At some point in the history of quantum mechanics, it was accepted that a single particle is described by a wavefunction which is a function of the position of the particle $\mathbf{r}$, denoted: $$\psi(\mathbf{r})\,.$$ At some (possibly later) point it was also accepted that two particles are described by a wavefunction which is a function of the positions of each one of the particles, $\mathbf{r}_1$ and $\mathbf{r}_2$, denoted: $$\psi(\mathbf{r}_1,\mathbf{r}_2)\,.$$ In other words, the Hilbert space describing the two-particle system is the tensor product of the Hilbert spaces describing the system of each particle.

  • When was this idea originated and how?

I understand that there are important consequences of the tensor product structure, mainly entanglement, but as far as I know these consequences were explored after this structure was already accepted.
For example in the EPR paper (1935) the tensor product structure was already taken for granted.

To make my question clearer, here is an alternative (false) way that could have been used to describe a system of two particles. If the particle is thought of as a "wave-like" object (like was the case in these days I think), why not describe two particles with a single wavefunction $\psi(\mathbf{r})$, which is normalized:

$$ \int d\mathbf{r} |\psi(\mathbf{r})|^2 =2~, $$

i.e. "double" the quantity of the particle. Of course a lot of information is missing in this description when compared to the correct description, but how did the originators of quantum mechanics know that this information is supposed to be there in the first place?
[Edit: As Luboš says in his answer, a better alternative description for the purpose of this question is to use two wavefunctions $\psi(\mathbf{r}_1)$ and $\psi(\mathbf{r}_2)$. Here too there is information missing, and the rest of the question is the same.]

  • Was there an experimental result at that time that could not have been explained by this alternative description?
  • Was it mere physical intuition that led to the tensor product structure?

2 Answers 2


This right description of multiparticle states via tensor product spaces may have been surprising for folks like Schrödinger and from the viewpoint of "wave mechanics", but it has been incorporated from the very beginning in "matrix mechanics", Heisenberg's and pals' approach to quantum mechanics.

After all, the wave functions for a single particle in 3 dimensions $$\psi(x,y,z)$$ are already elements of the tensor product of 3 copies of spaces of wave functions for a particle in 1 dimension $$\psi(x).$$ Now, if the number of coordinates in the configuration space is enhanced from 1 ($x$) not to 3 ($x,y,z$) but to $3N$, we clearly need a wave function that depends on all the variables, e.g. $$\psi(x_1,y_1,z_1,x_2,y_2,z_2)$$ because all these coordinates are equally good coordinates on the configuration space and we already know that the wave function should be a function defined on the whole configuration space. It would be inconsistent to treat $3N$ particles differently. Abstract mechanics – even abstract classical mechanics – doesn't care whether the six coordinates belong to one or two particles, it's just our way to think about these degrees of freedom, not an essential qualitative property of the theory.

There isn't any intuition or experiment needed here. What you need to understand is that observables become operators and the commutators of the observables etc. were determined by Heisenberg from the very beginning and they immediately imply the tensor structure. If all the coordinates in $\vec r_1$ and $\vec r_2$ commute with each other, they must act on independent "directions" of a space where the wave function is defined, so the whole space must be 6-dimensional. There's no guess involved.

Again, if one tries to think in Schrödinger's picture and give various wrong materialist interpretations to the wave function, he could end up with different guesses – such as $N$ independent waves in 3 dimensions – but if one actually does the "quantization" of the previously classical system systematically, according to the universal rules, and demands that the observables become linear operators whose commutators are the right ones, the whole theory is completely determined.

There is nothing to be adjusted about the quantum description of a system of particles that previously exist in classical mechanics.

Your single wave function in 3 dimensions normalized so that its (squared) norm is 2 instead of 1 is completely equivalent to the wave function for a single particle. Just take a $\psi$ normalized to unity; $\sqrt{2}\psi$ is then normalized in your way. The theory is clearly equivalent and it describes 1 particle, not 2 particles. A more "promising" attempt would be to consider 2 wave functions of 3 variables for 2 particles. However, observables such as $\vec r_1$ and $\vec r_2$ are operators and they must be operators acting on a vector space associated with a single theory. It would make no sense to consider operators acting on different wave functions – that would be like adding apples and oranges and one could define things such as products (compositions) of these operators.

