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% Dirac notation
# Dirac notation
*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:
$$\begin{aligned}
\bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
&= U\big[f(x) \: G[w(x)]\big]
= U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
\\
&= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
\\
&= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
\\
&= \braket{u}{f} \braket{g}{w}
\end{aligned}$$
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