There is a nice answer to this question by @joshphysics but it applies to multiple quantum systems combination. I was wondering whether this could also be applied to a single quantum system? For example, the Hilbert space $S$ of states for a single particle (the space abstraction including its spin, position, momentum, charge, energy, ...) is the tensor product of the Hilbert spaces generated by the basis (eigen-) vectors of each observable considered in the $S$ space.
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1$\begingroup$ I don't understand what you mean. The basis vectors of any observable always span the entire space - that's what being a basis means! Can you be a bit more explicit/mathematical about what you're imagining here? $\endgroup$– ACuriousMind ♦Apr 4, 2020 at 10:51
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$\begingroup$ @ACuriousMind Let's take a basic example. A particle with only 2 observables: its spin and its 3D-position. Any state that is generated with the spin operator (let's use the x component of the spin to simplify things here) can be expressed as a linear combination of 2 x-spin basis vectors, which happen to be normalized eigenvectors of the x-spin operator. But what you are saying is that ANY state, including those with more info than just spin (e.g.position info), can be expressed in the same basis, using only 2 vectors with 2 complex components ? Please see my next comment below $\endgroup$– CarlaApr 4, 2020 at 13:41
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$\begingroup$ My initial question was asking if it was not rather the tensor product of the space generated by the spin eigenvectors by the space generated by the position eigenvectors, which would then become the Hilbert space of all states owning ALL the information about the particle. $\endgroup$– CarlaApr 4, 2020 at 13:41
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1$\begingroup$ No, I'm not saying what you think I'm saying. You're simply misusing terminology - the basis of common eigenvectors $\lvert x, s_x\rangle$ of position and spin in x-direction are all eigenvectors of $\hat{x}$ and $\hat{S}_x$. There is no "subset" here of eigenvectors of $S_x$. I think I understand why you are trying to say, though - for that Valter's answer is correct. $\endgroup$– ACuriousMind ♦Apr 4, 2020 at 13:46
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$\begingroup$ @ACuriousMind I agree that "subset" is incorrect and misleading. I changed the tile of this post accordingly. Thank you for your comment $\endgroup$– CarlaApr 5, 2020 at 8:10
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The answer is negative if I correctly understand the question. There are observables which do not commute, whereas if each of them could be viewed in its own Hilbert space and the whole Hilbert space were the tensor product of all those Hilbert spaces, these observables would commute.
What it is true is that there is a maximal set of pairwise commuting observables whose common eigenvectors span the whole Hilbert space (the rigorous mathematical statement is a bit different and uses the direct integral, but the physical substance is that). This fact can be reformulated in terms of tensor product if necessary, even if the use of the tensor product is not necessary...
answered Apr 4, 2020 at 11:04
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$\begingroup$ OK. Thanks. So to make sure I understand (and validate your answer), in that particular case, the $\lvert x, s_x\rangle$ are elements of the tensor product, since x and Sx commute, correct? $\endgroup$– CarlaApr 4, 2020 at 14:16
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1$\begingroup$ Yes, they can be considered as elements of a tensor product. $\endgroup$ Apr 4, 2020 at 14:19