In my course we have just derived the momentum operator in the position basis to be the first-order derivative of the position. This was achieved by using a first-order taylor expansion:

$$ \psi(\boldsymbol r-d\boldsymbol r)\approx\psi(\boldsymbol r)-\nabla\psi(\boldsymbol r)\cdot d\boldsymbol r $$

The same derivation can also be found on wikipedia.

Now my question is whether the final result, $\boldsymbol p = -i\hbar\nabla$, is an approximation? What if we were to use a second order term in the taylor expansion, would that change anything?

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    $\begingroup$ In this sense, momentum generates infinitesimal changes in position. This is not an approximation - the approximation comes in when you want to relate two positions that are not infinitesimally close to each other by only considering the momentum operator in the Taylor series. $\endgroup$ Nov 18, 2021 at 16:08

1 Answer 1


No, it's not an approximation. There are a number of ways in which one might define the momentum operator, and one of those ways is as follows. In the following I set $\hbar = 1$ for convenience.

If you have a family of unitary operators $U_\lambda$ indexed by a continuous variable $\lambda\in \mathbb R$ which has the properties that $U_0 = \mathbb I$ (the identity operator) and that $U_\lambda \circ U_\rho = U_{\lambda + \rho}$, then it turns out that you can write $U_\lambda = \exp[-i \lambda A]$ where $A$ is a self-adjoint operator which is called the infinitesimal generator of the family of operators $U_\lambda$. In order to extract $A$ from $U_\lambda$, we use a difference quotient$^1$: $$A\psi(x) := i\lim_{\lambda\rightarrow 0} \frac{\big(U_\lambda \psi\big)(x)- \psi(x)}{\lambda}$$

In this particular case, we let $\big(U_\lambda\psi\big)(x) = \psi(x-\lambda)$, which means that $U_\lambda$ are the spatial translation operators$^2$ which shift the wavefunction by $\lambda$; we then find that $$A\psi(x) := i\lim_{\lambda\rightarrow 0} \frac{\psi(x-\lambda) - \psi(x)}{\lambda} = -i \psi'(x) \tag{$\star$}$$ and therefore that $A = -i \frac{d}{dx}$.

This approach is formally equivalent to writing $U_\epsilon \approx 1 -i \epsilon A$ and expanding $U_\epsilon \psi$ to first order in $\epsilon$. This yields $$ \big(U_\epsilon \psi\big)(x) = \psi(x-\epsilon)$$ $$\implies \big([1 - i\epsilon A]\psi\big)(x) + \mathcal O(\epsilon^2)=\psi(x) - \epsilon \psi'(x) + \mathcal O(\epsilon^2) $$ $$\implies \big(A\psi\big)(x) = -i \psi'(x)\tag{$\star\star$}$$ just as before.

The main point here is the relationship between families of unitary operators $U_\lambda$ and their corresponding self-adjoint infinitesimal generators $A$. Given a family $U_\lambda$, $A$ is technically defined via a difference quotient as in $(\star)$, but can also be obtained by expanding $U_\lambda$ to first order, as in $(\star\star)$. In the latter case, one might say that $A$ is defined to be the operator in the first-order term when one expands $U_\epsilon$ about $\epsilon=0$; though the first order expansion is an approximation to $U_\epsilon$, the identification of $A$ as the coefficent of the first-order term is not.

$^1$It's important to note that for a generic $\psi$, this difference quotient does not necessarily converge to a well-defined limit; in such cases, the domain of $A$ is defined to be those $\psi$ in the Hilbert space such that the difference quotient does have a well-defined limit.

$^2$In Hamiltonian mechanics, the momentum observable is the infintesimal generator of spatial translations (see e.g. here). Therefore, we define the momentum operator in quantum mechanics to be the infinitesimal generator of spatial translations by way of analogy.


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