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author | Prefetch | 2021-02-20 20:06:00 +0100 |
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committer | Prefetch | 2021-02-20 20:06:00 +0100 |
commit | a121b6e4affdf4e4ad26d179fbf2f6b8c6070b11 (patch) | |
tree | eed02b142dfb0aa724b493243094a372c03c6740 /latex/know/concept/pauli-exclusion-principle/source.md | |
parent | 549b1e963183b6284aefbc52b9167009fefa551d (diff) |
Add "Pauli exclusion principle"
Diffstat (limited to 'latex/know/concept/pauli-exclusion-principle/source.md')
-rw-r--r-- | latex/know/concept/pauli-exclusion-principle/source.md | 112 |
1 files changed, 112 insertions, 0 deletions
diff --git a/latex/know/concept/pauli-exclusion-principle/source.md b/latex/know/concept/pauli-exclusion-principle/source.md new file mode 100644 index 0000000..0a35869 --- /dev/null +++ b/latex/know/concept/pauli-exclusion-principle/source.md @@ -0,0 +1,112 @@ +% Pauli exclusion principle + + +# Pauli exclusion principle + +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: + +$$\begin{aligned} + \hat{P} \ket{a}\!\ket{b} = \ket{b}\!\ket{a} +\end{aligned}$$ + +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 +$\lambda = -1$. + +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: + +$$\begin{aligned} + \hat{P}_f \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = - \ket{a}\!\ket{b} + \qquad + \hat{P}_b \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = \ket{a}\!\ket{b} +\end{aligned}$$ + +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 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: + +$$\begin{aligned} + \ket{\Psi(a, a)}_f + = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big) + = 0 +\end{aligned}$$ + +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. |