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Let us consider three spin-1/2 particles and only focusing on their intrinsic spin $S$. The Hilbert space has then to be $\mathcal H = ℂ^2 ⊗ ℂ^2 ⊗ ℂ^2$. The spin can be described by $V ∈ \text{SU(2)}$ and the fundamental representation $\mathcal D_{1/2}$ with $$\vec{S} = \hbar\vec{M} = \frac{1}{2}\hbar\vec{\sigma}.$$ Let us choose for the base of $ℂ^2$ (1 particle): $$\left\lvert\frac{1}{2},\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(1)\\(0)\end{array}\right)≡\lvert\vec{e}_3\rangle, \quad \left\lvert\frac{1}{2},-\frac{1}{2}\right\rangle = \left(\begin{array}{cc}(0)\\(1)\end{array}\right)≡\lvert-\vec{e}_3\rangle. $$ Furthermore according to the Clebsch-Gordan series one gets: \begin{align} \mathcal D_{1/2}⊗\mathcal D_{1/2}⊗\mathcal D_{1/2} & = \mathcal D_{1/2} ⊗ (\mathcal D_1 ⊕ \mathcal D_0) \\ & = (\mathcal D_{1/2}⊗\mathcal D_1) ⊕ (\mathcal D_{1/2}⊗\mathcal D_0) \\ & = \mathcal D_{3/2} ⊕ \mathcal D_{1/2} ⊕ \mathcal D_{1/2}. \end{align} So we are left with 8 states in the combined 3-particle system.

Questions:

  1. If one would simply consider the direct sum of the three particles, i.e. $ℂ^2 ⊕ ℂ^2 ⊕ ℂ^2$ we would only have 6 states, correct?
  2. What is the simplest picture to see the consequences of doing this instead of taking the tensor product?
  3. Maybe one could also give me a good (physical) example for the difference of $ℝ^3 ⊗ℝ^2$ versus $ℝ^3 ⊕ℝ^2$ (phase space?).

I have searched for similar questions and found some stuff. However I am not in particular interested in the look of the outcoming states (i.e. their spin momentum) but more in the differences if one were not considering the tensor product.

Red Pencil
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  • tex tips: \mathbb{C} has got to be faster than finding a ℂ character to copy and paste, and similarly for \otimes, \oplus, \equiv, and \in. | doesn't always have the same spacing rules as \lvert, only the latter of which is an actual delimiter, and \hbar and \sigma typeset differently from (i.e. better than) ħ and σ in MathJax. (+1 on the content though.) –  Dec 30 '14 at 23:36
  • Related: Should it be obvious that independent quantum states are composed by taking the tensor product? –  Dec 30 '14 at 23:38
  • Thanks for your remarks.
    Sidenote: In my case it is the laziness of not being forced to copy such character nor typing in latex-style text due to using a tool named Neo (keyboard layout, see: http://neo-layout.org/index_en.htmlhttp://neo-layout.org/index_en.html).
    However I realized that the outcome sometimes can mess up the layout such that I will have to adapt to keep it reader friendly. So I appreciate corrections (edits) to learn.     – Red Pencil Dec 30 '14 at 23:43

2 Answers2

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The direct sum of Hilbert spaces is not a "good" notion when talking about spaces of states.

The "actual" states are elements of the projective Hilbert space, where every ray is shrunk to a point to reflect that phases and normalization do not alter the state a vector in the Hilbert space is supposed to represent.

Now, on the projective spaces, the direct sum of the original spaces simply does not produce something that does behave correctly when considering this: Observe that, for states $\phi\in\mathcal{H}_1$ and $\psi\in\mathcal{H}_2$, $(c\phi,\psi) \in\mathcal{H}_1\oplus\mathcal{H}_2$ is not on the same ray as $(\phi,\psi)\in\mathcal{H}_1\oplus\mathcal{H}_2$ for $c \in \mathbb{C} - \{1\}$. So the direct sum does not respect the nature of quantum states, it is simply the wrong notion of product to consider.

On the other hand, the bilinearity of the tensor product means that $(c\phi)\otimes\psi = c \cdot(\phi \otimes \psi)$, so the tensor product does indeed map the product of elements of rays to the same ray in the product space, regardless of which representant we choose.

Therefore, the proper quantum notion of product is the tensor product, not the direct sum. (see also this question, and, as another interesting note, "entangled states" arise simply as the failure of a map from the Cartesian product of the projective spaces to the tensor product to be surjective, since the latter is bigger)

The frequent occurence of direct sums is now because representations of groups have the notion of irreducible representations - representations that do not decompose as the direct sum of other representations.

