# Bra-Ket and inner products

We denote a scalar product of two vectors $$a, b$$ in Hilbert space $$H$$ as $$(a,b)$$ or $$\langle a, b\rangle$$.

In Bra-Ket notation, we denote a vector $$a$$ in Hilbert space as $$|a\rangle$$. Also, we say that bras such as $$\langle b|$$ belong to the dual space $$H^*$$ . So Bras are linear transformations that map kets to a number.

Then it isn't always true that $$\langle a \mid b\rangle=(a,b)$$.

In quantum mechanics the above equality holds, is it that we define bra in such a way so that the above equality holds? Are there cases when the above equation is not true?

• I'm not sure about what you ask. $\forall T \in \mathcal{H}^{*}, !\exists \, x_{T} \, / \, T(y) = \left\langle x_{T}, y \right\rangle \, \forall y \in \mathcal{H}$. en.wikipedia.org/wiki/Riesz_representation_theorem Oct 1, 2021 at 6:35
• I'm sorry i am not able to put it clearly. I'm saying is <a| |b> = (a, b) always? How is that so. Oct 1, 2021 at 8:03
• It is so, it is just notation. Oct 1, 2021 at 8:06

It's just notation! $$\langle \psi|\chi\rangle \equiv (\langle \psi|,|\chi\rangle )$$ Dirac denoted the number resulting from the pairing of the covector $$\langle \psi|$$ with the vector $$|\chi\rangle$$ by the "bra-c-ket" symbol $$\langle \psi|\chi\rangle$$.
Furthermore, We can regard the dagger map as either determining the inner-product on $$V$$ via $$\langle |\psi\rangle ,|\chi\rangle \rangle \equiv (|\psi\rangle ^\dagger,|\chi\rangle )=(\langle \psi|,|\chi\rangle )\equiv \langle \psi|\chi\rangle$$ or being determined by it as $$|\psi\rangle ^\dagger\equiv \langle |\psi\rangle,\ \ \rangle \equiv \langle \psi|$$