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I was looking at the hamiltonian of a particle confined to the $x$-$y$ plane when it has mass $m$ and charge $q$ coupled to the electromagnetic field. My question is actually a very simple one. During the calculation of the hamiltonian I get $$1/2m (p_x ^2 +(p_y -qBx)^2) $$where the $x$ is NOT an operator and $p_y$ and $p_x$ are the momentum operators along the y and x directions. I only care about focusing on this particular part here $(p_y-qBx)^2$ because this equals $p_y^2 -qBxp_y-p_y(qBx)+(qBx)^2$ and the interesting part are the two middle terms where my lecturer has simplified that bit into $-2qBxp_y$ but that would assume $p_y(x)=xp_y$ but in terms of a partial derivative (since the momentum operator is a differential operator) shouldn’t $p_y(x)$ actually be 0? I’m guessing either a) the lecturer is wrong or b) it’s not 0 because it should be thought of as an operator

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    $\begingroup$ You are doing quantum mechanics here, right? That's what the post tag says. In that case, how come you write "x is NOT an operator". How can that be? $\endgroup$
    – Marius Ladegård Meyer
    Commented Nov 17, 2021 at 8:22
  • $\begingroup$ @MariusLadegårdMeyer This is because our magnetic field=curl of vector potential where the vector potential is denoted by A. Now A is defined as the vector (0, xB, 0) with x being a variable rather than an operator at least that’s how I see the question because the lecturer did not write a hat symbol above the x but confusingly at the same time x operator is just defined as x in quantum mechanics. $\endgroup$
    – Captain HD
    Commented Nov 17, 2021 at 8:26
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    $\begingroup$ Okay, this seems like a quantum Hall type question, in which both x and p are certainly operators that satisfy the usual commutation relations. So the x should be an operator. $\endgroup$
    – Marius Ladegård Meyer
    Commented Nov 17, 2021 at 9:17
  • $\begingroup$ @MariusLadegårdMeyer Even if we assume x is an operator the main confusion here is whether or not x and p_y commute. $\endgroup$
    – Captain HD
    Commented Nov 17, 2021 at 9:56
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    $\begingroup$ They do. Remember, the whole operator is supposed to act on a wave function, and $(\partial / \partial_y) x \psi(x,y) = x (\partial / \partial_y) \psi(x,y)$. $\endgroup$
    – Marius Ladegård Meyer
    Commented Nov 17, 2021 at 10:12

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