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author | Prefetch | 2022-10-14 23:25:28 +0200 |
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committer | Prefetch | 2022-10-14 23:25:28 +0200 |
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diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md new file mode 100644 index 0000000..56af39b --- /dev/null +++ b/source/know/concept/pauli-exclusion-principle/index.md @@ -0,0 +1,119 @@ +--- +title: "Pauli exclusion principle" +date: 2021-02-22 +categories: +- Quantum mechanics +- Physics +layout: "concept" +--- + +In quantum mechanics, the **Pauli exclusion principle** is a theorem with +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 can occupy the same two allowed +states $a$ and $b$. We then define the permutation operator $\hat{P}$ as +follows: + +$$\begin{aligned} + \hat{P} \Ket{a}\Ket{b} = \Ket{b}\Ket{a} +\end{aligned}$$ + +That is, it swaps the states of the particles. Obviously, swapping the +states twice simply gives the original configuration again, so: + +$$\begin{aligned} + \hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b} +\end{aligned}$$ + +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 $\lambda = -1$: + +$$\begin{aligned} + \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b} +\end{aligned}$$ + +As it turns out, in nature, each class of particle has a single +associated permutation eigenvalue $\lambda$, or in other words: whether +$\lambda$ is $-1$ or $1$ depends on the type of particle that $x_1$ +and $x_2$ are. 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: + +$$\begin{aligned} + \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b} + \qquad + \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b} +\end{aligned}$$ + +Another fundamental fact of nature is that identical particles cannot be +distinguished by any observation. Therefore it is impossible to tell +apart $\Ket{a}\Ket{b}$ 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: + +$$\begin{aligned} + \Ket{\Psi(a, b)} + = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a} +\end{aligned}$$ + +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: + +$$\begin{aligned} + \hat{P} \Ket{\Psi(a, b)} + &= \lambda \alpha \Ket{a}\Ket{b} + \lambda \beta \Ket{b}\Ket{a} + \\ + \hat{P} \Ket{\Psi(a, b)} + &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b} +\end{aligned}$$ + +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 $\alpha = \beta$: + +$$\begin{aligned} + \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big) +\end{aligned}$$ + +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 $\alpha = -\beta$: + +$$\begin{aligned} + \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big) +\end{aligned}$$ + +This state is 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 $C$: + +$$\begin{aligned} + \Ket{\Psi(a, a)}_b + = C \Ket{a}\Ket{a} +\end{aligned}$$ + +However, for fermions, the state is unnormalizable and thus unphysical: + +$$\begin{aligned} + \Ket{\Psi(a, a)}_f + = C \big( \Ket{a}\Ket{a} - \Ket{a}\Ket{a} \big) + = 0 +\end{aligned}$$ + +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. |