Lecture10 - TheInniteSquareWell WedFeb2,2011 Lecture10 0 if...

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1/26/11 1 PHYS 360 Quantum Mechanics Wed Feb 2, 2011 Lecture 10: How do the boundary condiCon alone give quanCzed energy levels? The Infinite Square Well V x ( ) = 0, if 0 x a, , otherwise 2 2 m d 2 ψ n dx 2 + V ψ n = E n ψ n We will solve this equaCon first outside the well, then inside, then match up the soluCons. 2 2 m d 2 ψ n dx 2 = E n ψ n d 2 ψ n dx 2 = k n 2 ψ n , where k n 2 mE n ψ n x ( ) = A sin k n x + B cos k n x Inside, V(x)=0, so…. Outside the well V(x)=∞, so the wave funcCon must be zero. ψ n ( x ) = 0 outside the well Need to match up these funcCons at the boundaries…. 1. Any physically valid wave funcCon must be conCnuous at a boundary. 2. If the potenCal V(x) is physically valid, then the slopes must be conCnuous at a boundary. HOWEVER, a potenCal jumping abruptly from zero to infinity is NOT physically valid (but a nice model anyway). ψ 0 ( ) = ψ a ( ) = 0 So, Note, ψ (0) = A sin0 + B cos0 = B = 0 Hence ψ n x ( ) = A sin k n x ψ n a ( ) = A sin k n a = 0 At x=a: k n a = 0, ± π , ± 2 π , ± 3 π ,...
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