# Normalizing ket vectors

So I've run into a bit of a notation problem on my coursework.

I have a vector, $$\vert A\rangle$$ expressed in the orthogonal basis $$\vert 1\rangle$$ and $$\vert 2\rangle$$ as \begin{align} \vert A\rangle &= \vert 1\rangle + i \vert 2\rangle \end{align}

I need to convert it into bra form to normalize it. Is the answer $$\langle A\vert = \vert 1\rangle - i\vert 2\rangle$$ or $$\langle A \vert = \langle 1\vert - i\langle 2\vert$$?

Basically, I get that to convert between bra and ket form, you take the complex conjugate and change a column vector to a row one, but I need to know if you convert the basis vectors into bra form as well.

• Dual vectors are linear functionals that also form a vector space, so every bra has to be expressed in terms of other bras (linear functionals). And having in mind that the relation between kets and bras are conjugate-linear, if you have $|\phi\rangle = \alpha |1\rangle + \beta |2\rangle$, its corresponding bra is: $\langle \phi | = \alpha^* \langle 1|+\beta^* \langle 2|.$ Feb 7 '16 at 11:31

## 1 Answer

The comment from @Ellie is what you need to know. This shows the steps. $$|A\rangle = |1\rangle + i|2\rangle$$ $$|A\rangle = 1|1\rangle + i|2\rangle$$ $$\langle A| = |A\rangle^*$$ $$|A\rangle^* = \alpha^* \langle 1 | + \beta^*\langle 2 |$$ $$|A\rangle^* = 1^* \langle 1 | + i^* \langle 2 |$$ $$|A\rangle^* = 1 \langle 1 | + -i \langle 2 |$$ $$|A\rangle^* = 1 \langle 1 | -i \langle 2 |$$ $$|A\rangle^* = \langle 1 | -i \langle 2 |$$ $$|A\rangle^* = \langle A | = \langle 1 | -i \langle 2 |$$

Apologies if it's a bit verbose, but it should be clear.