Pauli s Derivation of the Hydrogen Spectrum with Matrix Representation

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1 Pauli s Derivation of the Hydrogen Spectrum with Matrix Representation Andrew MacMaster August 23, Motivation At the time Pauli s derivation was published, the primary method of calculating the hydrogen spectrum was via polar coordinates. While this method was successful in deriving the spectrum, it was necessary to add extra conditions to the derived quantum numbers s, m. If this is not done, the math predicts nonphysical results. Specifically for the m = 0 case, as polar coordinates do not have a bijective map to R 3, the old quantum mechanics predict the electron falls through the nucleus of the atom. Matrix representation bypasses these issues altogether, and as a result the hydrogen spectrum can be derived directly without such additional conditions. Pauli s proof of this fact was a necessary step in the adoption of the new quantum mechanics, and was instrumental in progressing our understanding of quantum mechanics. 2 Hydrogen Spectrum 2.1 Physical Assumptions Consider a single electron in orbit about a single proton, and assume we have three time-independent vector operators L, M, and E where L is the angular momentum about the origin, M is the perturbation in the single electron case due to Coulomb forces, and E is the diagonal matrix that corresponds to the Hamiltonian. We define them as follows: L = mr v 1 M = 2Ze 2 m 0 (L v v L + r r ) E = 1 2m 0 p 2 Ze2 r In defining our operators lie this, we can eliminate Cartesian coordinates in favor of these vector matrices. We also assume the following relations derived from physics. I L L = h 2πi L 1

2 II M x P x = P x M x M x P y P y M x = P x M y M y P x = h 2πi M z L M = M L = 0 III M M = h 2πi 2 m 0Z 2 e 4 EL IV 1 M 2 = 2 m 0Z 2 e 4 E(L 2 + h2 4π 2 ) From the existence of time-independent vector matrix M, we can infer that an atom with a single electron constitutes a degenerate system. This is because, from these relations, ML 2 L 2 M cannot vanish. As ME EM does vanish, we can conclude that for every value of E there is not just one value of L 2. In general, time-independent matrices need not be diagonal. Non-zero elements can occupy positions (n, m) which correspond to vanishing frequencies ν n m = En Em h = 0. However, if we force our matrices to be diagonal, the degeneracy disappears and we are able to find special solutions to equations (I) to (IV ) that is adapted to the fine structure of the hydrogen atom. 2.2 Pauli s Derivation Let P z, L 2 be diagonal matrices, and for a given L 2 let values of P z = mh 2π such that m. Also let partial vibrations of L, which belong to a change in m by ±1, be left- and right-circular in the (x, y)-plane. Then It then follows from (I) that P y ±1 2 = P x ±1 2 = P y ±1 = ±ip x ±1 h2 [( +1) m(m 1)] = h2 16π2 (L 2 ) ( + 1) 4π2 Now we define the components of M so that = h2 M y,m±1 = ±im x,m±1, ( = + 1 or 1) ( ±m)( +1 m) 16π2 M y ±1 2 = M x ±1 2 = 1 4 C+1 ( m)( m + 1) If we replace m with m ± 1, it further follows that M y ±1 2 = M x ±1 2 = 1 4 C +1( ± m + 1)( ± m + 2) M z 2 = C +1 [( + 1) 2 m 2 ] where C +1 is an as yet undetermined normalized non-negative function of that satisfies the symmetry relation C +1 = C+1. If we assume P x, M z are both positive and real, then M x is real and has its sign determined by whether or not m and are moving in the same direction. It was shown by Born, Heisenberg, and Jordan[1] that if L 2 and P z are assumed to be diagonal, then the expression chosen for L and M necessarily follows from (I) and (II). 2

