# In Dirac notation, why do we have $\langle cf|g\rangle = c^*\langle f\vert g\rangle$?

A Hilbert Space has this property

$$\langle cf,g\rangle=c\langle f,g\rangle$$

where $$f$$ and $$g$$ are the vectors in the Hilbert Space and $$c$$ is a complex number.

In Dirac Notation,

$$\langle cf|g\rangle = c^*\langle f\vert g\rangle$$

I am confused about why the Dirac notation takes complex conjugate on $$c$$. I know $$\langle f \vert$$ in the Dirac notation is in the Dual Hilbert space. Is this the reason? Do I need to define a Hilbert Space as

$$\langle cf,g\rangle=c\langle f,g\rangle$$

I feel it is equivalent if I define as

$$\langle cf,g\rangle = c^*\langle f,g\rangle$$

Am I right? Thanks a lot!

• There are actually two conventions for which argument of the inner product should be linear vs conjugate linear. The one you give at the start is often called the "mathematics convention" and the other one the "physics convention". However, many mathematics books also use the "physics convention", as it works better with the GNS construction. Feb 9, 2020 at 9:36

Inner products on Hilbert spaces are linear in their first argument and conjugate linear in their second argument. So Hilbert spaces also have the property $$\langle f, \, cg\rangle = c^\ast \langle f, \, g \rangle.$$
As you said, a bra represents the dual space to a ket. So to move to an inner product in Dirac notation simply note that for $$c_1, \, c_2 \in \mathbb{C}$$ and $$f, \, g \in \mathcal{H}$$ $$\langle c_1 f , \, c_2 g \rangle = \langle c_2 g \vert c_1 f \rangle = c_1 \, c_2^\ast \langle f, \, g \rangle.$$
• Shouldn't be $\langle c_2 g \vert c_1 f \rangle$ complex conjugated as $\langle f \vert g \rangle = \langle g \vert f \rangle^*$? Feb 9, 2020 at 7:20
• @MartinVesely It should, but in the final line above only the middle expression is in Dirac notation, the first and third expressions are in mathematics notation for Hilbert space inner products. The conventions are opposite as to whether the conjugate-linear argument is on the right (mathematics, comma-delimited) or left (Dirac, vertical bar-delimited). In other words $$\langle f, \, g \rangle = \langle g, \, f \rangle^\ast = \langle g \vert f \rangle = \langle f \vert g \rangle^\ast.$$ Feb 9, 2020 at 12:31