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diff --git a/content/know/category/quantum-mechanics.md b/content/know/category/quantum-mechanics.md index d61f03f..12c62d6 100644 --- a/content/know/category/quantum-mechanics.md +++ b/content/know/category/quantum-mechanics.md @@ -13,7 +13,7 @@ Alphabetical list of concepts in this category. * [Dirac notation](/know/concept/dirac-notation/) ## P -* [Pauli exclusion principle](/known/concept/pauli-exclusion-principle/) +* [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/) * [Probability current](/know/concept/probability-current/) ## T diff --git a/content/know/concept/index.md b/content/know/concept/index.md index 43f5928..4171da7 100644 --- a/content/know/concept/index.md +++ b/content/know/concept/index.md @@ -13,7 +13,7 @@ Alphabetical list of concepts in this knowledge base. * [Dirac notation](/know/concept/dirac-notation/) ## P -* [Pauli exclusion principle](/known/concept/pauli-exclusion-principle/) +* [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/) * [Probability current](/know/concept/probability-current/) ## T diff --git a/latex/know/concept/pauli-exclusion-principle/source.md b/latex/know/concept/pauli-exclusion-principle/source.md index 0a35869..e9e4d42 100644 --- a/latex/know/concept/pauli-exclusion-principle/source.md +++ b/latex/know/concept/pauli-exclusion-principle/source.md @@ -7,24 +7,24 @@ 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 +$\ket*{x_1}\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}$, 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 $[\hat{P}, \hat{P}^2] = 0$, $\ket{a}\!\ket{b}$ +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 $\lambda = -1$. @@ -38,14 +38,14 @@ 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 $\ket{b}\!\ket{a}$, +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 @@ -55,7 +55,7 @@ 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 @@ -64,10 +64,10 @@ 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 @@ -76,7 +76,7 @@ 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 @@ -84,7 +84,7 @@ Where $C$ is a normalization constant. As expected, this state is 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 $b$ @@ -95,14 +95,14 @@ 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: $$\begin{aligned} \ket{\Psi(a, a)}_f - = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big) + = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big) = 0 \end{aligned}$$ diff --git a/static/know/concept/pauli-exclusion-principle/index.html b/static/know/concept/pauli-exclusion-principle/index.html index 74a0954..5389c02 100644 --- a/static/know/concept/pauli-exclusion-principle/index.html +++ b/static/know/concept/pauli-exclusion-principle/index.html @@ -51,53 +51,53 @@ <hr> <h1 id="pauli-exclusion-principle">Pauli exclusion principle</h1> <p>In quantum mechanics, the <em>Pauli exclusion principle</em> is a theorem that has profound consequences for how the world works.</p> -<p>Suppose we have a composite state <span class="math inline">\(\ket*{x_1}\!\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\)</span>, where the two identical particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> each have the same two allowed states <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>. We then define the permutation operator <span class="math inline">\(\hat{P}\)</span> as follows:</p> +<p>Suppose we have a composite state <span class="math inline">\(\ket*{x_1}\ket*{x_2} = \ket*{x_1} \otimes \ket*{x_2}\)</span>, where the two identical particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> each can occupy the same two allowed states <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span>. We then define the permutation operator <span class="math inline">\(\hat{P}\)</span> as follows:</p> <p><span class="math display">\[\begin{aligned} - \hat{P} \ket{a}\!\ket{b} = \ket{b}\!\ket{a} + \hat{P} \ket{a}\ket{b} = \ket{b}\ket{a} \end{aligned}\]</span></p> <p>That is, it swaps the states of the particles. Obviously, swapping the states twice simply gives the original configuration again, so:</p> <p><span class="math display">\[\begin{aligned} - \hat{P}^2 \ket{a}\!\ket{b} = \ket{a}\!\ket{b} + \hat{P}^2 \ket{a}\ket{b} = \ket{a}\ket{b} \end{aligned}\]</span></p> -<p>Therefore, <span class="math inline">\(\ket{a}\!\ket{b}\)</span> is an eigenvector of <span class="math inline">\(\hat{P}^2\)</span> with eigenvalue <span class="math inline">\(1\)</span>. Since <span class="math inline">\([\hat{P}, \hat{P}^2] = 0\)</span>, <span class="math inline">\(\ket{a}\!\ket{b}\)</span> must also be an eigenket of <span class="math inline">\(\hat{P}\)</span> with eigenvalue <span class="math inline">\(\lambda\)</span>, satisfying <span class="math inline">\(\lambda^2 = 1\)</span>, so we know that <span class="math inline">\(\lambda = 1\)</span> or <span class="math inline">\(\lambda = -1\)</span>.</p> +<p>Therefore, <span class="math inline">\(\ket{a}\ket{b}\)</span> is an eigenvector of <span class="math inline">\(\hat{P}^2\)</span> with eigenvalue <span class="math inline">\(1\)</span>. Since <span class="math inline">\([\hat{P}, \hat{P}^2] = 0\)</span>, <span class="math inline">\(\ket{a}\ket{b}\)</span> must also be an eigenket of <span class="math inline">\(\hat{P}\)</span> with eigenvalue <span class="math inline">\(\lambda\)</span>, satisfying <span class="math inline">\(\lambda^2 = 1\)</span>, so we know that <span class="math inline">\(\lambda = 1\)</span> or <span class="math inline">\(\lambda = -1\)</span>.