# PS1sol - Solutions to Problem Set 1 Physics 341 by Callum...

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Unformatted text preview: Solutions to Problem Set 1 Physics 341 by: Callum Quigley 1 Commuting Hermitian Operators To begin, we know that A is a Hermitian operator and so there exists a complete basis for V such that A is diagonal in this basis. Call this basis A = {| a i n i , n ∈ N , i = 1 ,...,g n } , where g n labels the degeneracy of the n-th eigenvalue. Then we have A | a i n i = a n | a i n i , ∀| a i n i ∈ A . (1.1) Before going further, let’s establish some notation. We’ll write V n for the g n-dimensional subspace of V with common eigenvalue a n , and denote the basis for this subspace as A n . Clearly, V = V ⊕ V 1 ⊕ ... with V i ∩ V j = | i for i 6 = j (1.2) A = A ∪ A 1 ∪ ... with A i ∩ A j = {∅} for i 6 = j. (1.3) Now fix a value of n . For any | a i n i ∈ A n , consider the vector B | a i n i . Because [ A,B ] = 0 it follows that A ( B | a i n i ) = BA | a i n i = a n ( B | a i n i ) (1.4) and so B | a i n i ∈ V n . Since this is true for all n , we’ve shown that B is block-diagonal in the basis A , namely BV n ⊆ V n . If g n = 1, then B is clearly diagonal in that 1 × 1 block. For g n > 1, B will, in general, not be diagonal within V n . However, we know that B is also Hermitian, so it too can be diagonalized. The key point is that within the subspace V n , A acts like the identity operator, A | V n = a n 1 , for any basis of V n and so it will always be diagonal. In particular, we can chose a new basis A n for each V n (while not disrupting the decomposition V = V ⊕ V 1 ⊕ ... ) such that B is diagonal in each subspace. In this way, we have constructed a basis for V such that both A and B are diagonal. 2 Unitary 2 × 2 Matrices Consider the general 2 × 2 matrix U with complex entries: U = a b c d , with a,b,c,d ∈ C . Unitarity requires that U- 1 = U † , and so this gives four constraints among the entries of U . Defining u = det U = ad- bc , these constraints are a = u d * b =- u c * (2.1) c =- u b * d = u a * . (2.2) 1 If we combine the first and last of these (or equivalently the second and third), we see that | u | 2 = 1 and so it is a pure phase. Let’s write this as u = exp(2 iφ ). For the moment, let’s consider the case where...
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## This note was uploaded on 02/27/2010 for the course PHYSICS 341 taught by Professor Sghy during the Spring '10 term at King's College London.

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PS1sol - Solutions to Problem Set 1 Physics 341 by Callum...

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