# What is an example of the need for complex numbers implied by the need of continuity of linear operators?

Not a duplicate. Question and answers given elsewhere do not answer this one.

Context. Scott Aaronson, lecture 2, course 6.896 Quantum Complexity Theory, september 2008 says that a matrix such as $$\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ maps the state $$\alpha|0\rangle + \beta|1\rangle$$ to $$\alpha|0\rangle - \beta|1\rangle$$.

What is in between these two states? To have a continuous unitary transformation between these two states requires the use of complex numbers.

Question. Could you elaborate this conclusion with an example? I can't quite see the deduction. I can see why we want it to be continuous: the Schrödinger equation is a differential equation with respect to time, which therefore requires time to be continuous. So if that matrix maps a state into another one, we may ask what happened in between the change of state. And how does that require complex numbers?

Source:

• Thank you @FDGod for showing me how to write matrices in LaTeX and how to use a proper right-angled bracket. I appreciate it. Nov 11, 2023 at 19:04

We are told that there is an operation $$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} \alpha \\ \beta \end{bmatrix}=\begin{bmatrix} \alpha \\ -\beta \end{bmatrix}.$$ We are now asked to speculate about whether there are intermediate operations, what that means for intermediate states, and whether we are forced to include complex numbers.
If operation is continuous, I can stop somewhere and later restart. So, my evolution would be described by two matrices $$M_1$$ and $$M_2$$. $$M_2M_1=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$$ I'm imposing that $$M_1$$ and $$M_2$$ are neither identity or the target. To simplify the maths, consider the specific possibility that we stopped half way, which would suggest we can set $$M_1=M_2$$. So, no you're asking what $$2\times 2$$ matrices $$M_1$$ there are such that $$M_1^2=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}.$$ Try solving it. You'll find the only solutions contain complex numbers. (If you're not so happy with the $$M_2=M_1$$ special case, you'll probably need to introduce the extra restriction that the matrices are unitary, so if you want them to be real, they'll look something like $$M_1=\begin{bmatrix} \sqrt{p} & \sqrt{1-p} \\ \pm\sqrt{1-p} & \mp\sqrt{p} \end{bmatrix},$$ and you'll still find there's no solution.