Answer to the Question 06/00

HYDROGEN ATOM

The question was:

Is it possible that two different transitions in hydrogen atom give the same frequency of radiation?

(10/00) We did not recieve a general solution of the problem. However, a satisfactory solution has been submitted by Jared D. Kaplan (9/7/2000) (e-mail jared@imsa.edu) and his friend Howard Liu, both of them high school students from the Illinois Mathematics and Science Academy. Their solution is presented at the bottom of this page. Iddo Ussishkin (27/7/2000) from the Department of Condensed Matter Physics at the Weizmann Institute of Science in Israel (e-mail feiddo@wicc.weizmann.ac.il) has correctly identified the "lowest possible" transition pair. A subset of solutions has also been identified by Ganesh Sundaram (28/9/00) (e-mail gsundaram@hotmail.com).


The answer: There are infinitely many such transitions.

The solution:

Since the energy of the n'th level is propotional to -1/n2 the transition frequency for transiiton from m'th level to n'th level is proportional to (1/n2-1/m2). So we need to find different pairs of integers for which the difference in inverse squares is the same. And this can beeasily done: E.g., the transition from n=35 to n=7 has the same energy difference as the transition from n=7 to n=5. Similarly, the transition from n=90 to n=9 has the same energy difference as the transition from n=6 to n=5. Below we present a large sequence of such pairs for n<200. (This sequence has been found "by inspection" of all possible transitions with relatively small n's. We do not know a general solution. However, see below.)

( 6 -> 5)==( 90 -> 9)
( 7 -> 5)==( 35 -> 7)
( 9 -> 5)==( 90 -> 6)
( 8 -> 6)==( 72 -> 9)
( 9 -> 6)==( 72 -> 8)
( 8 -> 7)==( 56 -> 14)
( 14 -> 7)==( 56 -> 8)
( 11 -> 10)==( 55 -> 22)
( 12 -> 10)==(180 -> 18)
( 14 -> 10)==( 70 -> 14)
( 18 -> 10)==(180 -> 12)
( 22 -> 10)==( 55 -> 11)
( 16 -> 12)==(144 -> 18)
( 18 -> 12)==(144 -> 16)
( 16 -> 14)==(112 -> 28)
( 18 -> 14)==( 63 -> 21)
( 21 -> 14)==( 63 -> 18)
( 28 -> 14)==(112 -> 16)
( 21 -> 15)==(105 -> 21)
( 22 -> 20)==(110 -> 44)
( 28 -> 20)==(140 -> 28)
( 44 -> 20)==(110 -> 22)
( 24 -> 21)==(168 -> 42)
( 42 -> 21)==(168 -> 24)
( 35 -> 25)==(175 -> 35)
( 36 -> 28)==(126 -> 42)
( 42 -> 28)==(126 -> 36)
( 33 -> 30)==(165 -> 66)
( 34 -> 30)==( 85 -> 51)
( 51 -> 30)==( 85 -> 34)
( 66 -> 30)==(165 -> 33)
( 44 -> 33)==(176 -> 48)
( 48 -> 33)==(176 -> 44)
( 42 -> 35)==(120 -> 56)
( 56 -> 35)==(120 -> 42)
( 45 -> 36)==( 80 -> 48)
( 48 -> 36)==( 80 -> 45)
( 40 -> 38)==(152 -> 95)
( 95 -> 38)==(152 -> 40)
( 42 -> 39)==(182 -> 91)
( 45 -> 39)==(117 -> 65)
( 65 -> 39)==(117 -> 45)
( 91 -> 39)==(182 -> 42)
( 54 -> 42)==(189 -> 63)
( 63 -> 42)==(189 -> 54)
( 70 -> 55)==(154 -> 77)
( 77 -> 55)==(154 -> 70)
( 68 -> 60)==(170 -> 102)
(102 -> 60)==(170 -> 68)
( 75 -> 65)==(156 -> 100)
(100 -> 65)==(156 -> 75)
( 90 -> 72)==(160 -> 96)
( 96 -> 72)==(160 -> 90)


Here is what Kaplan and Liu wrote (7/00) about this problem (slightly edited):

The energy (and, consequently, the radiation frequency) released when electrons jump between energy levels a and b (integers) is proportional to (1/a2-1/b2). So the problem is equivalent to finding solutions to the diophantine equation:
1/a2 - 1/b2 = 1/c2 - 1/d2, where a != c and b != d.
Which we can simplify to:
(b2-a2)/(ab)2 = (d2-c2)/(cd)2
Our method of finding solutions here was to find solutions to:
b2-a2 = d2-c2
and then to multiply a and b by some number x and c and d by some number y so that the original expression works. We noticed that (52-12)=(72-52)=24, so we just need to find the constants x and y:
x2(52) = y2(5272)
So obviously we can let x = 7 and y = 1, to get the solution:
(a,b,c,d) = (7,35,5,7), meaning that a jump from energy level 35 to energy level 7 is the same as a jump from 7 to 5. There are obviously an infinite number of solutions of the form (7k, 35k, 5k, 7k), where k is a positive integer. I see no reason why there shouldn't be an infinity of other solutions, but so far I haven't proven this.

