Fund Quantum Mechanics Lect &amp; HW Solutions 33

# Fund Quantum Mechanics Lect &amp; HW Solutions 33 -...

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Unformatted text preview: 2.6. HERMITIAN OPERATORS 2.6.5 15 Solution herm-e 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 herm-f 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 skew-Hermitian 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.

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