Quantum Mechanics Solution Manual |
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© Leon van Dommelen |
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2.6.6 Solution herm-f
Question:
Show that the operator is not a Hermitian operator, but is, assuming that the functions on which they act vanish at the ends of the interval on which they are defined. (Less restrictively, it is only required that the functions are periodic
; they must return to the same value at that they had at .)
Answer:
You first need to show that
is not the same as
in order for not to be a Hermitian operator.
By definition,
You can use integration by parts,
[1, p. 64], to move the derivative from to :
(the differentiation can be moved inside the complex conjugate since it is a real operation.) Since the functions and are the same at the end points and you have
This makes a skew-Hermitian operator, rather than a Hermitian one: flipping over the operator to the other side changes the sign of the inner product.
To get rid of the change of sign, you can add a factor to the operator, since the adds a compensating minus sign when you bring it inside the complex conjugate:
This makes a Hermitian operator.