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This is your regularly scheduled reminder that the author of these papers (https://arxiv.org/abs/2101.10873 https://arxiv.org/abs/2101.10873, https://arxiv.org/
by eigenket 4y ago
This is your regularly scheduled reminder that the author of these papers (https://arxiv.org/abs/2101.10873 https://arxiv.org/abs/2101.10873, https://arxiv.org/abs/2111.15128 https://arxiv.org/abs/2111.15128), which this article is based on know very well that complex numbers have real matrix representations.
What they add which your comment discounts is the locality structure of quantum mechanics, i.e. what happens when you combine multiple quantum systems. Specifically if the state of one system lives in A, and the state of another system lives in B then the state of a combined system lives in the tensor product A ⊗ B.
If you do the trick where you replace complex numbers by real matrices what you end up is having the state of the first system living in A = X ⊗ A', and the second is in B = X ⊗ B', where A' and B' are real and the X subsystem is the degree of freedom you're using to "fake" the complex numbers.
Then if you try to combine the two systems you end up with a state that lives in A ⊗ B = (X ⊗ A') ⊗ (X ⊗ B') but what you need in order to get the behavior you want is actually for the combined system to live in X ⊗ (A'⊗ B').
What they do is a bit more complicated because they show that all ways to fake complex numbers using real numbers breaks this way of combining systems, but this is the gist.
- civilized 4y agoThanks, this is very helpful. When I took undergrad quantum physics, we saw that the Schrödinger equation can be represented without complex numbers. But that's for one particle, and it sounds like you're saying this somehow breaks down when you have multiple systems interacting, due to these tensor products not working as we think they should?
- eigenket 3y agoyes, specifically doing things without complex numbers breaks if you require composing systems to work like they do in normal quantum mechanics (composing with the tensor product). If you drop that assumption about how things compose then you can use something like the real matrix representation where 1 is a 2x2 identity matrix and i is some anti-symmetric 2x2 matrix.
- civilized 3y agoHelp a lapsed mathematician out here... the complex numbers can be represented as skew-symmetric 2x2 real matrices, right? And this is an isomorphism? So if the 2x2 real skew-symmetric matrices are "the same" as the complex numbers, how do we end up having this issue that comes up in quantum physics where you can only use the actual complex numbers and not the representation?
- eigenket 3y agoAny explanation I can give in a comment here will be pretty flawed, if you want more detail than my previous comment I think your best bet is to read the papers at the arxiv links I posted above.