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Consider a tight-binding square lattice in 2D. This lattice has two different nearest neighbor tunneling rates along the x and y directions; call them $J_{x}$ and $J_{y}$. All longer range tunneling rates are zero. How can I compute the response of the expected momentum of a very low momentum particle to a force applied along a vector n? Is the rate of change of momentum parallel to n? Where is the excess momentum going?

I think the excess momentum is being lost to heating the environment.

I think the rate of change of momentum IS parallel to the force applied along the vector n.

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up vote 1 down vote accepted

It is not parallel, the excess momentum can't go into heat, because heat is energy not momentum, but it does go into deflecting the atomic crystal as a whole a negligible amount.

The description of this is by the discrete Schrodinger equation with a different value for the x and y mass:

$$ H= - {\partial_x^2\over 2m_x} - {\partial_y^2\over 2m_y} - F_x x - F_y y $$

The deflections can be found by linearly rescaling x and y to turn it into the ordinary rotationally invariant Schrodinger equation, and using the ordinary Newtonian limit (or you can use the short-wavelength geometric optics approximation directly, or you can note that it factorizes into independent Schrodinger equations in x and y of a standard form that you can take a classical limit on). The form of Newton's laws here is:

$$ a_x = {F_x \over m_x} $$

$$ a_y = {F_y \over m_y} $$

This means that the force is not in the direction of the acceleration when the two hopping parameters are unequal.

This doesn't violate any conservation law, the extra momentum is absorbed by the lattice--- you must remember that the effective theory is not translationally invariant, so doesn't conserve momentum.

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I'll have to think about this, but by inspection you seem to have hit it right on the head. Have class today, might ask some Q's later. Thanks Ron! – Dylan Sabulsky Oct 29 '12 at 13:21

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