In quantum mechanics, the Pauli exclusion principle is a theorem that has profound consequences for how the world works.
-Suppose we have a composite state \(\ket*{x_1}\!\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\), where the two identical particles \(x_1\) and \(x_2\) each have the same two allowed states \(a\) and \(b\). We then define the permutation operator \(\hat{P}\) as follows:
+Suppose we have a composite state \(\ket*{x_1}\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\), where the two identical particles \(x_1\) and \(x_2\) each can occupy the same two allowed states \(a\) and \(b\). We then define the permutation operator \(\hat{P}\) as follows:
\[\begin{aligned} - \hat{P} \ket{a}\!\ket{b} = \ket{b}\!\ket{a} + \hat{P} \ket{a}\ket{b} = \ket{b}\ket{a} \end{aligned}\]
That is, it swaps the states of the particles. Obviously, swapping the states twice simply gives the original configuration again, so:
\[\begin{aligned} - \hat{P}^2 \ket{a}\!\ket{b} = \ket{a}\!\ket{b} + \hat{P}^2 \ket{a}\ket{b} = \ket{a}\ket{b} \end{aligned}\]
-Therefore, \(\ket{a}\!\ket{b}\) is an eigenvector of \(\hat{P}^2\) with eigenvalue \(1\). Since \([\hat{P}, \hat{P}^2] = 0\), \(\ket{a}\!\ket{b}\) must also be an eigenket of \(\hat{P}\) with eigenvalue \(\lambda\), satisfying \(\lambda^2 = 1\), so we know that \(\lambda = 1\) or \(\lambda = -1\).
+Therefore, \(\ket{a}\ket{b}\) is an eigenvector of \(\hat{P}^2\) with eigenvalue \(1\). Since \([\hat{P}, \hat{P}^2] = 0\), \(\ket{a}\ket{b}\) must also be an eigenket of \(\hat{P}\) with eigenvalue \(\lambda\), satisfying \(\lambda^2 = 1\), so we know that \(\lambda = 1\) or \(\lambda = -1\).
As it turns out, in nature, each class of particle has a single associated permutation eigenvalue \(\lambda\), or in other words: whether \(\lambda\) is \(-1\) or \(1\) depends on the species of particle that \(x_1\) and \(x_2\) represent. Particles with \(\lambda = -1\) are called fermions, and those with \(\lambda = 1\) are known as bosons. We define \(\hat{P}_f\) with \(\lambda = -1\) and \(\hat{P}_b\) with \(\lambda = 1\), such that:
\[\begin{aligned} - \hat{P}_f \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = - \ket{a}\!\ket{b} + \hat{P}_f \ket{a}\ket{b} = \ket{b}\ket{a} = - \ket{a}\ket{b} \qquad - \hat{P}_b \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = \ket{a}\!\ket{b} + \hat{P}_b \ket{a}\ket{b} = \ket{b}\ket{a} = \ket{a}\ket{b} \end{aligned}\]
-Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart \(\ket{a}\!\ket{b}\) and the permuted state \(\ket{b}\!\ket{a}\), regardless of the eigenvalue \(\lambda\). There is no physical difference!
+Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart \(\ket{a}\ket{b}\) and the permuted state \(\ket{b}\ket{a}\), regardless of the eigenvalue \(\lambda\). There is no physical difference!
But this does not mean that \(\hat{P}\) is useless: despite not having any observable effect, the resulting difference between fermions and bosons is absolutely fundamental. Consider the following superposition state, where \(\alpha\) and \(\beta\) are unknown:
\[\begin{aligned} \ket{\Psi(a, b)} - = \alpha \ket{a}\!\ket{b} + \beta \ket{b}\!\ket{a} + = \alpha \ket{a}\ket{b} + \beta \ket{b}\ket{a} \end{aligned}\]
When we apply \(\hat{P}\), 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:
\[\begin{aligned} \hat{P} \ket{\Psi(a, b)} - &= \lambda \alpha \ket{a}\!\ket{b} + \lambda \beta \ket{b}\!\ket{a} + &= \lambda \alpha \ket{a}\ket{b} + \lambda \beta \ket{b}\ket{a} \\ \hat{P} \ket{\Psi(a, b)} - = \alpha \ket{b}\!\ket{a} + \beta \ket{a}\!\ket{b} + &= \alpha \ket{b}\ket{a} + \beta \ket{a}\ket{b} \end{aligned}\]
This gives us the equations \(\lambda \alpha = \beta\) and \(\lambda \beta = \alpha\). In fact, just from this we could have deduced that \(\lambda\) can be either \(-1\) or \(1\). In any case, for bosons (\(\lambda = 1\)), we thus find that \(\alpha = \beta\):
\[\begin{aligned} - \ket{\Psi(a, b)}_b = C \big( \ket{a}\!\ket{b} + \ket{b}\!\ket{a} \!\big) + \ket{\Psi(a, b)}_b = C \big( \ket{a}\ket{b} + \ket{b}\ket{a} \big) \end{aligned}\]
Where \(C\) is a normalization constant. As expected, this state is symmetric: switching \(a\) and \(b\) gives the same result. Meanwhile, for fermions (\(\lambda = -1\)), we find that \(\alpha = -\beta\):
\[\begin{aligned} - \ket{\Psi(a, b)}_f = C \big( \ket{a}\!\ket{b} - \ket{b}\!\ket{a} \!\big) + \ket{\Psi(a, b)}_f = C \big( \ket{a}\ket{b} - \ket{b}\ket{a} \big) \end{aligned}\]
This state called antisymmetric under exchange: switching \(a\) and \(b\) causes a sign change, as we would expect for fermions.
Now, what if the particles \(x_1\) and \(x_2\) are in the same state \(a\)? For bosons, we just need to update the normalization constant \(C\):
\[\begin{aligned} \ket{\Psi(a, a)}_b - = C \ket{a}\!\ket{a} + = C \ket{a}\ket{a} \end{aligned}\]
However, for fermions, the state is unnormalizable and thus unphysical:
\[\begin{aligned} \ket{\Psi(a, a)}_f - = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big) + = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big) = 0 \end{aligned}\]
At last, 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 an atom only fit a limited number of electrons, since each must have a different quantum number.
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