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Complex numbers! I didn't expect complex numbers to have a deeper relation to quantum mechanics. It feels like negative probabilities is just the "sqrt(-1) does

Complex numbers! I didn't expect complex numbers to have a deeper relation to quantum mechanics. It feels like negative probabilities is just the "sqrt(-1) doesn't make sense" topic all over again. Yeah, why is it Quantum Mechanics taught on how it was historically sequentially discovered instead of reorganizing it in a clear manner for future students? This may be long overdue and might even yield new perspectives on quantum mechanics from future researchers.

layer8 2y ago
Quantum states are represented as vectors of length 1 (unit vectors), or corresponding complex numbers, so really just positions on a circle or n-dimensional sphere. To me that makes more sense than talking about roots of negative numbers or negative probabilities, and complex numbers are just a mathematically convenient representation.
- elashri 2y ago
  Yes, it is a very convenient representation, and it is possible to formulate quantum mechanics without using complex numbers at the fundamental level of the mathematical representation. In standard quantum mechanics, the state of a quantum system is represented by a complex-valued wave function or a vector in a complex Hilbert space. Observables are represented by self-adjoint operators acting on this complex Hilbert space. The complex nature of the wave function gives rise to the phenomenon of quantum interference and the probabilistic interpretation of quantum mechanics. However, in this real quantum mechanics, the complex Hilbert space is now replaced by a real Hilbert space, and the complex wave function is replaced by a real-valued wave function or a vector in this real Hilbert space. The observables are represented by self-adjoint operators acting on the real Hilbert space. One way to achieve this is by using real algebra or real matrix representations. Now we let the complex numbers represented by 2x2 real matrices of the form: a + ib = [a -b] [b a] now a and b are real and the imaginary unit i is represented by the matrix: i = [0 -1] [1 0] One also can use quaternions, but now it will be more complicated without much to gain.
  - itishappy 2y ago
    Complex numbers are just one example of more general Clifford Algebras. Matrices, quaternions, and bivectors can all be used to represent the same stuff.
    - nh23423fefe 2y ago
      quotient of polynomial ring R[x]/(x^2+1)
- moefh 2y ago
  For that to make sense, the n-dimensional sphere must not live in an Euclidean n-dimensional space, but a space over the complex numbers (a Hilbert space) -- that is, coordinates for the space are complex numbers. Also note that (this is a little beside the point) the n-dimensional sphere picture only works for quantum states that can be described with a finite number of dimensions (like in quantum computing): to talk about even simple things like position and momentum of particles, states must live in a infinite-dimensional space. More importantly (as is pointed out in the lecture), every time-evolution of a quantum state from A to B can be represented as a unitary operator, and if you want to take the square root any such operator, you must in general use complex numbers (or something equivalent). Taking the square root like that is a very simple operation that answers the question "what is the operator for the evolution to the point halfway between A and B?"
dboreham 2y ago
Every time you're tempted to freak out because complex numbers, reflect that you should also be freaking out about negative numbers. Those don't exist either.
- andoando 2y ago
  Negative numbers are very natural to me. Math isnt just about counting real objects, but also transformations through space/time. The notion of duality or things going in reverse is extremely natural. -5 is just the inverse of 5, such that if you go 5 and go back 5, you end up at the same place.
  - anyfoo 2y ago
    Complex numbers you just go sideways as well, and if you multiply you not only stretch or squish your path to the origin, you also change your angle to it by adding both angles together. Exponentiation is then doing this continuously, so you end up walking in a circle. That’s it.
- anyfoo 2y ago
  Besides, they’re really not that scary once you get to know them well. Take a deep dive into digital signal processing for example, and you become well comfortable with them, and understand how “the real world” fits into them quite neatly.
  - woopsn 2y ago
    Or differential equations -- that grueling and irrelevant subject physical scientists waste so much time on. :)
- mr_mitm 2y ago
  What does it even mean for a number to exist? In math we are happy if the things we talk about are well defined. Numbers are just mathematical objects which satisfy a few interesting properties that lead to very rich structures we can study. Some of those structures map onto concepts in nature, which makes them actually useful. Instead of real and imaginary or complex we could have chosen entitirely different adjectives. They're completely arbitrary. They don't matter.
unbrewn 2y ago
> Complex numbers! I didn't expect complex numbers to have a deeper relation to quantum mechanics. Can you elaborate more on what you mean? Why are complex numbers “unexpected”? So much of contemporary physics is about distillation. You care more about things like consistency and simplicity over readability. Our current methods of solving these equations, vector calculus, requires certain operations, specifically, square roots and derivatives. To be able to utilize the well studied methods of linear algebra we need to make ample use of the Pythagorean theorem that ultimately uses a square root. To ensure that our model works in all domains we have to use a form of the square root that allows negative numbers. Complex numbers. It’s why we use e^(iHt/h_bar) in the Schrödinger equation. Rather than see it as “nature works by using exponentiation!” I see it as “exponentiation is ‘stable’ in the face of complex derivatives”, ie the derivative of the exponential function is the exponential function f(x)=e^x == f’(x)=e^x. So this formulation simplifies our calculations. These things can be calculated with any coordinate and numbering system, the ones we choose just help make the problems more tractable for us.
nathan_compton 2y ago
I would argue complex numbers don't really have a deeper relation to quantum mechanics. Quantum Mechanics is all about squaring the fact that we must find a representation for a system which simultaneously supports an expression of a probability distribution over measurement outcomes while also providing sufficient representational complexity to support all the symmetries that are found in the kinematic space of the system. Complex numbers happen to serve well for a lot of systems but when you have spinors, for example, you can think of them as just another sort of number the use of which is suggested by the symmetries in (or redundancies in the description of) the underlying system. In other words, the appearance of complex numbers is no more or less mysterious than the allowed transformations of the system itself. At least that is how I've come to understand the question.

