I'd been studying introductory quantum mechanics from J.J. Sakurai's Modern Quantum Mechanics, 3rd Edition. I was reading chapter 5, especially about time-independent perturbation theory, non-degenerate case. This book used normalization convention as $$\langle n^0|n(\lambda)\rangle=1,$$ rather than $\langle n(\lambda)|n(\lambda)\rangle=1$. (Here $\lambda$ is perturbation parameter: full hamiltonian is given as $H^0+\lambda H'$)

At page 295-296, this convention is released by calculation constant for renormalization/normalization: $$|n\rangle _N = Z^{1/2}_n|n\rangle$$ and wavefunction renormalization $$Z_n=\langle n|n\rangle ^{-1}.$$ Up to 2nd order, since $\langle n^0|n^1\rangle=\langle n^0|n^2\rangle=0$, this constant is easily calculated as: $$\langle n|n\rangle = \langle n^0|n^0\rangle+\lambda^2\langle n^1|n^1\rangle+O(\lambda^3)=1+\lambda^2\sum_{k\neq n}\frac{|\langle k^0|H'|n^0\rangle|^2}{(E^0_n-E^0_k)^2}+O(\lambda^3)$$ $$Z_n=\langle n|n\rangle^{-1}=1-\lambda^2\sum_{k\neq n}\frac{|\langle k^0|H'|n^0\rangle|^2}{(E^0_n-E^0_k)^2}+O(\lambda^3)$$

On the other hand, Energy correction is easily calculated as: $$E_n=E^0_n+\lambda\langle n^0|H'|n^0\rangle+\lambda^2\sum_{k\neq n}\frac{|\langle k^0|H'|n^0\rangle|^2}{E^0_n-E^0_k}+O(\lambda^3)$$

Thus the following equality holds up to 2nd order:

$$Z_n=\frac{\partial E_n}{\partial E^0_n}.$$

J.J. Sakuarai says in his book that this equality is a general result for higher order.

  1. I'd checked it for 3nd order, and also tried to prove it by manipulating iterative equation given in (5.5.34) and (5.5.35), but failed.

  2. Is this a famous relation in renormalization theory or something?


1 Answer 1


We denote the perturbed Hamiltonian as $H=H_0+\lambda H'=\sum_{k}{E_{k}^{0}|k^0\rangle \langle k^0|}+\lambda H'$, and to show the equation, we start from the identity $$\langle n| H |n \rangle=E_n\langle n|n\rangle$$ Partially differentiate both sides with respect to $E_{n}^{0}$, and we have $$\bigg({\partial \langle n| \over \partial E_{n}^{0}}\bigg) H |n \rangle+\langle n| {\partial H \over \partial E_{n}^{0}} |n \rangle+\langle n| H \bigg({\partial |n \rangle \over \partial E_{n}^{0}}\bigg)={\partial E_n \over \partial E_{n}^{0}}\langle n|n\rangle+E_n{\partial \langle n|n\rangle \over \partial E_{n}^{0}}$$ Since $H|n\rangle=E_n|n\rangle$, and ${\partial H \over \partial E_{n}^{0}}=|n^0\rangle \langle n^0|$, the equation is simplified to $$1+E_n{\partial \langle n|n \rangle \over \partial E_{n}^{0}}={\partial E_n \over \partial E_{n}^{0}}\langle n|n\rangle+E_n{\partial \langle n|n\rangle \over \partial E_{n}^{0}}$$ Therefore, $$Z_n={\partial E_n \over \partial E_{n}^{0}}$$ The proof involves only algebraic calculation without any spirit of renormalization like scaling and renormalization group, so they may not be related. Also, I even doubt renormalization can be useful in the proof since the partial derivative is restricted to $E_{n}^{0}$ while we shall expect the scaling of complete energy spectrum if renormalization is used.

If you are asking whether the equation is useful in quantum field theory and renormalization theory, my answer is no. The purpose of renormalization in quantum field theory is to solve divergence when path integrals of closed loops are calculated, and as far as I can see, the equation is unrelated to renormalization, but I may be wrong since I am not an expert in quantum field theory. Please correct me if there is some advanced insight relating the two.


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