Bra-ket

Now let this be a bra-ket (get it?)

⟨ϕ∣ψ⟩=⟨ϕ∣∣ψ⟩=[c∗d∗][ab]\begin{gather*} \left\langle \phi|\psi \right\rangle=\left\langle \phi \right\rvert\left\lvert \psi \right\rangle=\begin{bmatrix}c^* & d^*\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}\quad \end{gather*}

This means we’re multiplying these two Matrix [c∗d∗],[ab]\begin{bmatrix}c^* & d^*\end{bmatrix},\begin{bmatrix}a\\b\end{bmatrix} together.

by doing matrix multiplication we get

⟨ϕ∣ψ⟩=⟨ϕ∣∣ψ⟩=(c∗d∗)(ab)=(c∗a+d∗b)=c∗a+d∗b\begin{gather*} \left\langle \phi|\psi \right\rangle\\=\left\langle \phi \right\rvert\left\lvert \psi \right\rangle\\=\begin{pmatrix}c^* & d^*\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\\=\begin{pmatrix} c^*a+d^*b\end{pmatrix}\\=c^*a+d^*b \end{gather*}

It always outputs a 1x1 matrix (i.e., a scalar value).

Note that

⟨x+∣x+⟩=⟨x−∣x−⟩=1⟨x+∣x−⟩=⟨x−∣x+⟩=0\begin{gather*} \left\langle x+|x+ \right\rangle=\left\langle x-|x- \right\rangle=1\\ \left\langle x+|x- \right\rangle=\left\langle x-|x+ \right\rangle=0 \end{gather*}

Where x+,x−x+,x- are orthonormal to each other. Another way we can say this is {∣x+⟩,∣x−⟩}\{\left\lvert x+ \right\rangle,\left\lvert x- \right\rangle\} forms an orthonormal basis.

a Set of Vector {v1,v2,...,vn}\{v_1,v_2,...,v_n\} form an orthonormal basis if and only if

where δij\delta_{ij} is the Kronecker Delta

⟨vi∣vj⟩=δij={1i=j0i≠j\begin{gather*} \left\langle v_i|v_j \right\rangle=\delta_{ij}=\begin{cases}1\quad i=j\\0\quad i\neq j\end{cases} \end{gather*}