Fund Quantum Mechanics Lect & HW Solutions 32

Fund Quantum Mechanics Lect & HW Solutions 32 - 14...

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Unformatted text preview: 14 CHAPTER 2. MATHEMATICAL PREREQUISITES (cos 45◦ , − sin 45◦ ). The dot product of (cos 45◦ , sin 45◦ ) and (cos 45◦ , − sin 45◦ ) is cos2 45◦ − sin2 45◦ . That is zero, because cos 45◦ = sin 45◦ , so the two eigenvectors are orthogonal. In linear algebra, you would write the relationship r2 = Ar out as: x2 y2 2.6.3 x y 11 11 = = x+y x+y Solution herm-c Question: Show that the operator 2 is a Hermitian operator, but i is not. Answer: By definition, 2 corresponds to multiplying by 2, so 2g is simply the function 2g . Now write the inner product f |2g and see whether it is the same as 2f |g for any f and g : f |2g = all x f ∗ 2g dx = all x (2f )∗ g dx = 2f |g since the complex conjugate does not affect a real number like 2. So 2 is indeed Hermitian. On the other hand, f |ig = all x f ∗ ig dx = all x −(if )∗ g dx = − if |g so i is not Hermitian. An operator like i that flips over the sign of an inner product if it is moved to the other side is called “skew-Hermitian”. An operator like 2 + i is neither Hermitian nor skew-Hermitian. 2.6.4 Solution herm-d Question: Generalize the previous question, by showing that any complex constant c comes out of the right hand side of an inner product unchanged, but out of the left hand side as its complex conjugate; f |cg = c f |g cf |g = c∗ f |g . As a result, a number c is only a Hermitian operator if it is real: if c is complex, the two expressions above are not the same. Answer: Since constants can be taken out of an integral: f |cg = cf |g = all x all x f ∗ cg dx = c (cf )∗ g dx = c∗ all x f ∗ g dx = c f |g all x f ∗ g dx = c∗ f |g . ...
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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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