Would it matter which inner product I choose in quantum mechanics?

[patric44]

TL;DR

it seems that quantum mechanics formalism is dependent on the inner product

hi guys
i was thinking about the inner product we choose in quantum mechanics to map the elements inside the hilbert space to real number which is given by :
$$\int^{∞}_{-∞}\psi^{*}\psi\;dV$$
or in some cases we might introduce a weight function dependent on the wave functions i have , it seems that if i had chosen another completely different inner product the formalism of quantum mechanics will be different ?! , is there a particular reason for that inner product that we use in QM or choosing adiffernt inner product will not change any thing and why .
thanks

 

[PeroK]

Summary:: it seems that quantum mechanics formalism is dependent on the inner product

hi guys
i was thinking about the inner product we choose in quantum mechanics to map the elements inside the hilbert space to real number which is given by :
$$\int^{∞}_{-∞}\psi^{*}\psi\;dV$$
or in some cases we might introduce a weight function dependent on the wave functions i have , it seems that if i had chosen another completely different inner product the formalism of quantum mechanics will be different ?! , is there a particular reason for that inner product that we use in QM or choosing adiffernt inner product will not change any thing and why .
thanks

The (position space) wave function and associated inner product may be derived from the general Dirac formalism:

We have a normalised state $|\alpha \rangle$ and the position eigenstates $|x \rangle$, which form a complete basis:$$\int dx \ |x \rangle \langle x| = I$$ Hence:
$$ 1 = \langle \alpha | \alpha \rangle = \int dx \ \langle \alpha|x \rangle \langle x|\alpha \rangle$$
We now define the position space wave-function:
$$\psi(x) = \langle x|\alpha \rangle$$ which leads to:
$$\int dx \ \psi(x)^*\psi(x) = 1$$ and all the other formulas involving the wave-function.

In that sense, the inner product is determined by the formalism.

 

[patric44]

thank you so much it becomes clear now

 

[dextercioby]

That the choice of inner product is limited is a consequence of the famous mathematical statement: All infinite-dimensional complex separable Hilbert space are isomorphic to $L^2 (\Omega)$, where $\Omega$ stands for an open subset of $\mathbb{R}^n$ endowed with a measure. Precisely because of this fact, QM works with separable Hilbert spaces (add to that the spectral theorem).

 

[atyy]

That the choice of inner product is limited is a consequence of the famous mathematical statement: All infinite-dimensional complex separable Hilbert space are isomorphic to $L^2 (\Omega)$, where $\Omega$ stands for an open subset of $\mathbb{R}^n$ endowed with a measure. Precisely because of this fact, QM works with separable Hilbert spaces (add to that the spectral theorem).

It's probably not that limiting, since it means even 3+1D relativistic QFT and quantum gravity (if they exist) have $L^2 (\Omega)$ as the Hilbert space.