0
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ What does $| a + b \rangle$ even mean? How are you defining it? $\endgroup$
    – Rammus
    Mar 13, 2023 at 15:48
  • $\begingroup$ I came across this question in a book, and wasn't sure what exactly this notation meant $\endgroup$
    – Saniaaa
    Mar 13, 2023 at 16:04

2 Answers 2

0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Got it! This question was given in a book as an exercise, I am guessing it was a misprint. Thanks for this explanation. $\endgroup$
    – Saniaaa
    Mar 13, 2023 at 16:05
1
$\begingroup$

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, $$|0> + |1>= \begin{bmatrix}1\\0\end{bmatrix} + \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.