Query in inner product axioms in QM

Question

In inner product spaces in $mathbb R$ we have an axiom stating that:
$$ langle x, xrangle geq 0  text{and}     langle x, xrangle = 0 iff x = 0$$
In Griffiths' textbook for Quantum Mechanics, they have stated that,
$$ langlealpha|alpharangle geq 0  text{and}     langlealpha|alpharangle = 0 iff x = 0$$
My question is that, since $alpha in mathbb C$ then why is true that $langlealpha|alpharangle geq 0$. the product of two complex number can be negative, right?

Gyro Gearloose · Accepted Answer

For complex numbers, the inner product is $sum_ibar {a_i} b_i$ rather than  $sum_i a_i b_i$, where $bar {a_i}$ denotes the complex conjugate.
Note that if $a=x+iy$ then $bar a cdot a=x^2+y^2ge0$

Prahar · Answer

$|alpharangle$ 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,
$$
langle alpha |alpha rangle geq 0 , qquad langle alpha |alpha rangle = 0 quad Longleftrightarrow quad |alpharangle = 0 . 
$$

Query in inner product axioms in QM

2 Answers

Add your own answers!

Ask a Question