Bra-Ket Matrix Notation

February 11, 2008

We’ll need a notation for matrices and vectors. The standard notation for a column vector is \mathbf{a}=\underline{a}=\left\{a_1,a_2,\ldots,a_n\right\}^T
whilst a row vector is written using transpose notation \underline{a}^T .

The standard notation for a matrix and a matrix composed of vectors is
\mathbf{A}=\underline{\underline{A}}=\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{m1} & \cdots & a_{mn} \end{pmatrix} = \begin{bmatrix} \underline{a}_1^T \\ \vdots \\ \underline{a}_m^T \end{bmatrix} = \begin{bmatrix} \underline{a}_1 \cdots \underline{a}_n \end{bmatrix}.

Unfortunately, this notation will be a bit awkward for my purposes. I will instead abuse Dirac’s bra-ket notation to write vectors and matrices in summation form as follows:

\mathbf{a}=\sum_{1 \le i \le n}\left\rvert \hat{\imath} \right\rangle a_i

\mathbf{a}^T=\sum_{1 \le j \le m} a_j \left\langle \hat{\jmath} \right\lvert

\mathbf{A}=\sum_{\substack{1 \le i \le n \\ 1 \le j \le m}} \left\rvert \hat{\imath} \right\rangle a_{ij} \left\langle \hat{\jmath} \right\lvert.

The rule for projecting unit vectors is:

\left\langle \hat{\imath} \right\lvert \left\rvert \hat{\jmath} \right\rangle =\delta_{ij} =\left\{ \begin{array}{ll} 0 & i \ne j \\ 1 & i = j \end{array} \right.

Bra \left\langle \right\lvert and ket \left\rvert \right\rangle do not commute with each other; though they do commute with numbers.

To illustrate, here is the multiplication of two matrices in this notation

\mathbf{A}=\sum_{\substack{1 \le i \le n \\ 1 \le j \le m}} \left\rvert \hat{\imath} \right\rangle a_{ij} \left\langle \hat{\jmath} \right\lvert \text{;} \mathbf{B}=\sum_{\substack{1 \le u \le m \\ 1 \le v \le p}} \left\rvert \hat{u} \right\rangle b_{uv} \left\langle \hat{v} \right\lvert

\mathbf{A}\times\mathbf{B} = \sum_{i,j} \left\rvert \hat{\imath} \right\rangle a_{ij} \left\langle \hat{\jmath} \right\lvert \times \sum_{u,v} \left\rvert \hat{u} \right\rangle b_{uv} \left\langle \hat{v} \right\lvert

= \sum_{i,j,u,v} \left\rvert \hat{\imath} \right\rangle a_{ij} \left\langle \hat{\jmath} \right\lvert \left\rvert \hat{u} \right\rangle b_{uv} \left\langle \hat{v} \right\lvert

= \sum_{i,j,u,v} \left\rvert \hat{\imath} \right\rangle a_{ij} \delta_{ju} b_{uv} \left\langle \hat{v} \right\lvert

= \sum_{i,j,v} \left\rvert \hat{\imath} \right\rangle a_{ij} b_{jv} \left\langle \hat{v} \right\lvert

= \sum_{i,v} \left\rvert \hat{\imath} \right\rangle \left\langle \hat{v} \right\lvert \sum_{j} a_{ij} b_{jv}

Here is the usual inner product between two vectors

\mathbf{a} \circ \mathbf{b} = \mathbf{a}^T \mathbf{b} = \sum_{i} a_i \left\langle \hat{\imath} \right\lvert \sum_{j} \left\rvert \hat{\jmath} \right\rangle b_j = \sum_{i,j} a_m \delta_{i,j} b_j = \sum_{i} a_i b_i

When there is no possibility for ambiguity I will use implicit summations. For example

\mathbf{A} = \left\rvert \hat{\imath} \right\rangle \left\langle \hat{\jmath} \right\lvert a_{ij} .