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Unformatted text preview: PHYS 507 Answers to Homework I 1. (a) h u 1  u 1 i = u † 1 u 1 = N * N 1 + i 2 1 i 2 =  N  2 ((1 + i )(1 i ) + 4) = 6  N  2 = 1 . ⇒  N  = 1 √ 6 This is the only thing we can say about the value of N , i.e., we can only determine its modulus,  N  , but not its phase. Luckily, no physical observable depends on the phase so that we can choose any value for the phase N . All of the following are valid choices. N = 1 √ 6 , N = 1 √ 6 , N = 1 √ 6 e i π 4 = 1 + i 2 √ 3 . We will use the first one above, because it is easier to write. (b) For u 2 , all we know is that it is orthogonal to u 1 (because eigenkets of an observable corresponding to different eigenvalues are orthogonal). Let u 2 = a b . Then h u 1  u 2 i = u † 1 u 2 = 1 √ 6 1 + i 2 a b = 1 √ 6 ((1 + i ) a + 2 b ) = 0 . ⇒ (1 + i ) a = 2 b (1 i )(1 + i ) a = 2(1 i ) b a = (1 i ) b ⇒ u 2 = a b = b 1 + i 1 Same as above, value of b cannot be calculated from the given information. We can determine its modulus,  b  , by requiring that it is normalized. Its phase has to be chosen again. We can find  b  = 1 / √ 3 by normalization and I choose b = 1 / √ 3. Then u 2 = 1 √ 3 1 i 1 . 1 (c) The observable A is a 2x2 matrix. From the given information, we know that u 1 is an eigenvector with eigenvalue a 1 = +1 and u 2 is an eigenvector with eigenvalue a 2 = 1. As a result we need to find a matrix satisfying Au 1 = a 1 u 1 and Au 2 = a 2 u 2 . Trying to find A by using the relations above is the longest way of solving this problem. There is a much shorter way. We have seen that A can be expressed as A = a 1 ...
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 Spring '11
 starg

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