WS state is a resonant state and quasi-stable.

Thanks a lot!

This sounds making sense.

but why?
 
In the above discussion,there exist Both Wannier state and Wannier-Stark state.But the two is different.Can you give a more detailed anwser?
 

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