Given another general state function f(x) = <x |f>
The overlap integral between the state functions |f> and |y > can be written:
ò <f | x ><x |y > dx = < f | y
>
The Dirac notation then shows clearly this is just the
generalization of the scalar product of two complex vectors to the infinite
continuous case. If the overlap integral is zero then we have:
< f
| y > = 0
And the
two state vectors are said to be orthogonal. (Perpendicular to each
other). It then follows from previous
definitions:
< f
| y > = < y | f >*
The form u n(x) can be used to denote the eigenvector of an operator A^ belonging to eigenvalue an and for which the general state function:
|f > = u n
Then also:
u n(x) = < x | an >
The applicable eigenvalue equation including the operator ( A^) can then be written:
A^ (x , ¶
/
¶ x) < x |
an > = an < x |
an >
Which
can be abbreviated to:
< A^ |
an > = < an | an >
The average value a f of repeated observations A^ on a normalized state | y > can be expressed:
a f = ò <y | x >
A^ (x , ¶
/
¶ x) <x |y > dx
= <y | A^ | y >
Suggested Problems:
1) Given eigenvalues an and eigenvectors un of the operator A^ find the eigenvectors if:
A^ is of the
diagonal form (e.g. for the matrix)
(a1 …….0)
(0…… a2)
2) In a quantum mechanics exam a student writes for the normalization condition:
< y
| y > = ò |< y | x >| dx = 1
Rewrite the above equation as it should be given
3) Prove that the eigenvalue a for an observable operator A^ is real.
4) Let a1 and a2 be different eigenvalues of A^ . Find:
A^ | a1 > and A^ | a2 > and thence show:
<a1| A^ |a2 > = a2 <a1 | a2 >
5) Show that u(x) = exp -(½x 2) is an eigenfunction of the operator:
A^ (x , ¶ / ¶ x) = ( ¶ 2 / ¶ x2 - x2 )
And find the corresponding eigenvalue
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