# Common observables and associated operators: operator momentum [duplicate]

Starting from my previous question Commutators in quantum mechanics and considering that the commutator

$$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar, \tag{1}$$ the associated linear operator momentum (for example the momentum $$p_x$$ of the $$x$$-axis) is:

$$p_x\longrightarrow -i\hbar\frac{\partial}{\partial x}=\frac{\hbar}{i}\frac{\partial}{\partial x} \tag{2}$$ The association of $$p_x$$ with $$-i\hbar\partial/\partial x$$ is it a postulate or exist a proof that $$\left[i\hbar\frac{\partial}{\partial x},x\right]=i\hbar\color{red}{\boldsymbol{\equiv}[p_x,x]\,\,?} \tag{3}$$

$$[\hat x, \hat{p}_x]=i\hbar$$

can be considered a postulate of quantum mechanics. In the position representation where wave functions are functions of position, and the position operator is just multiplication by $$x$$, the momentum operator then can be chosen as

$$\hat{p}_x=-i\hbar\frac{\partial}{\partial x}$$

in order to satisfy the commutation relation.

A more physical way to think about this choice for the momentum operator is to consider a plane wave, $$e^{i(k_xx-\omega t)}$$. Operating on this with a momentum operator should give the momentum egenvalue $$\hbar k_x$$, and it does with that choice for $$\hat{p}_x$$.

• Thank you very very much for your patience with me. Where is the into my question the minus sign error? Thank you again. – Sebastiano Aug 8 '19 at 11:34
• What you have in red would be $[-p_x,x]$ (or $[x,p_x]$ like I wrote). – G. Smith Aug 8 '19 at 17:07
• With honesty the doubt came to me (I do not want to refer to notes of any kind) otherwise we would not explain the (2) with the (3). The (2) has a minus sign, the (2) no. Always thank you and hi. Thank you for your friendliness. – Sebastiano Aug 8 '19 at 19:17