How to write completeness of wavefunctions without bra ket notation? In the quantum textbook I'm currently working from, the completeness relation is written as:
$$
\sum_i |\psi_i \rangle \langle \psi_i| = \mathbb{1}.
$$
But this seems to specifically require knowledge of individual bra and ket vectors. I know wavefunctions are supposed to satisfy both orthogonal and completeness relations, but I thought wavefunctions were written as coefficients of vectors $ \langle x | p \rangle = \psi(x) $ rather than the vectors themselves. Is there a way of writing the completeness relations if we're only given wavefunctions rather than bra or ket vectors? For example, the orthgonal relation for normalized wavefunction $ \psi_i $ and $ \psi_j $ is:
$$
\int_{-\infty}^\infty \psi^*_i(x) \psi_j(x) dx = \delta_{ij}.
$$
But I'm unsure of the equivalent for the completeness relation. Any insight would be greatly appreciated!
 A: The completeness relation reads:
$$\sum_{I=1}^{\infty} \psi^{*}_{I}(x') \psi_{I}(x)=\delta(x'-x
).$$
Proof:
Suppose any wave function $\Psi$ can be expanded using the $\psi_i$:
$$ \Psi(x)=\sum_{I=1}^{\infty} c_I \psi_I(x).$$
Taking inner product with $\psi_j$:
$$\langle \psi_j, \Psi \rangle=\sum_{I=1}^{\infty} c_I  \underbrace{\langle \psi_j, \psi_I \rangle}_{=\delta_{ij}} \implies \langle \psi_j,\Psi \rangle=\sum_{I=1}^{\infty} c_I \delta_{Ij}=c_j .$$
Plugging back into our expression, we obtain:
\begin{equation}
\Psi(x)=\sum_{I=1}^{\infty} \langle \psi_I,\Psi \rangle \psi_i(x).
\end{equation}
Using the definition of the inner product:
\begin{equation}
\Psi(x)=\sum_{I=1}^{\infty} \left(\int_{x'} \psi^{*}_{I}(x') \Psi(x') dx' \right) \psi_I(x).
\end{equation}
Exchanging the integral and the sum, ignoring the mathematical subtleties of why we are allowed to do so, we obtain the desired relation:
$$\Psi(x)=\int_{x'} \left(\sum_{I=1}^{\infty} \psi^{*}_{I}(x') \psi_{I}(x) \right) \Psi(x') dx' \implies \sum_{I=1}^{\infty} \psi^{*}_{I}(x') \psi_{I}(x)= \delta(x'-x).$$
EDIT: Since someone was not satisfied with the answer, I give an alternative proof using only the expressions you mentioned. We start with the completeness relation in bra-ket notation:
$$\sum_{I} |\psi_i \rangle \langle \psi_i| =\mathbb{1}.$$
Sandwich it between $|x' \rangle$ and $\langle x|$ to obtain:
$$\sum_{i} \langle x|\psi_i \rangle \langle \psi_i| x' \rangle=\langle x| x' \rangle.$$
Using the properties of the inner product and that position eigenstates are orthonormal yields:
$$\sum_{i}\langle x| \psi_i \rangle \left( \langle x'| \psi_i \rangle) \right)^{*}=\delta(x-x').$$
Finally, using the equation from the post that $\langle x| \psi_i\rangle=\psi_i(x)$, we have, as desired:
$$\sum_{i} \psi_i(x) \psi_i(x)^{*}=\delta(x-x').$$
A: One way you can show the completeness relation without bra-ket notation is just
$$\sum_{i} \langle \psi_i , v \rangle \psi_i  = v \qquad \forall v\in\mathcal{H},$$
where $\mathcal{H}$ is the Hilbert space in question, and $\langle \cdot,\cdot\rangle$ is the corresponding inner product.
Or you can say that the linear operator
$$\begin{cases}
T:\mathcal{H} \rightarrow \mathcal{H}\\[7pt]
v \mapsto T(v)=\sum_{i} \langle \psi_i , v \rangle \psi_i\end{cases}
$$
is the identity operator $T = 1\!\!1$.
