# On the completeness relation in Quantum Mechanics

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.

• @AccidentalFourierTransform thanks for helping, the formula is here : jpoffline.com/physics_docs/y3s6/mm_summary.pdf check equation (1.5) the second one. – TheQuantumMan Mar 5 '16 at 20:20
• Related : Take a look how this useful relation is used for the derivation of the propagator $\:K(\mathbf{x},t \; \boldsymbol{;} \;\mathbf{x}_{0},t_{0})\:$ in my answer here : Show that Propagator satisfies Schrödinger equation, equations from (sk-12) to (sk-18). Equation (sk-17) therein is identical to that of the question. – Frobenius Jun 25 '18 at 22:55

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.

• Your answer, written at no-mathematical-rigor level, starts off with the assumption that the "kets" are genuine vectors in a Hilbert space (notice that the completeness relation you started with involves a sum, i.e. $n$ takes values in a subset of $\mathbb N$), to end up with a discrete sum of functions in a realization of the Hilbert space you started with by a set of (classes of equivalence of) functions which will not longer add up to unit operator (number 1), but to a distribution. Therefore, there is a logical disconnection somewhere... – DanielC Nov 9 '17 at 22:31
• @DanielC Yes, the "logical disconnection" is that position kets are terrible objects and if you want rigor you should avoid them at all costs ;) I can't say anything in defense of this answer, mathematically, this is undoubtedly nonsense: I start with an expression that lives in the Hilbert space, then apply two bras/kets to it that are not in the Hilbert space, and end up with a sum over functions converging...somehow. But note that the question is the same kind of nonsense to begin with: The sense of convergence in which the sum is supposed to go to the distribution is not specified. – ACuriousMind Nov 9 '17 at 23:19
• Thank you for your comment. Actually, it should be emphasized at all costs in an answer to a mathematically ill-posed (or perhaps incompletely stated) problem that this is the case and that your answer would not pass a rigor validation. This is my problem with „elementary" or typically textbook-level physics. It strays from mathematics without saying (a disclaimer) it specifically. Even "X Theory for Mathematicians" books fall in the same trap, I am afraid... I have not been able to wrap my head around the full mathematical treatment of Dirac's bra/ket formalism, so I avoid it completely. – DanielC Nov 9 '17 at 23:31

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.

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.

• 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 '16 at 20:25