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Unformatted text preview: 2.6. HERMITIAN OPERATORS 2.6.5 15 Solution herme Question: Show that an operator such as x2 , corresponding to multiplying by a real function,
is an Hermitian operator.
Answer: If the operator corresponds to multiplying by a real function of x, call it r(x), then
f rg = all i f ∗ rg dx = all i (rf )∗ g dx = rf g since the complex conjugate does not aﬀect a real function. 2.6.6 Solution hermf Question: Show that the operator d/dx is not a Hermitian operator, but id/dx is, assuming
that the functions on which they act vanish at the ends of the interval a ≤ x ≤ b on which
they are deﬁned. (Less restrictively, it is only required that the functions are “periodic”; they
must return to the same value at x = b that they had at x = a.)
Answer: You ﬁrst need to show that
f d
g
dx is not the same as d
fg
dx
in order for d/dx not to be a Hermitian operator.
By deﬁnition, b
dg
d
g=
dx.
f∗
dx
dx
a
You can use “integration by parts,” [1, p. 64], to move the derivative from g to f : f b
d
f
g = f ∗g −
dx
a b
a df ∗
g dx = f ∗ (b)g (b) − f ∗ (a)g (a) −
dx b
a df
dx ∗ g dx (the diﬀerentiation can be moved inside the complex conjugate since it is a real operation.)
Since the functions f and g are the same at the end points a and b you have
d
g =−
f
dx b
a df
dx ∗ g dx = − d
fg
dx This makes d/dx a skewHermitian operator, rather than a Hermitian one: ﬂipping over the
operator to the other side changes the sign of the inner product. ...
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This note was uploaded on 01/06/2012 for the course PHY 3604 taught by Professor Dr.danielarenas during the Fall '11 term at UNF.
 Fall '11
 Dr.DanielArenas
 mechanics

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