I'm having troubles with the assertion "(normalizable) wave-functions constitutes (projective) Hilbert space".

The standard argument I find for this seems to go as following: say $\Psi(\vec{x},t)$ is a quantum state of a spinless particle. Born rule asserts that for a given $\vec{x}_0, t_0$, the number $|\Psi(\vec{x}_0, t_0)|^2$ should be interpreted as the probability of detecting the particle in position $\vec{x}_0$ at time $t_0$. So $\Psi$ induces a probability density, and we should have $\int_V{|\Psi(\vec{x},t_0)|^2}dV=1$. So we're dealing with square integrable functions (w.r.t. to the Lebesgue measure over $R^3$), that form a Hilbert space (with the inner product $\langle\Psi_1,\Psi_2\rangle=\int_V{\Psi_1^*\Psi_2}dV$).

But... all that was for a fixed $t_0$. Actually, the result of an "inner product" $\langle\Psi_1(\vec{x},t),\Psi_2(\vec{x},t)\rangle$ is a function $t\mapsto z\in C$. In what sense do the "full" quantum states ($\Psi(\vec{x},t)$, as a functions of both space and time), form a Hilbert space?

It doesn't seem to me that considering direct-sums $\oplus_{t\ge0}\mathscr{H}_t$ or tensor-products $\otimes_{t\ge0}\mathscr{H}_t$ (where $\mathscr{H}_t$ is the Hilbert space of all the possible quantum states at time $t$) leads anywhere (the temporal restriction imposed over $\Psi$ by the Schrödinger equation complicates things, and the inner-products make no physical-sense to me in this context).

The approach I find somewhat sensible, is to rely on the fact that the said function $t\mapsto z\in C$ is actually constant (I think so at least, based on the continuity equation for probability currents), so we can construct an inner product by canonically assigning this constant complex value to $\langle\Psi_1(\vec{x},t),\Psi_2(\vec{x},t)\rangle$). But I couldn't find a hint for such construction anywhere, which leads me to the conclusion that I'm missing something very basic somewhere. What am I missing? What is, explicitly, the Hilbert space to which quantum states belong?

  • $\begingroup$ Have a look at this answer of mine. You don't need a different Hilbert space for every time step, you give the time evolution in the Schrödinger picture as a map $\mathbb{R}\to\mathcal{H}$ into the space of time-independent states. If you wish, the space of time-dependent states is then $C^\infty(\mathbb{R},H)$. $\endgroup$
    – ACuriousMind
    Oct 15, 2015 at 11:08

1 Answer 1


The time evolution isn't random, it is unitary in the already given inner product.

So that time dependent function is already constant in the original inner product.

As an example, the Schrödinger equation evolution is unitary.

What is, explicitly, the Hilbert space to which quantum states belong?

A Hilbert Space is different than a projective Hilbert space. You can start with $\mathbb C^k$ or with functions from $\mathbb R^{3n}$ into the tensor product of the individual single particle spin states. Then put an inner product on the spin states and then restrict to functions that are square integrable in the sense that the inner product on the slin states is a probability density on the configuration space $\mathbb R^{3n}$. That's the Hilbert space (equivalence classes of functions whose difference has L2 norm of zero).

OK, then you make a projective Hilbert space by doing one more quotient, first, remove zero, then say that two elements are in the same equivalence class if one is a nonzero complex scalar multiple of the other. And that set of equivalence class is the projective Hilbert Space.


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