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$\left\{ e _ { i } \right\}$ is an orthonormal basis which has the orthonormal condition as following: $$e _ { i } ^ { T } \cdot e _ { j } = \delta _ { i j }$$ In Dirac Notation where $| i \rangle = | e _ { i } \rangle$ the condition can be written as $$\langle i | j \rangle = \left( e _ { i } ^ { T } \right) ^ { * } \cdot e _ { j } = \delta _ { i j }$$

Does that mean $\left( e _ { i } ^ { T } \right) ^ { * } = e _ { i } ^ { T }$? which could only be true when $\left\{ e _ { i } \right\}$ are made up of real vectors, isnt it?

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    $\begingroup$ Orthonormality for complex vectors is defined as $e _ { i } ^ { \dagger } \cdot e _ { j } = \delta _ { i j }$, where $\dagger$ means Hermitian conjugate. For vectors, this translates as the transpose of the complex conjugate. $\endgroup$ – TheAverageHijano Feb 18 at 8:10
  • $\begingroup$ so can i conclude $e _ { i } ^ { \dagger } = e _ { i }$ ? since the general form, like @InitialObserver said, is $e _ { i } \cdot e _ { j } = \delta _ { i j }$ $\endgroup$ – Jung Feb 18 at 20:36
  • $\begingroup$ No, the dagger means transpose and complex conjugate, a vector can't be equal to its Hermitian conjugate unless it has dimension one (a it is a real vector). $e_i$ is a $n\times 1$ vector while $e_i^\dagger$ is $1\times n$ vector. Take this vector as an example: $e_1=\begin{bmatrix} i\\ 0 \end{bmatrix}$, $e_1^\dagger=\begin{bmatrix} -i & 0 \end{bmatrix}$, so $e_1^\dagger e_1=1$ $\endgroup$ – TheAverageHijano Feb 18 at 20:45
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$\{ e_i \}$ is an orthonormal basis which has the orthonormal condition as following $$ e^T_i \cdot e_j = \delta_{ij} $$

Not quite. The reason I say this because this statement isn't the most general statement you could make about orthonormal bases for vector spaces (or at least it's not precise enough)


At any rate, the correct general statement is as follows.

Let $\{ e_i\}$ be a discrete (possibly infinite) orthonormal basis for a complex vector space $V$ equipped with an inner product " $\cdot$ ". Then, for any $i,j$ we have that

$$ e_i \cdot e_j = \delta_{ij}. \tag{a} $$

Wait a second. Shouldn't that be $(e_i^*)^T \cdot e_j$? No, and this is because the " $\cdot$" denotes an inner product between two vectors $v, w$---whatever that may be. That is, the dot does not refer to matrix multiplication necessarily, but rather refers to the inner product between two vectors $v,w\in V$, which is a more abstract notion than matrix multiplication.


Note that the case $[e_i]^T \cdot [e_j] = \delta_{ij}$ is a special case of (1) when we are representing the basis vectors in a matrix representation and (2) when it is a real vector space, since then $e_i^\dagger = e_i$ (in the abstract sense, not the matrix sense--if you find this notation confusing don't worry it is, which is why Dirac notation better as it distinguishes between vectors and their "dual"). It’s precisely when both (1) and (2) are true that

$$e_i \cdot e_j\ \dot{=} \ [e_i]^T \cdot [e_j] = \delta_{ij}$$

where the square brackets explicitly denote the matrix representation of the basis vectors, and the $\dot{=}$ means "is represented by".


Lastly, Dirac notation is just the notational change

\begin{align} & e_i \to | e_i \rangle\\ & e_i^\dagger \to \langle e_i|\\ & v \cdot w \to \langle v |w \rangle. \tag{b} \end{align}

Note that in pure mathematics, another common notation for $(b)$ is $v\cdot w \to \langle v, w \rangle$. While this notation is better than the dot-notation, Dirac notation is, in my opinion still better for any purpose I can readily think of.

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  • $\begingroup$ Does that mean $\langle i | j \rangle = \left( e _ { i } ^ { T } \right) ^ { * } \cdot e _ { j } = \delta _ { i j }$ can only be true when $i$ and $j$ live in the real vector space? $\endgroup$ – Jung Feb 18 at 18:35
  • $\begingroup$ It’s always true for two orthonormal vectors, but the conjugate is meaningless when it’s a reap vector space. $\endgroup$ – InertialObserver Feb 18 at 19:00

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