1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
|
% Bloch's theorem
# Bloch's theorem
In quantum mechanics, **Bloch's theorem** states that,
given a potential $V(\vec{r})$ which is periodic on a lattice,
i.e. $V(\vec{r}) = V(\vec{r} + \vec{a})$
for a primitive lattice vector $\vec{a}$,
then it follows that the solutions $\psi(\vec{r})$
to the time-independent Schrödinger equation
take the following form,
where the function $u(\vec{r})$ is periodic on the same lattice,
i.e. $u(\vec{r}) = u(\vec{r} + \vec{a})$:
$$
\begin{aligned}
\boxed{
\psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}}
}
\end{aligned}
$$
In other words, in a periodic potential,
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 $\psi(\vec{r})$ and $\psi(\vec{r} + \vec{a})$
are eigenstates with the same energy:
$$
\begin{aligned}
\hat{H} \psi(\vec{r}) = E \psi(\vec{r})
\qquad
\hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a})
\end{aligned}
$$
Now define the unitary translation operator $\hat{T}(\vec{a})$ such that
$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$.
From the previous equation, we then know that:
$$
\begin{aligned}
\hat{H} \hat{T}(\vec{a}) \psi(\vec{r})
= E \hat{T}(\vec{a}) \psi(\vec{r})
= \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big)
= \hat{T}(\vec{a}) \hat{H} \psi(\vec{r})
\end{aligned}
$$
In other words, if $\hat{H}$ is lattice-periodic,
then it will commute with $\hat{T}(\vec{a})$,
i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$.
Consequently, $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
$$
\begin{aligned}
\hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
\qquad
\hat{T}(\vec{a}) \:\psi(\vec{r}) = \tau \:\psi(\vec{r})
\end{aligned}
$$
Since $\hat{T}$ is unitary,
its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
Therefore a translation by $\vec{a}$ causes a phase shift,
for some vector $\vec{k}$:
$$
\begin{aligned}
\psi(\vec{r} + \vec{a})
= \hat{T}(\vec{a}) \:\psi(\vec{r})
= e^{i \theta} \:\psi(\vec{r})
= e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
\end{aligned}
$$
Let us now define the following function,
keeping our arbitrary choice of $\vec{k}$:
$$
\begin{aligned}
u(\vec{r})
= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
\end{aligned}
$$
As it turns out, this function is guaranteed to be lattice-periodic for any $\vec{k}$:
$$
\begin{aligned}
u(\vec{r} + \vec{a})
&= e^{- i \vec{k} \cdot (\vec{r} + \vec{a})} \:\psi(\vec{r} + \vec{a})
\\
&= e^{- i \vec{k} \cdot \vec{r}} e^{- i \vec{k} \cdot \vec{a}} e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
\\
&= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
\\
&= u(\vec{r})
\end{aligned}
$$
Then Bloch's theorem follows from
isolating the definition of $u(\vec{r})$ for $\psi(\vec{r})$.
|