  • $\begingroup$ Indeed using two wavefunctions of 3 variables is a better alternative, I added that to the question. $\endgroup$
    – Joe
    Commented Feb 4, 2013 at 18:04
  • $\begingroup$ Yes it seems completely natural when you put it like that. $\endgroup$
    – Joe
    Commented Feb 4, 2013 at 18:14
  • $\begingroup$ What's amazing to me is this: when quantum mechanics was discovered it was first described for systems with continuous variables, that have infinite dimensional Hilbert spaces. This has the potential of complicating things, today one might want to start teaching QM from systems with finite Hilbert spaces, which still have most of the quantum features, but are simpler. However, the tensor product structure looks over-complicated in the setting of finite dimensional Hilbert spaces, so if QM would have started in this setting who knows how long it would have taken to discover entanglement... $\endgroup$
    – Joe
    Commented Feb 4, 2013 at 18:15
  • $\begingroup$ Dear Joe, I think that the tensor product structure of the composite system is simple even for systems with finite-dimensional Hilbert spaces. It's still true that the operators from the subsystems have to commute with each other... Quantum mechanics started with continuous variables such as $x$ because continuous variables are needed to isolate the classical limit. When the states are discrete, like for $J_z$ in a discrete set, the quantum effects are always of order one. Heisenberg et al. wanted just a deformation of classical physics, so choices with $\Delta x \Delta p\gg\hbar$ had to exist $\endgroup$ Commented Feb 4, 2013 at 21:32
  • $\begingroup$ But quantum mechanics where the quantum effects are of order 100%, like the case of the spin for spin-1/2 particles, was understood within months as well. So the delay wasn't as significant as you suggest. ... Also, Heisenberg and his school - which was also closer to Bohr etc. - was thinking in terms of matrix mechanics and "matrices" are sometimes defined with the extra conditions of being of finite size. Of course that he also meant matrices (operators) that are allowed to be infinite in size. But the basic logic is really the same in both cases, just the value of $\dim H$ differs $\endgroup$ Commented Feb 4, 2013 at 21:34

I think that Joe wondering why wave functions of two particles are not "interfere" or "mixing" like the case with the usual electromagnetic/mechanical waves that will lead to the need for just a one wave function $\psi(\vec{r})$ as he mentioned.

First of all I will assume that you understand clearly that wave functions are not describing real waves, their name is misleading, secondly please note that by definition, wave function should have all needed information (in other words, all quantum numbers) that will guarantee "individuality" of the particles, so even if we have two particles, to describe them (we suppose that we are not dialing with statistical mechanics or Bose-Einstein condensate) has to be done through two wave functions, if we will try to do that by only one wave function, this will lead to lose of their "individuality", which is not acceptable (for example, from Paule exclusion principle prospective).

For that, we write that the wave function of the system, can be written as a superposition of all possible eigenstates of both particles, in sense if first particle can take the states $1,2,3...$ while the second one the states $a,b,c...$ then our system can be in one of the states $1a,1b,...2a,2b...$ (where $2b$ stands for that first particle is at the state $1$ and in the same time second particle is at the state $b$), so the intuition here is very obvious I think, and those 1a,2b... states are simply called tensor product in mathematics, so there is no need even for experiments to prove that (because we have really no other way to describe the state of the system if we require that our description will preserve individuality of the particles), with one condition which is we assume as a given thing that all our operators are linear, otherwise putting $1a,1b...$ as a possible states of the system will be not enough, because then a structures like (formally writing) $1b^2, 2^3c...$ should be included

  • $\begingroup$ Sorry but I think you completely missed the point of the question. This is not a question about quantum mechanics, but about the history of quantum mechanics. $\endgroup$
    – Joe
    Commented Feb 4, 2013 at 18:21
  • $\begingroup$ History? you asked about the intuition that leading to this and the experiments (your last two questions), and I explained that there is no need for such experiments and explained the intuition, even so I ignored the first historical question because it was answered above by Lubos. $\endgroup$
    – TMS
    Commented Feb 4, 2013 at 19:50
  • $\begingroup$ I didn't ask about intuition and experiments that lead to this, I asked whether it was intuition or experiments that historically led the originators of quantum mechanics. $\endgroup$
    – Joe
    Commented Feb 4, 2013 at 19:59
  • $\begingroup$ I see no difference, if you were familiar with the intuition behind it, you wouldn't wonder about the experiments, anyway, most important thing that you get the answer you wanted :) $\endgroup$
    – TMS
    Commented Feb 4, 2013 at 20:04

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