When we go from the classical many-body hamiltonian

$$H = \sum_i \frac{\vec{p}_i^2}{2m_e} - \sum_{i,I} \frac{Z_I e^2 }{|\vec{r}_i - \vec{R}_I|} + \frac{1}{2}\sum_{i,j} \frac{ e^2 }{|\vec{r}_i - \vec{r}_j|} + \sum_I \frac{\vec{p}_I^2}{2M_I}+ \frac{1}{2}\sum_{I,J} \frac{Z_IZ_J e^2 }{|\vec{R}_I - \vec{R}_J|}$$

to the quantum many-body hamiltonian

$$H = -\sum_i \frac{\hbar^2}{2m_e}\nabla_i^2 - \sum_{i,I} \frac{Z_I e^2 }{|\vec{r}_i - \vec{R}_I|} + \frac{1}{2}\sum_{i,j} \frac{ e^2 }{|\vec{r}_i - \vec{r}_j|} - \sum_I \frac{\hbar^2}{2M_I} \nabla_I^2+ \frac{1}{2}\sum_{I,J} \frac{Z_IZ_J e^2 }{|\vec{R}_I - \vec{R}_J|}$$

only the kinetic energy parts turn into operators. I mean the other parts are also operators but merely numbers.

Why is this the case? My guess is, it has to be with the representation we are working with, but that's as far as I go, I don't know how it affects.

Can someone give a heuristic explanation also?

  • 3
    $\begingroup$ The potential operator is actually: $\hat{V} \equiv V(x)\hat{I}$ with $\hat{I}$ the identity operator for the 1D case here. Often you omit the $\hat{I}$. $\endgroup$ Mar 11, 2019 at 14:57

2 Answers 2


Actually the potential is also an operator. It just so happens that, in the position representation, $\hat x\psi(x)=x\psi(x)$, so that the potential energy operator $V(\hat x)$ acts by multiplication: $V(\hat x)\psi(x)=V(x)\psi(x)$.

Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator.

In the momentum representation $\hat x$ acts by differentiation so in this case the potential energy operator becomes a (usually quite complicated) differential operator since one needs to use the formal expansion of potential to convert it to a polynomial.

  • 1
    $\begingroup$ Pseudodifferential operators are not complicated, since they are conveniently defined by Fourier transform. And acting on $L^2$, as in quantum mechanics, they can even be defined directly by spectral calculus. No power expansion (formal or not) of the function is needed, only measurability (essentially with respect to the Lebesgue measure). $\endgroup$
    – yuggib
    Mar 11, 2019 at 12:30
  • $\begingroup$ "Even though it acts “by simple multiplication”, it doesn’t commute with the momentum, a sign that it is still a legitimate operator." is incorrect, e.g. the identity operator is an operator and commutes with any other operator you like. You are confusing the operator with it's action on some specific basis, and these are different things. $\endgroup$ Jan 3, 2020 at 19:28
  • $\begingroup$ @user2820579 the point is perfectly clear and the basis is implied in the question. Yes of course if you want to be really formal about everything your argument is right on point but it’s difficult to imagine the OP had in mind such nuances as you highlighted when he or she formulated the question. $\endgroup$ Jan 3, 2020 at 22:00

OP's hunch is exactly right: In QM they are all operators. However each operator may become a multiplication operator in certain representations. E.g. position operators become multiplication operators in the Schrödinger position representation, while momentum operators become multiplication operators in the momentum representation, and so forth.


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