Question on Time-Independent Perturbation Theory

[cwill53]

Homework Statement

I just need a quick check on something from Appendix A in "Nanostructures and Nanotechnology" by Douglas Natelson.

Relevant Equations

$$H|\psi \rangle=E|\psi \rangle$$
$$H^0|\psi^0 \rangle=E^0|\psi^0 \rangle$$
$$E_j=E^0_j+\sum_{i=1}^{n}\lambda ^iE^i_j$$
$$|\psi _j\rangle=|\psi^0 _j\rangle+\sum_{i=1}^{n}\lambda ^i|\psi^i _j\rangle$$

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.

 

[Grelbr42]

I think (A.4) does not follow from the previous equations. It's just the usual orthonormal conditions on a basis set.

 

[cwill53]

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.

 

[kuruman]

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.