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 V\ket{V} from the Hilbert space H\mathbb{H} and bras V\bra{V} from its dual space H\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 VW\inprod{V}{W}, which returns a scalar, and the outer product VW\ket{V} \bra{W}, which returns a linear operator that maps kets V\ket{V} to other kets V\ket{V'}. Recall that by definition the Hilbert inner product must satisfy:

VW=WV\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 NN-dimensional columns vectors, the corresponding bras are given by the kets’ adjoints, i.e. their transpose conjugates:

V=[v1vN]    V=[v1vN]\begin{aligned} \ket{V} = \begin{bmatrix} v_1 \\ \vdots \\ v_N \end{bmatrix} \qquad \implies \qquad \bra{V} = \begin{bmatrix} v_1^* & \cdots & v_N^* \end{bmatrix} \end{aligned}

The inner product VW\inprod{V}{W} is then just the familiar dot product VWV \cdot W:

VW=[v1vN][w1wN]=v1w1+...+vNwN\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 VW\ket{V} \bra{W} creates an N×NN \cross N matrix, which can be thought of as applying an operation to any vector it multiplies:

VW=[v1vN][w1wN]=[v1w1v1wNvNw1vNwN]\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)f(x) of x[a,b]x \in [a, b], then the bras are functionals F[u(x)]F[u(x)] that take an arbitrary function u(x)u(x) as an argument and return a scalar:

f=f(x)    f=F[u(x)]=abf(x)u(x)dx\begin{aligned} \ket{f} = f(x) \qquad \implies \qquad \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:

fg=F[g(x)]=abf(x)g(x)dx\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:

fg=f(x)G[u(x)]=f(x)abg(ξ)u(ξ)dξ\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 u\bra{u} and a ket w\ket{w} and rearrange:

u ⁣( ⁣fg ⁣) ⁣w=U[f(x)G[w(x)]]=U[f(x)abg(ξ)w(ξ)dξ]=abu(x)f(x)(abg(ξ)w(ξ)dξ)dx=(abu(x)f(x)dx)(abg(ξ)w(ξ)dξ)=ufgw\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.