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Unformatted text preview: PHYS 507 Answers to Homework I (Fall ’05) 1. (a) We know that u 2 has to be orthogonal to u 1 , i.e., h u 1  u 2 i = 0. Using this, we can completely determine u 2 . Let u 2 = • a b ‚ . Then h u 1  u 2 i = 1 √ 6 £ 2 1 + i / • a b ‚ = 2 a + (1 + i ) b √ 6 = 0 . From here we find a = (1 + i ) b/ 2 and therefore u 2 = • a b ‚ = b • (1 + i ) / 2 1 ‚ . Here, b can be any complex number, i.e., the equation h u 1  u 2 i = 0 is satisfied by any complex value of b . However, in applications, especially in the computation of probabilities, it is advantageous to have u 2 to be normalized to unity, i.e., h u 2  u 2 i = 1. This condition reads, h u 2  u 2 i =  b  2 1 2 + 1 ¶ = 3  b  2 2 = 1 . As a result, we have  b  = p 2 / 3. Normalization gives only the value of modulus of b and not its phase. There is no way to determine its phase and moreover we don’t need to because it is just an overall phase factor for u 2 . For this reason, we choose the phase angle of b to be anything we like. For example, b = p 2 / 3 is one such possible choice. In that case we have u 2 = 1 √ 6 • (1 + i ) 2 ‚ ....
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This note was uploaded on 02/11/2012 for the course MATH 435 taught by Professor Starg during the Spring '11 term at Al Ahliyya Amman University.
 Spring '11
 starg

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