Can complex vectors be orthogonal even if none of the components are?

Question

I tried researching this on my own but I couldn't find a concise answer that didn't bog me down with terms and math I don't understand. I'm attempting to read Leonard Susskind's Theoretical Minimum book on Quantum Mechanics and I'm just getting into ket-vectors, inner products, etc. but I had a question I couldnt wrap my head around. (Just assume I know very little and go from there, also I have no idea how to properly format the mathematical expressions so I'll use plain English).

If the definition of orthogonal is that the inner product of two vectors is 0, there are hypothetically multiple ways this could be achieved, in my mind at least. What I want to know is, which of these is actually true.

The first case is if every component (B* times A) is 0, then the sum of all those 0's is of course 0.

What if two components are positive and negative pairs? Say +1, -1, and 0? The sum is 0 but some or all components are non-zero.

The third case, based on the previous one, is what if there are no +/- pairs but the sum is 0? Say -0.5, -0.5, and +1. Can either of these last two cases actually happen?

Is there something I'm missing that makes my interpretation too simplistic to be accurate? Again, I don't have a strong math or physics background so I wouldn't understand complicated mathematical proofs or explanations, just a simple answer (if that's even possible).

Answer 1

You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $a\cdot b = 0$ without all $a_ib_i = 0$.

Examples in 3D matching your exact example numbers-

$$ \begin{pmatrix} 1 \\ 1\\ 0\end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1\\ 0\end{pmatrix} = 0, \text{ with one} \pm \text{pair}, \quad \begin{pmatrix} 1 \\ 1\\ 1\end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1/2\\ -1/2\end{pmatrix} = 0, \text{ with no} \pm \text{pair}$$

Answer 2

To illustrate the flexibility of orthogonality: Given any numbers $a_1,\ldots,a_n$ and $b_1,\ldots,b_{n-1}$, as long as $a_n\neq0$, there is a number $b_n$ such that the vectors $$ A=(a_1,\ldots,a_n)\\ B=(b_1,\ldots,b_n) $$ are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.

Answer 3

Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$\mathbf{v}_1 = (1,0,0)\;\;\;\mathbf{v}_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$\mathbf{v}_1 \cdot \mathbf{v}_2 = (1\times0) + (0\times0) + (0\times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero