# Why does $\langle x' | x \rangle$ give the Dirac delta distribution? [duplicate]

I'm having difficulty understanding why the following is true:

$$\int_\mathbb{R} \langle x' | x \rangle dx = \int_ \mathbb{R} \delta(x-x')dx$$

where $$\delta(x)$$ is the delta distribution. Are we taking this to be true because we require a normalized basis for the Hilbert space? Or can this relation be shown to be true from some other postulate in QM?

I understand that if we assume it's true then the following works out:

$$\psi(x) = \langle x|\psi\rangle = \int_\mathbb{R} \langle x | x'\rangle\langle x' | \psi\rangle dx' = \int_\mathbb{R} \delta(x'-x)\psi(x') dx' = \psi(x)$$

## marked as duplicate by Qmechanic♦ quantum-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 24 '18 at 3:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.