Fund Quantum Mechanics Lect &amp; HW Solutions 27

# Fund Quantum Mechanics Lect &amp; HW Solutions 27 - x A...

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2.3. THE DOT, OOPS, INNER PRODUCT 9 2.3.3 Solution dot-c Question: Find the inner product of the functions sin( x ) and cos( x ) on the interval 0 x 1. Answer: a sin( x ) | cos( x ) A = i 1 0 sin( x ) cos( x ) d x = 1 4 cos(2 x ) v v v v 1 0 = 1 4 p 1 cos(2) P 2.3.4 Solution dot-d Question: Show that the functions sin( x ) and cos( x ) are orthogonal on the interval 0 x 2 π . Answer: They are by de±nition orthogonal if the inner product is zero. Check that: a sin( x ) | cos( x ) A = i 2 π 0 sin( x ) cos( x ) d x = 1 4 cos(2 x ) v v v v 2 π 0 = 1 4 p 1 cos(4 π ) P = 0 2.3.5 Solution dot-e Question: Verify that sin( x ) is not a normalized function on the interval 0 x 2 π , and normalize it by dividing by its norm. Answer: || sin( x ) || = r a sin( x ) | sin(
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Unformatted text preview: x ) A = R i 2 π sin 2 ( x ) d x = √ π Since || sin( x ) || is not one, sin( x ) is not a normalized function on 0 ≤ x ≤ 2 π . If you divide by its norm, i.e. by √ π , however, || sin( x ) / √ π || = R i 2 π (sin( x ) / √ π )(sin( x ) / √ π ) d x = r π/π = 1 2.3.6 Solution dot-f Question: Verify that the most general multiple of sin( x ) that is normalized on the interval ≤ x ≤ 2 π is e i α sin( x ) / √ π where α is any arbitrary real number. So, using the Euler formula,...
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