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If I use British money the amounts I can have are isomorphic to $\mathbb{Z}_{\geq0}$ (in pennies). If I also use Australian money, if I want to think about the amount I have in total, I can use addition and make a vector space. The possible amounts I can have are isomorphic to $\mathbb{Z}_{\geq0} \times \mathbb{Z}_{\geq0}$ (in pennies and cents).

$\times$ fairly clearly describes how to build the more complicated space out of the simpler ones. It also carries across the structure and even semantics (if I use vector space $\times$ rather than set $\times$). Is there an equivalent operator for quantum spaces? ie an operator that if given two singletons/single points/values returns a Bloch sphere, and so forth.

It would be perfect if there is an operator like this specifically for quantum mechanics. But if there is only a mathematical one that lacks semantics or some structure it would still be great to learn of it. I exclude $\otimes^\text{note 1}$ on Hilbert spaces as an answer because as far as I understand this includes unphysical degrees of freedom and I am looking for one that captures the construction exactly.

Edit - note 1: to make sense in the context of this question $\otimes$ should be replaced with $\oplus$.

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    $\begingroup$ What's wrong with $\otimes$, the tensor product of the Hilbert spaces of the two subsystems you're joining? Or are you envisioning the 'state space' of the subsystems as something other than a vector space? If so, what? $\endgroup$ Aug 12, 2018 at 12:06
  • $\begingroup$ @EmilioPisanty Thanks for the comment Emilio. I exclude Hilbert spaces because as I understand it they include unphysical degrees of freedom. I believe all physics can be recreated after restricting to projective Hilbert space for example. I may prefer talking about currency in $\mathbb{Z}_{\geq0}$ rather than $\mathbb{R}$ for a similar reason. I should have written $\otimes$ rather than $\times$ to make this clearer and I will update my answer to reflect this. $\endgroup$
    – user183966
    Aug 12, 2018 at 12:13
  • $\begingroup$ By "unphysical degrees of freedom", do you simply mean a global phase and a normalization constant? If so, it would be good to name them explicitly (so as to separate them e.g. from gauge degrees of freedom). $\endgroup$ Aug 12, 2018 at 12:27
  • $\begingroup$ @EmilioPisanty Unfortunately I'm not sure what I mean in terms of the usual formalism. In part this question is intended to help with that. I had hoped my example case would imply a unique answer but I guess not. I have not studied gauge theories. If there are theories which reproduce the same observable predictions without gauge degrees of freedom them I exclude them, otherwise I include them. At the risk of being cheeky, an amazing answer could give the operator for constructing spaces with gauge degrees of freedom and the one without them. $\endgroup$
    – user183966
    Aug 12, 2018 at 12:39
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    $\begingroup$ Gauge degrees of freedom are a whole other kettle of fish, and they make very little sense in the context of your question. I point them out explicitly as one way in which your question admits readings which make no sense and which it would be good if you can clear out explicitly by being clear about what "unphysical degrees of freedom" you perceive. (In particular, note that "degree of freedom" is normally used in the dynamical sense, say, to count spatial dimensions in an oscillator, and it would not normally include the normalization that you seem to be worried about here.) $\endgroup$ Aug 12, 2018 at 13:18

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I exclude $\otimes$ on Hilbert spaces as an answer

That's a shame, because the tensor product of the underlying Hilbert spaces really is the correct constructor for the joint state space of quantum systems.

because as far as I understand this includes unphysical degrees of freedom

To the degree that it does $-$ i.e. in that the Hilbert space includes a global phase and a normalization constant $-$ those degrees of freedom are trivial to add back in. More technically, the 'true' state space of a $d$-dimensional quantum system is the complex projective space $$ \mathbb{C}\mathbf{P}^{d-1} = (\mathbb{C}^{d}\setminus\{0\})/\mathbb{C}^\times $$ of all complex $d$-ples modulo a complex amplitude (i.e. modulo a global phase and normalization). If you have two independent quantum systems, with state spaces $\mathbb{C}\mathbf{P}^{d_1-1}$ and $\mathbb{C}\mathbf{P}^{d_2-1}$, and you want to find the space of states that they can occupy jointly (i.e. including entangled states; if you don't want to include them then you just do a set-theoretic $\times$ and you're done) then

  • first you recover the $\mathbb C^d$ structure, which you can always do automatically,
  • you then form the tensor product $\mathbb C^{d_1}\otimes \mathbb C^{d_2}\cong \mathbb C^{d_1+d_2}$,
  • and finally you re-take the projection to $\mathbb{C}\mathbf{P}^{d_1+d_2-1} = (\mathbb C^{d_1}\otimes \mathbb C^{d_2}\setminus\{0\})/\mathbb{C}^\times$ to get the joint state space.

For this construction to make sense, there are two key things to note:

  1. You don't actually need a unique way to get from the projective space up $\mathbb{C}\mathbf{P}^{d-1}$ to the base Hilbert space $\mathbb{C}^{d}$: the former already comes equipped with the latter, and you don't need to do anything with it.
  2. What you do need to do, in order to deal with the unphysical re-specification of the normalization and phase, is to show that you do have a well-defined way to take two states of the subsystems, $[u]\in \mathbb{C} \mathbf{P}^{d_1-1}$ and $[v]\in\mathbb{C}\mathbf{P}^{d_2-1}$, and get a uniquely-defined state of the joint system.

