It's part of the weird effect of quantum mechanics.
To simplify, let's suppose you have two electrons that can jump between 4 Hydrogen atoms. Let's call the ("1s" orbitals) in the atoms 1, 2, 3, 4.
Now the 2 electrons can be in the atoms 1 and 2. The official notation is |12> and it's essentially equal to |21> because the wo electrons are exactly equal.
The two electrons can be in |13>, |14>, |23>, |24> or |34>, so there are in total 6 possibilities.
For a calculation of the state with minimal energy, you must find a combination of them, i.e.
where the Greek letters are real number, and the square of them sum 1.
With 3 electrons in 6 Hydrogen atoms you have 20 possibilities, so you must calculate 20 real numbers.
With 4 electrons in 8 Hydrogen atoms you have 70 possibilities, so you must calculate 70 real numbers.
(It's more usual to consider that the number of electrons is equal to the number of Hydrogens, but the electrons in the Hydrogens can have spin up or down. So you have to double the numbers in the previous examples.)
The general calculation for N Hydrogens is Combinatorial(N, N/2) (or Combinatorial(2N, N)), and it is exponential.
With more complex atoms, you have more orbitals and more electrons, so it's even bigger.
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As a side effect, if you can modify the global state of one of this systems with N atoms, you are operating on an exponential number of states. This is the main idea of quantum computers, and why they can go faster than a classical computer. They use better system than my examples, so the calculation are simpler, but it is the same "exponential" magic/curse. There are a lot of problems to build a "good" system, so the current quantum computer have a small amount of atoms.
To simplify, let's suppose you have two electrons that can jump between 4 Hydrogen atoms. Let's call the ("1s" orbitals) in the atoms 1, 2, 3, 4.
Now the 2 electrons can be in the atoms 1 and 2. The official notation is |12> and it's essentially equal to |21> because the wo electrons are exactly equal.
The two electrons can be in |13>, |14>, |23>, |24> or |34>, so there are in total 6 possibilities.
For a calculation of the state with minimal energy, you must find a combination of them, i.e.
|state> = alpha |12> + beta |13> + gamma |14> + delta |23> + epsilon |24> + dseta |34>
where the Greek letters are real number, and the square of them sum 1.
With 3 electrons in 6 Hydrogen atoms you have 20 possibilities, so you must calculate 20 real numbers.
With 4 electrons in 8 Hydrogen atoms you have 70 possibilities, so you must calculate 70 real numbers.
(It's more usual to consider that the number of electrons is equal to the number of Hydrogens, but the electrons in the Hydrogens can have spin up or down. So you have to double the numbers in the previous examples.)
The general calculation for N Hydrogens is Combinatorial(N, N/2) (or Combinatorial(2N, N)), and it is exponential.
With more complex atoms, you have more orbitals and more electrons, so it's even bigger.
---
As a side effect, if you can modify the global state of one of this systems with N atoms, you are operating on an exponential number of states. This is the main idea of quantum computers, and why they can go faster than a classical computer. They use better system than my examples, so the calculation are simpler, but it is the same "exponential" magic/curse. There are a lot of problems to build a "good" system, so the current quantum computer have a small amount of atoms.