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I'm reading notes from a friend of mine taking a quantum mechanics class, and I see something I don't quite get.

$$\left<x_i|x_j\right> = \delta_{ij}.$$

The notes say this implies orthogonality. Generally, the dot product of two orthogonal vectors is just zero, yes? So delta here = 0, but what exactly does $\delta_{ij}$ represent? Thanks in advance for any help and feel free to ask for more details. $x_i$ and $x_j$ are elementary basis vectors.

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  • $\begingroup$ So that you know, that symbol is called the kronecker delta $\endgroup$
    – Triatticus
    Commented Jun 29, 2020 at 22:09

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$\delta_{ij}$ is the Kronecker Delta. It's $0$ when $i \neq j$, and 1 when $i=j$. In other words, the "dot product" of $|x_i\rangle$ and $|x_j\rangle$ is zero unless $i=j$, in which case it's 1, just like with "usual" vectors.

EDIT: Of course, as @user12262 points out, the dot product of two vectors is only 1 when the vectors themselves have a length of 1 each. (We then say that such vectors are "normalised"). If the vectors have arbitrary length, then the condition for normalisation is actually:

$$\langle x_i|x_j\rangle = \delta_{ij} \sqrt{\langle x_i|x_i\rangle} \sqrt{\langle x_j|x_j\rangle} ,$$

where the terms on the right-hand-side are the "lengths" of each of the vectors.

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  • $\begingroup$ Philip: "[...] just like with "usual" vectors." -- The given formula $$\langle x_j | x_k \rangle = \delta_{jk}$$ also implies that any and all $|x_j\rangle$ (of index $j$ range under consideration) are separately normalized; i.e. $$\langle x_j | x_j \rangle = 1.$$ Such orthonormal vectors are of course often convenient to use. (@Tolemus M´s OP calls them "elementary basis vectors".) Orthogonality without implying normalization: $$\langle a_j | a_k \rangle = \delta_{jk}\,\sqrt{\langle a_j|a_j \rangle \, \langle a_k|a_k \rangle}.$$ $\endgroup$
    – user12262
    Commented Jul 1, 2020 at 2:54
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    $\begingroup$ @user12262 You're right, of course. I was just keeping it simple since the OP's question seemed more conceptual, but you're right that it's a little sloppy. I'll edit to include it :) Thanks! $\endgroup$
    – Philip
    Commented Jul 1, 2020 at 4:38
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    $\begingroup$ Technically what you edited in isn't equivalent to @user12262's equation; consider the case where one or both vectors vanish. $\endgroup$
    – J.G.
    Commented Jul 1, 2020 at 6:53
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    $\begingroup$ @J.G. That's true. I've taken them over to the other side :) $\endgroup$
    – Philip
    Commented Jul 1, 2020 at 6:55

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