In a recent lecture we were told that $\langle x'|\hat{O}|x \rangle = O(x,x') = O(x)\delta(x-x')$ "due to the locality of quantum mechanical observables".

I have no idea what this is supposed to mean any comments or references to a book from which this, or something similar, originates would be much appreciated


2 Answers 2


Okay, $\delta$ is the Dirac $\delta$-function; it is an infinitely thin infinitely tall spike at $0$ where the limit is taken such that the area of the spike remains a constant $1.$ In general there are lots of function families $f_k$ which satisfy the essential property $\lim\limits_{k\to0}\int dx~f_k(x) ~g(x) = g(0),$ and for some of these families every $f_k$ is a smooth function, so you can always essentially pretend that $\delta(x)$ is smooth. One such choice is $k = \sigma$ in the Gaussians of unit integral, $(2\pi\sigma^2)^{-1/2} \exp\big[-x^2/(2\sigma^2)\big].$ That is valid if $x$ is defined in the position-space $\mathbb R$, but there are certainly other Dirac-$\delta$ functions for higher-dimensional spaces, where generally $\int dx_1\int dx_2\dots \int dx_n~\delta_n(\vec x) ~g(\vec x) = g(\vec 0).$

The basis vectors $|x\rangle$ and $|x'\rangle$ are being written in the "position basis", which is not actually 100% well-defined. The general idea is that $\int dx~|x\rangle\langle x|$ is the identity operator $\hat 1$ (when integrated over $(-\infty, \infty)$), with the "orthonormality" relation $\langle m|n\rangle = \delta_{mn}$ using the Kronecker $\delta$ being replaced by the Dirac $\delta$ as $\langle x' | x\rangle = \delta(x - x').$

The fact that you cannot really find a set of wavefunctions which do this perfectly is viewed as a side-note to a valid way of managing the algebra involved. Yes, basically $|x_0\rangle$ is essentially supposed to be a square-root-of-$\delta$ function, so that in 1D we're supposed to consider $\psi_{x_0}(x)$ as the limit of $(2\pi\sigma^2)^{-1/4} \exp\big[-(x - x_0)^2/(4\sigma^2)\big]$ in the limit (after the integrals have been all performed) as $\sigma\to 0.$ But more importantly: the math works at the bra- and ket- levels, so why not use it?

The claim being offered here is that a quantum mechanical observable must be local, presumably in the sense that it must not connect two different positions at the same time. Therefore probably $\vec x$ lives in Minkowski space and the $\delta$ functions we're meant to use are 4D ones, though you could also retrieve the 1D case if you are making the statement that $\hat O$ is a measurement that is performed instantaneously rather than by evolving the system with some Hamiltonian first and then performing your instantaneous projective measurement.

The claim probably requires that $O(x)$ remain a differential operator, because locality should really only demand that an observable depend on a wavefunction and its derivatives at some point. In particular I'd very much struggle to see how the momentum operator in 1D would look if not $O(x) = -i\hbar \partial_x.$


Since having asked this question I know feel I understand the meaning:

Firstly the matrix element can only really be understood within a corresponding matrix equation such as:

$$\langle \psi |\hat{O}|\psi\rangle=\int \mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle \langle x' |\hat{O}|\,x\rangle \langle x\, |\,\psi\rangle $$

Now in non relativistic QM $\hat{O}=f(\hat{p},\hat{x})$ i.e. the operator can be written as some function of momentum and position operators. Firstly the simplest case is if $\hat{O}=f(\hat{x})$ then we have

\begin{align}&=\int \mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle \langle x' |\,f(\hat{x})|\,x\rangle \langle x\, |\,\psi\rangle \\&= \int \mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle f(x)\langle x' |\,x\rangle \langle x\, |\,\psi\rangle \\ &= \int \mathrm dx \int\mathrm dx' \langle \psi |\,x'\rangle f(x)\delta(x-x') \langle x\, |\,\psi\rangle \end{align}

Next we can consider the case when $\hat{O}=f(\hat{p})$:

\begin{align}&=\int\mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle \langle x' |\,f(\hat{p})|\,x\rangle \langle x\, |\,\psi\rangle\\ & = \int\mathrm dp \int\mathrm dx \int\mathrm dx' \langle \psi |\,x'\rangle f(p)\langle x' |\,p\rangle\langle p\, |x\rangle \langle x\, |\,\psi\rangle \\& = \int \mathrm dp \int\mathrm dx \int\mathrm dx' \langle \psi |\,x'\rangle f(p)e^{-ip(x'-x)/\hbar} \langle x\, |\,\psi\rangle \\& = \int \mathrm dp \int \mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle f\left(-i\hbar\frac{\partial}{\partial x'}\right)e^{-ip(x'-x)/\hbar} \langle x\, |\,\psi\rangle \\ &= \int \mathrm dx \int \mathrm dx' \langle \psi |\,x'\rangle f\left(-i\hbar\frac{\partial}{\partial x'}\right)\delta(x-x') \langle x\, |\,\psi\rangle \\&= \int \mathrm dx' \langle \psi |\,x'\rangle f\left(-i\hbar\frac{\partial}{\partial x'}\right) \langle x'\, |\,\psi\rangle \end{align}

Hence we have again related the matrix form of the operator to its corresponding differential operator in the position representation. This process can be generalised for any product of operators $\hat{p}$ and $\hat{x}$ and so in general we see that:

$$\langle x' |\hat{O}|\,x\rangle=\langle x' |f(\hat{x},\hat{p})|\,x\rangle=f\left(x',-i\hbar\frac{\partial}{\partial x'}\right)\delta(x-x')\,.$$

  • $\begingroup$ Independent of this mathematically Is there any physical interpretation for $<x'|O|x>$ ?? $\endgroup$
    – asaa
    Commented Feb 10, 2016 at 10:11
  • $\begingroup$ @asaa Physical interpretation: maybe but meaning I would say no, which was the point I was making at the start it only really even has mathematical meaning as part of an integral. Only observable quantities have meaning and it is most certainly not an observable quantity. $\endgroup$
    – J.L.
    Commented Feb 10, 2016 at 11:12
  • 1
    $\begingroup$ if it was observable at was? $\endgroup$
    – asaa
    Commented Feb 10, 2016 at 11:20

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