91Ó°ÊÓ

Consider a three-dimensional vector space spanned by an Orthonormality basis $|1âŸ\copyright ,|2âŸ\copyright ,|3âŸ\copyright$.

Kets localid="1658312431665" $|\mathrm{Î\pm }âŸ\copyright$and $|\mathrm{β}âŸ\copyright$are given by

$$\begin{gathered} |\mathrm{Î\pm }âŸ\copyright =i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright , \\ |\mathrm{β}âŸ\copyright =i|1âŸ\copyright +2|3âŸ\copyright . \end{gathered}$$

Here we consider the following two kets:

$$\begin{gathered} |\mathrm{Î\pm }âŸ\copyright =i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright \\ |\mathrm{β}âŸ\copyright =i|1âŸ\copyright +2|3âŸ\copyright \end{gathered}$$

In order to construct bras $⟨\mathrm{Î\pm }|$and $⟨\mathrm{β}|$we need to take a complex conjugate of each factor and substitute kets for bras.

Following the above recipe we have for $⟨\mathrm{Î\pm }|$and $⟨\mathrm{β}|$:

$$\begin{gathered} ⟨\mathrm{Î\pm }|=-i⟨1|-2⟨2|+i⟨3| \\ ⟨\mathrm{β}|=-i⟨1|+2⟨3|. \end{gathered}$$

The inner product$⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}âŸ\copyright$ is now given by

$⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}âŸ\copyright =(-i⟨1|-2⟨2|+i⟨3|\left)\right(i|1âŸ\copyright +2|3âŸ\copyright )$.

Term by term multiplication yields

$⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}âŸ\copyright =-i^2⟨1âˆ\pounds 1âŸ\copyright -2i⟨1âˆ\pounds 3âŸ\copyright -2i⟨2âˆ\pounds 1âŸ\copyright -4⟨2âˆ\pounds 3âŸ\copyright +i^2⟨3âˆ\pounds 1âŸ\copyright +2i⟨3âˆ\pounds 3âŸ\copyright$

This is an Orthonormal basis which means that

$⟨iâˆ\pounds jâŸ\copyright =\left\{\begin{array}{l}1,i=j\\ 0,i≠j\end{array}\right.$

Applying this to the above equation we find

$$\begin{gathered} ⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}âŸ\copyright =-i^2â‹\dots 1-2iâ‹\dots 0-2iâ‹\dots 0-4â‹\dots 0+i^2â‹\dots 0+2iâ‹\dots 1 \\ ⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}âŸ\copyright =1+2i \end{gathered}$$

Similarly, we have

$$\begin{gathered} ⟨\mathrm{β}âˆ\pounds \mathrm{Î\pm }âŸ\copyright =(-i⟨1|+2⟨3\left|\right)\left(i\right|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright ) \\ ⟨\mathrm{β}âˆ\pounds \mathrm{Î\pm }âŸ\copyright =-i^2â‹\dots 1+2iâ‹\dots 0+i^2â‹\dots 0+2iâ‹\dots 0-4â‹\dots 0-2iâ‹\dots 1 \\ ⟨\mathrm{β}âˆ\pounds \mathrm{Î\pm }âŸ\copyright =1-2i \end{gathered}$$

We now also see that

$⟨\mathrm{β}âˆ\pounds \mathrm{Î\pm }âŸ\copyright =⟨\mathrm{Î\pm }âˆ\pounds \mathrm{β}{âŸ\copyright }^{*}$.

We now consider the operator$\stackrel{Áåœ}{A}=|\mathrm{Î\pm }âŸ\copyright ⟨\mathrm{β}|$.

Its matrix elements are given by

$A_{ij}=⟨iâˆ\pounds \mathrm{Î\pm }âŸ\copyright ⟨\mathrm{β}âˆ\pounds jâŸ\copyright$
Let us now compute the first row of the matrix A:

$$\begin{gathered} A_{11}=⟨1\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 1âŸ\copyright \\ {A}_{11}=(i+0+0)(-i+0) \\ {A}_{11}=1 \\ {A}_{12}=⟨1\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 2âŸ\copyright \\ {A}_{12}=(i+0+0)(0+0) \\ {A}_{12}=0 \end{gathered}$$

$$\begin{gathered} A_{13}=⟨1\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 3âŸ\copyright \\ {A}_{13}=(i+0+0)(0+2) \\ {A}_{13}=2i \end{gathered}$$

Similarly, other elements are

$$\begin{gathered} A_{21}=⟨2\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 1âŸ\copyright \\ {A}_{21}=2i \end{gathered}$$

$$\begin{gathered} A_{22}=⟨2\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 2âŸ\copyright \\ {A}_{22}=0 \end{gathered}$$

$$\begin{gathered} A_{23}=⟨2\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 3âŸ\copyright \\ {A}_{23}=-4 \end{gathered}$$

$$\begin{gathered} A_{31}=⟨3\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 1âŸ\copyright \\ {A}_{31}=-1 \end{gathered}$$

$$\begin{gathered} A_{32}=⟨3\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 2âŸ\copyright \\ {A}_{32}=0 \end{gathered}$$

$$\begin{gathered} A_{33}=⟨3\left|\right(i|1âŸ\copyright -2|2âŸ\copyright -i|3âŸ\copyright )(-i⟨1|+2⟨3\left|\right)âˆ\pounds 3âŸ\copyright \\ {A}_{33}=-2i. \end{gathered}$$

We can now write out the full matrix explicitly:

$A=\left(\begin{array}{ccc}1& 0& 2i\\ 2i& 0& -4\\ -1& 0& -2i\end{array}\right)$

This is not a hermitan matrix (note, for example, that $A_{21}≠A_{21}^{*}$).