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Why does $$ \sum_n \Phi^{\ast}_n(x)\Phi_n(r)=\delta(x−r) $$ represents a completeness relation? Or, put differently, why does it imply completeness?

Is there any way to see it intuitively? Maybe an intuition concerning vector analogy? It seems to me that we are summing up the product of the component of a vector along the $x$ axis with its component along the $r$ axis (roughly speaking) of $n$ vectors but I can not quite see through the relation and I can not see why completeness holds through it.

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@AccidentalFourierTransform thanks for helping, the formula is here : check equation (1.5) the second one. – TheQuantumMan Mar 5 at 20:20
up vote 3 down vote accepted

A "completeness relation" for a set of vectors $\lvert \psi_n \rangle$ is that the sum of the projectors onto them is the identity since that assures use there is no basis vector "missing", i.e. $$ \sum_n \lvert \psi_n\rangle\langle \psi_n \lvert = \mathbf{1}$$ and your relation is this evaluated in position space: Apply $\langle x \rvert$ from the left and $\lvert x' \rangle$ from the right to obtain $$ \sum_n \langle x \lvert \psi_n \rangle\langle \psi_n \vert x' \rangle = \langle x \vert x' \rangle$$ and since the wavefunction is defined by $\psi_n(x) := \langle x \vert \psi_n\rangle$ this gives $$ \sum_n \psi_n(x)\psi_n^\ast(x') = \delta(x -x')$$ so your equation is the completeness relation for the kets $\lvert \psi_n \rangle$ expressed in position space.

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Let's do a finite example. Suppose we have a vector space $V$ with a subset of vectors $E=\{\mathbf{e}_1,\mathbf{e}_2,\dots\}$. Also assume that $V$ has an inner product, and that this set of vectors is orthonormal. That is $$\langle\mathbf{e}_i,\mathbf{e}_j\rangle = \delta_{ij}$$

If the span of these vectors is all of $V$ (they form a basis for $V$) then we say that the set is a complete orthonormal basis for $V$. ("Complete" and "basis" pretty much mean the same thing here. There might be some nuance in the infinite dimensional case, but the intuition is the same)

Given $E$, we can construct a special linear operator that acts on $V$. We'll call it $P_E$ and its action on an arbitrary vector $v\in V$ is given by $$P_E(v) = \sum_{n}\langle \mathbf{e}_n,\mathbf{v}\rangle \mathbf{e}_n$$ $P_E(v)$ is a linear combination of $\mathbf{e}$'s, so it is a linear map that takes a vector from $V$ and returns a vector in $\operatorname{span}(E)$.

What if $\operatorname{span} E = V$? Well, then we know we can write every vector $\mathbf{v}$ as $\sum_i v^i\mathbf{e}_i$, so $$P_E(v) = \sum_{n,i}\langle \mathbf{e}_n,\mathbf{e}_i\rangle v^i \mathbf{e}_{n} =\sum_{n,i}\delta_{ni} v^i \mathbf{e}_{n}= \sum_{i}v^i \mathbf{e}_{i} = \mathbf{v}$$ Hence, if $E$ is a complete, orthonormal basis for $V$, then $P_E = I$, the identity operator on $V$. If it is orthonormal but not complete, $P_E$ sends $V$ to a restricted subspace of $V$.

To address your question all we have to do is notice that $\sum |\Phi_n\rangle \langle \Phi_n |$ is just a convenient way of writing $P_\Phi$ and that $\delta(x-r)$ is the identity operator on the Hilbert space of square-integrable functions.

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In the finite dimensional case completeness can be represented by $$I= \sum_{i=0}^n |i\rangle\langle i|, $$ where the kets are typically orthogonal. The relation in the OP is for continuous variables, and establishes that counting everything, the entirety is present somewhere.

It's often called the resolution of the identity, and it means that the expression accounts for all of the elements of probability; thus probability is conserved.

Here is an entire lecture on the topic:

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Hello and thanks for the answer, but why does your formula and my formula(which are the same) imply completeness? Is there a way to see through it and understand it? – TheQuantumMan Mar 5 at 20:25

protected by Qmechanic Mar 5 at 23:14

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