Later (8/00) Kaplan expanded the solution he suggested. His line of reasoning is quite complicated and tedious, so we only will give you the final answer which you can verify directly:
a = (k3+k6)/2
b = (k6-k3)/2
c = k2(k+1)(k4+k2+1)/2
d = k2(k-1)(k4+k2+1)/2


Ussishkin made few interesting and important remarks (7/00):

1. The l and m quantum numbers should be chosen to agree with the transition selection rules.
2. On the practical level, observing these transitions may be very difficult, as they involve high lying levels, and various deviations from the "ideal" hydrogen may become important.

(4/0) We recieved quite a general method for generating pairs, triples, etc. of pairs of states producing identical frequencies. The solution has been sent by L.F.Perondi - a researcher at Instituto Nacional de Pesquisas Espaciais - SP, Brazil (e-mail perondi@directnet.com.br). While his method does not produce all the possible states, it certainly very broad. The method is as follows:

We give a procedure for finding a set of pairs (x1,y1;x2,y2;.....;xn,yn), with xi and yi integers, which satisfy the equation

(1) ... 1/x12 - 1/y12 = 1/x22 - 1/y22 = ... = 1/xn2 - 1/yn2 = R,
where R is a rational number.

For a particular pair of integers (xi,yi), we may write
(2) ... 1/xi2 - 1/yi2 = (yi-xi) * (yi+xi) / (xi2 * yi2) = m1 * m2 * ... * mk / D2,
D = d1* d2* ... * dl,
where mi and di denote the prime factors of the numerator and the denominator of Eq. (2), respectively. Let {mi} and {mj} denote a partition of {m1,m2,...,mk} into two groups, such that if C1 and C2 are the integers formed by the multiplication of all elements in {mi} and {mj}, respectively, then C1*C2 = m1*m2*...*mk. Making use of these definitions, Eq. (2) may be rewritten as

(3) ... m1 * m2 * ... * mk / D2 = ( (C1 + C2)/(2 * D) )2 - ( (C1 - C2)/(2 * D) )2.

From Eq. (3), we observe that if D is chosen so as to contain the prime factors appearing in either (C1+C2)/2 and (C1-C2)/2, then this equation assumes the form given on the left-hand side of Eq. (2), with

(4) ... xi = 2 * D / ( C1 + C2 ) , yi = 2 * D / ( C1 - C2) .

Hence, departing from a given set of integers {m1,m2,..., mk}, we may "construct" as many solutions to Eq. (2) as there are partitions of {m1,m2,...,mk} in the sense described above.

Without any loss of generality , it may be easily shown that one may drop the factor 2 appearing in Eqs. (3) and (4) (it will appear multiplying m1 * m2 * ... * mk) and still get correct solutions to Eq. (1) through the same procedure described above.

Taking what has been described so far, we may summarise the procedure for generating solutions to Eq. (1), as follows:

- choose a set {m1,m2,...,mk}; for each partition, find the prime factors in the corresponding numbers ( C1 + C2 ) and ( C1 - C2 );
- form D in such a way that it contains all prime factors emerging from the above step, (D is given by the minimum common multiple of the set of the ( C1 + C2 ) and ( C1 - C2 ) factors);
- compute the pairs (xi,yi) from xi = D / ( C1 + C2 ) and yi = D / ( C1 - C2).

A few examples will help illustrating the procedure outlined above.

Let m1=3, m2 =5 (evidently, 1 is always a member of the set); consider the partitions:

2,3
(C1 + C2) = 5
(C1 - C2 ) = 1

6,1 (C1 + C2) = 7
(C1 - C2) = 5 ;
hence, D = 5 * 7, and

x1 = D / 5 = 7
y1 = D / 1 = 35

x2 = D / 7 = 5
y2 = D / 5 = 7.
Now let us consider an example with three pairs. Let m1 = 22, m2 =3

4,3
(C1 + C2) = 7
(C1 - C2 ) = 1

6,2
(C1 + C2) = 8 = 2^3
(C1 - C2 ) = 4 = 22

12,1
(C1 + C2) = 13
(C1 - C2 ) = 11;
hence, D = 23 * 7 * 11 * 13, and

x1 = D / 7 = 1,144
y1 = D / 1 = 9,152

x2 = D / 23 = 1,001
y2 = D / 22 = 2,002

x3 = D / 13 = 616
y3 = D / 11 = 728.

From the above example, by taking only the first two partitions we find another pair of solutions: now D = 23 * 7 and the corresponding two pairs are (8, 56) and (7, 14).

Concluding, we may say the number of pairs that satisfy equations of the form of Eq. (1) may be made as large as one pleases, through a suitable choice of R. The procedure given above allows one to build solutions with arbitrary number of pairs.

Y. Kantor: We are still waiting for a general solution of the problem.
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