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Time independent perturbation theory
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[QUOTE="barefeet, post: 4993396, member: 490981"] [h2]Homework Statement [/h2] The following text on the time independent perturbation theory is given in a textbook: [tex] \hat{H} = \hat{H}_0 + \alpha \hat{H'} [/tex] We expand its eigenstates [itex] \mid n \rangle [/itex] in the convenient basis of [itex] \mid n \rangle^{(0)} [/itex] [tex] \mid n \rangle = \sum_m c_{nm} \mid m \rangle^{(0)} [/tex] The Schrodinger equation in these notations becomes [tex] \left\{ E_n(\alpha) - E_m^{(0)} \right\}c_{nm} = \alpha \sum_p c_{np} M_{mp} [/tex] With [tex] M_{nm} = \langle n \mid \hat{H'} \mid m \rangle [/tex]I don't understand how the second last equation is derived and I don't know how the Schrodinger equation is used [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] The only thing I can think of is to use the first equation and let both sides be sandwiched between an eigenstate [itex] \mid n \rangle [/itex] of the operator [itex] \hat{H} [/itex] [tex]\langle n \mid \hat{H} \mid n \rangle = \langle n \mid \hat{H_0} \mid n \rangle + \alpha \langle n \mid \hat{H'} \mid n \rangle [/tex] [tex] \langle n \mid E_n(\alpha) \mid n \rangle = \sum_m c_{nm}^* \langle m \mid^{(0)} \hat{H_0} \mid \sum_k c_{nk} \mid k \rangle^{(0)} + \alpha \langle n \mid \hat{H'} \mid \sum_p c_{np} \mid p \rangle^{(0)} [/tex] [tex] E_n(\alpha) = \sum_m c_{nm}^*c_{nm} E_m^{(0)} + \alpha \sum_p c_{np} \langle n \mid \hat{H'} \mid p \rangle^{(0)} [/tex] [tex] E_n(\alpha) - \sum_m |c_{nm}|^2 E_m^{(0)} = \alpha \sum_p c_{np} M_{np} [/tex] And here I am stuck: - [itex] E_n(\alpha) [/itex] doesn't have a factor [itex] c_{nm} [/itex] - [itex] E_m^{(0)} [/itex] is still a summation and has a factor of [itex] |c_{nm}|^2 [/itex] instead of [itex] c_{nm}[/itex] - I have [itex] M_{np} [/itex] instead of [itex] M_{mp} [/itex] - The p's are eigenstates of [itex] H_0 [/itex] and not of [itex] H [/itex] [/QUOTE]
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Time independent perturbation theory
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