5 ms·

This is your regularly scheduled reminder that complex numbers have (real) matrix representations, and what matters in any model is the properties it has not it

This is your regularly scheduled reminder that complex numbers have (real) matrix representations, and what matters in any model is the properties it has not its identity as an object.

sebzim4500 4y ago
Did someone claim otherwise?
- whatshisface 4y ago
  "Quantum physics falls apart without imaginary numbers." >Marco made a curious face, so Toni posed the question: “Can standard quantum theory work without imaginary numbers?”
  - sebzim4500 4y ago
    By that logic, it doesn't even need real numbers. Just do everything with cauchy sequences of rationals.
    - whatshisface 4y ago
      But you can't really disagree with that. It's wrong to say that physics "needs" any one thing in particular when you can construct it from other things, and use them instead.
    - deleted 4y ago
      [deleted]
    - jiggawatts 4y ago
      Something I've always wondered is: what is the weakest algebra that could be used to model physics? E.g.: Are nationals sufficient? Integers? Finite integers?
      - Koshkin 4y ago
        Sure, but like what they say about Lisp, you are then bound to reinvent the complex numbers, poorly.
- contravariant 4y ago
  If the title doesn't count then we need to question what it even means for a quantum theory to 'use' imaginary numbers. Because if real skew symmetric matrices count, then a harmonic oscillator also inescapably uses imaginary numbers.
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 ago
  Thanks, 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 4y ago
    yes, 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 4y ago

This is your regularly scheduled reminder that complex numbers have (real) matrix representations, and what matters in any model is the properties it has not its identity as an object.

sebzim4500 4y ago

Did someone claim otherwise?

whatshisface 4y ago
"Quantum physics falls apart without imaginary numbers." >Marco made a curious face, so Toni posed the question: “Can standard quantum theory work without imaginary numbers?”
- sebzim4500 4y ago
  By that logic, it doesn't even need real numbers. Just do everything with cauchy sequences of rationals.
  - whatshisface 4y ago
    But you can't really disagree with that. It's wrong to say that physics "needs" any one thing in particular when you can construct it from other things, and use them instead.
  - deleted 4y ago
    [deleted]
  - jiggawatts 4y ago
    Something I've always wondered is: what is the weakest algebra that could be used to model physics? E.g.: Are nationals sufficient? Integers? Finite integers?
    - Koshkin 4y ago
      Sure, but like what they say about Lisp, you are then bound to reinvent the complex numbers, poorly.
contravariant 4y ago
If the title doesn't count then we need to question what it even means for a quantum theory to 'use' imaginary numbers. Because if real skew symmetric matrices count, then a harmonic oscillator also inescapably uses imaginary numbers.

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 ago
Thanks, 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 4y ago
  yes, 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 4y ago