There are nice and fairly detailed discussions of your question at the first two links below (the third link deals with the mathematical part of the story).

http://physics.stackexchange.com/questions/53039/when-and-how-did-the-idea-of-the-tensor-product-originate-in-the-history-quantum

https://www.quora.com/What-is-a-tensor-product-in-quantum-mechanics

https://en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces

There are nice and fairly detailed discussions of your question at the first two links below (the third link deals with the mathematical part of the story).

http://physics.stackexchange.com/questions/53039/when-and-how-did-the-idea-of-the-tensor-product-originate-in-the-history-quantum

https://www.quora.com/What-is-a-tensor-product-in-quantum-mechanics

https://en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces

just think you do not know about spin (but it exists). Logically that the space you work on is a sub-variety of the full space (to you not known). So when you do discover a new quantity, you are simply accessing the dimensions till then mute.

The more difficult question is if you have a scalar product for each dimension, to prove that the global scalar prod. is the product of the individual dimensions' ones.

Suppose H_1 is a Hilbert space, and H_2 another. Suppose each space has a basis: H_1 has phi_i and H_2 psi_j, where i runs from 1 to m and j from 1 to n. Then you can construct a new vector space using the pairs phi_i cross psi_j as a basis. That means that

psi_1 cross phi_1 + psi_2 cross phi_2

cannot in any way be expressed in terms of other elements. This is to  be contrasted, say, with the direct sum, in which such vectors are summed component by component.

Since the phi_i cross psi_j are all quite independent, we have that the dimension of the tensor product is m times n.

What happens, now, in fucntion spaces? For example, the functions on the [0, 2 pi] interval can ll be expressed as a sum of functions of the type exp(i m x). This is one Hilbert space. Now consider another one. To distinguish the functions, let me assume they depend on the variable y. They are then generated by exp(i m y). The basis for the tensor product  is then simply

exp(i m x) exp(i n y)

which generates arbitrary functions psi(x, y) of two variables. So if the two Hilbert spaces are function spaces, the tensor product is simply the one that describes functions depending on both arguments.

Now a quantum mechanical state for one system assigns an amplitude to each configuration x of the system. A compound state of two suystems assigns an amplitude to the combined values of the configurations of bothsystems. It is therefore part of the tensor product of the spqces to which the states of either system belong.

... good question; and F. Leyvraz I think I can say good answer, but in total honesty it is not totally clear: you saying that a many-particle state is described by a tensor product because that is the one that describes functions depending on all arguments but when you say this do you mean that the tensor product on n objects is the most general way in which a function depending on n variables can be written?

And a more general treatment, I hope I am doing well when I re-interpret Oliver Passon's question in this way: suppose we know nothing about quantum physics, would it still be possible on grounds of generality to infer that we must employ tensor product to describe such systems?

For example, even if we knew nothing about general relativity, by implementing a principle that requires quantities to give rise to equations that are independent on the system of coordinates in the most general case, then we are automatically given the concept of tensor... so is there an analogous argument to require the same in quantum physics?

Why are many variable probability distribution described by a Tensor product? Many quantum formulae correspond to formulae of probability theory with probabilities replaced by probability amplitudes.

... right, but still: why?

One need a sum of vector spaces to describe something which are superpositions of "this or that or ..."

And a product of vector spaces to describe something which are               "(this1 and this2 and ...) or (that1 and that2 and ...) or ..."

Count how many independent quantum numbers you need to specify a state completely, that give you the number of subspace products. Count of many possible values each quantum number can have, that give you the dimension of each subspace.

Then, for a more interesting life, there may be some restrictions (like those coming from absolute identity of identical particles).

Yes, cf. http://ejde.math.txstate.edu/conf-proc/04/w1/wightman.pdf

Dear Colleagues,

many thanks to all of these remarks and answers! The remark of Luiz Botelho concerning the probability preservation appears of particular importance to me.

also: first need to know what a wavefunction is. "Silly question", right ?... well, maybe not:

http://www.nature.com/nphys/journal/v8/n6/full/nphys2309.html

http://www.nature.com/news/quantum-physics-what-is-really-real-1.17585

and from there on proceed on higher dimensionality of such.

(NB - to me adding orthogonal vectors in an embedding, or doing the tensor product of 2 sub-spaces seems the same thing)

First of all, you need no tensor products in quantum physics: note, there is - up to isomorphism - only 1 Hilbert space. A nxn matrix may be regarded as well as tensor and as n² dimensional vectorspace. the writing (aik) is only book keeping. The same happens with n-fold tensor products, as long n is finite. In physics, they don't consider infinite tensor products.

In Schrödinger picture you have then 3n coordinates, and in Heisenberg?

Considerably clearer, also for the references

If system 1 has n possible states and system 2 has m possible states then the combination has n*m possible states. This determines the dimensionality of the combined space. If probability of state 1 being i is pi and state 2 being j is pj, and both probs are independent then prob of state i AND state j is pi*pj. Tensor products give new vectors that have these properties. 

Tell me the equation for this tensor. Is it the Schrödinger's? I mean, where you put your tensor into?

Schrodinger's eqn is for a wave-function which is a simple scalar product. Of course you must use a 2-particle Hamiltonian.

Dear brother in love, Mike York! And you can show me the sane math of it? The two particles Hamiltonian is H = Alembert + q1 q2/(x1 - x2), where Alembert is the differential operator, the q1, q2 are the electric charges of particles, x1, x2 are coordinates of particles, which are on the x-axes of space. The state-tensor of the system is the direct product of the state w1 of first particle and state w2 of the second. So: W=w1 w2. Hereby the w1 is found from H w1= E1 w1, so the w1 contains the x2 as the parameter. The w2 is found from H w2 = E2 w2, so the w2 contains the x1 as the parameter. Let us apply the Hamiltonian: H W = (Alembert + q1 q2/(x1 - x2)) w1 w2. It is not the H W = w2 H w1 + w1 H w2 = (E1+E2) w1 w2 = (E1+E2) W. Latter we can not produce. So the N. Bohr theory is wrong? Perhaps one shall study the David Bohm's theory?

