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| author | Prefetch | 2026-09-05 21:55:33 +0200 |
|---|---|---|
| committer | Prefetch | 2026-09-05 21:55:33 +0200 |
| commit | 5cacf4ffaf3a9621ab536195f6469f98a420f054 (patch) | |
| tree | 317b734468a287c50d2403fdfafa45434f538150 /source/know/concept/blochs-theorem/index.md | |
| parent | 29b49508a751649310173e592b63415dbf563a2a (diff) | |
Diffstat (limited to 'source/know/concept/blochs-theorem/index.md')
| -rw-r--r-- | source/know/concept/blochs-theorem/index.md | 25 |
1 files changed, 15 insertions, 10 deletions
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md index d7fcf90..c6278f3 100644 --- a/source/know/concept/blochs-theorem/index.md +++ b/source/know/concept/blochs-theorem/index.md @@ -12,14 +12,14 @@ 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, +to the time-independent Schrödinger equation take the following form, where the function $$u(\vb{r})$$ is periodic on the same lattice, i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$: $$\begin{aligned} \boxed{ - \psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}} + \psi(\vb{r}) + = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}} } \end{aligned}$$ @@ -33,9 +33,11 @@ then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$ are eigenstates with the same energy: $$\begin{aligned} - \hat{H} \psi(\vb{r}) = E \psi(\vb{r}) - \qquad - \hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a}) + \hat{H} \psi(\vb{r}) + = E \psi(\vb{r}) + \qquad \qquad + \hat{H} \psi(\vb{r} + \vb{a}) + = E \psi(\vb{r} + \vb{a}) \end{aligned}$$ Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that @@ -52,18 +54,21 @@ $$\begin{aligned} In other words, if $$\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})$$: +Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ +must share eigenstates $$\psi(\vb{r})$$: $$\begin{aligned} - \hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r}) + \hat{H} \psi(\vb{r}) + = E \psi(\vb{r}) \qquad \qquad - \hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r}) + \hat{T}(\vb{a}) \psi(\vb{r}) + = \tau \psi(\vb{r}) \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}$$: +so there exists a vector $$\vb{k}$$ such that: $$\begin{aligned} \psi(\vb{r} + \vb{a}) |
