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!


2 Answers 2


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 $T = 1\!\!1$.


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: $$\sum_{i} \psi_i(x) \psi_i(x)^{*}=\delta(x-x').$$


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