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-rw-r--r--source/know/concept/blochs-theorem/index.md42
1 files changed, 21 insertions, 21 deletions
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index e049a71..6f445f1 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -8,14 +8,14 @@ layout: "concept"
---
In quantum mechanics, **Bloch's theorem** states that,
-given a potential $V(\vb{r})$ which is periodic on a lattice,
-i.e. $V(\vb{r}) = V(\vb{r} + \vb{a})$
-for a primitive lattice vector $\vb{a}$,
-then it follows that the solutions $\psi(\vb{r})$
+given a potential $$V(\vb{r})$$ which is periodic on a lattice,
+i.e. $$V(\vb{r}) = V(\vb{r} + \vb{a})$$
+for a primitive lattice vector $$\vb{a}$$,
+then it follows that the solutions $$\psi(\vb{r})$$
to the time-independent Schrödinger equation
take the following form,
-where the function $u(\vb{r})$ is periodic on the same lattice,
-i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$:
+where the function $$u(\vb{r})$$ is periodic on the same lattice,
+i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
$$
\begin{aligned}
@@ -30,8 +30,8 @@ the solutions are simply plane waves with a periodic modulation,
known as **Bloch functions** or **Bloch states**.
This is suprisingly easy to prove:
-if the Hamiltonian $\hat{H}$ is lattice-periodic,
-then both $\psi(\vb{r})$ and $\psi(\vb{r} + \vb{a})$
+if the Hamiltonian $$\hat{H}$$ is lattice-periodic,
+then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
are eigenstates with the same energy:
$$
@@ -42,8 +42,8 @@ $$
\end{aligned}
$$
-Now define the unitary translation operator $\hat{T}(\vb{a})$ such that
-$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$.
+Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
+$$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$$.
From the previous equation, we then know that:
$$
@@ -55,10 +55,10 @@ $$
\end{aligned}
$$
-In other words, if $\hat{H}$ is lattice-periodic,
-then it will commute with $\hat{T}(\vb{a})$,
-i.e. $[\hat{H}, \hat{T}(\vb{a})] = 0$.
-Consequently, $\hat{H}$ and $\hat{T}(\vb{a})$ must share eigenstates $\psi(\vb{r})$:
+In other words, if $$\hat{H}$$ is lattice-periodic,
+then it will commute with $$\hat{T}(\vb{a})$$,
+i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
+Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
$$
\begin{aligned}
@@ -68,10 +68,10 @@ $$
\end{aligned}
$$
-Since $\hat{T}$ is unitary,
-its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
-Therefore a translation by $\vb{a}$ causes a phase shift,
-for some vector $\vb{k}$:
+Since $$\hat{T}$$ is unitary,
+its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
+Therefore a translation by $$\vb{a}$$ causes a phase shift,
+for some vector $$\vb{k}$$:
$$
\begin{aligned}
@@ -83,7 +83,7 @@ $$
$$
Let us now define the following function,
-keeping our arbitrary choice of $\vb{k}$:
+keeping our arbitrary choice of $$\vb{k}$$:
$$
\begin{aligned}
@@ -92,7 +92,7 @@ $$
\end{aligned}
$$
-As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$:
+As it turns out, this function is guaranteed to be lattice-periodic for any $$\vb{k}$$:
$$
\begin{aligned}
@@ -108,4 +108,4 @@ $$
$$
Then Bloch's theorem follows from
-isolating the definition of $u(\vb{r})$ for $\psi(\vb{r})$.
+isolating the definition of $$u(\vb{r})$$ for $$\psi(\vb{r})$$.