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In Heisenberg's The Physical Principles of the Quantum Theory he has a section of Perturbation Theory, where he develops Perturbation theory on the Matrix Theory he's developed in the earlier sections of the book. It goes accordingly:

There is the Hamiltonian $H$ and a matrix $S$, which transforms the Hamiltonian into a coordinate system in which the Hamiltonian is diagonal, in which it is represented by $W$. As such: $$S^{-1}HS=W$$ or $$HS=SW$$ He then proceeds to develop the perturbed Hamiltonian, Energy and Transformation Matrices, with $\lambda$ as a dummy variable $$H=H_0+\lambda H_1+\lambda^2H_2+...$$ $$W=W_0+\lambda W_1+\lambda^2W_2+...$$ $$S=S_0+\lambda S_1+\lambda^2S_2+...$$ He then says that by replacing them in $HS=SW$ and setting equal the terms with equal $\lambda$ coefficients he obtains $$\begin{matrix} H_0S_0=S_0W_0\\ H_0S_1-S_1H_0=S_0W_1\\ H_0S_2-S_2H_0+H_2S_2-S_1W_1=S_0W_2\\ \vdots \end{matrix} $$ But these are not the results I obtain. The results I obtain are $$\begin{matrix} H_0S_0=S_0W_0\\ H_0S_1+H_1S_0=S_0W_1+S_1W_0\\ H_0S_2+H_2S_0+H_1S_1=S_0W_2+S_2W_0+S1W_1\\ \vdots \end{matrix}$$ Please help.

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  • $\begingroup$ What do you obtain? If you post that, it can be corrected $\endgroup$ – Greg.Paul Aug 2 '18 at 20:39
  • $\begingroup$ Show some of your work. This is rather lazy way of asking questions. $\endgroup$ – Jaswin Aug 2 '18 at 20:40
  • $\begingroup$ Yes, my apologies. I have edited it. $\endgroup$ – user140323 Aug 2 '18 at 20:47
  • $\begingroup$ Indeed, the missing $H_1$ on the l.h.side of the first order is a typo, corrected in the next edition. Set $S_0=1\!\!1$ as he ends up doing, to make life easier on yourself. H actually restructured this discussion in the next edition of the book.... $\endgroup$ – Cosmas Zachos Aug 3 '18 at 14:52
  • $\begingroup$ Well thank you! Do you happen to have a link to the next edition? $\endgroup$ – user140323 Aug 3 '18 at 15:53
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The other answer is correct, of course, but here is the 1941 edition streamlined version: $$H=H_0+\lambda H_1+\lambda^2H_2+...$$ $$W=W_0+\lambda W_1+\lambda^2W_2+...$$ $$S=1\!\! 1+\lambda S_1+\lambda^2S_2+...\qquad \Longrightarrow S^{-1}=1\!\! 1-\lambda S_1+\lambda^2(S_1^2-S_2)+... $$ So that $S^{-1} H S= W$ resolves to $$ H_0 = W_0\\ H_0S_1-S_1H_0 +H_1=W_1\\ H_0S_2-S_2H_0+H_2 +S_1^2H_0-S_1H_0S_1+H_1S_1-S_1H_1=W_2\\ =H_0S_2 -S_2 W_0 +H_2+H_1 S_1-S_1W_1,\\ \vdots $$ consistent with your calculation.

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  • $\begingroup$ Thank you! Where have you read this 1941 edition? $\endgroup$ – user140323 Aug 3 '18 at 19:35
  • $\begingroup$ I own it. As I indicate, I set $S_0=1$ right from the start to simplify things... $\endgroup$ – Cosmas Zachos Aug 3 '18 at 20:56
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In Heisenberg scenario, $H_0=W_0$ because he thought the $H_0$ should be some simple Hamiltonian that already been diagonalized.

I don't see any reason for my answer being downvoted. After all, if you look one page more, you will observe it's an obvious typo as indicated just above (58), after you take $W_0$ to $H_0$ as I said.

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  • $\begingroup$ You are right; it might be helpful to the OP if you added more detail. Heisenberg corrected the 1930 edition typo in the 1941 edition. Of course $H_1$ must be present to first order. In fact, you might as well set $S_0=1$ ab initio, to clear the underbrush. He much simplified all this in the next, 1941, edition. H, spectacular thinker as he was, was irredeemably sloppy when it came to such "details" ... he was used to talking to crackerjack types that quickly caught on to minor technical slipups.... $\endgroup$ – Cosmas Zachos Aug 3 '18 at 14:49
  • $\begingroup$ @CosmasZachos sure you're right. your answer should be helpful to OP :) $\endgroup$ – RoderickLee Aug 7 '18 at 22:50

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