Therefore, if we wish to get a combined quantum system, we first take the tensor product of their spaces of states (which usually carry representations of the symmetry groups of the theory), and then decompose these tensor products into the direct sum of representations, because the irreducible ones are easier to work with, and, in the case of spin, directly correspond to the total spin of a state in that representation.

(To answer your questions literally and directly:

  1. Yes, correct.
  2. You have a less dimensional space that doesn't behave properly w.r.t. the projective structure.
  3. $\mathbb{R}^3\otimes\mathbb{R}^2 = \mathbb{R}^6$, but $\mathbb{R}^3 \oplus \mathbb{R}^2 = \mathbb{R}^5$ )
Community
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ACuriousMind
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  • So far, I did not think about an argumentation from the original Hilbert Space.
    However to fully follow your argument, shouldn't I calculate the probabilities (i.e. like measuring and normalization) and check that it does really not cancel out in the case of using the direct sum.
    I mean for a non-composite system I have seen this at the beginning of my QM lectures, but here I don't see it.     – Red Pencil Dec 31 '14 at 01:19
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    @RedPencil: I'm not sure what you are asking: Look at $\phi$ and $c \cdot \phi$, which are the same state in $\mathcal{H}_1$ (since they differ only by multiplication by a complex number!). Now look at the tupels $(\phi,\psi)$ and $(c\cdot\phi,\psi)$. These should represent the same state, yet there is no complex number $d$ such that $d\cdot(\phi,\psi) = (c\cdot\phi,\psi)$ if $c$ wasn't 1 already, so these tupels do not represent the same state. This shows the direct sum does not respect the projective nature of states. – ACuriousMind Dec 31 '14 at 01:31
  • This is exactly what I have meant: Right, $d (φ,ψ) = (cφ,ψ)$ can't hold if we exclude $1$. However what I was thinking about is, that the stuff with $λψ ~ ψ$ is due to Born's interpretation of QM ($\text{Pr}(ψ) = |\langleψ|φ\rangle|^2 /\langleψ|ψ\rangle \langle φ|φ\rangle$). And then how do I conclude that $Pr((cφ,ψ)) ≠ Pr((φ,ψ))$? – Red Pencil Dec 31 '14 at 02:09
  • @RedPencil: Observe that the inner product of $(c\phi,\psi)$ with $(\phi,\psi)$ is $c\langle\phi\vert\phi\rangle + \langle\psi\vert\psi\rangle$. Assume (for simplicity only!) that $\phi,\psi$ are normalised, so the inner product is just $c + 1$. The denominator of the Born rule applied to the two states is $2(c^2 + 1)$, and its numerator is $(c+1)^2$. Now, if the two things were the same state, the Born rule should yield 1. Demanding that this actually happens gives as the only solution $c=1$. One can trust the projective spaces, they are the Born rule in disguise ;) – ACuriousMind Dec 31 '14 at 02:44
  • I appreciate your comments. It is a really lovely page with helpful human beings. Yeah it is a curse that I don't get the mathematical concepts so quickly which would allow me spending more time with physical consequences... Cheers! – Red Pencil Dec 31 '14 at 02:52
  • @L.Su: Yes, there is still a problem with this: First, it maps normalized to unnormalized states, and second, this map does not descend to a proper map on the projective Hilbert spaces: $\phi$ and $c\phi$ are the same point in the projective space, but if $(\phi,\psi) = (\phi,0) + (0,\psi)$ and $(c\phi,\psi)$ are not the same state, then $(\phi,0)$ and $(c\phi,0)$ are not the same state, and hence we do not get a well-defined map on the projective spaces - another hint that this is the wrong notion to consider! – ACuriousMind Jan 02 '15 at 13:42
  • @ACuriousMind I guess we need to identify $c(\phi,\psi)~(c\phi,\psi)~(\phi,c\psi)$. Then the definition above works for finite dimension Hilbert spaces because we have the isomorphism between $H_1 \oplus H_2$ and $H_1 \times H_2$. Using the definition of tensor product together with the identifications, we may prove the equivalence. But then everything seems cumbersome... – DKS Jan 02 '15 at 14:00
  • @L.Su: Then we are identifying arbitrarily states in the Hilbert space which are not on the same rays, meaning the actual space of states is no longer the projective Hilbert space. Also, you are making the direct sum, which is already smaller than the tensor product, even smaller with such identification, so this cannot be the same as simply taking the tensor product. The tensor product is the correct (categorial) notion of product in the category of projective spaces, and the direct sum isn't - there's no way to "fix" this. – ACuriousMind Jan 02 '15 at 14:04
  • @ACuriousMind What's in my mind is the isomorphism between direct sum and direct product of finite dimensional spaces. The vector space spanned by $(e_1,e_2)$ with $e_1,e_2$ in the bases of $H_1$ and $H_2$, respectively, is no longer smaller than the tensor product. There is also no identification of different rays because this construction produces exactly a tensor product. – DKS Jan 02 '15 at 14:16
  • @L.Su: You cannot really construct the tensor product in this way, since $(c\phi,c\psi) = c(\phi,\psi)$ in the direct sum/product, and hence $c(\phi,\psi) = (c\phi,\psi) = (c\phi,c\psi)$, which is false in the tensor product ($(c\phi)\otimes(c\psi) = c^2(\phi\otimes\psi)$) Anyway, it's a bit needlessly convoluted - by the universal property of the tensor product, you already always have a (bilinear!) map from the direct product/sum into the tensor product. I guess I've not understood what you are trying to do. – ACuriousMind Jan 02 '15 at 14:27
  • I have deleted the previous wrong statements and conclude here: I need to add two more identifications as we can do for constructing a tensor product from two vector spaces. What I did was actually reconstructing tensor product from direct sum. As you said, it is convoluted. Let's stick to your answer. – DKS Jan 02 '15 at 15:02
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    @ACuriousMind Dear friend, could you please add your discussion with Red Pencil to the answer, because I think the arguments you gave, were very crucial. I didn't really understand how you calculated $2(c^2+1)$ could you maybe show that in your answer as well? (and maybe the correct version of the same calculation obtained with $\otimes$ product). Thank you so much in advance. – user929304 Jan 06 '15 at 08:56
  • This is a good answer, but it seems to me to simply push the question back to "why are physical states represented by lines in Hilbert space, instead of individual vectors in Hilbert space?" After all, phase differences between components within a single pure-state vector have physically observable consequences, so it's certainly logically conceivable that phase differences between different subsystems could as well. I think at the end of the day, we just have to resort to the experimental evidence for entanglement. – tparker Aug 06 '17 at 08:26
  • Note that combining subsystems via direct sums instead of tensor products wouldn't break the local predictiveness of QM, because just like entanglement in the usual picture, the phase differences between subsystems could only be measured by a joint measurement on both subsystems. So as long as you only ever measure one subsystem, you're free to rescale that subsystem's state vector by any constant without affecting observables. – tparker Aug 06 '17 at 08:31
  • Indeed, under this rule for combining subsystems together, we would be completely justified in thinking of a single spin-1/2 as a "joint system" of two "subsystems" $|\uparrow\rangle$ and $|\downarrow\rangle$! – tparker Aug 06 '17 at 08:35
  • Hi @tparker, I also reached this idea when thinking "ok, but what are the consequences of taking the direct sum? What if indeed we lose the property that two vectors inside a ray of a subsystem leads to two vectors in the same ray in a joint system?", then it means that by considering a larger system, one would be able to distinguish between the two state vectors, i.e to observe their phase difference by measuring them as part of the composite system. But Nature is not like this, and preserves the rays of subsystems. – Voidt Apr 24 '23 at 15:06
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This is mainly about your question 2.

