We rewrite 2 2 2 2 2 1 2 2 2 2 2 2 2 The nonrelativistic limit can also be

We rewrite 2 2 2 2 2 1 2 2 2 2 2 2 2 the

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? , ? ? ] = ? ? ? ? ? ? ? ? = ?? ( 𝜕 ? ? ? ) ?? ( 𝜕 ? ? ? ) We rewrite ? ? ? ? : ? ? ? ? = ? ?? ? ? ? ? = 𝜕 ? 𝜕 ? + ?? (( 𝜕 ? ? ? ) + 2 ? ? 𝜕 ? ) ? 2 ? ? ? ? = = 𝜕 ? 𝜕 ? + ?? (( 𝜕 0 ? 0 ) + 2 ? 0 𝜕 0 + ( 𝜕 ? ? ? ) + 2 ? ? 𝜕 ? ) ? 2 ( ? 0 ? 0 + ? ? ? ? ) = = 𝜕 ? 𝜕 ? + ? 1 ? 2 𝜕? 𝜕? + 2 ? ? ? 2 𝜕 𝜕? + ?? ( 𝜕 ? ? ? ) + 2 ??? ? 𝜕 ? ? 2 ? 2 ? 2 ? ? ? ? The nonrelativistic limit can also be applied directly to the Klein-Gordon equation: 0 = (¯ 2 ? 2 ? ? ? ? + ? 2 ? 4 ) ? = = (︂ ¯ 2 ? 2 𝜕 ? 𝜕 ? + ? 𝜕? 𝜕? + 2 ?? 𝜕 𝜕? + ? ¯ ℎ?? 2 ( 𝜕 ? ? ? ) + 2 ? ¯ ℎ?? 2 ? ? 𝜕 ? ? 2 ? 2 ? 2 ? ? ? ? + ? 2 ? 4 )︂ ? ? ¯ ?? 2 ? 𝜙 = = (︂ ¯ 2 𝜕 2 𝜕? 2 ? 2 ¯ 2 2 + 2 ?? 𝜕 𝜕? + ? 𝜕? 𝜕? + ? ¯ ℎ?? 2 ( 𝜕 ? ? ? ) + 2 ? ¯ ℎ?? 2 ? ? 𝜕 ? ? 2 ? 2 ? 2 ? ? ? ? + ? 2 ? 4 )︂ ? ? ¯ ?? 2 ? 𝜙 = = ? ? ¯ ?? 2 ? (︂ ¯ 2 ( ? ¯ ?? 2 + 𝜕 𝜕? ) 2 ¯ 2 ? 2 2 + 2 ?? ( ? ¯ ?? 2 + 𝜕 𝜕? ) + ? 𝜕? 𝜕? + ? ¯ ℎ?? 2 ( 𝜕 ? ? ? ) + 2 ? ¯ ℎ?? 2 ? ? 𝜕 ? ? 2 + ? 2 ? 2 ? ? ? ? + ? 2 ? 4 )︀ 𝜙 = = ? ? ¯ ?? 2 ? (︂ 2 ? ¯ ℎ?? 2 𝜕 𝜕? + ¯ 2 𝜕 2 𝜕? 2 ? 2 ¯ 2 2 + 2 ? ? ? 2 ¯ + 2 ?? 𝜕 𝜕? + ? 𝜕? 𝜕? + ? ¯ ℎ?? 2 ( 𝜕 ? ? ? ) + 2 ? ¯ ℎ?? 2 ? ? 𝜕 ? ? 2 + ? 2 ? 2 ? ? ? ? )︀ 𝜙 = = 2 ?? 2 ? ? ¯ ?? 2 ? (︂ ? ¯ 𝜕 𝜕? + ¯ 2 2 2 ? ? 1 2 ?? 2 𝜕 2 𝜕? 2 ? 2 ?? 2 𝜕? 𝜕? + ? 2 2 ?? 2 ?? ?? 2 𝜕 𝜕? + ? ¯ ℎ? 2 ? 𝜕 ? ? ? ? ¯ ℎ? ? ? ? 𝜕 ? + ? 2 2 ? ? ? ? ? )︂ 𝜙 Taking the limit ? → ∞ we again recover the Schrödinger equation: ? ¯ 𝜕 𝜕? 𝜙 = (︂ ¯ 2 2 2 ? + ? + ? ¯ ℎ? 2 ? 𝜕 ? ? ? + ? ¯ ℎ? ? ? ? 𝜕 ? ? 2 2 ? ? ? ? ? )︂ 𝜙 , we rewrite the right hand side a little bit: ? ¯ 𝜕 𝜕? 𝜙 = (︂ ¯ 2 2 ? ( 𝜕 ? 𝜕 ? + ? ¯ ?𝜕 ? ? ? + 2 ? ¯ ?? ? 𝜕 ? ? 2 ¯ 2 ? ? ? ? ) + ? )︂ 𝜙 , ? ¯ 𝜕 𝜕? 𝜙 = (︂ ¯ 2 2 ? ( 𝜕 ? + ? ¯ ?? ? )( 𝜕 ? + ? ¯ ?? ? ) + ? )︂ 𝜙 , 334 Chapter 8. Quantum Field Theory and Quantum Mechanics