3 We must now determine the normalization of and m and the function C+1. To do this we mae use of formula (III), and it suffices to just use the z component, M x M y M y M x = h 2 2πi m 0 Z 2 e 4 EP z Now we form the expression P y (M x M y M y M x ) (M x M y M y M x )P y = h 2 2πi m 0 Z 2 e 4 E(P yp z P z P y ) Using (I) and (II), we find an equation that matches the x-component of (III). Similarly, the y-component of (III) follows from the z-component of this vector equation and equations (I) and (II). If we form the element of the previous equation which occupies the (, m) position in the diagonal series, we obtain on the left hand side, (M x M y M y M x ) = 2i( M x 1 2 M x M x 1,m 1 2 M x 1,m+1 2 ) = im( (2 + 3)C +1 + (2 1)C 1) Noting that E has a negative sign, and introducing the Rydberg constant R = 2π2 e 4 m 0 h 3 together with our constraint on P z, we can derive the condition m{ (2 + 3)C +1 + (2 1)C 1} = E RhZ 2 m To mae sense of this condition, we will consider the extremes of. When is minimum, C 1 will vanish. Then the condition forces m = 0, as the coefficient m on the left hand side cannot be positive, but the coefficient m on the right hand side must be non-negative. Also, because of our earlier condition on P z, must also vanish. Otherwise m would be allowed to tae non-zero values. Thus and m are necessarily integers, and taes values = 1, 2,..., n where n is the maximum value that can tae for a given E. Now our relation implies (2 1)C 1 (2 + 3)C +1 = E for = 1, 2,..., n RhZ2 If we consider = n, then we must set C n +1 n = 0 as the contribution from transitioning + 1 vanishes for = n. So by starting at = n and iterating down stepwise, we can successfully calculate the values of Thus C +1 C n n 1, C n 1 n 2,..., C1 0 can be stated explicitly by the formula 3

4 C +1 = E n (n + 2) ( + 2) RhZ 2 (2 + 1)(2 + 3) By substituting 1 for, we get C 1 = = E (n )(n + + 2) RhZ 2 (2 + 1)(2 + 3) E n (n + 2) ( 1)( + 1) RhZ 2 (2 1)(2 + 1) = E (n + 1)(n + + 1) RhZ 2 (2 1)(2 + 1) With these expressions, we can directly confirm that our relations from earlier are satisfied. Now in order to derive the energy expression, we must use (IV ). First, we determine the value of M 2 at the (, m) position of the diagonal series. We ultimately obtain (M 2 ) = 2 M x M x M z 2 +2 M x 1,m M x 1,m M x 1,m 2 = ( + 1)(2 + 3)C +1 + (2 1)C 1 which, by substituting our expressions for C +1 and C 1, yields (M 2 ) = E RhZ 2 [n 2 + 2n ( + 1)] Now our full expressions for M 2 and L 2 can be plugged into (IV ), which gives us and thus 1 = E RhZ 2 (n 2 + 2n + 1) = E RhZ 2 (n + 1) 2 E = RhZ2 (n + 1) 2 ) = RhZ2 n 2, where n + 1 = n. This characterizes an n th order quantum state, and correctly predicts the Balmer series with weight n 2. 3 Mathematical Justification An important result of Pauli s paper is the generalization of the Runge-Lenz (RL) vector to the observable operator M. The reason this was so important is because it allowed Pauli to obtain an so(4) symmetry of the bound states. The classical Poisson bracets involving the classical quantities l and m generalize to the commutator relations [L i, L j ] = i ɛ ij L 4

5 [L i, M j ] = i ɛ ij M [M i, M j ] = ( 2H 0 µ ) i ɛ ijl where µ is the reduced mass. From these, if we restrict ourselves to the Hilbert space spanned by bound states with negative energy, we can define M = ( µ 2H 0 ) 1/2 M Because L and M commute with H 0, the algebra becomes [L i, L j ] = i ɛ ij L [L i, M j] = i ɛ ij M [M i, M j] = i ɛ ij L which is recognizable as an so(4) algebra with generators L and M. We can further define generators I = 1 2 (L + M ) K = 1 2 (L M ) The new commutation rules are lie so: [I i, I j ] = i ɛ ij I [K i, K j ] = i ɛ ij K [I i, K j ] = 0 The operators I generate an so(3) I Lie algebra, and the operators K generate an so(3) K Lie algebra. Further these algebras are commuting, and their Casimir operators are respectively I 2 and K 2. This correspondence between L, M and I, K gives a change of basis that displays the isomorphism so(4) so(3) I so(3)k While the full exploration of this is beyond the scope of this paper, by extending quantum mechanics into so(4), Pauli was able to tae advantage of the symmetry of such spaces. In so doing, he found an elegant solution to a difficult problem. References [1] M. Pauli Jr, On the Hydrogen Spectrum from the Standpoint of the New Quantum Mechanics. Sources of Quantum Mechanics, edited by B. L. van der Waerden ( Dover, New Yor, 1968), pp [2] Galliano Valent, The hydrogen atom in electric and magnetic fields : Pauli s 1926 article arxiv:quant-ph/