# Why is the wave-function defined in abstract configuration space, as opposed to real space $\mathbb{R}^3$?

I've just began studying QM so I apologise in advance if this question is silly or badly put.

My initial understanding of the wave-function $$\psi(x,t)$$ was that it exists in real 3-dimensional space $$\mathbb{R}^3$$, with $$x$$ being a position in said space. I've come to understand that this understanding is wrong and that $$\psi$$ is actually an abstract function defined on an abstract "configuration-space" whose points are possible states of the system under consideration, and not actually physical locations.

This distinction is important as, for example, I initially thought that in order to address a system of two particles we'd have to compose the wave functions of each particle in a fashion similar to classical waves, i.e: $$\psi(x,t)=\psi_1(x,t)+\psi_2(x,t)$$ But in actuality the proper treatment would be to consider a higher-dimensional configuration space. That is, the composite function would have to be of the form: $$\psi(x_1,x_2,t)$$

My question is how come physicists decided to describe quantum mechanics with the (seemingly more complicated) idea of abstract configuration-space functions instead of wave-functions defined in physical space?

I understand that the configuration-space approach is more sensible considering Born's interpretation of the wave-function, but my lecturer cited the fact that $$\psi$$ exists in configuration-space as one of the reasons that Born's interpretation was preferred over alternative approaches (for example Schrödinger's interpretation).

Were there any experiments that motivated physicists to describe $$\psi$$ as a configuration-space function? For instance, was there an experiment that discredited the former approach I presented to two-particle quantum systems?

• Physicsts had been using the notion of configuration space in classical mechanics for over a century before the development of quantum mechanics. Are you interested how to understand configuration space generally, or are you interesting in why the wavefunction for a compound quantum system should have the form that it does? – By Symmetry Mar 19 at 17:55
• For one thing, you don’t get any entanglement with that scheme, directly contradicting hundreds or perhaps thousands of experiments. – knzhou Mar 19 at 22:32
• The first equation looks like a direct sum, while the latter could be a more general mapping. – Emil Mar 19 at 22:47
• @BySymmetry I am interested in the latter. You could also say I'm trying to understand why a theory where the wave function exists in physical space (whatever that wave function may mean physically) was rejected so early in the development of QM. – Bar Alon Mar 20 at 11:25
• @BarAlon Different interpretations of QM describe what's happening in different ways but make the same predictions -- and you simply do not have enough information to get the right predictions if you don't use configuration space. For example, even in Bohmian mechanics, the guiding wave is a wave on configuration space. (There are many oversimplified pop-sci presentations that pretend it is a wave in physical space, like that annoying bouncing droplet thing, but it's simply not true -- the Bohmian guiding wave has the exact same value and behavior as what we usually call the wavefunction.) – knzhou Mar 20 at 11:45

Representation in $$\mathbb{R}^3$$ is often convenient, but the quantum state $$|\psi\rangle$$ itself lives out there in Hilbert space $$\mathcal{H}$$, an abstract, complete vector space that has a well-defined inner product. Establishing this framework allows us to analyze the state in whatever the most convenient basis is for the problem at hand. Specifically, for any set of complete eigenstates of your Hamiltonian, we can represent the wavefunction in that space by sticking in completeness. Whether you have a continuum or discretization of these states dictates the form of your completeness. For example, completeness of discrete energy eigenstates $$|n\rangle$$ looks like

$$\sum_{n}|n\rangle\langle n| = 1,$$

whereas completeness of position eigenstates is represented as

$$\int d^3\vec{r}\:'\:|\vec{r}\:'\rangle\langle \vec{r}\:'| = 1.$$

Representing your quantum state in the space of a specific complete set of eigenfunctions is as simple as hitting it with completeness. Let's use the above completeness to represent $$|\psi\rangle$$ in this space:

$$|\psi(t)\rangle = \int d^3\vec{r}\:'\:|\vec{r}\:'\rangle\langle \vec{r}\:'|\psi(t)\rangle = \int d^3\vec{r}\:'\:|\vec{r}\:'\rangle \psi(\vec{r}\:',t).$$

A key property from the completeness of your representation space is orthogonality. In position space, this looks like $$\langle \vec{r}|\vec{r}\:'\rangle = \delta^3(\vec{r}-\vec{r}')$$. So let's see what this means for our representation in $$\mathbb{R}^3$$:

$$\langle \vec{r}|\psi(t)\rangle = \int d^3\vec{r}\:'\:\langle\vec{r}\:|\vec{r}\:'\rangle \psi(\vec{r}\:',t) = \int d^3\vec{r}\:'\:\delta^3(\vec{r}-\vec{r}') \psi(\vec{r}\:',t) = \psi(\vec{r},t).$$

