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authorPrefetch2026-09-05 21:55:33 +0200
committerPrefetch2026-09-05 21:55:33 +0200
commit5cacf4ffaf3a9621ab536195f6469f98a420f054 (patch)
tree317b734468a287c50d2403fdfafa45434f538150 /source/know/concept/blochs-theorem/index.md
parent29b49508a751649310173e592b63415dbf563a2a (diff)
Improve knowledge baseHEADmaster
Diffstat (limited to 'source/know/concept/blochs-theorem/index.md')
-rw-r--r--source/know/concept/blochs-theorem/index.md25
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})