Recently, I have just learnt about the concept of a Hilbert space. As far as I can understand, a Hilbert space is a generalized Euclidean space.

When talking about an Euclidean space $E$, indeed there must be a mapping from $E \times E$ to the scalar field, called the "inner product". Back then, when I started to study linear algebra, I studied about the Euclidean space on $\mathbb{R}$. The inner product there has this property $$\langle x,y \rangle = \langle y,x \rangle $$ But, when expanded to a Hilbert space on $\mathbb{R}$ or $\mathbb{C}$, that property has been changed into $$\langle x,y \rangle = \overline{\langle y,x \rangle} $$ Indeed, when our scalar field is $\mathbb{R}$, nothing have changed. But what I'm concerning is: Why do we need the inner product of $x$ and $y$ to be the conjugate of the inner product of $y$ and $x$ when the scalar field is $\mathbb{C}$? Why is it neccesary to define that property like that, while we can just simply keep the commutativity like when the scalar field is $\mathbb{R}$?

Sorry if I asked something stupid. Thank you.


It's not a stupid question at all!

One reason is that using conjugate symmetry instead of symmetry allows for a norm to be defined on the space, as with real inner product spaces. We define $$\|x\| = \sqrt{\langle x, x \rangle},$$ and expect such a thing to be well-defined, real, and non-negative so as to measure distance. Having conjugate symmetry means that, $$\langle x, x \rangle = \overline{\langle x, x \rangle},$$ making the inner product of $x$ with itself real (of course, further axioms are required to make it non-negative). It better serves a geometric purpose than the slightly more obvious generalisation.

  • $\begingroup$ Your explaination cleared my mind, but there must be other reasons, isn't it? Because, to ensure that $\langle x, x \rangle$ is a real number, we can just define that the inner product map to be from $H \times H$ to $\mathbb{R}$. Please enlighten me more, sir. $\endgroup$
    – ElementX
    Apr 21 '18 at 13:11
  • 1
    $\begingroup$ That's true; you can always consider the space over $\mathbb{R}$ and put an inner product on it that way. Putting the complex inner product also makes the inner product play more nicely with eigenvalues, such as Hermitian matrices of operators under orthonormal bases having real eigenvalues (which is not always true of symmetric complex matrices). More generally, these operators form a $C^*$ algebra, which is what allows us to plug operators into analytic functions, like the complex exponential or the square root function. We need the conjugate to tap into complex analysis in order to do this. $\endgroup$ Apr 21 '18 at 16:27

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