By UriAceves


2019-03-11 09:00:50 8 Comments

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?

2 comments

@Qmechanic 2019-03-11 10:07:12

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.

@ZeroTheHero 2019-03-11 10:06:57

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.

@yuggib 2019-03-11 12:30:53

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).

Related Questions

Sponsored Content

0 Answered Questions

1 Answered Questions

[SOLVED] Dimensions in the Second Quantization of an Operator

0 Answered Questions

1 Answered Questions

[SOLVED] The central field approximation and the quantum number $n$

0 Answered Questions

0 Answered Questions

Notation in a paper on quantum mechanics and gravitation (2)

Sponsored Content