layer8 2y ago

Quantum states are represented as vectors of length 1 (unit vectors), or corresponding complex numbers, so really just positions on a circle or n-dimensional sphere. To me that makes more sense than talking about roots of negative numbers or negative probabilities, and complex numbers are just a mathematically convenient representation.

elashri 2y ago
Yes, it is a very convenient representation, and it is possible to formulate quantum mechanics without using complex numbers at the fundamental level of the mathematical representation. In standard quantum mechanics, the state of a quantum system is represented by a complex-valued wave function or a vector in a complex Hilbert space. Observables are represented by self-adjoint operators acting on this complex Hilbert space. The complex nature of the wave function gives rise to the phenomenon of quantum interference and the probabilistic interpretation of quantum mechanics. However, in this real quantum mechanics, the complex Hilbert space is now replaced by a real Hilbert space, and the complex wave function is replaced by a real-valued wave function or a vector in this real Hilbert space. The observables are represented by self-adjoint operators acting on the real Hilbert space. One way to achieve this is by using real algebra or real matrix representations. Now we let the complex numbers represented by 2x2 real matrices of the form: a + ib = [a -b] [b a] now a and b are real and the imaginary unit i is represented by the matrix: i = [0 -1] [1 0] One also can use quaternions, but now it will be more complicated without much to gain.
- itishappy 2y ago
  Complex numbers are just one example of more general Clifford Algebras. Matrices, quaternions, and bivectors can all be used to represent the same stuff.
  - nh23423fefe 2y ago
    quotient of polynomial ring R[x]/(x^2+1)
moefh 2y ago
For that to make sense, the n-dimensional sphere must not live in an Euclidean n-dimensional space, but a space over the complex numbers (a Hilbert space) -- that is, coordinates for the space are complex numbers. Also note that (this is a little beside the point) the n-dimensional sphere picture only works for quantum states that can be described with a finite number of dimensions (like in quantum computing): to talk about even simple things like position and momentum of particles, states must live in a infinite-dimensional space. More importantly (as is pointed out in the lecture), every time-evolution of a quantum state from A to B can be represented as a unitary operator, and if you want to take the square root any such operator, you must in general use complex numbers (or something equivalent). Taking the square root like that is a very simple operation that answers the question "what is the operator for the evolution to the point halfway between A and B?"

dboreham 2y ago

Every time you're tempted to freak out because complex numbers, reflect that you should also be freaking out about negative numbers. Those don't exist either.

andoando 2y ago
Negative numbers are very natural to me. Math isnt just about counting real objects, but also transformations through space/time. The notion of duality or things going in reverse is extremely natural. -5 is just the inverse of 5, such that if you go 5 and go back 5, you end up at the same place.
- anyfoo 2y ago
  Complex numbers you just go sideways as well, and if you multiply you not only stretch or squish your path to the origin, you also change your angle to it by adding both angles together. Exponentiation is then doing this continuously, so you end up walking in a circle. That’s it.
anyfoo 2y ago
Besides, they’re really not that scary once you get to know them well. Take a deep dive into digital signal processing for example, and you become well comfortable with them, and understand how “the real world” fits into them quite neatly.
- woopsn 2y ago
  Or differential equations -- that grueling and irrelevant subject physical scientists waste so much time on. :)
mr_mitm 2y ago
What does it even mean for a number to exist? In math we are happy if the things we talk about are well defined. Numbers are just mathematical objects which satisfy a few interesting properties that lead to very rich structures we can study. Some of those structures map onto concepts in nature, which makes them actually useful. Instead of real and imaginary or complex we could have chosen entitirely different adjectives. They're completely arbitrary. They don't matter.

unbrewn 2y ago

> Complex numbers! I didn't expect complex numbers to have a deeper relation to quantum mechanics. Can you elaborate more on what you mean? Why are complex numbers “unexpected”? So much of contemporary physics is about distillation. You care more about things like consistency and simplicity over readability. Our current methods of solving these equations, vector calculus, requires certain operations, specifically, square roots and derivatives. To be able to utilize the well studied methods of linear algebra we need to make ample use of the Pythagorean theorem that ultimately uses a square root. To ensure that our model works in all domains we have to use a form of the square root that allows negative numbers. Complex numbers. It’s why we use e^(iHt/h_bar) in the Schrödinger equation. Rather than see it as “nature works by using exponentiation!” I see it as “exponentiation is ‘stable’ in the face of complex derivatives”, ie the derivative of the exponential function is the exponential function f(x)=e^x == f’(x)=e^x. So this formulation simplifies our calculations. These things can be calculated with any coordinate and numbering system, the ones we choose just help make the problems more tractable for us.

nathan_compton 2y ago

I would argue complex numbers don't really have a deeper relation to quantum mechanics. Quantum Mechanics is all about squaring the fact that we must find a representation for a system which simultaneously supports an expression of a probability distribution over measurement outcomes while also providing sufficient representational complexity to support all the symmetries that are found in the kinematic space of the system. Complex numbers happen to serve well for a lot of systems but when you have spinors, for example, you can think of them as just another sort of number the use of which is suggested by the symmetries in (or redundancies in the description of) the underlying system. In other words, the appearance of complex numbers is no more or less mysterious than the allowed transformations of the system itself. At least that is how I've come to understand the question.