Dirac bra-ket notation

The Dirac bra-ket notation is a concise way to represent objects in a complex vector space $\mathbb{C}^{n}$ .

A ket, denoted by $\vert \textbf{\textit{v}} \rangle$ , is a vector $\textbf{\textit{v}}$ . Since a linear operator $\hat{O}$ maps a vector to another vector, we have $\hat{O}\vert \boldsymbol{\mathit{v_{1}}}\rangle=\vert \boldsymbol{\mathit{v_{2}}} \rangle$ .

A bra, denoted by $\langle\boldsymbol{\mathit{u}}\vert$ , is often associated with a ket in the form of an inner product, denoted by $\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle$ . If a ket is expressed as a column vector, the corresponding bra is the conjugate transpose of its ket, i.e. $\langle\boldsymbol{\mathit{u}}\vert=\vert\boldsymbol{\mathit{u}}\rangle^{\dagger}$ . The inner product can therefore be written as the following matrix multiplication:

$\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle=\begin{pmatrix} \mathit{u_{1}^{*}}&\mathit{u_{2}^{*}}& \cdots & \mathit{u_{N}^{*}}\end{pmatrix}\begin{pmatrix} \mathit{v_{1}}\\\mathit{v_{2}} \\ \vdots \\\mathit{v_{N}} \end{pmatrix}=\sum_{i=1}^{N}\mathit{u_{i}^{*}}\mathit{v_{i}}$

or in the case of functions:

$\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v}}\rangle=\int_{-\infty}^{\infty}f_{u}^{*}(x)f_{v}(x)dx$

Since a linear operator acting on a ket is another ket, we can express an inner product as:

$\langle\phi_{i}\vert\phi_{k}\rangle=\langle\phi_{i}\vert\hat{O}\vert\phi_{j}\rangle=\int \phi_{i}^{*}\hat{O}\phi_{j}d\tau$

where $\hat{O}\vert\phi_{j}\rangle=\vert\phi_{k}\rangle$ .

If $i=j$ , $\langle\phi\vert\hat{O}\vert\phi\rangle$ is the expectation value (or average value) of the operator $\hat{O}$ .

As mentioned above, bras and kets can be represented by matrices. Therefore, the multiplication of a bra and a ket that involves a linear operator is associative, e.g.:

$\langle\boldsymbol{\mathit{u}}\vert(\hat{O}\vert\boldsymbol{\mathit{v}}\rangle)=(\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle\equiv\langle\boldsymbol{\mathit{u}}\vert\hat{O}\vert\boldsymbol{\mathit{v}}\rangle$

$(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)\vert\boldsymbol{\mathit{w}}\rangle=\vert\boldsymbol{\mathit{u}}\rangle(\langle\boldsymbol{\mathit{v}}\vert\boldsymbol{\mathit{w}}\rangle)$

$(\hat{O}\vert\boldsymbol{\mathit{u}}\rangle)\langle\boldsymbol{\mathit{v}}\vert=\hat{O}(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)\equiv\hat{O}\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert$

You can verify the above examples using a 2×2 matrix with complex elements to represent the operator acting on a vector in $\mathbb{C}^{2}$ . The three examples reveal that:

$\hat{O}\vert\boldsymbol{\mathit{v}}\rangle$ produces another ket.
$\langle\boldsymbol{\mathit{u}}\vert\hat{O}$ results in another bra. This is because $(\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle=\langle\boldsymbol{\mathit{u}}\vert(\hat{O}\vert\boldsymbol{\mathit{v}}\rangle)=\langle\boldsymbol{\mathit{u}}\vert\boldsymbol{\mathit{v'}}\rangle=c$ , where $c$ is a scalar; and if $(\langle\boldsymbol{\mathit{u}}\vert\hat{O})\vert\boldsymbol{\mathit{v}}\rangle=c$ , the only possible identity of $(\langle\boldsymbol{\mathit{u}}\vert\hat{O})$ is a bra.
$\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert$ , which is called an outer product, is an operator because $(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)\vert\boldsymbol{\mathit{w}}\rangle=\vert\boldsymbol{\mathit{u}}\rangle(\langle\boldsymbol{\mathit{v}}\vert\boldsymbol{\mathit{w}}\rangle)=c\vert\boldsymbol{\mathit{u}}\rangle$ , i.e. $\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert$ maps the ket $\vert\boldsymbol{\mathit{w}}\rangle$ to another ket $c\vert\boldsymbol{\mathit{u}}\rangle$ . In other words, the operator $\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert$ transforms the vector $\vert\boldsymbol{\mathit{w}}\rangle$ in the direction of the vector $\vert\boldsymbol{\mathit{u}}\rangle$ , i.e. $\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert$ projects $\vert\boldsymbol{\mathit{w}}\rangle$ onto $\vert\boldsymbol{\mathit{u}}\rangle$ .
The product of two linear operators is another linear operator: $\hat{O}\hat{O'}=\hat{O}(\vert\boldsymbol{\mathit{u}}\rangle\langle\boldsymbol{\mathit{v}}\vert)=(\hat{O}\vert\boldsymbol{\mathit{u}}\rangle)\langle\boldsymbol{\mathit{v}}\vert=\vert\boldsymbol{\mathit{u'}}\rangle\langle\boldsymbol{\mathit{v}}\vert=\hat{O''}$ .

Dirac bra-ket notation

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