2
$\begingroup$

$\langle p | x \rangle=e^\frac{ipx}{\hbar}$ can pretty easily be derived from wave mechanics. But how can it be derived without resorting to it? I've seen it be derived from the translation operator $\hat T$. It is shown that for an infinitesimal displacement $\epsilon$ the translation operator in the position eigenbasis is: $$\langle x |\hat T |\psi\rangle=\left( 1+\dfrac{\partial}{\partial x}\right)\langle x | \psi\rangle=\left( 1+\dfrac{\partial}{\partial x}\right)\psi_x$$ In which $\psi_x$ is the state vector's representation in position space, such that $$|\psi\rangle=\int^{\infty}_{-\infty}\psi_x |x\rangle dx$$ And it then assumes that because momentum generates translation, $\frac{\partial}{\partial x}=\frac{i}{\hbar}\hat p$.

Now this leaves me rather unsatisfied. Is there any way to prove the form of the momentum operator, and from it prove $\langle p | x \rangle=e^\frac{ipx}{\hbar}$ using, for example, the matrix representation of quantum mechanics?

$\endgroup$

marked as duplicate by Chris, Cosmas Zachos, Art Brown, caverac, glS Feb 19 '18 at 10:23

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.