Non relativistic quantum mechanics assumes that a composite system should be described with the tensor product of the component systems. This is the tensor product postulate of quantum mechanics.

I think that the postulate originated in wave mechanics due to the following isomorphism: $\ L^2 \ (R\times R)=\ L^2 (R)\otimes \ L^2 (R) $. The lhs of the former equation is quite intuitive, the geometry of Hilbert spaces do the rest.

Now, disregarding position representation and taking for example two simple quantum systems (e.g a qubit and a qutrit) why, in principle, should we describe the composite system according to the tensor product postulate?

  • $\begingroup$ Because the results of calculations using this formalism agree with experiments and there is no known exception (within the limits of non-relativistic QM, of course)? Are you looking for a mathematical proof that nature has to behave this way? Any such "proof" would merely replace one set of "postulates" (not a good word to use in physics, by the way) that are required and sufficient to describe/generate the entire theory with an equivalent one. $\endgroup$ – CuriousOne Oct 26 '15 at 0:00
  • $\begingroup$ No. My question is how do i define a composite systems in quantum theory? Why, in principle, the tensor product rule must be used? Why it is so different from the classical case? Is there any physical principle ruling this description? $\endgroup$ – A433dd541iti Oct 26 '15 at 7:04
  • $\begingroup$ The rule "must" be used because it agrees with experiments. That's the principle behind every scientific theory. Newton's laws must be used because they agree well with experiments and observations. GR must be used because it agrees well with other observations. Maxwell must be used because it agrees with other observations and experiments, yet, again. And so on... QM is no different, whatsoever. Now, you can ask what the key observations for this are and my best answer would be that atomic and molecular spectroscopy are probably the most forceful experimental arguments, by far. $\endgroup$ – CuriousOne Oct 26 '15 at 7:28
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    $\begingroup$ Possible duplicate of Should it be obvious that independent quantum states are composed by taking the tensor product? $\endgroup$ – ACuriousMind Oct 26 '15 at 13:09

It comes down to linearity, and to be clear life isn't just a tensor product of two systems, it lives inside the tensor product.

If you think it reasonable to have a single particle state like $\psi_1$ for part of the system and to have a single particle state like $\psi_2$ for part of the system then it might seem reasonable to have a multiparticle state like $\psi_1\otimes\psi_2$ and then by linearity you would be forced to have at least the whole tensor product.

This really comes from having a basis for the tensor product made out of products of basis elements from the base spaces.

Now let's come down to reality. We merely said that it might seem reasonable to have a multi-particle state like $\psi_1\otimes\psi_2$ but if the states describe identical particles then $\psi_2\otimes\psi_1$ would be experimentally indistinguishable except in as far as this is a subsystem of a larger system. And what its also possible that nature could impose a superselection rule such as that a base state look like $\psi_1\otimes\psi_2+\psi_2\otimes\psi_1$ or $\psi_1\otimes\psi_2-\psi_2\otimes\psi_1$ and in fact the kicker is that nature does impose such superselection rules the former for bosons and the latter for fermions, and every particle is one or the other.

And then linearity gives you something provably smaller than the tensor product. So there isn't a tensor product rule. There is linearity and symmetry superselection and the former could have given you the full tensor product, but because of the latter it doesn't.


From a formal point of view, the reason for the tensor product rule is as follows:

For any quantum system the Hilbert space of states $\mathcal{H}$ is defined in terms of a complete set of degrees of freedom $S_1, S_2,…, S_n$ that can be measured simultaneously. The corresponding quantum states, characterized by sets of values $\{s_1,s_2,…,s_n\}$ and denoted $| s_1, s_2,…, s_n \rangle$, define both a basis in $\mathcal{H}$, and a complete set of mutually commuting, self-adjoint observables $\hat{S}_1$, $\hat{S}_2$, ..., $\hat{S}_n$, that have the states $| s_1, s_2,…, s_n \rangle$ as common eigenstates and the values $s_1, s_2,…, s_n$ as respective eigenvalues. The Hilbert space $\mathcal{H}$ so defined is isomorphically equivalently to the direct product $\mathcal{H}_1\otimes \mathcal{H}_2\otimes … \otimes \mathcal{H}_n$, where each $\mathcal{H}_i$ is spanned by states $| s_i\rangle$ corresponding to a single degree of freedom, and $| s_1, s_2,…, s_n \rangle$ is isomorphically equivalent to $\otimes_i| s_i\rangle$.

A similar construction applies when the total system is composed of a number of independent subsystems or different particles. In this case Hilbert spaces $\mathcal{H}_i^{(\alpha)}$ corresponding to different degrees of freedom of a subsystem $\alpha$ are naturally factors of the subsystem's Hilbert space $\mathcal{H}^{(\alpha)} = \otimes_i{\mathcal{H}_i^{(\alpha)}}$, and the total Hilbert space amounts to a direct product over subsystem spaces, $\mathcal{H} = \otimes_\alpha \mathcal{H}^{(\alpha)}$.

The case of identical particles is no different in this respect, although the set of states allowed is further restricted as required by indistinguishability and statistics. Supersymmetry does not supersede the tensor product rule, but reinforces it.

  • $\begingroup$ You are saying that IF the algebra of observables has a "tensor structure", THEN there's an induced tensor product structure on the underlying space. This is quite good. But then, why the algebra of observable should has that structure? $\endgroup$ – A433dd541iti Oct 26 '15 at 7:19
  • $\begingroup$ That's a description of the structure of quantum theory, not a reason why it "has to be this way". It doesn't have to be this way, it simply is this way. $\endgroup$ – CuriousOne Oct 26 '15 at 7:30
  • $\begingroup$ @Additi No, I did not make any mention of an algebra of observables here, and an algebra structure is not in fact necessary. No algebra composition laws are defined yet, however easy it is to define them. There is just a set of operators representing observables. The set is defined in terms of a common eigenbasis that implicitly defines the Hilbert space, and the respective eigenvalues. The nature of the common eigenbasis is what induces the tensor product isomorphism and does so at the level of the Hilbert space, not at the level of an algebra on the Hilbert space. $\endgroup$ – udrv Oct 26 '15 at 8:05
  • $\begingroup$ @CuriousOne Well, if the structure of quantum theory doesn't have to be this way please state the viable alternatives. Whatever you have any in mind, you ought to make sure that those are not isomorphic to the current one that "simply is this way". And much as you can think of it as a description, this is how it actually was set up and why it was set up this way way back when by Dirac, in his Principles of Quantum Mechanics. $\endgroup$ – udrv Oct 26 '15 at 8:15
  • $\begingroup$ Physics is an empirical science. It does not have to and it does not ask for "viable alternatives" to reality. All it asks for is sufficiently precise descriptions of empirical measurements. Since quantum mechanics as is explains all observations of the microscopic world and no deviation from the theoretical prediction has been observed, so far, we are done for now. As soon as somebody does observe deviations we have to go back to the drawing board, of course. My point is that the only place one can look for "why something is as it is" in physics is in empirical data. $\endgroup$ – CuriousOne Oct 26 '15 at 8:21

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