@ChrisWhite I just posted a quick (probably wrong) answer since I thought OP was confusing the dimensionality of a single state with that of the set of all states
user54412
Yeah it's a bit more subtle. I think deleting is good (at least until it's sorted out) - you'd be surprised how common it is for big names around here to speak too quickly and subsequently delete answers.
Heh, I guess since you can see deleted posts you should know ;)
user54412
The answer is the full tensor product space is bigger than the space of "pure" tensor products - i.e. there are sums \sum c_ij e_{1,i} \otimes e_{1,j} that cannot be expressed as psi_1 \otimes \psi_2
user54412
but I've always been really hesitant on this sort of thing - I've never found a good intuition for the difference between H1 \oplus H2 and H1 \otimes H2
I don't really understand what notion of dimensionality Nakahara is using for "the set of separable states", since the sum of two separable states need not be separable, so it is not a subvectorspace of the tensor product, if we "naively" work with normal Hilbert spaces
If we go rigorous, we should take the spaces as projective spaces and look at the Segre embedding of the direct sum/cartesian product into the projective space of the tensor product, and then we can talk about the $dim(H_1) + dim(H_2)$ dimensional subvariety of the tensor product corresponding to the separable states.
But that's not dimension in the vector space sense, so I'm not really sure what Nakahara wants to tell us
@ChrisWhite I think the best intuition is to think in terms of bases - $\oplus$ generates the space with the basis of $H_2$ added (i.e. set-theoretic union) to the basis of $H_1$, while $\otimes$ generates the space with all possible products of the basis vectors of $H_1$ and $H_2$ as basis. Since you now can pair every vector in $H_1$ with every vector in $H_2$, it is quite natural that this space is larger than just "pouring the bases together", I think.
user54412
1:48 PM
Do you have any intuition as to why, e.g., we can't use $\mathcal{H}_1 \oplus \mathcal{H}_2$ to accomplish the same effect? I think this can be very subtle and confusing, especially once students start doing Clebsch-Gordan stuff and both methods of constructing combined spaces are in play. — Chris WhiteFeb 24 '13 at 2:26
user54412
@ACuriousMind perhaps your intuition extends to that question I asked long ago
I think it does - when we create the combined state of two "independent systems", we create the smallest Hilbert space that contains all possible states. So, when I know that the first system is in a state $\psi$, that doesn't tell me anything about what the second should be - so I can pair every $n$ basis states of the first system with any $m$ basis states of the second system to get a valid basis state of the combined system, and this is precisely how the tensor product is created
You also need every linear operator on the original spaces to extend to a linear operator on the combined space that leaves the "other system" untouched. The tensor product, by its universal property, lends itself naturally to that - just extend $A$ as $A \otimes 1$, while the direct sum has no such property.
user54412
That makes sense, and I suppose I agree that combining two systems leads to the tensor product of Hilbert spaces.
user54412
But what physically gives rise to a direct sum of Hilbert spaces?
user54412
For instance, consider the cutesy "2 \otimes 2 = 3 \oplus 1" saying my particle physics prof loved to throw around. Sure, two spin-1/2 particles' Hilbert spaces combine via a tensor product, and I can represent this as a spin-1 system summed with a spin-0 one. But what about going the other way?
user54412
2:01 PM
When could I have a system with Hilbert space H_1, and other with H_2, and say something physically meaningful about H_1 \oplus H_2?
@ChrisWhite We don't go the other way, physically. The $\oplus$ here is merely the group theoretic decomposition of the tensor product of reps into the irreducible reps, and the latter has no intrinsic physical motivation, I think - it's just that we understand the irreducible rep better than general ones, so it is often mathematically convenient to decompose the physical tensor product into the sum of irreps
user54412
Direct sum spaces just seem too pretty to have no direct physical interpretation. But maybe that's the way it is.
@ChrisWhite Perhaps one could tell some kind of physical story at least for one particle and angular momentum: Angular momentum tells us in which rep of SO(3) are particle transforms, and these are labeled by integers. So, we can have no a.m., which is the trivial rep V_0. But we can also have $\hbar$ a.m., so we add V_1, getting the full space V_0 + V_1. But we can also have 2$\hbar$ a.m, so we add...
