# Distributive property of Ket over addition

If:

$$|a> = \binom{-4i}{2}$$

$$|b> = \binom{1}{-1+i}$$

Then what will the value of $$|a + b>$$ be?

That is, is addition distributive over the ket notation?

• What does $| a + b \rangle$ even mean? How are you defining it? Commented Mar 13, 2023 at 15:48
• I came across this question in a book, and wasn't sure what exactly this notation meant Commented Mar 13, 2023 at 16:04

As far as I'm aware, writing something like $$\lvert a+b\rangle$$ does not make sense (is not defined). Here's a quote from the Wiki page on bra-ket notation:
Symbols, letters, numbers, or even words—whatever serves as a convenient label—can be used as the label inside a ket, with the $$\lvert\rangle$$ making clear that the label indicates a vector in vector space. In other words, the symbol $$\lvert A\rangle$$ has a specific and universal mathematical meaning, while just the "$$A$$" by itself does not. For example, $$\lvert 1\rangle + \lvert 2\rangle$$ is not necessarily equal to $$\lvert 3\rangle$$. Nevertheless, for convenience, there is usually some logical scheme behind the labels inside kets, such as the common practice of labeling energy eigenkets in quantum mechanics through a listing of their quantum numbers.
What we put inside of the $$\lvert \rangle$$ is, essentially, just a name for a vector, and this name may be a number, like $$\lvert 0\rangle$$ (which we use to denote the first standard basis vector of $$\mathbb C^2$$), or it could be some other symbol like $$\lvert +\rangle$$ (used to denote the first element of the Hadamard basis) or $$\lvert \uparrow\rangle, \lvert\downarrow\rangle$$ (used to denote a "spin up" or "spin down" state). In the latter case, it becomes more clear that what you're asking is not really well-defined. What would $$\uparrow + \downarrow$$ even mean?
If what the book meant was |a> + |b>; I believe it would just be element-wise addition. i.e. |a>+|b> = $$\begin{bmatrix}a_1+b_1\\a_2+b_2\end{bmatrix}$$
As well, to be more explicit, $$|0> + |1>= \begin{bmatrix}1\\0\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$$