Lect33_LatticeSymmetry - Discrete Translation Symmetry...

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Lecture 33: Symmetry IV: Lattice Symmetry PHY851 Quantum Mechanics I Fall, 2008 M.G. Moore Discrete Translation Symmetry Suppose the potential is periodic Let a be the period: Formally this means: In terms of the shift operator, this means: Since we have the more general case: It Follows that: Which is clearly a ‘discrete symmetry’ V ( x ) a ) ( ) ( x V a x V = + V ma VT ma T = ! ) ( ) ( 2 2 ) ( ) ( P d T P d T = ! H ma HT ma T = ! ) ( ) ( K , 2 , 1 , 0 ± ± = m
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Simultaneous Eigenstates The basic requirement for a periodic symmetry is: So clearly simultaneous eigenstates of H and T ( a ) exist Lets try to find these simultaneous eigenstates… As usual, we begin by writing equations which define the states we want to find: [ ] 0 ) ( , = a T H t E t t E a T t E E t E H , , ) ( , , = = T ( a ) eigenvalue equation Hitting the T ( a ) equation from the left with a k - state (momentum eigenstate) gives: Recall that: Which leads to: This requires: All eigenvalues of T ( a ) are pure phase factors: We can define the ‘Bloch Vector’ as: Also known as the ‘quasi momentum’ t E k t t E k e ika , , = ! iaK e a T ! = ) ( t E k t t E a T k , , ) ( = ika e t ! = ! i e t " = " # < $ a q = a q a " < # h P K =
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Bloch Theorem Thus we can re-label our simultaneous eigenstates as: The wavefunctions must then satisfy: Multiplying both sides by
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Lect33_LatticeSymmetry - Discrete Translation Symmetry...

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