A decomposition of the Hilbert space into a direct sum $\mathcal H = \mathcal H_1 \oplus \mathcal H_2$ represents that the system can be in a state in $\mathcal H_1$, or a state in $\mathcal H_2$ (or a superposition, of course). In your example, $\mathcal H = \mathcal D_{3/2} \oplus \mathcal D_{1/2} \oplus \mathcal D_{1/2}$, the system can be in states of spin $\frac 3 2$, or spin $\frac 1 2$ (and superpositions thereof).

A decomposition of the Hilbert space into a tensor product $\mathcal H = \mathcal H_1 \otimes \mathcal H_2$ represents that the system is described by a state from $\mathcal H_1$ and a state from $\mathcal H_2$ (and superpositions of such states). This is naturally the case for multi-particle systems: a description of the system means giving a state for each particle.

Robin Ekman
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    If the Hilbert space admits a direct sum decomposition, there is no way of forming a superposition of two states from different invariant subspaces. This sort of phenomenon leads to the so-called superselection sectors. I think the argument of ACuriousMind's answer somehow hints to this. So the direct sum is not the correct way of composing independent components of a quantum system. – Phoenix87 Dec 31 '14 at 00:05
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    @Phoenix87: A space of two spin-1/2 particles can be decomposed into a direct sum of a spin 1 and a spin 0 space, and still, $\lvert\uparrow\downarrow\rangle$ is a valid state, and a superposition of spin 0 and spin 1. Superselection rules require that the density matrices and observables commute with a symmetry, i.e., have a direct sum structure. – Norbert Schuch Jan 03 '15 at 15:36