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# set3 - Energy eigenvalues For a par1cle in a ring we have...

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Energy’ eigenvalues’ For’a’par1cle’in’a’ring’we’have’ ±’ d 2 d φ 2 ψφ ( ) = 2 IE 2 ( ) = k 2 ( ) But from the boundary condition k is a whole number E = 2 k 2 2 I k = 0,1,2,3,. . with a wavefunction (eigenfunction) ( ) = A sin k + δ ( ) where A and are determined from the boundary condition We define a node of function as a position in which the eignefnction becomes zero. How can we determine the number of nodes for a particle in a ring?

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Nodes’for’a’par1cle’in’a’ring’ 2’ 0’ 2π’ Sin( ϕ )’ 0’ 2π’ Sin(2 ϕ )=0’ ϕ =0,90,±80,270’ Sin( ϕ )=0’ ϕ =0,±80’
Number’of’nodes’ vs’Energy’ In’one’dimensional’system’the’number’of’ nodes’is’growing’monotonically’with’energy.’ The’more’nodes’we’have’the’higher’is’the’ energy.’ The’nodes’are’simple’qualita1ve’measure’of’ the’energy’of’an’ eignefunc1on.’The’more’ nodes’the’ eigenfunc1on’has’the’more’ energe1c’it’is’likely’to’be.’ 3’

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Angular’momentum’ The’Hamiltonian’is’the’operator’of’energy.’The’ energy’is’quantum’mechanics’is’discrete.’ When’we’discussed’the’Bohr’model’we’also’ men1oned’that’the’angular’momentum’ obtained’discrete’values’too’ What’is’the’quantum’mechanics’operator’for’ angular’momentum?’ 4’ J = n J = i ∂φ
The’ wavefucn1on’needs’to’be’an’ eigenfunc1on’

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set3 - Energy eigenvalues For a par1cle in a ring we have...

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