2009.02.06-ECE659-L11-notes

2009.02.06-ECE659-L11-notes - Example : k x k y y x k and v...

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ECE 659 Quantum Transport: Atom to Transistor Lecture 11: Semi classical Dynamics Supriyo Datta Spring 2009 Notes prepared by Samiran Ganguly
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S D L Vd μ2 μ1 Newton’s Law: dx v dt dp F dt = = Bandstructure E E 2 2 2 2 2 p k E m m = = ( 29 f f df df dz dz λ + - + - - = = - k k Hamilton’s form of Newton’s Law: dx E dt p dp E dt x = = - ( 29 2 2 p E U x m = + , dx p dp U dt m dt x = = -
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Schrodinger Equation: ( 29 2 2 d i d dx i U x dt m ψ = + If U(x) is constant, we can solve this by: iEt ikx e e - If U is space variant (say in x) but time invariant: 1 1 dx E A dt k dk E B dt x = - = - - eqn. B can be shown to be true in the following way: ( 29 , 0 0 0 dE x k dt E x E k x t k t E k v x t = + = ∂ ∂ + = ( 29 ' ' iEt i dx k x e e - ( 29 2 2 2 k E U x m = + In 3D: 1 1 k dx E dt dk E dt = = - h h k x k y ( 29 2 2 2 2 x y k k E m + = dx k dt m = h h Which gives: Hence, v and k are collinear Example :
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Unformatted text preview: Example : k x k y y x k and v are collinear depending upon E-k diagram If E goes down with k (e.g. in valence band) v will be in opposite direction ( 29 ( 29 2 k qA k qA E qV m--= + h h h h x z x z y A A B z x =- x y z y A B z A B x = = -( 29 ( 29 2 2 2 x x z z k qA k qA E m-+-= Effect of Magnetic Field v 2 2 2 k E qV m U V B A = + = - = h h h x z z z z dx k qA k dt m m dk k qA A q dt m x v -= =- = h h h h dHDH We can show: ( 29 dk q v B dt = h h h ( 29 dx k dt m dk q v B dt = = h h h h h Hence:...
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2009.02.06-ECE659-L11-notes - Example : k x k y y x k and v...

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