By Rajasekar, Shanmuganathan; Velusamy, R
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Extra info for Quantum Mechanics II: Advanced Topics
Also ﬁnd ψ|H|ψ . Hd3 express H We obtain H H d3 x = = 1 2 = 1 (2π)6 π2 + (∇ψ)2 + m2 ψ 2 d3 x d3 x ′ d3 p′ ei(p+p )·r Ep Ep′ ap − a†−p 4 × + = d3 p −p · p′ + m2 4 Ep Ep′ 1 (2π)3 ap′ − a†−p′ ap + a†−p d3 p Ep a†p ap + ap′ + a†−p′ 1 ap , a†p 2 . 139) Ep d3 p . 140) Simplifying the right-side of the above equation we get H = 1 (2π)3 d3 p Ep a†p ap + 1 2 The expectation value 0|H|0 = (1/2) Ep d3 p is inﬁnite. This is because there is an oscillator for each value of momentum. Hence, contribution from all the oscillators is inﬁnite.
59) respectively and with the transformation ∂ψ ∂t → −i ∇ψ → i ∇ψ ∗ → i ∂ − eφ ψ , ∂t e −i ∇ − A ψ , c e i ∇ − A ψ∗ . 89c) The resulting L is L = ψ∗ i 1 e e ∂ − eφ ψ − i ∇ − A ψ ∗ · −i ∇ − A ψ ∂t 2m c c −V ψ ∗ ψ + 1 8π 1 ∂A + ∇φ c ∂t 2 − 1 (∇ × A)2 . 91a) e e 1 e 1 ψ∗ −i ∇ψ − Aψ − ψ i ∇ψ ∗ − Aψ ∗ 2mi −i c i c 2 e e [ψ ∗ ∇ψ − ψ∇ψ ∗ ] − Aψ ∗ ψ . 91b) 2mi mc The momentum canonically conjugate to ψ is i ψ ∗ and that of A is π= 1 4πc 1 ∂A + ∇φ c ∂t . 92) The Hamiltonian H is given by H = = ∂ψ ∂A +π· d3 x − L ∂t ∂t 1 e e i ∇ − A ψ ∗ · −i ∇ − A ψ + eφψ ∗ ψ 2m c c 1 +V ψ ∗ ψ + 2πc2 π2 + (∇ × A)2 − c π · ∇φ d3 x .
68a) we get [Ax (r, t), ∇′ · π(r′ , t)] = i ∂ 3 δ (r − r′ ) . 72c) This inconsistency arises since A is not experimentally measurable and hence it is not a physical quantity. This inconsistency will be removed if the commutation relations are given in terms of the experimentally measurable quantities E and H. 57b) we get ∇×∇×A+ 1 ∂2A =0. 73) 18 Quantum Mechanics II: Advanced Topics Since ∇ × ∇ × A = ∇(∇ · A) − ∇2 A, in the gauge ∇ · A = 0, Eq. 73) gives the d’Alembert equation ∇2 A = 1 ∂2A . 74) If we assume that the electromagnetic ﬁeld is conﬁned to a large box of size of volume V with side length L and that it satisﬁes the periodic boundary conditions with period L then the general free solution to Eq.
Quantum Mechanics II: Advanced Topics by Rajasekar, Shanmuganathan; Velusamy, R