Practice Dirac Notation

Determine size and transpose

$${\left[\begin{array}{ccc}1& 0& 0\end{array}\right]}_{1\times 3}^T=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]3\times 1$$

b)

$${\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]}_{2\times 2}^T={\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]}_{2\times 2}$$

c)

$${\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]}_{3\times 1}^T={\left[\begin{array}{ccc}0& 0& 1\end{array}\right]}_{1\times 3}$$

Express $|\psi ⟩=\left[\begin{array}{c}1\\ i\end{array}\right]$ using Dirac notation

$$=1\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]+i\left[\begin{array}{c}0\\ 1\end{array}\right]=|0⟩+i|1⟩$$

6.3) Compute $v^{†}$ for $v=\left[\begin{array}{c}1\\ i\\ 2+3i\end{array}\right]$

$$v^{†}=(v^{\ast }{)}^T=⟨v|$$ $$=\left[\begin{array}{ccc}(1{)}^{\ast }& (i{)}^{\ast }& (2+3i{)}^{\ast }\end{array}\right]$$ $$=\left[\begin{array}{ccc}1& -i& 2-3i\end{array}\right]$$

Reminder: conjugate is written as $(a+bi{)}^{\ast }=a-bi$, we flip the sign. A valid state must have $\Vert \psi \Vert =1$.