Categories: Physics, Quantum mechanics.

Pauli exclusion principle

In quantum mechanics, the Pauli exclusion principle is a theorem with profound consequences for how the world works.

Suppose we have a composite state ∣x1⟩∣x2⟩=∣x1⟩⊗∣x2⟩\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}, where the two identical particles x1x_1 and x2x_2 each can occupy the same two allowed states aa and bb. We then define the permutation operator P^\hat{P} as follows:

P^∣a⟩∣b⟩=∣b⟩∣a⟩\begin{aligned} \hat{P} \Ket{a}\Ket{b} = \Ket{b}\Ket{a} \end{aligned}

That is, it swaps the states of the particles. Obviously, swapping the states twice simply gives the original configuration again, so:

P^2∣a⟩∣b⟩=∣a⟩∣b⟩\begin{aligned} \hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b} \end{aligned}

Therefore, ∣a⟩∣b⟩\Ket{a}\Ket{b} is an eigenvector of P^2\hat{P}^2 with eigenvalue 11. Since [P^,P^2]=0[\hat{P}, \hat{P}^2] = 0, ∣a⟩∣b⟩\Ket{a}\Ket{b} must also be an eigenket of P^\hat{P} with eigenvalue λ\lambda, satisfying λ2=1\lambda^2 = 1, so we know that λ=1\lambda = 1 or λ=−1\lambda = -1:

P^∣a⟩∣b⟩=λ∣a⟩∣b⟩\begin{aligned} \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b} \end{aligned}

As it turns out, in nature, each class of particle has a single associated permutation eigenvalue λ\lambda, or in other words: whether λ\lambda is −1-1 or 11 depends on the type of particle that x1x_1 and x2x_2 are. Particles with λ=−1\lambda = -1 are called fermions, and those with λ=1\lambda = 1 are known as bosons. We define P^f\hat{P}_f with λ=−1\lambda = -1 and P^b\hat{P}_b with λ=1\lambda = 1, such that:

P^f∣a⟩∣b⟩=∣b⟩∣a⟩=−∣a⟩∣b⟩P^b∣a⟩∣b⟩=∣b⟩∣a⟩=∣a⟩∣b⟩\begin{aligned} \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b} \qquad \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b} \end{aligned}

Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart ∣a⟩∣b⟩\Ket{a}\Ket{b} and the permuted state ∣b⟩∣a⟩\Ket{b}\Ket{a}, regardless of the eigenvalue λ\lambda. There is no physical difference!

But this does not mean that P^\hat{P} is useless: despite not having any observable effect, the resulting difference between fermions and bosons is absolutely fundamental. Consider the following superposition state, where α\alpha and β\beta are unknown:

∣Ψ(a,b)⟩=α∣a⟩∣b⟩+β∣b⟩∣a⟩\begin{aligned} \Ket{\Psi(a, b)} = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a} \end{aligned}

When we apply P^\hat{P}, we can “choose” between two “intepretations” of its action, both shown below. Obviously, since the left-hand sides are equal, the right-hand sides must be equal too:

P^∣Ψ(a,b)⟩=λα∣a⟩∣b⟩+λβ∣b⟩∣a⟩P^∣Ψ(a,b)⟩=α∣b⟩∣a⟩+β∣a⟩∣b⟩\begin{aligned} \hat{P} \Ket{\Psi(a, b)} &= \lambda \alpha \Ket{a}\Ket{b} + \lambda \beta \Ket{b}\Ket{a} \\ \hat{P} \Ket{\Psi(a, b)} &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b} \end{aligned}

This gives us the equations λα=β\lambda \alpha = \beta and λβ=α\lambda \beta = \alpha. In fact, just from this we could have deduced that λ\lambda can be either −1-1 or 11. In any case, for bosons (λ=1\lambda = 1), we thus find that α=β\alpha = \beta:

∣Ψ(a,b)⟩b=C(∣a⟩∣b⟩+∣b⟩∣a⟩)\begin{aligned} \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big) \end{aligned}

Where CC is a normalization constant. As expected, this state is symmetric: switching aa and bb gives the same result. Meanwhile, for fermions (λ=−1\lambda = -1), we find that α=−β\alpha = -\beta:

∣Ψ(a,b)⟩f=C(∣a⟩∣b⟩−∣b⟩∣a⟩)\begin{aligned} \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big) \end{aligned}

This state is called antisymmetric under exchange: switching aa and bb causes a sign change, as we would expect for fermions.

Now, what if the particles x1x_1 and x2x_2 are in the same state aa? For bosons, we just need to update the normalization constant CC:

∣Ψ(a,a)⟩b=C∣a⟩∣a⟩\begin{aligned} \Ket{\Psi(a, a)}_b = C \Ket{a}\Ket{a} \end{aligned}

However, for fermions, the state is unnormalizable and thus unphysical:

∣Ψ(a,a)⟩f=C(∣a⟩∣a⟩−∣a⟩∣a⟩)=0\begin{aligned} \Ket{\Psi(a, a)}_f = C \big( \Ket{a}\Ket{a} - \Ket{a}\Ket{a} \big) = 0 \end{aligned}

And this is the Pauli exclusion principle: fermions may never occupy the same quantum state. One of the many notable consequences of this is that the shells of atoms only fit a limited number of electrons (which are fermions), since each must have a different quantum number.