Let $\vec{A}$ be the magnetic vector potential and $\vec{p}$ be momentum.

$$ \vec{p} \cdot \vec{A} \psi = (\vec{p} \cdot \vec{A}) \psi + \vec{A} \cdot (\vec{p} \psi) $$ $$ \vec{A} \cdot \vec{p} \psi = (\vec{A} \cdot \vec{p}) \psi $$

This makes sense in coordinate representation, since $\vec{p}$ is differentiation which has a chain rule. $\vec{A}$ is dependent only on position, so it does not have a chain rule.

What if I decided to change my representation of the operators to momentum representation? Then $\vec{p}$ would not follow a chain rule, and instead I would find that the position dependent $\vec{A}$ would now have a chain rule (since it depends on derivatives of momentum).

This seems really weird to me, since the algebra of my operators (the chain rule in this example) now depends on my representation.

Perhaps this has always been the case and I just never realized it. Or it's due to messy notation. Does anyone know a truly representation independent way of expressing the above relations? I'm interested in this since I want to expand the Pauli equation into its standard form with a magnetic field (the Pauli equation actually contains cross products, but I'd like to stick with dot products here for simplicity).

  • $\begingroup$ No, I think you already have the representation independent way. The equations you wrote at the top are representation independent -- they are the 'algebra' of the operators. You can derive them in either position or momentum space, and the steps in the derivation will look different, but you'll get the same result. $\endgroup$
    – knzhou
    Commented Dec 3, 2015 at 22:28

1 Answer 1


The correct way is to consider $\vec A$ a function of the operator $\hat{\vec x}$. Then we have the ("abstract") operator $\hat{\vec p} \cdot \vec A(\hat{\vec x})$. (Operators with hats represent operators in the "abstract" Hilbert space).

Without concern for the messy details of the position and momentum operators (which technically do not have eigenfunctions, as their "eigenfunctions" are not in the Hilbert space), a function of an operator is defined as follows: $$ f(\hat{A}) = \sum_\lambda f(\lambda) \left|\lambda\right>\left<\lambda\right|. $$ where $\hat{A}\left|\lambda \right> = \lambda\left|\lambda\right>$.

Now there remains no problem, you can simply represent the equations in position space: \begin{align*} \langle x \rvert \hat{\vec p} \cdot \vec A(\hat{\vec x}) \lvert \psi \rangle &= \int d^3x'\, \langle x \rvert \hat{\vec p} \lvert \vec x' \rangle \cdot \langle \vec x' \rvert \vec A(\hat{\vec x}) \lvert \psi \rangle \\ &= \int d^3x'\, \langle x \rvert \hat{\vec p} \lvert \vec x' \rangle \cdot \vec A(\vec x') \langle \vec x'\vert \psi \rangle \\ &= \int d^3x'\, \delta(\vec x - \vec x') (-i \hbar \nabla) \cdot \vec A(\vec x') \langle \vec x'\vert \psi \rangle \\ &= -i\hbar \nabla \vec A(\vec x) \psi(\vec x) = \vec p \cdot \vec A \psi(\vec x). \end{align*} Here we used, that the position representation of a state vector is given by $\psi(x) = \langle x \vert \psi \rangle$ and of an operator by $\langle x' \vert \hat O \vert x \rangle = \delta(x - x') O$ (all operators are local).

It also works without problem in momentum space: \begin{align*} \langle \vec p \rvert \hat{\vec p} \cdot \vec A(\hat{\vec x}) \lvert \psi \rangle &= \vec p \cdot \langle \vec p \rvert \vec A(\hat{\vec x}) \lvert \psi \rangle \\ &= \vec p \cdot \int d^3p'\, \langle \vec p \rvert \vec A(\hat{\vec x})\lvert \vec p' \rangle\langle \vec p'\rvert \psi \rangle. \end{align*} Here it is more difficult to evaluate the matrix element, one way to approach this is to use the above formula for functions of operators: \begin{align*} \langle \vec p \rvert \vec A(\hat{\vec x})\lvert \vec p' \rangle &= \int d^3x \, \langle \vec p \vert \vec x \rangle \vec A(x) \langle \vec x \vert \vec p' \rangle \\ &= \frac{1}{2\pi\hbar} \int d^3x\, e^{i \vec x \cdot (\vec p' - \vec p)} A(\vec x) \\ &= \frac{1}{2\pi\hbar} \tilde A(\vec p - \vec p'). \end{align*} So the matrix element is the Fourier transform of the potential at the argument $\vec p - \vec p'$. This gives the result for the above term: $$ \langle \vec p \rvert \hat{\vec p} \cdot \vec A(\hat{\vec x}) \lvert \psi \rangle = \vec p \cdot \int \frac{d^3p'}{2\pi\hbar} \tilde{\vec A}(\vec p - \vec p') \psi(\vec p'). $$ In other words: In momentum representation the term gets a non-local convolution integral. (One can also rewrite this equation to make $\vec A$ a function of the operator $\vec x$ in momentum representation, which is a derivative in $\vec p$, this is however even less useful for explicit calculation).

Note, that the original question (associativity and product rule behaviour) of the operators simply vanishes in the abstract notation (there are to vectors operators whose scalar product is calculated and the result is applied to a state vector).


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