0
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Alfred Centauri, Danu, Kyle Kanos, Jim
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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:

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

Indeed:

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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