# Pauli's exclusion principle

• Gravitonion
In summary, the principle stating that no two fermions can have the same atomic number means that identical fermions cannot exist in the same state. This is important because it allows for the creation of the periodic table and explains the increase in size of atoms as electrons are added. This principle is based on the use of anti-commutators to construct fermionic creation and annihilation operators, which dictate that two fermions cannot occupy the same state. However, they can occupy the same state if they have different positions in momentum space.

#### Gravitonion

I know the principle says that no two fermions canNot have the same atomic number, but could you explain that in detail?

It means two identical fermions (say, any two electrons) cannot be in exactly the same state. So if you have a discrete way of labeling the states, say by atomic quantum numbers, then you can only have one electron in each of those states (or two if you are not counting the spin in the quantum numbers). One reason this is really important is that it gives us the periodic table-- atoms get a lot larger when you "break into a new valence shell" as you add electrons, and that's only because the new electrons can't go into the states that the previous electrons are already in.

Gravitonion said:
I know the principle says that no two fermions canNot have the same atomic number, but could you explain that in detail?
Are you familiar with the harmonic oscillator?

Fermionic creation and annihilation operators are constructed using anti-commutators (instead of commutators

$$\{b_s,b^\dagger_{s^\prime}\} = \delta_{s\,s^\prime}$$

$$\{b_s,b_{s^\prime}\} = \{b^\dagger_s,b^\dagger_{s^\prime}\} = 0$$

The last equation is important b/c for s=s' it can be rewritten as :

$$(b^\dagger_s)^2 = 0$$

s,s' represent the quantum numbers

You can construct a states where each s, s', s'', ... is occupied by either zero or one particle. But if you want to construct a state with two particles sitting in state s you get

$$|2s\rangle = (b^\dagger_s)^2|0\rangle = 0$$

due to the anticommutator.

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They can be in the same state as long as they have different positions though right?

Runner 1 said:
They can be in the same state as long as they have different positions though right?
Sorry, I forgot to explain that creation and annihilation operators act in momentum space and that this means that s is a collection of momentum, spin, isospin, etc.}. Therefore they are not located anywhere in position space.