Query in inner product axioms in QM

Question

In inner product spaces in $\mathbb R$ we have an axiom stating that: $$ \langle x, x\rangle \geq 0\ \ \text{and} \ \ \langle x, x\rangle = 0 \iff x = 0$$

In Griffiths' textbook for Quantum Mechanics, they have stated that,

$$ \langle\alpha|\alpha\rangle \geq 0\ \ \text{and} \ \ \langle\alpha|\alpha\rangle = 0 \iff x = 0$$

My question is that, since $\alpha \in \mathbb C$ then why is true that $\langle\alpha|\alpha\rangle \geq 0$. the product of two complex number can be negative, right?

Answer 1

For complex numbers, the inner product is $\sum_i\bar {a_i} b_i$ rather than $\sum_i a_i b_i$, where $\bar {a_i}$ denotes the complex conjugate.

Note that if $a=x+iy$ then $\bar a \cdot a=x^2+y^2\ge0$

Answer 2

$|\alpha\rangle$ is a state, not a number. The rules of inner product state that $$ \langle \beta |\alpha \rangle \in {\mathbb C} , \qquad \langle \beta |\alpha \rangle^* = \langle \alpha |\beta \rangle. $$ It follows that $$ \langle \alpha |\alpha \rangle^* = \langle \alpha | \alpha \rangle \quad \implies \quad \langle \alpha | \alpha \rangle \in {\mathbb R}. $$ We can therefore assign a sign to this quantity. Another rule for QM then requires that the inner product be positive semi-definite, $$ \langle \alpha |\alpha \rangle \geq 0 , \qquad \langle \alpha |\alpha \rangle = 0 \quad \Longleftrightarrow \quad |\alpha\rangle = 0 . $$