Hermitian inner product in C3

The square of the norm is the sum of the squared moduli of the components:

$$\norm{u}^2=|1|^2+|i|^2+|1|^2=3,\qquad \norm{v}^2=|i|^2+|1|^2+|1-i|^2=1+1+2=4 ,$$

and hence $\norm u=\sqrt3$ and $\norm v=2$. To compute the inner product, the components of the first vector must be conjugated, so $\bra u=(1,\,-i,\,1)$:

$$\braket uv=1\cdot i+(-i)\cdot1+1\cdot(1-i)=i-i+1-i=1-i .$$

Similarly, with $\bra v=(-i,\,1,\,1+i)$,

$$\braket vu=(-i)\cdot1+1\cdot i+(1+i)\cdot1=-i+i+1+i=1+i .$$

We find that $\braket vu=\braket uv^*$: this is the Hermitian symmetry of the inner product.

We have $|\braket uv|=|1-i|=\sqrt2\simeq1{,}41$, and $\norm u\,\norm v=2\sqrt3\simeq3{,}46$. The inequality is satisfied, and it is strict: the two vectors are not collinear. Recall that equality holds only if one of the vectors is a scalar multiple of the other.

The vector $\lambda\ket u=i\ket u$ has components $(i,\,-1,\,i)$, and its bra is $(-i,\,-1,\,-i)$. Thus

$$\braket{iu}{v}=(-i)\cdot i+(-1)\cdot1+(-i)(1-i)=1-1-i-1=-1-i=-i\,\braket uv .$$

By contrast, $\braket{u}{iv}=i\braket uv=i(1-i)=1+i$. The inner product is linear in the second vector but antilinear in the first: $\braket{\lambda u}{v}=\lambda^*\braket uv$. This is the convention used by physicists.

We seek $\ket w=(w_1,w_2,w_3)$ such that $\braket uw=0$ and $\braket vw=0$, that is, using the bras computed above,

$$w_1-iw_2+w_3=0,\qquad -iw_1+w_2+(1+i)w_3=0 .$$

The first equation gives $w_1=iw_2-w_3$. Substituting this into the second gives

$$-i(iw_2-w_3)+w_2+(1+i)w_3=w_2+iw_3+w_2+(1+i)w_3=2w_2+(1+2i)w_3=0,$$

and hence $w_2=-\frac{1+2i}{2}w_3$. Choosing $w_3=2$ gives $w_2=-1-2i$ and $w_1=i(-1-2i)-2=-i$:

$$\ket w=\begin{pmatrix}-i\\ -1-2i\\ 2\end{pmatrix}.$$

Let us verify this: $\braket uw=-i+(-i)(-1-2i)+2=-i+i-2+2=0$, and $\braket vw=(-i)(-i)+(-1-2i)+(1+i)\cdot2=-1-1-2i+2+2i=0$. The vector $\ket w$ is determined up to a complex factor; its norm is $\sqrt{1+5+4}=\sqrt{10}$.

We find $\sum_iu_i^2=1+i^2+1=1$, whereas $\norm u^2=3$. Without complex conjugation, the resulting quantity is not always real or positive: for $(1,i)\in\C^2$, it is $1+i^2=0$, even though the vector is non-zero. It therefore cannot define a norm; the components of the first vector are conjugated precisely to ensure that $\braket uu=\sum_i|u_i|^2\geq0$.