2 1 4 D O C U M E N T 1 9 8 F E B R U A R Y 1 9 2 6 If , then according to the canonical transformation, , , which you propose, this does not agree with the formula given by you. (It is, of course, .) But the canonical trans- formation chosen by you is still somewhat impractical insofar as P is no longer a Hermitian matrix. That is why I would suggest[4] . Then . If these examples are treated as perturbation calculations, then, as far as I can see, the old energy matrix always results.[5] But perhaps I am mistaken in any event I would be very thankful if you would very briefly retrace the course of your calculations for me. For the treatment of point (a) (1), on the rotator, I hope that Born has sent you the correction proofs of his and Wiener’s paper.[6] The problem of the rotator really is very characteristic of the whole theory but I will write down all I know about it in detail. First, the calculation:[7] Let r be the constant distance between the particles, m the mass, ϕ the angle, hence, . , and the energy as the commutation condition. Then the equa- tions of motion yield , and the whole problem is, as you say, identical to that of the translation thus it seems as if there were no quantization here. However, now one can ask, how the radiation or how x and y look as a function of time. (I now always denote as the k 1 2 -- - p2 ω0 2q2) + ( = p P 2λPQ … + + = q Q λQ2 – … + = H 1 2 --{ - P2 ω0 2Q2 2λ( P2Q PQP) + 2λQ3ω0 2} – + + = p2 P 2λPQ)2 + ( P2 2λ( P PQ PQ P)… ⋅ + ⋅ + + = = q Q λQ2 – … + = p P λ( PQ QP) + … + + = Q q λq2 += P p λ( pq qp) + – … + = H 1 2 -- - P2 ω0 2Q2 λ( P2Q 2PQP GP2) + + 2λQ3ω0 2} – + + { = x rcosϕ = y rsinϕ = p mr2ϕ · = H p2 2mr2 ------------ = p ϕ ϕ p ⋅ – ⋅ h 2πi -------- = ϕ · ∂p ∂H const. = = td d