Question on Time-Independent Perturbation Theory

AI Thread Summary
The discussion focuses on the derivation of Equation (A.4) in time-independent perturbation theory, specifically regarding the normalization of the eigenstate ##|\psi_j\rangle##. Participants express confusion about how the equation follows from taking the inner product of the eigenstate with itself, particularly questioning the assumption of known perturbative corrections. There is a consensus that the textbook's approach seems to presuppose normalization without adequately justifying why certain cross terms must vanish. The need for a clear argument demonstrating that the summations on the right side equal zero is emphasized. Overall, the conversation highlights the complexities and potential gaps in the explanation of perturbation theory in the text.
cwill53
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Homework Statement
I just need a quick check on something from Appendix A in "Nanostructures and Nanotechnology" by Douglas Natelson.
Relevant Equations
$$H|\psi \rangle=E|\psi \rangle$$
$$H^0|\psi^0 \rangle=E^0|\psi^0 \rangle$$
$$E_j=E^0_j+\sum_{i=1}^{n}\lambda ^iE^i_j$$
$$|\psi _j\rangle=|\psi^0 _j\rangle+\sum_{i=1}^{n}\lambda ^i|\psi^i _j\rangle$$
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I'm currently reading this passage to review perturbation theory. Just before Equation (A.4), this passage tells me to take the inner product of the proposed eigenstate ##|\psi _j\rangle## with itself. Writing this out, I got:

$$1=\left \langle \psi _j| \psi _j\right \rangle=\left ( |\psi^0 _j\rangle+\sum_{k=1}^{n}\lambda ^k|\psi^k _j\rangle \right )^\dagger\left ( |\psi^0 _j\rangle+\sum_{i=1}^{n}\lambda ^i|\psi^i _j\rangle \right )$$

$$= \left ( \langle\psi^0 _j|+\sum_{i=1}^{n}\left (\lambda ^i \right )^*\langle\psi^i _j| \right )\left ( |\psi^0 _j\rangle+\sum_{k=1}^{n}\lambda ^k|\psi^k _j\rangle \right )$$

$$=\left \langle \psi^0 _j| \psi^0 _j\right \rangle+\sum_{i=1}^{n}\lambda ^i\left \langle \psi^0 _j| \psi^i _j\right \rangle+\sum_{i=1}^{n}\left (\lambda ^i \right )^*\langle\psi^i _j|\psi^0 _j\rangle+\sum_{i=1}^{n}\sum_{k=1}^{n}(\lambda ^i)^*\lambda ^k\left \langle \psi^i _j| \psi^k _j\right \rangle$$

I'm not sure how Equation (A.4) follows from this though.
 
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I think (A.4) does not follow from the previous equations. It's just the usual orthonormal conditions on a basis set.
 
Grelbr42 said:
I think (A.4) does not follow from the previous equations. It's just the usual orthonormal conditions on a basis set.
Okay, so basically you’re saying we impose the first equation in (A.4) and that’s what yields the second one? The reason that it’s confusing to me is that when the passage asks us to take the inner product, it makes it seems as if we know something about the perturbative corrections to the state function a priori.
 
cwill53 said:
I'm not sure how Equation (A.4) follows from this though.
I'm with you on this. I think the textbook's author begs the question. He says "let's check normalization" and then apparently uses normalization to claim that the cross terms on the right hand side must vanish because the leading term on the right is equal to 1. One should consider $$\begin{align} \left \langle \psi _j| \psi _j\right \rangle & =\left \langle \psi^0 _j| \psi^0 _j\right \rangle+\sum_{i=1}^{n}\lambda ^i\left \langle \psi^0 _j| \psi^i _j\right \rangle+\sum_{i=1}^{n}\left (\lambda ^i \right )^*\langle\psi^i _j|\psi^0 _j\rangle+\sum_{i=1}^{n}\sum_{k=1}^{n}(\lambda ^i)^*\lambda ^k\left \langle \psi^i _j| \psi^k _j\right \rangle \nonumber \\
& = 1+\sum_{i=1}^{n}\lambda ^i\left \langle \psi^0 _j| \psi^i _j\right \rangle+\sum_{i=1}^{n}\left (\lambda ^i \right )^*\langle\psi^i _j|\psi^0 _j\rangle+\sum_{i=1}^{n}\sum_{k=1}^{n}(\lambda ^i)^*\lambda ^k\left \langle \psi^i _j| \psi^k _j\right \rangle \nonumber \end{align}$$and argue why the summations on the right hand side add up to give zero. I don't believe there is such an argument. See discussion here. Note the explicit use of the normalization constant ##N(\lambda)## in the end.
 
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