While this may not seem like anything beyond what you already know, it is how we should be thinking of the representation that you've chosen, and shows everything is nice and consistent. Following this same procedure for any set of complete eigenstates allows you to represent your function in that space. For example, let's see how we would go from a state represented in $$\mathbb{R}$$ to its representation in momentum space:

$$\psi(p,t) = \langle p|\psi(t)\rangle = \int dx \langle p|x\rangle\langle x|\psi(t)\rangle = \int dx \langle p|x\rangle\psi(x,t).$$

The tricky thing here is the $$\langle p|x\rangle$$, which is the position eigenstate represented in momentum space. However, it's also equivalent to $$(\langle x|p\rangle)^*$$, the momentum eigenstates represented in position space, which are plane waves $$e^{ikx} = e^{ipx/\hbar}$$. Completeness ends up giving us a prefactor $$1/\sqrt{2\pi \hbar}$$ on this, so lets see what this tells us about the above:

$$\psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int \psi(x,t)e^{-ipx/\hbar}dx \\ \psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int \psi(p,t)e^{ipx/\hbar}dp.$$

Where the second one follows from starting with $$\langle x|\psi(t)\rangle$$ in the above steps. This shows us that representations in $$x$$ and $$p$$ are Fourier transforms of one another, and all of this can be immediately extended to $$\mathbb{R}^3$$. Now there is nothing exclusive or special to $$x$$ or $$p$$ about any of this, all we need is completeness of eigenstates to represent a state in that space.

Now, representing multiple particles with a single wavefunction requires us to extend our Hilbert space into its higher dimensional equivalent: Fock space. Refraining from going into way too much detail, I'll just motivate this space with the intro definition on Wiki:

Fock space is the direct sum of tensor products of copies of a single-particle Hilbert space $$\mathcal{H}$$ $$\mathcal{F} = \mathbb{C}\oplus\mathcal{H}\oplus(S_\nu(\mathcal{H}\otimes\mathcal{H}))\oplus...$$ where $$S_\nu$$ is the operator which symmetrizes or antisymmetrizes a tensor, depending on whether the Hilbert space describes particles obeying bosonic ($$\nu=+$$) or fermionic ($$\nu=-$$) statistics.

Your multiple particle state $$\psi$$ must characterize the positions of each separately, thus explicitly depending on both $$\vec{r}_1$$ and $$\vec{r}_2$$. Perhaps this can be most intuitively understood by asking the question "What would the probability be that we measure particle 1 in a volume $$d^3\vec{r}_1$$ and particle 2 in volume $$d^3\vec{r}_2$$?" The answer would be $$|\psi(\vec{r}_1,\vec{r}_2,t)|^2d^3\vec{r}_1d^3\vec{r}_2$$, and normalization would require

$$\int|\psi(\vec{r}_1,\vec{r}_2,t)|^2d^3\vec{r}_1d^3\vec{r}_2 = 1.$$

If you have indistinguishable particles, you need to construct a wave function that is non-committal as to which particle is in which state. With the symmetry requirements, you're two-particle position wavefunction would need to satisfy $$\psi(\vec{r}_1,\vec{r}_2,t) = \pm\psi(\vec{r}_2,\vec{r}_1,t)$$. If you know the two occupied states, call them $$\psi_\alpha$$ and $$\psi_\beta$$, then this is satisfied with

$$\psi(\vec{r}_1,\vec{r}_2,t) = \psi_\alpha(\vec{r}_1,t)\psi_\beta(\vec{r}_2,t)\pm\psi_\alpha(\vec{r}_2,t)\psi_\beta(\vec{r}_1,t),$$

where the $$+$$ is for bosons, and $$-$$ for fermions. A nice way of seeing this association of signs is by looking at the case $$\psi_\alpha = \psi_\beta$$. This gives you zero wavefunction for the fermions, no state at all! This is Pauli's exclusion principle.

• Thank you for the detailed response. I am still slightly confused. Was it ever historically the case that the QM wavefunction was thought of as a physical wave propagating through space (similar to how EM radiation is described and how EM interference is explained)? Or was the wavefunction always thought of as a state-vector living in a Hilbert space? – Bar Alon Apr 8 at 11:25