@ACuriousMind @ChrisWhite Isn't the whole point of the sum that it's just geometrically combining the two spaces? This is how Reed&Simon seem to be presenting it, with a figure showing a 3D-coordinate system where one can decompose a vector in the usual way to show that the 3D is 1D+1D+1D
At the same time, I feel like this should be a product instead of a sum (I'm so confused)...
@Danu Why not? It is perfectly right - direct sum means "pouring the base vectors together", so the n-fold sum of R is just a vector space with n basis vectors, aka R^n
@Danu It's more intuitive, yes. But, I'd say, not more physical in the context of Hilbert spaces, since the tensor product is the natural operation of combining the spaces of states of seperate systems - but the direct sum is no such natural operation on the spaces of states, as far as I can see
@Jim ...they're nice because I know what the 'grand overview' of my work should look like ;)
@ACuriousMind I would love to know what it means! :) I've seen the wikipedia article and the nCatLab (or whatever it's called) page but of course we all know those are terrible places to start
And I'm actually not at all opposed to doing pure mathematics either, I just know it's probably not my specialty - the way of thinking isn't quite natural to me.
Also, does anyone know a good reason why the generalization of a sequence should be called a 'net'?
@Jim That's...not really what should go into an intro, is it?
@Danu Uhhh...another terminology question I never thought about :D And, no, I cannot come up with a good reason to call it a net rather than "abstract sequence" presently
@ACuriousMind Yes and no. I have the intro about my work and the rest of the intro about what I've learned (like as in from classes). It's sort of relevant, but the main reason is that this thesis isn't going before a panel. It's going before just my supervisor and program director because I'm the first cosmology student in our university
and because I started 2 years ago without a background in the subject, actually a lot of competing and complicated reasons have made my thesis not need to have a defense
@ACuriousMind As I said, complicated process. Very long and boring story. But now I'm here. Instead of defending, I have to publish a paper is an insufficient summary of it
@alemi so I have to ask. How springy are your socks if you thought they might be a suitable replacement for a hookian spring?
@Jim I was just hoping they would have a measurable hookian regime over the span of a couple pounds. Also, my socks are quite nice, wool with a nice amount of elasticity.
@alemi I think there's more risk of harm. What if you fall and can't get up because without socks the floor is too slippery? Nobody around to save you and we're all too far away
@Danu Seen it - I think he's saying that the partial order is "closer to the top", but I don't see either why we should delete the element by which we picked up the net. Also, this would be an argument to call posets nets, not the nets.
Sometimes I wonder if there's some kind of drug all people with "math intuition" like this produce so that it makes sense to them - and if there's a way to get some doses of that ;)
@Danu: Well, it is not wrong what he says - deleting the maximal element leaves behind a cofinal subset, but why this should be a better picture for the net than the whole net isn't clear to me either. (Perhaps because the absence of a global maximal element is a bit less restrictive)
@Jim Hmm...if it is, it definitely doesn't work that way with all people - I know some people on that stuff who definitely do not develop intuitive math skills from it
Ah - I wouldn't call that an upper bound, but rather the idea that there should be some element that is comparable (and larger) to any pair of elements.
A user recently asked if a certain computation she/he saw in Weinberg's QM was in error:
Rate of change of a normalised wave function
I've found other typos/errors in the same book, and I think having a single question in which users write "answers" in which they point out an apparent error (an...
@ACuriousMind ah yes had seen it earlier today, he didn't seem satisfied with your initial reply if I recall correctly. Going to read your full answer now
@Phonon As I say, I've no idea if that is what Nakahara is talking about, so I cannot begrudge the OP hoping for something else
I suspect it isn't what Nakahara means, since he talks about ordinary Hilbert spaces (I think), and there the dimensionality statement don't seem to make much sense
@Phonon I also destroyed this guy physics.stackexchange.com/a/133506/23473 that was claiming the Earth will start spinning the other way because the end of the universe is coming
I never know if people posting such answers have some private reason to believe what they write, if they are just confused about what we really know (and really think that is scientifically backed up), or if they are trolling
@ACuriousMind I would say it's more because most people are just clueless and feel entitled to comment on area of physics without any considerations whatsoever, rather than being trolls
Oh no's! a couple more votes and I hit the rep cap. But only 2 more hours until that is reset. Come on random internet people, wait to vote for a couple of hours
A user recently asked if a certain computation she/he saw in Weinberg's QM was in error:
Rate of change of a normalised wave function
I've found other typos/errors in the same book, and I think having a single question in which users write "answers" in which they point out an apparent error (an...
