2016-08-09 09:48:15 8 Comments

This question is about non-normalisable states in quantum mechanics, e.g. the eigenstates of the position operator $|x\rangle$ which are defined by the eigenvalue equation

$$\tag{1} \hat{X}|x\rangle = x|x\rangle.$$

Since the spectrum of the operator is continuous, the states are not square normalisable. Instead one has an equation such as

$$\tag{2} \langle x'|x\rangle = \delta(x-x')$$

which gives the overlap meaning as a distribution. However it is clear that there is still at least an ambiguity in choosing a constant, i.e. why no have

$$\langle x'|x\rangle = a \times \delta(x-x')$$

where $a$ is $a \in \mathbb{R} $.

So I went to see how Equation (2) is derived. In Schleich's textbook on quantum optics in phase space I found the following, very short and in my opinion not clear/complete derivation.

From Equation (1) which can be taken to be the definition of the position operator and the assumption that the operator is **hermitian** which is a fundamental postulate of quantum mechanics one obtains:

$$\tag{3} \left( x - x' \right) \langle x'|x\rangle = 0.$$

That was easy so far, but then Schleich jumps immediately to equation (2). But the solutions for $\langle x'|x\rangle$ of (3) are all functions/generalised functions that are non-zero everywhere except at $x=x'$. So my question is: **Why is (2) the unique solution? Where does the additional information come from?**

My suspicion is that we impose another requirement, which is probably the completeness of the states, which would make (2) clear since it is the identity element in the position representation.