D sin n n cos n not an eigenfunction i d i how

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Unformatted text preview: lar momentum? J= i ∂φ 4 The wavefucn1on needs to be an eigenfunc1on of both angular momentum and the Hamiltonian d H =− 2 2 I dφ d J= i dφ 2 2 How about sin(nϕ)? d sin ( nφ ) = n cos ( nφ ) not an eigenfunction! i dφ i How about exp(inϕ)? d exp ( inφ ) = n exp ( inφ ) eigenfunction!! (eigenvalue n ) i dφ 2 d 2 2n2 n22 − exp ( inφ ) = exp ( inφ ) eigenfunction!! (eigenvalue ) 2 I dφ 2 2I 2I 5 How to normalize the new eigenfunc1on? 2π ⎣ ⎦ ∫ ⎡ a ⋅ exp (inφ )⎤ * a exp ( inφ ) dφ = 1 determine a 0 2π a* exp ( −inφ ) a exp ( inφ ) dφ = ∫ 0 2π 2π ∫ a a exp ( −inφ + inφ ) dφ = a ∫ dφ = a 2 * 0 a =1 2 2π = 1 0 2π We cannot determine the phase of a and make the choice of posi1ve real number 6 Par1cle in a box X V=∞ 0 V=0 L V=∞ 7 More Mathema1cally d H =− + V ( x) 2 2 m dx 2 2 ⎧∞ x<0 ⎪ V ( x) = ⎨ 0 L ≥ x ≥ 0 ⎪∞ x>L ⎩ ⎫ ⎪ ⎬ ⎪ ⎭ When the poten1al is infinite the eigenfunc1on must be zero ψ ( x = 0) = ψ ( x = L ) = 0 8 The Schroedinger eq for a par1cle in a box 2 d 2 H =− + V ( x) 2 2 m dx ⎧ 0 x<0 ⎪ ⎪ 2 d 2 Hψ ( x ) = ⎨ − ψ n ( x ) = Enψ n ( x ) L ≥ x ≥ 0 2 ⎪ 2 m dx ⎪ 0 x>L ⎩ ⎛ nπ x ⎞ ψ n ( x ) = c sin ⎜ n = 1, 2, 3, ... sin ( nπ ) = 0 ⎟ ⎝L⎠ ⎫ ⎪ ...
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This document was uploaded on 03/04/2014 for the course CH 354L at University of Texas.

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