Postulate 3 Preliminaries

Before proceeding with the third postulate of quantum theory, which will involve using matrices to transform and evolve quantum states, it's best to get familiar with matrices in bra-ket notation first.

For example, consider the matrix:

$$A = \begin{bmatrix} 1 - i & 4 \\ 2 & -3 \end{bmatrix}.$$

Then, in index notation, we would say:

$$\begin{align*} A_{11} \text{ or } a_{11} &= 1 - i, \\ A_{12} \text{ or } a_{12} &= 4, \\ A_{21} \text{ or } a_{21} &= 2, \\ A_{22} \text{ or } a_{22} &= -3. \end{align*}$$

In quantum mechanics, however, it''s common to start indexing from 0, that is, the first row and first column are labeled as the 0th row and 0th column, respectively. For the example above, we actually have:

$$\begin{align*} A_{00} &= a_{00} = 1 - i, \\ A_{01} &= a_{01} = 4, \\  A_{10} &= a_{10} = 2, \\ A_{11} &= a_{11} = -3. \end{align*}$$

We also sometimes express matrix entries using bras and kets in the following way. The $ij$-th entry of the matrix $A$ is $\langle i | A | j \rangle$. For our example above:

$$\begin{align*} \langle 0 | A | 0 \rangle &= 1 - i, \\ \langle 0 | A | 1 \rangle &= 4, \\ \langle 1 | A | 0 \rangle &= 2, \\  \langle 1 | A | 1 \rangle &= -3. \end{align*}$$

Notice this is not simply another way of writing matrix entries; it is also an explicit way to calculate them.

We can see this for the matrix above by explicitly performing the calculations $\langle 0 | A | 0 \rangle$, $\langle 0 | A | 1 \rangle$, $\langle 1 | A | 0 \rangle$, $\langle 1 | A | 1 \rangle$, that is:

\begin{align*} \langle 0 | A | 0 \rangle &= \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 - i & 4 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1-i \\ 2 \end{bmatrix} = 1\cdot(1 -i) + 0\cdot 2 = 1-i, \\  \langle 0 | A | 1 \rangle &= \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 1 - i & 4 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} 4 \\ -3 \end{bmatrix} = 1\cdot 4 + 0\cdot(-3) = 4, \\ ...