$\DeclareMathOperator{\tr}{tr}$I am rather troubled by how BW perturbation is derived, i.e., my main concern is the assumption of intermediate normalization, i.e., $\langle \psi_0|\psi \rangle=1$, so I attempted to rederive it rigorously for the non-degenerate case (I'm not sure if anyone else has done this before, so it would be great if someone has references), and I arrived at the following formula, i.e., if $H_\lambda = H_0 +\lambda V$ is the perturbed (finite-dim) Hamiltonian, and $E_\lambda$ represents the ground state eigen-energy of $H_\lambda$, then $$ \frac{1}{n!} \frac{d^n}{d\lambda^n} E_\lambda = \frac{1}{n} \frac{1}{2\pi i} \oint_\Gamma dz \tr \left (V \frac{1}{z-H_\lambda}\right)^n $$ where $\Gamma$ is a contour in $\mathbb{C}$ containing only the ground state eigen-energy $E_0$ and none of the other eigen-energies of $H_0$. By Weyl's inequality, for sufficiently small $\lambda$, the contour $\Gamma$ must also only enclose the ground state eigen-energy $E_\lambda$ and none of the other eigen-energies of $H_\lambda$.

I will provide a sketch of the proof later on, but let me try to convince you that this is correct. Indeed, if we plug in $\lambda =0$, the equation tells us the $n$th order correction to energy $E_\lambda$ (Taylor series). Let $P^k=P^k(\lambda=0),k\ge 0$ denote the projection onto the distinct eigenspaces of $H_0$ with energies $E^k=E^k(\lambda=0)$ so that if $n=1$, then the RHS is equal to \begin{align} \left.\frac{d}{d\lambda}\right|_{\lambda=0} E_\lambda &=\frac{1}{2\pi i} \tr\left( \oint_\Gamma V \frac{1}{z-H_0} dz \right) \\ &= \frac{1}{2\pi i} \tr\left( \oint_\Gamma V P^k\frac{1}{z-E^k} dz \right) \\ &= \tr (VP^0) =\langle 0|V |0\rangle \end{align} where I used the fact that $\Gamma$ only encloses $E^0$ and not the other eigen-energies $E^k,k\ne 0$.

QUESTION. For second order $n=2$ perturbation, you can do something similar. Indeed, \begin{align} \frac{1}{2!}\left.\frac{d^2}{d\lambda}\right|_{\lambda=0} E_\lambda &= \frac{1}{2}\frac{1}{2\pi i} \sum_{k,l} \tr\left( V P^k V P^l \oint_\Gamma \frac{1}{(z-E^k)(z-E^l)} dz \right) \end{align} Since $\Gamma$ only encloses $E^0$, we see that the contour integral is nonzero only if one of $E^k,E^l$ is equal to $E^0$ but not both ($(z-E^0)^{-2}$ has a pole of order 2). Therefore, if we write $$ R^0=\sum_{k\ne 0} \frac{P^k}{E^0 -E^k} $$ Then we have \begin{align} \frac{1}{2!}\left.\frac{d^2}{d\lambda}\right|_{\lambda=0} E_\lambda &=\frac{1}{2}\tr\left( V P^0 V R^0+VR^0 VP^0 \right) \\ &= \tr(P^0VR^0 V)=\langle 0|VR^0 V|0\rangle \end{align} where I used the commutation invariance of the trace function. Now you can see that the formula is starting to look like the BW perturbation, and you would assume that the trend goes on, i.e., $n=3$ order perturbation would be $\langle 0| VR^0 V R^0 V|0\rangle$ and so on. However, this doesn't seem to be true. Indeed, let us look at the $n=3$ contour integral $$ \propto \tr\left( VP^a VP^b VP^c\oint_\Gamma \frac{dz}{(z-E^a)(z-E^b)(z-E^c)} \right) $$ Since $\Gamma$ only encloses $E^0$, we see that the contour integral is nonzero only if one of $E^a,E^b,E^c$ is equal to $E^0$ and all of them are distinct, e.g., $E^a=E^0$ and $E^0\ne E^b\ne E^c \ne E^0$. However, it's not too hard to see that in order for the 3rd order perturbation to be of the form $P^0 V R^0 V R^0 V$, we instead should be summing over something like $E^a=0$, but $E^b,E^c\ne E^0$ where $E^b,E^c$ are allowed to be equal.

With this in mind, I'm having trouble believing in the BW perturbation, and it would be greatly appreciated if someone could point out the flaw in my logic.

APPENDIX. If you don't believe in the equation written down, let me provide a sketch of the proof. Notice that we can write the projection $P_\lambda$ onto the eigenspace $E_\lambda$ of $H_\lambda$ as a contour integral (Riesz projector) $$ P_\lambda = \frac{1}{2\pi i} \oint_\Gamma \frac{dz}{z-H_\lambda} $$ Also notice that since the energy state $E_0$ is non-degenerate, by Weyl's theorem, so is $E_\lambda$ for sufficiently small $\lambda$. Hence, we can write $$ E_\lambda = \tr (P_\lambda H_\lambda) $$ We can then calculate the derivative of $E^\lambda$ by calculating that of $P_\lambda$ and $H_\lambda$ separately. Indeed, one can easily write down $$ \frac{1}{\epsilon} (P_{\lambda+\epsilon} - P_\lambda) $$ And take $\epsilon \to 0$.

  • $\begingroup$ Is this from a reference? Related: physics.stackexchange.com/q/717102/2451 $\endgroup$
    – Qmechanic
    Commented Dec 29, 2023 at 0:48
  • $\begingroup$ @Qmechanic I'm not sure if there's a reference for the way I wrote it here. I've heard that Kato has some nice text on rigorous perturbation theory, but unfortunately, I never checked. $\endgroup$ Commented Jan 3 at 1:33

1 Answer 1


I realized that the contour integral $$ \oint_\Gamma \frac{dz}{(z-E^a)(z-E^b)(z-E^c)} $$ is nonzero if and only if one of $E_a,E_b,E_c$ is $=E_0$ and the remaining are $\ne E_0$, e.g., $E_a=E_0, E_b,E_c\ne E_0$, that is, $E_b, E_c$ are allowed to be equal. Then there's nothing wrong with the BW perturbation. In fact, you can do something similar with the degenerate case, i.e., take derivative of $P_\lambda H_\lambda P_\lambda$ and project onto the the ground state, e.g., first order perturbation would be $$ P_0 \left. \frac{d}{d\lambda} \right|_{\lambda=0}(H_\lambda P_\lambda) P_0 = P_0 VP_0 $$


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