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Theoretical Physics Reference, Release 0.5 ? ¯ 𝜕 𝜕? 𝜙 = (︂ 1 2 ? ¯ 2 ? ? ? ? + ? )︂ 𝜙 , Using (see the appendix for details): ¯ 2 ? ? ? ? = ¯ 2 ? ?? ? ? ? ? = ¯ 2 (︂ ? ¯ ( p ? A ) )︂ 2 = ( p ? A ) 2 we get the usual form of the Schrödinger equation for the vector potential: ? ¯ 𝜕 𝜕? 𝜙 = (︂ ( p ? A ) 2 2 ? + ? )︂ 𝜙 . A little easier derivation: 0 = (¯ 2 ? 2 ? ? ? ? + ? 2 ? 4 ) ? = = (¯ 2 ? 2 ? 0 ? 0 + ¯ 2 ? 2 ? ? ? ? + ? 2 ? 4 ) ? = = 2 ?? 2 (︂ ¯ 2 2 ? ? 0 ? 0 + ¯ 2 2 ? ? ? ? ? + 1 2 ?? 2 )︂ ? = = 2 ?? 2 (︂ ¯ 2 2 ? (︂ 𝜕 0 + ? ¯ ?? 0 )︂ (︂ 𝜕 0 + ? ¯ ?? 0 )︂ + 1 2 ?? 2 + ¯ 2 2 ? ? ? ? ? )︂ ? ? ¯ ?? 2 ? 𝜙 = = 2 ?? 2 (︂ ¯ 2 2 ? (︂ 𝜕 0 + ? ¯ ?? 0 )︂ ? ? ¯ ?? 2 ? (︂ 𝜕 0 ? ¯ ?? + ? ¯ ?? 0 )︂ + 1 2 ?? 2 + ¯ 2 2 ? ? ? ? ? )︂ 𝜙 = = 2 ?? 2 ? ? ¯ ?? 2 ? (︂ ¯ 2 2 ? (︂ 𝜕 0 ? ¯ ?? + ? ¯ ?? 0 )︂ (︂ 𝜕 0 ? ¯ ?? + ? ¯ ?? 0 )︂ + 1 2 ?? 2 + ¯ 2 2 ? ? ? ? ? )︂ 𝜙 = = 2 ?? 2 ? ? ¯ ?? 2 ? (︂ ¯ 2 2 ? 𝜕 0 𝜕 0 1 2 ?? 2 ? 2 ? 0 ? 0 2 ? + ??? 0 + ¯ 2 ? ? ¯ ? ( 𝜕 0 ? 0 + ? 0 𝜕 0 ) ? ¯ ℎ?𝜕 0 + 1 2 ?? 2 + ¯ 2 2 ? ? ? ? ? )︂ 𝜙 = = 2 ?? 2 ? ? ¯ ?? 2 ? (︂ ? ¯ 𝜕 𝜕? + ¯ 2 2 ? ? ? ? ? + ??? 0 + ¯ 2 2 ?? 2 𝜕 2 𝜕? 2 ? 2 ? 2 2 ?? 2 + ?? ¯ ?? 2 ( 𝜕 𝜕? ? + ? 𝜕 𝜕? ) )︂ 𝜙 = = 2 ?? 2 ? ? ¯ ?? 2 ? (︂ ? ¯ 𝜕 𝜕? + ( p ? A ) 2 2 ? + ?? + ¯ 2 2 ?? 2 𝜕 2 𝜕? 2 ? 2 ? 2 2 ?? 2 + ?? ¯ ?? 2 ( 𝜕 𝜕? ? + ? 𝜕 𝜕? ) )︂ 𝜙 and letting ? → ∞ we get the Schrödinger equation: ? ¯ 𝜕 𝜕? 𝜙 = (︂ ( p ? A ) 2 2 ? + ?? )︂ 𝜙 8.4. Quantum Mechanics 335
