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My knowledge of linear algebra is limited and my physics knowledge mostly comes from high school and Youtube so please bear with me.

In the equation $$|x\rangle = a|0\rangle+b|1\rangle,$$ I read that $a$ and $b$ can be complex. Does that mean that they can be imaginary? If so, does something like that have any sort of non-abstract meaning? Would it be some sort of imaginary vector that can't actually be visualized?

  • sure - let a=1 and b=i; then the two components are out of phase by 90 degrees. The same technique is used in electrical engineering, but they prefer the letter j. – Peter Diehr Mar 08 '16 at 16:42

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A complex number is equivalent to a vector in $\mathbb{R}^2$. Therefore you can visualize these vectors very easily. Of course, once you want to visualize a two-dimensional vector of complex numbers, you'd need four real dimensions, which is hard to "visualize" in three dimensions. However, for qubits, you don't really have four dimensions: A global phase leaves the state invariant and the state needs to be normalized, which means that any pure state is given by only two real dimensions. This leads naturally to the visualisation of a qubit on the Bloch sphere.

The term "imaginary" does not mean that it can't be visualised - it is a bit of an unfortunate name, because there is nothing imaginary about it. Do you need complex numbers in physics? Yes and no. See this question and answer for more details: QM without complex numbers

One thing you need to understand though is that the mathematical desciption of an object is always an abstraction. After all, the vector lies in a "Hilbert space", a purely mathematical concept that you cannot "see" in the real world.

Martin
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