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Unformatted text preview: 16 CHAPTER 2. MATHEMATICAL PREREQUISITES To get rid of the change of sign, you can add a factor i to the operator, since the i adds a compensating minus sign when you bring it inside the complex conjugate: (bigg f vextendsingle vextendsingle vextendsingle vextendsingle i d d x g )bigg = − integraldisplay b a parenleftBigg − i d f d x parenrightBigg ∗ g d x = (bigg i d d x f vextendsingle vextendsingle vextendsingle vextendsingle g )bigg This makes id / d x a Hermitian operator. 2.6.7 Solution herm-g Question: Show that if A is a Hermitian operator, then so is A 2 . As a result, under the conditions of the previous question, − d 2 / d x 2 is a Hermitian operator too. (And so is just d 2 / d x 2 , of course, but − d 2 / d x 2 is the one with the positive eigenvalues, the squares of the eigenvalues of id / d x .) Answer: To show that A 2 is Hermitian, just move the two operators A to the other side of the inner product one by one. As far as the eigenvalues are concerned, each application of A to one of its eigenfunctions multiplies by the eigenvalue, so two applications of A multiplies by the square eigenvalue. 2.6.8 Solution herm-h Question: A complete set of orthonormal eigenfunctions of − d 2 / d x 2 on the interval 0 ≤ x ≤ π that are zero at the end points is the infinite set of functions sin( x ) radicalBig π/ 2 , sin(2 x ) radicalBig π/ 2 , sin(3 x ) radicalBig π/ 2 , sin(4 x ) radicalBig π/ 2 , . . . Check that these functions are indeed zero at x = 0 and x = π , that they are indeed orthonor- mal, and that they are eigenfunctions of − d 2 / d x 2 with the positive real eigenvalues 1 , 4 , 9 , 16 , . . ....
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This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.
- Fall '11