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?

  • 3
    $\begingroup$ 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 $\endgroup$
    – Gilgamesh
    Commented Oct 1, 2021 at 6:35
  • $\begingroup$ I'm sorry i am not able to put it clearly. I'm saying is <a| |b> = (a, b) always? How is that so. $\endgroup$
    – Kashmiri
    Commented Oct 1, 2021 at 8:03
  • 1
    $\begingroup$ It is so, it is just notation. $\endgroup$
    – ohneVal
    Commented Oct 1, 2021 at 8:06

1 Answer 1


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 $$|\psi\rangle ^\dagger\equiv \langle |\psi\rangle,\ \ \rangle \equiv \langle \psi|$$


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