# Doubt in the completeness of wave function

The energy eigenstates are complete in the sense that any reasonable wave function $$\psi(x)$$ can be expanded as a superposition of energy eigenstates.
$$\psi(x)=\sum_{n=1}^\infty b_n\psi_n(x),\;\;b_n\in\mathbb{C}\tag{1}$$
Assume that the wavefunction at time equal zero is the $$\psi(x)$$ above. Then we have
$$\psi(x,t=0)=\sum_{n=1}^\infty b_n\psi_n(x),\;\;b_n\in\mathbb{C}\tag{2}$$
If this function is normalized then
$$\sum_{n=1}^\infty|b_n|^2=1\tag{3}$$
As $$\psi(x,t)$$ is a stationary state, so $$\psi(x,t)=e^\frac{-iEt}{\bar{h}}\psi(x)$$
So, $$\psi(x)=\sum_{n=1}^\infty b_ne^\frac{-iEt}{\bar{h}}\psi_n(x)\tag{4}$$
$$b_n=\int_{-\infty}^{\infty}dx\psi_n^*(x)\psi(x)\tag{5}$$
Substitute $$b_n$$ in $$(1)$$
$$\psi(x)=\sum_{n=1}^{\infty}\Big(\int_{-\infty}^{\infty}dx'\psi_n^*(x')\psi(x')\Big)\psi_n(x)$$
$$\implies\psi(x)=\int_{-\infty}^{\infty}dx'\Big(\sum_{n=1}^{\infty}\psi_n^*(x')\psi_n(x)\Big)\psi(x')\tag{6}$$
The above equation is of the form
$$f(x)=\int_\infty^\infty dx'K(x',x)f(x')\tag{7}$$
Take $$f(x)=\delta(x-x_o)$$
$$\delta(x-x_o)=\int dx'K(x',x)\delta(x'-x_o)=K(x_o,x)\tag{8}$$
So, $$K(x',x)=\delta(x-x')$$
Thus, $$\boxed{\text{Completeness}:\;\sum_{n=1}^{\infty}\psi_n^*(x')\psi_n(x)=\delta(x-x')}\tag{9}$$

I have the following doubts -

i) I am not able to understand why we have put $$f(x)=\delta(x-x_o)$$ in $$(7)$$ to get $$(9)$$. What is the point in doing so?

ii) What is the equation $$(9)$$ actually telling us. It follows from equation $$(1)$$. Equation $$(1)$$ is the superposition of wavefunction. If $$\psi_n$$ are eigen vectors corresponding to $$E_n$$ then their superposition is also a solution to time dependent Schrodinger equation. But I am not able to understand what is the meaning of completeness and what is the interpretation of equation $$(9)$$.

• @Jakob, yes you are correct.
– Iti
Jun 10 at 17:16
• It would be good if you have also met the general idea of a complete set of vectors in the sense of spanning a vector space, and then the complete set of functions idea in general. After that it is easier to see what it is doing in quantum theory. Jun 10 at 17:17
• @Jakob, here $\psi_n(x,t)$ is a stationary state meaning $\psi(x,t)exp(-iEt/\bar{h})\psi(x)$. IF $\psi(x,0)$ is a solution of time independent Schrodinger equation then $\psi(x,t)$ is a solution of time dependent Schrodinger equation and so is their superpositions. Now am I correct?
– Iti
Jun 10 at 17:25
• Equation (2) looks incorrect. Why is there $\psi(x)$ in front of the sum? Jun 10 at 18:32

Your notes are trying to say that when you have $$\begin{equation} \psi(x) = \int_{-\infty}^{\infty} dx K(x,x^\prime) \psi(x^\prime) \end{equation}$$ for any reasonable function $$\psi(x)$$, the only possibility is that $$K(x, x^\prime) = \delta(x - x^\prime)$$. They "prove" this by considering the special case where $$\psi(x)$$ is a Dirac delta itself. (Presumably, this is because the space of "reasonable functions" includes the distributions needed to make the Fourier transform an automorphism of $$L^2$$.)

If you have doubts about this analysis which is for an infinite dimensional Hilbert space, a great remedy is to consider the analogous formula for a finite dimensional Hilbert space. Then, instead of $$\begin{equation} \sum_{n = 1}^\infty \psi_n(x) \psi_n^*(x^\prime) = \delta(x - x^\prime) \end{equation}$$ you will have $$\begin{equation} \sum_{n = 1}^N \textbf{e}_n \textbf{e}^T_n = I_N. \end{equation}$$ Here, $$\textbf{e}_1, \dots, \textbf{e}_N$$ are any set of length 1 orthogonal vectors which form a basis for $$\mathbb{C}^N$$. It is easy to convince yourself of this formula and moreover to prove it.

Now we can consider what happens when we "remove" one of the basis vectors from this set. Without loss of generality, we will have $$\textbf{e}_1, \dots, \textbf{e}_{N - 1}$$. These are still linearly independent but they fail the other criterion for being a basis: there are no longer enough of them to span $$\mathbb{C}^N$$. I.e. they are not complete. This is obvious in the finite dimensional case because we can just count them and see that there are less than $$N$$. But to generalize to infinite dimensions, it is much better to diagnose this by adding up the outer products as above and checking whether we get the identity $$I_N$$.

So that is the intuitive meaning of equation (9). It means you have enough $$\psi_n(x)$$ to get any wavefunction you want by taking linear combinations. Removing one element from this infinite set will still leave you with infinitely many. So counting is no longer a valid way to check whether a linearly independent set is complete. The condition (9) is what replaces it.

• ,I have not studied about Hilbert pace. I am beginner to quantum mechanics. May you please answer accordingly.
– Iti
Jun 10 at 17:45
• Do you know how to find an orthonormal basis for $\mathbb{C}^N$ (the space of $N$ component vectors with complex entries) at least? The most important part is being able to do this exercise and then realizing that wavefunctions are somewhat analogous (they satisfy superposition and therefore form a vector space). Jun 10 at 17:57

I'm not sure if your result for completeness is correct. The completeness is given by $$\sum_n|n\rangle \langle n|=\mathbb{1}$$ On position basis $$\sum_n\iint \langle x'|n\rangle \langle n|x\rangle|x'\rangle\langle x| dxdx' =1$$ $$\rightarrow \sum_n\iint \psi_n^*(x')\psi_n(x)|x'\rangle\langle x|dx'dx=1$$ The interpretation for completeness can be found in this answer.

• In Dirac notation, the last equation reads $\sum\limits_n \psi^*_n(x^\prime) \, \psi_n(x) = \sum\limits_n \langle n|x^\prime\rangle \langle x|n\rangle = \langle x|\sum\limits_n |n\rangle \langle n| |x^\prime\rangle = \langle x|x^\prime \rangle = \delta(x-x^\prime)$. Jun 10 at 17:07
• And so what I do? Jun 10 at 17:35
• You said: 'I'm not sure if your result for completeness is correct'... Jun 10 at 17:44
• $\langle x|x'\rangle =\delta(x-x')$ is orthonormality relation, Not a completeness. That's what I mean, I'm not sure if it's right to call it completeness. Jun 10 at 18:01