Two Remarks on Schrodingers Quantum Theory

PROC. N. A. S.

PHYSICS: P. S. EPSTEIN

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components have not been resolved, and its wider separation is determined by the measurements. Thus this work confirms the theory of Heisenberg in its essential points. In this way the experimental work, by showing that, the helium spectrum contains 'singlets and triplets, confirms the theoretical predictions and thus removes this spectrum from its anomalous position. * NATIONAL RuSuARCa F}uLOW. 1 Nutting, Astroph. J., 23, 64, 1906; Lohman, Zeit. Wiss. Phot., 6, 1 and 41, 1908; R. Brunetti, Accad. Lincei, Atti., 33, ii, 413, 1924; Ruark, Foote and Mohler, Opt. .Soc. Amer. J., 8, 17, 1924; MacNair and McCurdy, Nature, 117, 159, 1926. 2 W. Heisenberg, Zeit. Physik, 39, 499, 1926. 3 Houston, Physic. Rev., March, 1927. 4 Merrill, Astroph. J., 46, 357, 1917. 6 D. Burger, Zeit. Physik, 38, 437, 1926. * McCurdy, Phil. Mag., 2, 529, 1926.

TWO REMARKS ON SCHRODINGER'S QUANTUM THEORY By PAUL S. EPSTZIN CALIjORNIA INSTITUTE or TSCHNOIOGY Communicated February 16, 1927

1. As our first remark we wish to point out that the method of reduction of Schrbdinger's equation which the author used in his treatment of the Stark effect' permits also an extremely simple treatment of the relativity effect in the neutral hydrogen atom. The Hamiltonian equation in this case has the form

Mc2 1 + (VS)2/m2c2 - E' + e2/r,

(1)

where e, m are the charge and mass of the electron, c the velocity of light, 't&d where the energy E' includes the inner potential energy Mc2 of the t1ectron (E' = E + mc2). Solving with respect to VS:

(VS)2 - [(E'

e2/r)2

_

m2c4]/C2

=

0.

(2)

We obtain the wave equation of the problem replacing V S by the operator (where K = h/27r) V 2/ + [(E' - e2/r)2 -m2c4]jP/K2c2 = 0. (3)

A14V

As in the non-relativistic case, we substitute y6 =

X(r)Pk_i(cos t) cos nip,

(4)

ToL. 13, 1927

PHYSICS: P. S. EPSTEIN

and obtain for x(r) the equation d2 d2 d2 + 2dX + A +2B+po

95

-k(k1)] x =,

(5)

(6) A = (E'2-m2c4)/K2c2, B = e2E'/K2c2, Po2 = e4/K2c2. This equation can be treated exactly in the same way which we have used in the other case. We put as there a2 = -A. (7) x = rk lexp(ar)M, This leads to the equation for M: k\d__ _ d2M dM + [2 ak+B+p]2 d 2 + 2 a+ (8) M=O.

r ra+-r type and M can be easily This equation is again of the hypergeometric

-+2

represented as a power series

M = Ear. (9) For the coefficients a, we obtain the relation = 0. (10) [52 + (2k - 1)5 + po2]aa + 2a[5 + k-1 + The coefficient of a, gives us the exponent of the first term of the expansion (9). The roots of 62 + (2k- 1)5 + po2 = 0 are

B/a]a8_.

50

=

i1

(

k

)) 11 -2 o-(-

We may use only the upper sign as the function x(r) must be finite in the point r = 0 and as the lower sign would violate this requirement. In the other terms of the sum (9) a will assume the values 5 = 5o + n, k and n being integers. The coefficient of aa_, then can be written as [So + B/a + n + k - 1]. This coefficient will vanish for a certain value of n, if and only if So + B/a + k is a negative integral number -1: So + B/a = -(k + 1). (12) On the other hand it is easy to see that, unless the series (9) breaks off, the function x(r) will be infinite for r = co. Relation (12) represents, therefore, the condition of finiteness. Substituting 5o, B, a from (12), (6), (7) and solving with respect to E, we have E = mc2[ 1 +

po2/(l

+

+ i (k +

-

Po2

c2

(13)

as found by Schrodinger by a different method. 2. The second remark bears on the problem of the relation of Schrodinger's equation to the theory of light quanta. The fundamental concept introduced by Louis de Broglie is that with motion of a mass there is always associated a wave motion.' Let the mass, m, be moving without forces in

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96

PRoc' N. A. S.

the direction x, with the velocity x. Denoting the momentum by p = mx we can write the energy equation as p2/2m = E. (14) The corresponding wave equation results then by Schrodinger's method as

62,/bX2 + 8X2mE4/h2 = 0. (15) Let us now reverse the problem: We consider a plane light wave as given, according to the electromagnetic theory, by the -equation (16) 624,/1x2 + kP% = 0, k = 7r/ = 2,x/c. We put the question: Will there be associated with this light wave the motion of a material particle and what will be the momentum and the kinetic energy of this particle? According to de Broglie the velocities x and V (of the wave) are related by the equation xV = c2. In the case of light waves we have* V = c, giving x = c. Our particles will move with the velocity of light, and this makes a change of .equation (15) necessary: For particles moving with the velocity c the energy is expressed as Mc2 and not mx2/2. The, momentum is p = mc and the energy equation can be written as mc2= p2/m = E. (17) The corresponding wave equation is 82+/8x2 + 47r2mE%&/h2 = 0. (18) This must be compared with the form (16). We see that we have to identify 47r2mE/h2 = k2 = 4r2V2/C2. (19) Making use of relation (17) we find E = hv. (20) We see that the particles associated with the motion of light waves have the energy hv and the momentum hv/c, i.e., they have exactly the characteristics which Einstein ascribed to the light quanta. Some interesting considerations on the connection of light quanta with Schrodinger's theory were published by L. de Broglie,3 but the aspect of the matter presented in the above lines seems to be new. Of course, we regard it as a first and incomplete attempt, as there remain still many difficulties to be elucidated. Especially it should be mentioned that Schr6dinger's functions represent a system of stationary waves, while in optics we are particularly interested in progressing waves. ' P. S. Epstein, Physic. Rev., 28, p. 695. 1926.

P. S. Epstein, Proc. Nat. A cad. Sci., 12, p. 629, 1926. ' L. de Broglie, Nature, 118, p. 441, 1926.

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