Fund Quantum Mechanics Lect & HW Solutions 35

Fund Quantum Mechanics Lect & HW Solutions 35 -...

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Unformatted text preview: 2.6. HERMITIAN OPERATORS 17 First check the end points. The graph of the sine function, [1, item 12.22], shows that a sine is zero whenever its argument is a whole multiple of . That makes both sin( k 0) = sin(0) and sin( k ) zero. So sin( kx ) / radicalBig / 2 must be zero at x = 0 and x = too. Now check that the norm of the eigenfunctions is one. First find the norm of sin( kx ) by itself: || sin( kx ) || = radicalBig ( sin( kx ) | sin( kx ) ) = radicalBigg integraldisplay sin( kx ) sin( kx ) d x. Since the sine is real, the complex conjugate does not do anything, and you get || sin( kx ) || = radicalBigg integraldisplay sin 2 ( kx ) d x = radicalBig / 2 using [1, item 18.26]. Dividing this by radicalBig / 2, the norm vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle sin( kx ) / radicalBig / 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle becomes one; every eigenfunction is normalized....
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