Understanding the Role of Tensor Product in Quantum Systems - Goldbeetle

[Goldbeetle]

Dear all,
why is it that the tensor product is used to describe two quantum system described by Hilbert spaces H1 and H2? What were the example systems or situations that were generalized and that led to this postulate?
Thanks.
Goldbeetle

 

[meopemuk]

In my opinion, these two papers provide a satisfactory answer

T. Matolcsi, "Tensor product of Hilbert lattices and free orthodistributive product of orthomodular lattices", Acta Sci. Math. (Szeged), 37 (1975), 263.


D. Aerts, I. Daubechies, "Physical justification for using the tensor product to describe two quantum systems as one joint system", Helv. Phys. Acta, 51 (1978) 661.

Eugene.

 

[Fredrik]

I read the Aerts and Daubechies article (ok, not the whole thing, but enough to get the general idea) after the last time you posted these references. I think it's a very well-written article, but I'm not so sure that it can give you a much deeper understanding of the answer to the OP's question than the simple argument:

Consider two systems that aren't interacting with each other. If system 1 is in state |\psi\rangle when we measure A, the probability of result a is

P(a)=|\langle a|\psi\rangle|^2

If system 2 is in state |\phi\rangle when we measure B, the probability of result b is

P(b)=|\langle b|\phi\rangle|^2

According to the standard rules for probabilities, the probability of getting both of these results is

P(a,b)=P(a)P(b)=|\langle a|\psi\rangle|^2|\langle b|\phi\rangle|^2=|\langle a|\otimes\langle b|\ |\psi\rangle\otimes|\phi\rangle|^2

This means that if we use the tensor product space to represent the states of the combined system, the Born rule will hold for that space too. Can you really look at this and think that we didn't choose to use tensor product to make sure that the probabilities assigned by the Born rule satisfy P(a,b)=P(a)P(b) when the systems aren't interacting?

I had to elaborate on what the point of my calculation was the next day:

I just proved that if we use the tensor product and the Born rule, we get P(a,b)=P(a)P(b) for non-interacting systems. (It would be a disaster to get P(a,b)≠P(a)P(b). If QM works for nuclei and electrons separately, it wouldn't work for atoms. But of course it wouldn't work for nuclei either...) The point is that quantum mechanics for individual systems, which by definition includes the Born rule, more or less forces us to us to use the tensor product to represent the states of the composite system.

At first it seemed to me that Aerts and Daubechies made no reference to probabilities at all, but that's not entirely true. Their starting point is the quantum logic approach to QM, in which we use a mathematical structure (which is isomorphic to the lattice of closed subspaces of a complex separable Hilbert space) to represent the set of statements of the form "if I measure the observable A, the result will be in the set E with probability 1". (An arbitrary member of this set can have a different A and a different E, but the probability is always 1). So what they're doing has something to do with probabilities as well. That's why it seems to me that even if I understood the details of their argument, it wouldn't give me a much deeper understanding of why we use the tensor product. (Let me know if you disagree).

It's still a very nice paper though, and I intend to return to it if I ever get around to study quantum logic seriously. Unfortunately that probably won't be until next year.

Goldbeetle, you might find this thread useful too. It explains the definition/construction of the tensor product pretty well.

 

[meopemuk]

I just proved that if we use the tensor product and the Born rule, we get P(a,b)=P(a)P(b) for non-interacting systems.

Fredrik,

You have proved that tensor product is sufficient to get P(a,b)=P(a)P(b). However, you haven't proved that tensor product is necessary to get P(a,b)=P(a)P(b). In other words, after your proof there is still a possibility that some other construction (not the usual tensor product) can be in agreement with all probability postulates.

Matolcsi and Aerts&Daubechies basically filled this gap. They formulated some postulates about probabilities in compound systems (like yours P(a,b)=P(a)P(b)) and then they proved that there are only two inequivalent ways to satisfy these postulates in quantum mechanics. One is the usual tensor product of component's Hilbert spaces H = H1 x H2. The other one is the tensor product of H1 with the dual Hilbert space of H2: H = H1 x H2*. I am not quite sure about the physical meaning of the second possibility. However, it remains there as a mathematical fact.

Eugene.

 

[strangerep]

[Matolcsi and Aerts&Daubechies] proved that there are only two inequivalent ways to satisfy these postulates in quantum mechanics. One is the usual tensor product of component's Hilbert spaces H = H1 x H2. The other one is the tensor product of H1 with the dual Hilbert space of H2: H = H1 x H2*. I am not quite sure about the physical meaning of the second possibility. However, it remains there as a mathematical fact.

I wonder why they make this distinction. I was under the impression that ordinary
Hilbert spaces are self-dual, i.e., H2 is isomorphic to H2*, so why bother?

 

[meopemuk]

I wonder why they make this distinction. I was under the impression that ordinary
Hilbert spaces are self-dual, i.e., H2 is isomorphic to H2*, so why bother?

Apparently, the two ways of building the compound Hilbert space are not isomorphic. Both papers claim the same result independently. I don't know what is the significance of this.

Eugene.

 

[Goldbeetle]

Thanks too all! Actually, I was looking for something else, namely some examples of compound systems (described for instance by a wave function) that can be interpreted as tensor products of hilbert spaces and that motivate the introduction of the generalization of the tensor product in conceptual framework of QM. The books I've consulted so far define more or less the tensor product and then use it, but lack any motivation.

 

[Fredrik]

The Aerts & Daubechies article has an example. See page 1 and the beginning of page 2.

 

[Goldbeetle]

Fredrik,
thanks. O, it's clear now for square-integrable wave function.

 

[strangerep]

Apparently, the two ways of building the compound Hilbert space are not isomorphic.

I guess it's because even though there's a one-to-one mapping between
H and H*, the mapping is antilinear and hence not strictly isomorphic since
it doesn't preserve linearity. Cf. Ballentine's footnote on p27, and his eqn (1.8).

I.e., if \langle F| \leftrightarrow |F\rangle , then \bar{c}\langle F| \leftrightarrow c|F\rangle , etc.

 

[meopemuk]

I guess it's because even though there's a one-to-one mapping between
H and H*, the mapping is antilinear and hence not strictly isomorphic since
it doesn't preserve linearity. Cf. Ballentine's footnote on p27, and his eqn (1.8).

I.e., if \langle F| \leftrightarrow |F\rangle , then \bar{c}\langle F| \leftrightarrow c|F\rangle , etc.

I guess you are right. The Aerts&Daubechies' proof requires use of structure-preserving maps between two quantum propositional systems. The analysis of such maps has been done in the preceding paper of the same authors:

D. Aerts and I. Daubechies, "About the structure-preserving maps of a quantum mechanical propositional system", Helv. Phys. Acta, 51 (1978), 637.

The theorem proved there is a generalization of the famous Wigner theorem about unitary/antiunitary operators. So, the two possibilities H1 x H2 (which is isomorphic to H1* x H2*) and H1 x H2* (which is isomorphic to H1* x H2) are related to two inequivalent (unitary and antiunitary) representations of structure-preserving maps.

Eugene.