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Diffstat (limited to 'source/know/concept/blochs-theorem')
| -rw-r--r-- | source/know/concept/blochs-theorem/index.md | 42 |
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})$$. |