    To do that, what you have to show is that the choice of representatives within the classes $[u]$ and $[v]$ do not matter: i.e. that if $u_1,u_2\in [u]$ and $v_1,v_2\in[v]$ are different representatives, with $u_2=\lambda u_1$ and $v_2=\mu v_1$, then the classes they give rise to, $[u_1\otimes v_1]$ and $$ [u_2 \otimes v_2] = [(\lambda u_1) \otimes (\mu v_1)] = [\lambda\mu (u_1\otimes v_1)] = [u_1\otimes v_1] $$ are the same $-$ which they are, because the combined multiplier $\lambda\mu$ gets ignored when we take the $/\mathbb C^\times$ equivalence in the final step.

On the other hand, it is important to realize that not all states in the joint state space are of this form, because entangled states are simply not separable as the tensor product of two individual system states. Those cannot be constructed explicitly within the joint state space given only the classes, i.e. the entangled state

the even superposition of system 1 in $[u]$ and system 2 in $[v]$ superposed with system 1 in $[w]$ and system 2 in $[x]$

is not sufficient as a description, because in forming the class $$\left[\frac{1}{\sqrt{2}}(u\otimes v + w\otimes x)\right]$$ the choices of representatives from $[u]$ and $[w]$ (resp. $[v]$ and $[x]$) do affect which class you end up in. This is, however, not an artifact of the math, and it has the physical backing that you've failed to specify a common phase reference for $[u]$ and $[w]$ (resp. $[v]$ and $[x]$) and you're not even able to form meaningful single-state superpositions $[u+w]$ or $[v+x]$ without that common phase reference.

What does that mean? Basically, that while the projective space $\mathbb{C} \mathbf{P}^{d-1}$ does trim out some 'unphysical' aspects of the description from the vector-space picture, that doesn't mean that you can discard the vector-space structure in $\mathbb{C}^{d}$, which is essential to formulate the full ontology of the system.


And finally, this brings me to my final comment on your question:

an operator that if given two singletons/single points/values returns a Bloch sphere

What you're describing there is not the joint state space of two systems whose individual description are singletons: if you have two systems whose state spaces $\mathbb{C} \mathbf{P}^{0} = \{[|\mathrm{s}⟩]\}$ only allow for a single state $|\mathrm{s}⟩$ on each system, then the only possible state for the system is $$ [|\mathrm{s}⟩\otimes|\mathrm{s}⟩], $$ i.e. the state in which each system is in its only allowed state; in other words, the joint state space is $$ (\mathbb C^{1}\otimes \mathbb C^{1}\setminus\{0\})/\mathbb{C}^\times = (\mathbb C^{1}\setminus\{0\})/\mathbb{C}^\times = \mathbb{C}\mathbf{P}^{0}, $$ which is also a singleton.

The process that you describe,

given two singletons/single points/values returns a Bloch sphere

does not correspond to the joint space state of two subsystems, but rather to the expanded state space of a single system which is allowed to inhabit superpositions of states from space 1 and from space 2. This is, physically, a very different situation, but your vague description of what actual physical situations you do (and don't) want to consider makes it very hard to comment more accurately.

(And no, I'm unlikely to expand this answer if you update your question to add in those specifications. You ask a flawed question, you get an imperfect answer.)

If what you want is not the joint states of two subsystems but rather the superposition states of two possible (sets of) states, then the operator you want is not the tensor product $\otimes$, but rather the direct sum $\oplus$ of the Hilbert spaces that underlie each state space. Here, however, it is important to note that, in contrast with the tensor-product structure, the de-projectivization procedure $$ \mathbb C\mathbf P^{d-1} \to \mathbb C^d, $$ together with its attendant addition of "unphysical" information which was not present in the original projective space, cannot be ignored as it was before.

The reason for this is that if you're taking superpositions, then the "unphysical" information contained in the "global" phase is no longer unphysical, because the phase is no longer global, and you need a way to specify the relative phase of the superpositions. Put another way, there is an infinity of possible ways to form superpositions of the physical states $[|0⟩]$ and $[|1⟩]$, depending on their relative phase; and, if you don't know the phase, then you've completely decohered and you're not doing quantum mechanics $-$ you're doing classical mechanics, and you're just forming a probability distribution over a larger set of states.


Further reading: They don't cover joint-space constructors in detail (I imagine because they consider it a bit too basic), but if what you're looking for is a formalized geometric approach to the states of quantum systems, the book you probably want to be reading is Geometry of Quantum States: An Introduction to Quantum Entanglement, by Ingemar Bengtsson and Karol Życzkowski (Cambridge University Press, 2017).

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  • $\begingroup$ Thank you, almost perfect answer! It was a silly mistake to give an example with $times$ when $+$ would have been more appropriate, although I am glad to see your explanation with both cases. One of the reasons I asked this question was because looking at higher dimensional Bloch spheres, and their great complexity, made me suspect they removed more information, very happy to know that's incorrect. Thank you very much for the book reference. A final confirmation: I assume since you have not given a name to the operations directly on complex projective spaces they do not have a standard name? $\endgroup$
    – user183966
    Aug 12, 2018 at 15:09
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    $\begingroup$ They may well have a standard name, but if they do I've never seen one. But it's not quite my community so there might be a specialized technical term that I've missed thus far. $\endgroup$ Aug 12, 2018 at 15:11

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