# Is there a proof or example to show that a noiseless subsystem is not necessarily closed under addition?

In a text (section 3.6 page 92) about noiseless subsystems by D. Lidar, it is mentioned:'A subsystem is a tensor factor in a tensor product, and this does not have to be a subspace (e.g., in general it is not closed under addition).'

I do not understand how if some quantum states $$\rho, \sigma \in \mathbb{C}^n \otimes \mathbb{C}^m$$ then how can their addition not be in $$\mathbb{C}^n \otimes \mathbb{C}^m$$?

Thank you in advance.

• Can you maybe provide a link to the text and explain where in the text this is mentioned? This might provide extra context which might make it easier to answer the question.
– JSdJ
Aug 14 '20 at 15:33

I do not understand how if some quantum states $$\rho, \sigma \in \mathbb{C}^n \otimes \mathbb{C}^m$$ then how can their addition not be in $$\mathbb{C}^n \otimes \mathbb{C}^m$$?

The author is not claiming that this is false. This is certainly true, you are right about that!

In a [text][1] (section 3.6 page 92) about noiseless subsystems by D. Lidar, it is mentioned:'A subsystem is a tensor factor in a tensor product, and this does not have to be a subspace (e.g., in general it is not closed under addition).'

The author is saying that there are non-subspace structures (i.e. tensor factors) which have the noiseless property. The point is kind of abstract but it comes from the notion that $$\mathbb{C}^n$$ isn't naturally a subspace of $$\mathbb{C}^n\otimes \mathbb{C}^m$$. However, this doesn't mean you can't think of a way to think of $$\mathbb{C}^n$$ as a subspace, it just means you need to fix something in $$\mathbb{C}^m$$ to do so.

More to the point, if you did want to think about $$\mathbb{C}^n$$ as a subspace of $$\mathbb{C}^n\otimes \mathbb{C}^m$$ you would need to fix a basis of $$\mathbb{C}^m$$ and identify how the basis of $$\mathbb{C}^n$$ gets mapped to the basis of $$\mathbb{C}^n\otimes \mathbb{C}^m$$. It's because of this "identication requirement" these noiseless subsystems are not technically subspaces.

Ultimately, the difference between noiseless subspace and noiseless subsystem is subtle, but I hope to illustrate the difference with an example. Encoding a single qubit into a noiseless subspace of a 2-qubit space can be given by the mapping $$\alpha|0\rangle+\beta|1\rangle \mapsto \alpha|0\rangle_1|0\rangle_2+\beta|1\rangle_1|1\rangle_2$$ this encoding can be seen as a linear (or subspace) endcoding as we are mapping a basis of $$\mathbb{C}^2$$ to a basis of $$\mathbb{C}^2\otimes \mathbb{C}^2\cong \mathbb{C}^4$$. Now, consider an encoding of a qubit of the form $$\alpha|0\rangle+\beta|1\rangle \mapsto (\alpha|0\rangle_1+\beta|1\rangle_1)\otimes |\psi\rangle_2$$ where $$|\psi\rangle_2$$ is an arbitrary qubit in the second space of $$\mathbb{C}^2\otimes \mathbb{C}^2$$. This encoding is into the first subsystem and is not a subspace encoding because we have not fixed the state of the second system in the tensor product (this makes the encoding non-linear and that's why its not the same as a subspace encoding). That being said, if we did fix $$|\psi\rangle_2$$ to be a specific element of $$\mathbb{C}^2$$ then we would get a subspace encoding like the first example, just in a different basis.

• Can you give a meaningful example in this context? Sep 1 '20 at 19:54
• @NorbertSchuch thanks, see revised answer. Sep 2 '20 at 13:30
• Hm, I see - this seems to follow my gut feeling. However, what this really means is that there is no way to think of this subsystem as any kind of subset of the full space, since there is no natural embedding (the embedding requires tensoring with some state, which is not canonical). So the real point to me seems not so much that it is not a subspace (i.e. does not have a sub-vector space structure), but that it is not a subset of the full space in a meaningful way to start with. If that's the way to look at it, then the quote from the book is certainly rather misleading. Sep 2 '20 at 15:43
• @NorbertSchuch yeah it does make the book quote seem misleading (or that I am missing something...), but perhaps his reasoning is that the book audience isn't familiar with the naturalness properties of vector spaces and tensor products. Btw my answer is based on the section of decoherence free subsystems in Dave Bacon's PhD thesis which I assume are the same thing as Lidar's noiseless subsystems (hopefully I am right about this)... Sep 2 '20 at 18:06
• What you say indeed fits with the feeling I had. I just feel that then the way it is phrased in the quote missing the point (or misleads about the actual point). Sep 2 '20 at 18:33