1. Introduction
For more than a century, quantum mechanics has been the fundamental theory that guides our understanding of how nature works at the scale of atoms and subatomic particles. The hydrogen atom is the simplest possible atom for theoretical investigation, consisting of only a single proton and a single electron orbiting around [
1]. While the hydrogen atom has been intensively studied since the dawn of quantum mechanics (
Figure 1), as a demonstration of what measurements can reveal about atoms, there are still surprises and hidden structures [
2]. One example is the emergence of equifrequency transitions, in which many distinctive jumps between atomic levels can radiate identical photon energy. This question was raised during a Ph.D. oral exam in 1997 at the University of Colorado Boulder [
3] and soon became well-known in the physics community, especially among graduate students. The answer is definitely yes, and an infinite number of transitions have been found [
4]; however, to the best of our knowledge, a generalization is still lacking.
Here, we show the connection between the above question and a Diophantine equation [
5], and present a general solution, i.e., how all equifrequency transition pairs can be obtained. While this finding might not address any foundational issue or important problem in quantum mechanics, it definitely provides us with a more complete understanding of the most popular atom in all quantum mechanics textbooks, and the relationship between atomic levels (disregarding degeneracies due to angular momentum and spin). It is also a simple illustration of how number theory can be of relevance to physics [
6], in a way that is accessible to non-experts.
2. A General Solution for All Equifrequency Transitions
In quantum mechanics, the
n-th energy level of a hydrogen atom is given by
, where
is a positive integer and
eV is the Rydberg energy [
7]. For simplicity, we will not consider any relativistic effects [
8] or other corrections (such as fine structure [
9,
10]) to this equation. The challenge is to find all transition pairs (
,
) with an equal radiation energy, which means:
Here, we will find a general solution to this equation, including trivial solutions where
and
.
Consider the Diophantine equation [
5] with a parameter
and unknowns
,
With any two solutions
and
to this equation, for any positive integer pair
that satisfies
a solution to Equation (
1) can be obtained:
which can be checked by direct substitution. To generate all solutions
to Equation (
3), we use any
and
,
where the operation
determines the greatest common divisor of
[
11].
We can prove that the above procedure comprises all solutions of Equation (
1). Starting from this equation, denote
and
. Write
. Note that
and
i.e.,
and
. Then, we rewrite (
1) as,
and put the fraction
into irreducible form
where
and both are non-zero,
Thus,
and hence there exists
such that,
Note here that condition (
7) is exactly Equation (
3) and condition (
9) provides us two solutions (
2). Combined with the above paragraph, we see that these two conditions (
7) and (
9) are both necessary and sufficient. This completes the proof.
To generate the set of all non-zero integer solutions
to Equation (
2), we need the set of all non-zero rational solutions
into their dehomogenized version (by dividing both sides of (
2) by
):
This equation is very similar to the Pell equation [
12], but can be solved using a much simpler method. By taking any
that satisfies (
10) and any
, we obtain all triples,
of (
2) where
.
The geometric way [
13] of dealing with Equation (
10) is to draw a line in the
-plane passing through
with a rational slope
; for example, the line
(see
Figure 2). For
, this line will cut the curve (
10) at another point,
and, more importantly, all solutions of Equation (
10) can be attained in this way by varying
q. Note that
gives
, and changing the sign of
q changes the sign of
. Hence, if we let
where
,
; then,
,
, where
The positive triple
can be obtained from (
11) with the correct sign choice.
In summary, we can generate a solution
to Equation (
2) with parameter
from any number
. Given the pair, we go through Equations (
13) and (), pick a value
and use Equation (
11) to arrive at
. Then, with two such solutions, say,
and
, we choose a value
and use Equation (
5) to obtain
before plugging in Equation (
4) to obtain a pair
. See
Figure 3 for a demonstration. The key difference in our approach compared to previous ones is using (
2), where we can generate all possible rational solutions, which enables us to find all possible solutions to the puzzle (
1).
3. Families of Equifrequency Transitions
Perondi [
4] found an infinite number of solutions to the generalization of (
1):
for any
. His approach is to start with a set of
k primes
, for some
, and then try to find an integer
, for which
has a solution. By partitioning [
14], the set of indices
into two sets
and denoting
, he found that if
and
is divisible by
and
, then
is a solution to
which, again, using the identity
. Now, by enlarging (or shrinking)
if necessary, and splitting the set of indices differently, he found
n distinct pairs
. Then, he chose a positive integer
, which is divisible by
n ’s, and found
n pairs
solution to (
16).
Similar to the above, we can find all solutions to the generalized Equation (
15) of Perondi by simply solving the first equation, which is equal to the
ith equation, for all
i, using the method we found in the previous section. First, choose
and
n distinct triple
, satisfying:
Then, we want to find
, such that
It suffices that
is divisible by
and the remaining
are deduced from (19). The final solution to the generalized Equation (
15) is: