# Requirements for the inner product in Hilbert space

I was reading the following text about the mathematical foundations of quantum mechanics when I stumbled upon the following conditions that the inner product must satisfy:

I woud like to understand why the above conditions must be satisfied, and resolve some seeming conflicts.

1. I always associated the inner product in quantum mechanics with an overlap integral, this notion would intuitively imply that it should be symmetric in exchange of $\psi$ and $\omega$ because the overlap of the two functions should not be a function of how I label them. Why do I require the complex part of the inner product to be anti-symmetric with respect to swapping of the arguments?

2. I understand that choosing it to be linear in its second argument is just a matter of convention and it makes the Dirac notation work nicer, as operators can then act on the elements to the right.

3. Again, this is not clear to me why I should require the inner product to be antilinear in its first argument. Is it just because it is a necessary requirement to satisfy the condition 4 below?

4. This makes sense.

I would be very glad if someone could explain the reason for the first condition to me, as it seems mystifying to me why it is necessary to impose it.

1. The inner product between $\psi(x)$ and $\phi(x)$ is $$\int \phi^\ast(x)\psi(x)\mathrm{d}x,$$ otherwise the inner product of $\psi$ with itself would not be $\int \lvert \psi\rvert^2$ but $\int \psi^2$ which would not necessarily be real. This is an "overlap integral", just not the naive $\int \phi\psi$ you were probably thinking of.
2. If you do not require anti-linearity, then for $I(\psi,\psi)$ real, $I(\alpha\psi,\alpha\psi)$ for some complex number $\psi$ would be $\alpha^2 I(\psi,\psi)$ instead of $\lvert \alpha \rvert^2 I(\psi,\psi)$, which again would not be necessarily real, but for an inner product we want the inner product of a vector with itself to be real.
• @Akerai I'm not sure what you mean - the first condition is also ensuring that the inner product of all vectors with themselves are real, since $I(\psi,\psi)^\ast = I(\psi,\psi)$ implies $I(\psi,\psi)$ is real. – ACuriousMind Jan 14 '18 at 14:32