Tensor product of two Hilbert spaces

Question

Does tensor product of two Hilbert spaces explains interactions between the systems also? Or is it just the way to represent two spaces in a combined fancy way? Let me make it more clear by an example. Consider two electrons in an atom. Can I explain whole system where electron are interacting by just tensor product of single electron solutions?

Answer 1

It is not fancy. Have you thought about the difference between the tensor product and the dirrect sum ?

For an object $x $ living in a space $X$ and $y$ an other object living in on the space $Y$, we can consider the product space $X \times Y$ for the configuration space of the pair of objects $(x,y)$. Now in the quatum world, we must replace $X$ by something like $L^2(X)$, $Y$ by $L^2(Y)$ and $X \times Y$ by $L^2(X \times Y)$. But this last space appears to be isomorphic to $L^2(X) \otimes L^2(Y)$ (can you show this for say $X = Y =\mathbb R $ ? ).

Hope to convice you that the tensor prodcut is not that "fancy". Tensor product represent in the quantum world the consideration of multiparticle states. These particles may or may not interact between themselves. But this will be encoded in the Hamiltonian.

Answer 2

Tensor product is a formalism that allows us to express states of a system in terms of its subsystems. Just like how Dirac’s bra-ket notation is a formalism to express quantum states. They themselves have no inherent physical implications. That is what the Hamiltonian enforces.

The job of a notation system is to be as clear as possible to implement the calculations as painlessly as possible. For example decimal place value notation is far superior to Roman numerals. All you need to do realise that is to try and add two numbers!

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Edit

Consider a Hamiltonian of the form: $$H=H_1+H_2+H_i$$ where $|n_1\rangle$ diagonalises $H_1$ and $|n_2\rangle$ diagonalises $H_2$. If there were no interaction term $H_i$, then we could diagonalise the two Hamiltonians independently and the states $|n_1\rangle\otimes|n_2\rangle$ would diagonalise the total $H$. But because of the interaction, the general eigenstates are linear combinations of the product states like: $$|\psi\rangle=\sum_{ij}c_{ij}|n_i\rangle\otimes|n_j\rangle$$

We can do so because the product states span the Hilbert space of the total system, ie, they form a basis.

Answer 3

Tensor product does not account for interactions arising due to composing two systems. Consider two particle example in mind. If Hamiltonian has extra term apart from two independent terms then that may be a difficult system to solve and in general can't be explained by two independent systems. However, quantum mechanics is unique as tensor product gives rise to some superposition states which are not there otherwise. Entangled states are the examples. Entanglement is present between two independent systems explained by tensor product. So in a sense there is some kind of pure quantum correlation which arises when we consider two independent(no classical interactions) systems together.