Dmitri, two-particle Schrodinger is in most text books. You can probably find it from Google or Wikipedia.

Loving brother in love, Mike. Thank you. But in the link below the wave function is not the tensor product of the two state vectors. But in this post-question of RG it must be the tensor product.

http://www.physicspages.com/2013/02/20/schrodinger-equation-for-2-particles-separation-of-variables/

Dmitri> the wave function is not the tensor product

This is painful to read.

The space of functions f(r1, r2) is the tensor product of the space of function g(r1) and the space of functions h(r2), just like the space of functions g(r) is the tensor product of the space of function gx(x), the space of function gy(y), and the space of function gz(z).

If you find this unfamiliar, consider expanding all functions in a discrete basis, f.i. harmonic oscillator eigenfunctions.

Dear brother in love, Olaussen. Thank you for attempt to put the scientific sense into by brain. I am adjusted to the tensor components. See: f^{i k} = g^i h^k. The f^{i k} is not the tensor f, but it is its components in some basis. Is it correct? P.S. discrete basis of functions do not describe any possible function. For example, any final sum of the trigonometric functions (2 sin(x) + 56 sin(2x)+...) can hardly be regarded as the exp(x).

Kare, the wave-function is a scalar. I am surprised you don't know this. Wave-functions are coefficients of base vectors. The tensor products of base vectors can be employed as base vectors for two-particle states. The product of the wave-functions is the (scalar) coefficient in the new space.

With all respect, the Mike could be more closer to the Truth, than the Kare. However (in my Tartu University class of vector and tensor analyses) the coefficients before the base vectors are not scalars. They are just functions. The scalar function do mean, what it does not change in value while any of the coordinate (the base vectors) transformation. 

Dmitri, if you apply a unitary transformation to a state vector, then the coefficients of the base vectors do change.

One Hilbert space for particle 1. 

One Hilbert space for particle 2.

.......

One Hilbert space for particle N. 

That is what they call tensor product. 

Fock space consists of N Hilbert spaces. 

It is tricky if the particle are indistinguishable. Entangled all over. 

Dear loved Mike, yes, they do. Therefore, they are not the scalars. Scalar function differs from just a function.

Dmitri, the coefficients of the base vectors are individually scalars just like any other vector components. A unitary transformation will create a different state vector and therefore a different set of coefficients of base vectors which translates to different probabilities of each eigenstate. 

It is still painful to read some of the things written here; only the brother Dmitri's remarks make it a bearable experience.

In general, abstract quantum mechanical states are vectors in their Hilbert spaces; this concept is defined by how its expansion coefficients transform under changes of basis. [Actually, since quantum states remain unchanged by multiplication with arbitrary phases (some will say with arbitrary non-zero finite complex numbers), they should actually be considered as rays. But this distracting fact is best saved to the back burner of your mind; otherwise you will live an unhappy life in permanent sorrow over the loss of your beloved superposition principle.]

One may also consider the explicit representation by ordinary Schrödinger wave functions, defined on (multi-particle) configuration space. How do they transform under change of coordinates? Not as scalar functions! Not as densities either! To mention two familiar objects which are not weighed down with the burden of carrying a bunch of indices. In faithful respect of the Born interpretation, the wave functions must transform like a half densities. If they are to represent the states of a spin-less particle: i.e., a scalar particle.

There are additional annoying details to struggle with when the particle(s) have spin, due to the more complicated (some will also say non-classical) structure of configuration space for such objects. Further, one way to consider identical particles is by identifying points of multi-particle configuration space (the cartesian product of individual-particle spaces) which are related by permutations. This gives a more interesting multi-connected topology to configuration space (if configurations with two or more identical particles at the same point are forbidden), leading to shivers of joy propagating the spine of true practitioners of quantum mechanics.

As to the question of whether identical particles are automatically entangled all over, this has been considerably discussed over the years. I proudly belong to the camp of sensible people which declares that particle statistics by itself do not imply entanglement. This is a natural standpoint at the viewpoint of quantum field theory, consistent with calculations of entanglement entropies.

Kåre, you here statistics when it is not said. There is no staistics in a Hydrogen molecule. Just two electrons. Undistinguished.  

Igor@ Yes, undistinguished, and sometimes unentangled.

so far I understand physics is unentangled with entanglement.

Why the tensor product method makes sense can be demonstrated even on an almost classical level. In order to 'quantize' the e.m. field in a mathematical rigorous way one considers a 'field packet' or 'wave packet' which is specified by some appropriate functions in a Hilbert space. Then, with starting from this Hilbert space the Fock space is constructed in the usual way, i.e. tensor products of H are used to describe N individual and independent 'wave packets', and the machinery of creation/destruction operators acting on Fock space can be defined (cf. "Relativist. Qu.theorie" by N. Straumann, Springer, 2005, ch.1, Quantentheorie der Strahlung). The fact that this mathematical machinery, which is traditionally called 'second quantization', just works and finally describes a "many-wavepacket-system" seems to be the most convincing 'physical' reason why the tensor product is useful. Proceeding in the same way, but using anticommutators instead of commutators, the Schrodinger field (considered as a 'classical' matter field) leads to the formalism of non.rel. many-particle-QM (without spin), and the Dirac field in conjunction with the e.m. field leads to QED.