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 it will commute with the unitary translation operator \(\hat{T}(\vec{a})\), i.e. \(\comm{\hat{H}}{\hat{T}(\vec{a})} = 0\). Therefore \(\hat{H}\) and \(\hat{T}(\vec{a})\) must share eigenstates \(\psi(\vec{r})\):
+This is suprisingly easy to prove: if the Hamiltonian \(\hat{H}\) is lattice-periodic, then it will commute with the unitary translation operator \(\hat{T}(\vec{a})\), i.e. \([\hat{H}, \hat{T}(\vec{a})] = 0\). Therefore \(\hat{H}\) and \(\hat{T}(\vec{a})\) must share eigenstates \(\psi(\vec{r})\):
\[
\begin{aligned}
\hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
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+ Dirac notation is a notation to do calculations in a Hilbert space without needing to worry about the space’s representation. It is basically the lingua franca of quantum mechanics. In Dirac notation there are kets \(\ket{V}\) from the Hilbert space \(\mathbb{H}\) and bras \(\bra{V}\) from a dual \(\mathbb{H}'\) of the former. Crucially, the bras and kets are from different Hilbert spaces and therefore cannot be added, but every bra has a corresponding ket and vice versa. Bras and kets can only be combined in two ways: the inner product \(\braket{V | W}\), which returns a scalar, and the outer product \(\ket{V} \bra{W}\), which returns a mapping \(\hat{L}\) from kets \(\ket{V}\) to other kets \(\ket{V'}\), i.e. a linear operator. Recall that the Hilbert inner product must satisfy: \[\begin{aligned}
+ \braket{V | W} = \braket{W | V}^*
+\end{aligned}\] So far, nothing has been said about the actual representation of bras or kets. If we represent kets as \(N\)-dimensional columns vectors, the corresponding bras are given by the kets’ adjoints, i.e. their transpose conjugates: \[\begin{aligned}
+ \ket{V} =
+ \begin{bmatrix}
+ v_1 \\ \vdots \\ v_N
+ \end{bmatrix}
+ \quad \implies \quad
+ \bra{V} =
+ \begin{bmatrix}
+ v_1^* & \cdots & v_N^*
+ \end{bmatrix}
+\end{aligned}\] The inner product \(\braket{V | W}\) is then just the familiar dot product \(V \cdot W\): \[\begin{gathered}
+ \braket{V | W}
+ =
+ \begin{bmatrix}
+ v_1^* & \cdots & v_N^*
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix}
+ w_1 \\ \vdots \\ w_N
+ \end{bmatrix}
+ = v_1^* w_1 + ... + v_N^* w_N
+\end{gathered}\] Meanwhile, the outer product \(\ket{V} \bra{W}\) creates an \(N \cross N\) matrix: \[\begin{gathered}
+ \ket{V} \bra{W}
+ =
+ \begin{bmatrix}
+ v_1 \\ \vdots \\ v_N
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix}
+ w_1^* & \cdots & w_N^*
+ \end{bmatrix}
+ =
+ \begin{bmatrix}
+ v_1 w_1^* & \cdots & v_1 w_N^* \\
+ \vdots & \ddots & \vdots \\
+ v_N w_1^* & \cdots & v_N w_N^*
+ \end{bmatrix}
+\end{gathered}\] If the kets are instead represented by functions \(f(x)\) of \(x \in [a, b]\), then the bras represent functionals \(F[u(x)]\) which take an unknown function \(u(x)\) as an argument and turn it into a scalar using integration: \[\begin{aligned}
+ \ket{f} = f(x)
+ \quad \implies \quad
+ \bra{f}
+ = F[u(x)]
+ = \int_a^b f^*(x) \: u(x) \dd{x}
+\end{aligned}\] Consequently, the inner product is simply the following familiar integral: \[\begin{gathered}
+ \braket{f | g}
+ = F[g(x)]
+ = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{gathered}\] However, the outer product becomes something rather abstract: \[\begin{gathered}
+ \ket{f} \bra{g}
+ = f(x) \: G[u(x)]
+ = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi}
+\end{gathered}\] This result makes more sense if we surround it by a bra and a ket: \[\begin{aligned}
+ \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
+ &= U\big[f(x) \: G[w(x)]\big]
+ = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
+ \\
+ &= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
+ \\
+ &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
+ \\
+ &= \braket{u | f} \braket{g | w}
+\end{aligned}\]
+Dirac notation
+
+© "Prefetch". Licensed under CC BY-SA 4.0.
+
+
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