One of the quantum mechanics postulates states that a composite system can be described with the tensor product of the component systems. I've read some rationalization about this fact in some post

Should it be obvious that independent quantum states are composed by taking the tensor product?

Tensor product postulate

As I understand the answers, they (the answers) refer to non interacting parts of the system or at least they propose the plausibility of associate a single particle state with a part of the system.

I was thinking about the case of a many electron system, e.g. an atom. I can not see any kind of separability (except because a mean field approach) of the wave function. For example, in an two electron system (neglecting the spin part) it is said that we have

$$ \psi(x_1,x_2) = \sum_{i,j} d_{ij}\,\phi_i(x_1) \phi_j(x_2)$$

(Please correct me if I am wrong, but I think that the validity of the equation above is because the knowledge that the state of the one particle system is in $L^2(\Omega_i,\mu_i)$, and the two particle system must be in $L^2(\Omega_1 \otimes \Omega_2, \mu_1 \otimes \mu_2)$. Knowing that $\{\phi_i\}$ is a complete orthonormal set in the one particle system space $L^2(\Omega_i,\mu_i)$, $\{\phi_i(x_1)\phi_j(x_2) \}$ is a complete orthonormal set in $L^2(\Omega_1 \otimes \Omega_2, \mu_1 \otimes \mu_2)$).

I have not found by intuition any physical sense to the one electron functions $\{\phi_i\}$ in the two particle system (because electron-electron repulsion), as I see it, it is not even necessary to use the solutions of the one particle system but just a basis set of that space:

Question 1: Leaving aside spin contribution, why we need tensor product if not as a method to recover the vectors in the many particle space in a simple way (that also is helpful to impose symmetry restrictions)?

Additional doubt: Is the equation above valid in a more general space of functions, or it is restricted to square-integrable functions?

  • $\begingroup$ Could you try clarifying your actual question a little? I think you're asking something along the lines of 'is the Hilbert space of an interacting many particle system encoded in the tensor product of the individual particles, or just the non interacting part?' Is that correct? $\endgroup$
    – Mason
    Commented Mar 1, 2016 at 19:18
  • $\begingroup$ @Mason, Thank you very much for your time. I mean, if I am right interpreting that {ϕi} functions doe not have physical sense in the two electron system, then why we need to speak about tensor product of one particle space? Leaving aside spin, I only find it useful to easily construct the two system square-integrable functions space. I am right? $\endgroup$ Commented Mar 1, 2016 at 20:06

2 Answers 2


I believe the problem is that you think that since two one particle systems do not behave like an interacting two particle system, that the Hilbert space of a two particle system is not the tensor product of the two individual systems. This is incorrect.

No matter if the particles interact non trivially or not, The Hilbert space is given by $$\mathcal{H}_{1,2}=\mathcal{H}_1 \bigotimes \mathcal{H}_2 $$ Solving for states in this tensor product of Hilbert spaces is in general, very difficult and in general cannot be done analytically which is why we need perturbation theory to describe the Helium atom. The states in this Hilbert space are NOT in general just the two single particle solutions superimposed on top of each other.

In the cases where the two particles do not interact, the space is separable and we can just say that the solutions to the product space are the solutions of the two single particle spaces.


To answer your actual question, when you get down to it, the reason why we use the tensor product to describe a multiparticle system is because the tensor product between Hilbert spaces is defined to be the product that will give you the properties you want in a multiparticle system. The tensor product is built as a way of combining spaces, so it logically follows that we'd use that product to find the Hilbert space describing two interacting particles.

We define the tensor product between two Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ as the tensor product between the two vector spaces described by $\mathcal{H}_1$ and $\mathcal{H}_2$ and then imbue that space with an inner product by defining $$\langle\phi_1 \otimes \phi_2 | \psi_1 \otimes \psi_2\rangle=\langle\phi_1| \psi_1\rangle_1\langle\phi_2|\psi_2\rangle_2 $$

So you see, we first axiomatically stated that our states live in Hilbert space and then we invented a product that lets us combine our Hilbert spaces in matches our expectations of how multiparticle states should behave. Somewhat unsurprisingly, this ends up working very well.

This makes your question hard to answer. I suppose you could conceivably cook up a new set of axioms describing quantum mechanics and make no reference to the 'tensor product' or Hilbert space. So, you could say such concepts are not theoretical necessities, but at the end of the day your theory if it matches experiment will have to be equivalent to standard quantum mechanics at some level so you will have to have some structure in some way equivalent to how we describe multiparticle states with tensor products.

  • $\begingroup$ Thank you very much for the answer. I understand that the Hilbert space of the two particle system MUST be the Hilbert space $$\mathcal{H}_{1,2}=\mathcal{H}_1 \bigotimes \mathcal{H}_2 $$. But, the only way that I can see it is that this space is the space of square-integrable functions of two space variables (suppose only one spatial dimension), I can't find anything more to require the usage of tensor product. The only advantage (but is not a theoretical necessity) I find is a clear framework that justify the usage of this basis set ($\{\phi_i(x_1)\phi_j(x_2) \}$)... $\endgroup$ Commented Mar 2, 2016 at 1:13
  • $\begingroup$ ... which is very useful. $\endgroup$ Commented Mar 2, 2016 at 1:15
  • 1
    $\begingroup$ So the question is if there is an ulterior theoretical necessity for the usage of tensor products. $\endgroup$ Commented Mar 2, 2016 at 1:16
  • $\begingroup$ Oh, I see. I'll edit my response to answer your question $\endgroup$
    – Mason
    Commented Mar 2, 2016 at 2:02

Physics is an experimental science; momenta adds like vectors because it's been experimentally verified that they do do - if it were otherwise, then some other description would be neccessary.

A composite description is described by a tensor product, as I imagine its been verified that in fact it does; the actual history of how this was decided and conjectured would be indeed interesting.

It may simply have been guessed as it's one of the few natural operations available in combining vector spaces; it's not been elaborated in the texts I've seen - so full marks for a difficult to answer, and interesting question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.