I am currently trying to understand a set of lecture notes, where the notation is very poorly defined, unfortunately. In a "proof" that canonical quantisation works, the following Hamiltonian (operator) is defined: \begin{equation} H = e\Phi + \frac{(\mathbf{p}-e\mathbf{A})\cdot (\mathbf{p}-e\mathbf{A})}{2m} \end{equation} where $\mathbf{A}$ is a vector potential and $\mathbf{p} = -i\hbar\nabla$. Then, Hamilton's equations are being applied: \begin{aligned} \dot{\mathbf{x}} &= \frac{\partial H}{\partial \mathbf{p}} \stackrel{(*)}{=} \frac{1}{m}(\mathbf{p}-e\mathbf{A}) \\ \dot{\mathbf{p}} &= -\frac{\partial H}{\partial \mathbf{x}} \stackrel{(*)}{=} -e\frac{\partial \Phi}{\partial \mathbf{x}} + \frac{e}{m}(\mathbf{p}-e\mathbf{A}) \cdot \frac{\partial}{\partial \mathbf{x}} \mathbf{A} \end{aligned}

where the dot product in the final term is between the outer two vectors, in a sad indictment of the notation.

I do not really understand the steps marked with (*).

For the first equation, I get how to do this if you treat the $\mathbf{p}$ as a scalar and the derivative as a "scalar" derivative. However, the result would be different if I treat it as a vector, and the derivative as the corresponding divergence. But even then, it is clearly an operator, so how do I think about this correctly? For the second equation, I don't really have an idea about what's going on.

I hope this question is suitable for Physics SE.

  • 4
    $\begingroup$ Are you sure this isn't supposed to take place prior to the quantization? It doesn't make a lot of sense to define ${\bf{p}}=-i\hbar \nabla$ and then to turn around and start talking about $\dot{\bf{p}}$. $\endgroup$
    – J. Murray
    May 11, 2018 at 13:24
  • $\begingroup$ The notes do talk about quantization first, and after that this calculation appears, without any notice that this happens prior to quantization. But it makes a lot more sense, you are right! Thanks! The issue with the second equation will probably be resolved using index notation in that case. $\endgroup$
    – scaphys
    May 11, 2018 at 13:45
  • $\begingroup$ Which lecture notes? $\endgroup$
    – Qmechanic
    May 11, 2018 at 15:52
  • $\begingroup$ They are unpublished lecture notes on Quantum Mechanics. $\endgroup$
    – scaphys
    May 11, 2018 at 16:00

1 Answer 1


Agreed, the notation is confusing.

Strictly speaking, there is no such a thing as a "derivative wrt an operator". Here the instruction is simply to take the derivative of $H=H(p ,q)$ as if it was a classical function of $p$ and $q$, and then set $p$ and $q$ equal to the corresponding operators in the Hilbert space.

Without entering into a boring discussion (basicly on notation), the physical message is that the time-derivatives of the quantum operators in the Heisenberg picture, that is: $$\dot q \equiv \frac{i}{\hbar}[H,q],\qquad \dot p \equiv \frac{i}{\hbar} [H,p],$$agree with the RHSs of the hamiltonian equations of motion. More precisely, one has the Ehrenfest theorem: $$m \ddot q = e [E+\frac{1}{2}(\dot q \wedge B - B\wedge \dot q)],$$ where all objects are operators.

A boring discussion. Now, the notation may be somewhat clarified as follows. If you consider an arbitrary polynomial $P(p)$ in the $p$s, then you can easily prove that: $$[q,P]=i\hbar\frac{\partial P}{\partial p},$$where again the RHS is defined by taking the usual derivative of $P$ with respect to $p$, and then setting $p$ equal to the momentum operator. This follows from the (trivial) identity: $$[A,BC]=[A,B]C+B[A,C].$$

To understand what's going on, notice that the operation $X\mapsto [A,X]$ satisfies the Leibnitz rule of ordinary differentiation:$$\text d(fg)=(\text d f)g+ f(\text d g).$$ In particular, for an operator $B$ satisfying $[A,X]=1$, this implies that $A$ acts on polynomials in $X$ as an ordinary derivative.

Things are however a bit more complicated for operators involving a product of $p$s and $q$s. Consider, for instance: $$F=pqp,\qquad G=qp^2.$$ Classically, these functions are the same, $F=G$. However: $$\frac{1}{i\hbar}[q,F]=qp+pq,\qquad \frac{1}{i\hbar}[q,G] = 2 qp.$$

It is clear that if you don't make any particular conventions for the ordering of $p$s and $q$s, you can't naively equate a commutator with the corresponding derivative.

  • $\begingroup$ Thank you very much! Even if J. Murray is probably right in his comment, this was helpful. $\endgroup$
    – scaphys
    May 11, 2018 at 14:53
  • $\begingroup$ Glad to help :-) If I understand correctly J. Murray's comment, he is pointing out that the position $\boldsymbol p = -i\nabla$ refers to a Schrödinger picture, while the time dependence in operators refers to the Heisenberg picture. To be precise, the Ehrenfest theorems refer to the operators: $$\mathbf x (t) = e^{iHt}\mathbf x e^{-iHt},\qquad \mathbf p (t) = e^{iHt}(-i\nabla)e^{-iHt}$$(I'm assuming that the electromagnetic potentials do not depend explicitly on time) $\endgroup$
    – pppqqq
    May 12, 2018 at 8:45
  • $\begingroup$ If I understood correctly, he meant that the Hamiltonian is supposed to be the Hamiltonian function of a classical particle, which is used to demonstrate that it leads to the correct equations of motion (i.e. Lorentz force, ...). This Hamiltonian can then be turned into a QM operator using canonical quantization. $\endgroup$
    – scaphys
    May 12, 2018 at 18:04

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