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Theoretical Physics Reference, Release 0.5 8.4.2 Perturbation Theory We want to solve the equation: ? ¯ d d ? | ? ( ? ) = ? ( ? ) | ? ( ? ) (8.9) with ? ( ? ) = ? 0 + ? 1 ( ? ) , where ? 0 is time-independent part whose eigenvalue problem has been solved: ? 0 | ? 0 = ? 0 ? | ? 0 and ? 1 ( ? ) is a small time-dependent perturbation. | ? 0 form a complete basis, so we can express | ? ( ? ) in this basis: | ? ( ? ) = ∑︁ ? ? ? ( ? ) ? ? ¯ ? 0 ? ? | ? 0 (8.10) Substituting this into ( 8.9 ), we get: ∑︁ ? (︂ ? ¯ d d ? ? ? ( ? ) + ? 0 ? ? ? ( ? ) )︂ ? ? ¯ ? 0 ? ? | ? 0 = ∑︁ ? (︀ ? 0 ? ? ? ( ? ) + ? 1 ? ? ( ? ) )︀ ? ? ¯ ? 0 ? ? | ? 0 so: ∑︁ ? ? ¯ d d ? ( ? ? ( ? )) ? ? ¯ ? 0 ? ? | ? 0 = ∑︁ ? ? ? ( ? ) ? ? ¯ ? 0 ? ? ? 1 | ? 0 Choosing some particular state | ? 0 of the ? 0 Hamiltonian, we multiply the equation from the left by ? 0 | ? ? ¯ ? 0 ? ? : ∑︁ ? ? ¯ d d ? ( ? ? ( ? )) ? ?? ?? ? ? 0 | ? 0 = ∑︁ ? ? ? ( ? ) ? ?? ?? ? ? 0 | ? 1 | ? 0 where ? ?? = ? 0 ? ? 0 ? ¯ . Using ? 0 | ? 0 = ? ?? : ? ¯ d d ? ? ? ( ? ) = ∑︁ ? ? ? ( ? ) ? ?? ?? ? ? 0 | ? 1 | ? 0 we integrate from ? 1 to ? : ? ¯ (( ? ? ( ? ) ? ? ( ? 1 )) = ∑︁ ? ∫︁ ? ? 1 ? ? ( ? ) ? ?? ?? ? ? 0 | ? 1 ( ? ) | ? 0 d ? Let the initial wavefunction at time ? 1 be some particular state | ? ( ? 1 ) = | ? 0 of the unperturbed Hamiltonian, then ? ? ( ? 1 ) = ? ?? and we get: ? ? ( ? ) = ? ?? ? ¯ ∑︁ ? ∫︁ ? ? 1 ? ? ( ? ) ? ?? ?? ? ? 0 | ? 1 ( ? ) | ? 0 d ? (8.11) This is the equation that we will use for the perturbation theory. In the zeroth order of the perturbation theory, we set ? 1 ( ? ) = 0 and we get: ? ? ( ? ) = ? ?? In the first order of the perturbation theory, we take the solution ?
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