# Need some help with Bra-Ket notation (specifically orthogonality in bra-ket notation)

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

$$\left = \delta_{ij}.$$

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.

• So that you know, that symbol is called the kronecker delta Jun 29, 2020 at 22:09

$$\delta_{ij}$$ 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.

• 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: $$\langle a_j | a_k \rangle = \delta_{jk}\,\sqrt{\langle a_j|a_j \rangle \, \langle a_k|a_k \rangle}.$$ Jul 1, 2020 at 2:54
• @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! Jul 1, 2020 at 4:38
• Technically what you edited in isn't equivalent to @user12262's equation; consider the case where one or both vectors vanish.
– J.G.
Jul 1, 2020 at 6:53
• @J.G. That's true. I've taken them over to the other side :) Jul 1, 2020 at 6:55