Categories: Physics, Quantum mechanics.

# Dirac notation

Dirac notation enables us to do calculations in a general Hilbert space without needing to worry about the space’s representation. It is 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 its dual space $\mathbb{H}'$. 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 be combined in two ways: the inner product $\inprod{V}{W}$, which returns a scalar, and the outer product $\ket{V} \bra{W}$, which returns a linear operator that maps kets $\ket{V}$ to other kets $\ket{V'}$. Recall that by definition the Hilbert inner product must satisfy:

\begin{aligned} \inprod{V}{W} = \inprod{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 $\inprod{V}{W}$ is then just the familiar dot product $V \cdot W$:

$\begin{gathered} \inprod{V}{W} = \begin{bmatrix} v_1^* & \cdots & v_N^* \end{bmatrix} \cdot \begin{bmatrix} w_1 \\ \vdots \\ w_N \end{bmatrix} = v_1^* w_1 + ... + v_N^* w_N \end{gathered}$

Meanwhile, the outer product $\ket{V} \bra{W}$ creates an $N \cross N$ matrix, which can be thought of as applying an operation to any vector it multiplies:

$\begin{gathered} \ket{V} \bra{W} = \begin{bmatrix} v_1 \\ \vdots \\ v_N \end{bmatrix} \cdot \begin{bmatrix} w_1^* & \cdots & w_N^* \end{bmatrix} = \begin{bmatrix} v_1 w_1^* & \cdots & v_1 w_N^* \\ \vdots & \ddots & \vdots \\ v_N w_1^* & \cdots & v_N w_N^* \end{bmatrix} \end{gathered}$

If the kets are instead represented by continuous functions $f(x)$ of $x \in [a, b]$, then the bras are functionals $F[u(x)]$ that take an arbitrary function $u(x)$ as an argument and return a scalar:

\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:

$\begin{gathered} \inprod{f}{g} = F[g(x)] = \int_a^b f^*(x) \: g(x) \dd{x} \end{gathered}$

However, the outer product is then rather abstract: a continuous analogue of a matrix:

$\begin{gathered} \ket{f} \bra{g} = f(x) \: G[u(x)] = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi} \end{gathered}$

This maybe makes more sense if we surround it by a bra $\bra{u}$ and a ket $\ket{w}$ and rearrange:

\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) \\ &= \inprod{u}{f} \inprod{g}{w} \end{aligned}

## References

1. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.