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 ofthe inner product one by one....
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- Fall '11