Fund Quantum Mechanics Lect & HW Solutions 28

Fund Quantum Mechanics Lect & HW Solutions 28 - 10...

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Unformatted text preview: 10 CHAPTER 2. MATHEMATICAL PREREQUISITES √ √ the following multiples of sin(x) are all normalized: sin(x)/ π , (for α = 0), − sin(x)/ π , (for √ α = π ), and i sin(x)/ π , (for α = π/2). Answer: A multiple of sin(x) means c sin(x), where c is some complex constant, so the magnitude is 2π c sin(x)|c sin(x) = ||c sin(x)|| = 0 (c sin(x))∗ (c sin(x)) dx You can always write c as |c|eiα where α is some real angle, and then you get for the norm: ||c sin(x)|| = 2π 0 (|c|e−iα sin(x)) (|c|eiα sin(x)) dx = 2π 0 √ |c|2 sin2 (x) dx = |c| π √ So for the multiple to be normalized, the magnitude of c must be |c| = 1/ π , but the angle α can be arbitrary. 2.3.7 Solution dot-g Question: Show that the functions e4iπx and e6iπx are an orthonormal set on the interval 0 ≤ x ≤ 1. Answer: You need to show that both functions are normalized, e4iπx = 1 and e6iπx = 1, and that they are mutually orthogonal, e4iπx |e6iπx = 0. Work each out in turn (don’t forget to take complex conjugate of the first function in the inner products): ||e4iπx || = e4iπx |e4iπx = ||e6iπx || = e6iπx |e6iπx = e4iπx |e6iπx = 1 0 1 0 1 0 Operators e−6iπx e6iπx dx = e−4iπx e6iπx dx = (Since the Euler formula shows that ei2π = 1.) 2.4 e−4iπx e4iπx dx = 1 0 e2iπx dx = 1 0 1 0 1 dx = 1 1 dx = 1 1 2iπx 1 =0 e 0 2iπ ...
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