QFT_solutions2 - Part III Quantum Field Theory Example Sheet 2 solutions DAMTP University of Cambridge Dated 11\/09 by Raquel H Ribeiro Please send any

QFT_solutions2 - Part III Quantum Field Theory Example...

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Part III - Quantum Field Theory - Example Sheet 2 solutions Dated: 11/09 DAMTP - University of Cambridge by Raquel H. Ribeiro Please send any corrections or comments to [email protected] 1 Quantisation of a set of HOs Given the classical Hamiltonian for a string and the canonical commutation relations, one may introduce the creation and annihilation operators as follows ˆ a n a n = ω n 2 q n + i 2 ω n p n and ˆ a n a n = ω n 2 q n - i 2 ω n p n , (1) which in turn should verify the usual commutation relations. These should be consistent with the canonical commutation relations. Indeed [ a n , a m ] = ω n 2 q n + i 2 ω n p n , ω m 2 q m + i 2 ω m p m = = i 2 ω n ω m [ q n , p m ] + i 2 ω m ω n [ p n , q m ] = 0 similarly = a n , a m and a n , a m = - i 2 ω n ω m [ q n , p m ] + i 2 ω m ω n [ p n , q m ] = = - i 2 ω n ω m ( nm ) + i 2 ω m ω n ( - nm ) = δ nm , (2) as required. Inverting (1): q n = 1 2 ω a n + a n and p n = i ω n 2 a n - a n . (3) Plugging into the expression for the Hamiltonian, we get H = + n =1 ω n 2 a n a n + a n a n as required . We acknowledge the existence of a ground state | 0 in the theory, which is annihilated by any destruction operator, a n | 0 = 0 , n . Using the non-vanishing commutation relations (2), we rewrite the Hamiltonian as H = + n =1 ω n 2 2 a n a n + δ nn = + n =1 ω n a n a n + + n =1 ω n 2 δ nn . (4) The last term is clearly divergent and it is due to the vacuum energy. Indeed, for the ground state 0 | H | 0 = + n =1 ω n 2 δ nn . 1
Now, usually one is interested in differences of energy 1 ; hence, one may define all energies with respect to the vacuum energy. Applying the prescription known as normal ordering , it then follows that : H := + n =1 ω n a n a n . (5) To confirm the way the creation operator acts on states, we compute the following commutation relation: H, a n = + m =1 ω m a m a m , a n = - + m =1 ω m a m ( - δ nm ) = ω n a n , where we have used the identity [ A, BC ] = [ A, B ] C + B [ A, C ]. This result is in agreement with the fact that when the creation operator acts on a given state, it increases its energy by a factor of ω : H a n | 0 = ω n a n | 0 . Also, 1 , 2 , . . . , N | H | 1 , 2 , . . . , N = 0 | ( a N ) N . . . ( a 1 ) 1 + m =1 ω m a m a m a 1 1 . . . a N N | 0 = = + m =1 m ω m , where we have used the fact that a m H a n p | 0 = a m a n Ha n p - 1 | 0 + ω m a m a n p | 0 = ( p ω n ) a m a n p | 0 . 2 Canonical quantisation relations From the Fourier decomposition of a real scalar field and its conjugate momentum in the Schr¨ odinger picture, one can get an expression for both the creation and the annihilation operators. Firstly d 3 x e - iq · x φ ( x ) = d 3 p (2 π ) 3 2 E p d 3 x a p e i ( p - q ) · x + a p e - i ( p + q ) · x = = d 3 p 2 E p a p δ (3) ( p - q ) + a p δ (3) ( p + q ) = = 1 2 E q a q + a - q , 1 except for situations which lie in a gavitational context where the energy-momentum tensor sources the Einstein equations.

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