With $[x,p_x]=i\hbar $, how to determine the form of the operator $p_x$?


closed as off-topic by Alfred Centauri, Danu, Kyle Kanos, Jim, user10851 Jun 23 '14 at 13:59

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Well, it doesn't need to have any form before you choose a representation for the operators.

In general, any Hermitian operator $Q$ satisfying $[Q,\psi]={\cal F}$, is the generator of the transformation $\psi \rightarrow\psi + i\epsilon{\cal F} $.

Let's try to figure this out. Consider the action of the unitary operator $U(\epsilon)=e^{i\epsilon{\cal Q}}$ on the operator $\psi$, $$U(\epsilon)\psi U^{\dagger}(\epsilon)=\psi+i\epsilon{\cal F}+{\cal O}(\epsilon^2) $$ Expanding and comparing terms with $\epsilon$, you will get $[Q,\psi]={\cal F}$. That's why $Q$ is the generator of the transformation $\psi \rightarrow\psi + i\epsilon{\cal F} $.

In your case, $[x,p_x]=i\hbar$ means that $p_x$ is the generator of the transformation $x\rightarrow x+\epsilon \hbar$. If one chooses the coordinate representation, any function $f$ of the coordinate $x$ transform in the way, $f(x)\rightarrow f(x+\epsilon\hbar) $

Thus you can write an explicit form of the transformation in this way, $$U(\epsilon)f(x) U^{\dagger}(\epsilon)=f(x+\epsilon\hbar)= e^{\epsilon\hbar\frac{d}{dx}}f(x)=e^{-i*i\epsilon\hbar\frac{d}{dx}}f(x) $$ where $U(\epsilon)=e^{i\epsilon p_x}$. In the last expression, you can identify $p_x$ in coordinate representation is $-i\hbar \frac{d}{dx}$.

Of course, $[x,-i\hbar \frac{d}{dx}]=i\hbar$.

I hope that this helps you understanding the operator language.

  • $\begingroup$ It cannot work since also $p_x = -i \hbar \frac{d}{dx} + g(x)$, with every given function $g$ verifies the initial commutation relation... $\endgroup$ – Valter Moretti Jun 23 '14 at 13:59

Using Fourier analysis, and setting $\hbar$ to 1 (I leave it to you to reintroduce it consistently using dimensional analysis), we have $$ f(x) =\int \tilde f(k) e^{ikx} dk\\ xf(x) = \int i \frac{\partial \tilde f(k)}{\partial k} e^{ikx} dk $$ where we used integration by part. Applying the commutation relation holds $$ i (\frac{\partial( p_x[\tilde f(k)])}{\partial k} -p_x[\frac{\partial \tilde f(k)}{\partial k}]) = i\tilde f(k) $$ where $p_x[]$ means the operator applied to the inside of the bracket. Here its representation in $k$ space. As the right hand side of this equation contains no derivatives and that $f$ is arbitrary, derivatives must cancel on the left hand side.

$p_x[g(k)]=kg(k)$ is obviously solution. This in turn implies its definition in real space:

$$ p_x f(x)=-i\frac{\partial }{\partial x} f(x) $$


$$ -i\frac{\partial }{\partial x} f(x) = \int k \tilde f(k) e^{ikx} dk $$

  • $\begingroup$ It cannot work since also $p_x = -i \hbar \frac{d}{dx} + g(x)$, with every given function $g$ verifies the initial commutation relation... $\endgroup$ – Valter Moretti Jun 23 '14 at 13:59
  • $\begingroup$ @V.Moretti Of course $[g(x),x]=0$, however this is only true for smooth, i.e. Taylor expandable $g$. I think dimensional consistency would require to introduce some length $L$, which contradicts the general postulate of commutation relations, i.e. the fact that they are problem independent. $\endgroup$ – Frédéric Jun 23 '14 at 14:21
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    $\begingroup$ Well, actually $[g(x),x]=0$ holds true for every measurable function, even if it is nowhere continuous. Moreover if one wants, as necessary, that $p_x$ is self-adjoint, he should study carefully its domain. For instance any space of smooth functions (not necessarily analytic [you are making confusion between analytic and smooth]) is not a self-adjointness domain of $p_x$ (with $g=0$). The self-adjointness domain, for $g=0$, is the Sobolev space $H^1$. The procedures you exploited in your answer and in mastrok's one have just an heuristic (however important) value. $\endgroup$ – Valter Moretti Jun 23 '14 at 15:01