</p> <p>As it turns out, in nature, each class of particle has a single associated permutation eigenvalue <span class="math inline">\(\lambda\)</span>, or in other words: whether <span class="math inline">\(\lambda\)</span> is <span class="math inline">\(-1\)</span> or <span class="math inline">\(1\)</span> depends on the species of particle that <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> represent. Particles with <span class="math inline">\(\lambda = -1\)</span> are called <em>fermions</em>, and those with <span class="math inline">\(\lambda = 1\)</span> are known as <em>bosons</em>. We define <span class="math inline">\(\hat{P}_f\)</span> with <span class="math inline">\(\lambda = -1\)</span> and <span class="math inline">\(\hat{P}_b\)</span> with <span class="math inline">\(\lambda = 1\)</span>, such that:</p> <p><span class="math display">\[\begin{aligned} - \hat{P}_f \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = - \ket{a}\!\ket{b} + \hat{P}_f \ket{a}\ket{b} = \ket{b}\ket{a} = - \ket{a}\ket{b} \qquad - \hat{P}_b \ket{a}\!\ket{b} = \ket{b}\!\ket{a} = \ket{a}\!\ket{b} + \hat{P}_b \ket{a}\ket{b} = \ket{b}\ket{a} = \ket{a}\ket{b} \end{aligned}\]</span></p> -<p>Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart <span class="math inline">\(\ket{a}\!\ket{b}\)</span> and the permuted state <span class="math inline">\(\ket{b}\!\ket{a}\)</span>, regardless of the eigenvalue <span class="math inline">\(\lambda\)</span>. There is no physical difference!</p> +<p>Another fundamental fact of nature is that identical particles cannot be distinguished by any observation. Therefore it is impossible to tell apart <span class="math inline">\(\ket{a}\ket{b}\)</span> and the permuted state <span class="math inline">\(\ket{b}\ket{a}\)</span>, regardless of the eigenvalue <span class="math inline">\(\lambda\)</span>. There is no physical difference!</p> <p>But this does not mean that <span class="math inline">\(\hat{P}\)</span> is useless: despite not having any observable effect, the resulting difference between fermions and bosons is absolutely fundamental. Consider the following superposition state, where <span class="math inline">\(\alpha\)</span> and <span class="math inline">\(\beta\)</span> are unknown:</p> <p><span class="math display">\[\begin{aligned} \ket{\Psi(a, b)} - = \alpha \ket{a}\!\ket{b} + \beta \ket{b}\!\ket{a} + = \alpha \ket{a}\ket{b} + \beta \ket{b}\ket{a} \end{aligned}\]</span></p> <p>When we apply <span class="math inline">\(\hat{P}\)</span>, 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> <p><span class="math display">\[\begin{aligned} \hat{P} \ket{\Psi(a, b)} - &= \lambda \alpha \ket{a}\!\ket{b} + \lambda \beta \ket{b}\!\ket{a} + &= \lambda \alpha \ket{a}\ket{b} + \lambda \beta \ket{b}\ket{a} \\ \hat{P} \ket{\Psi(a, b)} - = \alpha \ket{b}\!\ket{a} + \beta \ket{a}\!\ket{b} + &= \alpha \ket{b}\ket{a} + \beta \ket{a}\ket{b} \end{aligned}\]</span></p> <p>This gives us the equations <span class="math inline">\(\lambda \alpha = \beta\)</span> and <span class="math inline">\(\lambda \beta = \alpha\)</span>. In fact, just from this we could have deduced that <span class="math inline">\(\lambda\)</span> can be either <span class="math inline">\(-1\)</span> or <span class="math inline">\(1\)</span>. In any case, for bosons (<span class="math inline">\(\lambda = 1\)</span>), we thus find that <span class="math inline">\(\alpha = \beta\)</span>:</p> <p><span class="math display">\[\begin{aligned} - \ket{\Psi(a, b)}_b = C \big( \ket{a}\!\ket{b} + \ket{b}\!\ket{a} \!\big) + \ket{\Psi(a, b)}_b = C \big( \ket{a}\ket{b} + \ket{b}\ket{a} \big) \end{aligned}\]</span></p> <p>Where <span class="math inline">\(C\)</span> is a normalization constant. As expected, this state is <em>symmetric</em>: switching <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> gives the same result. Meanwhile, for fermions (<span class="math inline">\(\lambda = -1\)</span>), we find that <span class="math inline">\(\alpha = -\beta\)</span>:</p> <p><span class="math display">\[\begin{aligned} - \ket{\Psi(a, b)}_f = C \big( \ket{a}\!\ket{b} - \ket{b}\!\ket{a} \!\big) + \ket{\Psi(a, b)}_f = C \big( \ket{a}\ket{b} - \ket{b}\ket{a} \big) \end{aligned}\]</span></p> <p>This state called <em>antisymmetric</em> under exchange: switching <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> causes a sign change, as we would expect for fermions.</p> <p>Now, what if the particles <span class="math inline">\(x_1\)</span> and <span class="math inline">\(x_2\)</span> are in the same state <span class="math inline">\(a\)</span>? For bosons, we just need to update the normalization constant <span class="math inline">\(C\)</span>:</p> <p><span class="math display">\[\begin{aligned} \ket{\Psi(a, a)}_b - = C \ket{a}\!\ket{a} + = C \ket{a}\ket{a} \end{aligned}\]</span></p> <p>However, for fermions, the state is unnormalizable and thus unphysical:</p> <p><span class="math display">\[\begin{aligned} \ket{\Psi(a, a)}_f - = C \big( \ket{a}\!\ket{a} - \ket{a}\!\ket{a} \!\big) + = C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big) = 0 \end{aligned}\]</span></p> <p>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.</p> |