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When we solve the single coordinate Schrödinger equation,

\begin{equation} i \hbar \partial_t \psi = - \frac{\hbar^2}{2 \ m} \ \nabla^2 \psi \ + \ V(x) \ \psi, \tag{1} \end{equation}

we imply the potential $V(x)$ is exactly defined everywhere on $x$ with no uncertainties, i.e. a classical $V(x)$, despite the fact that the equation we are solving is quantum. It seems, there is no inconsistency in doing so as long as we can assume by definition of the problem that the $V(x)$ in the problem is classical.

But when we turn to the two-body problem, $V(x)$ is generated by a particle which admits an uncertainty in its position. In this case we should no longer be able to use the classical potential as in (1).

Yet in textbooks the Hamiltonian for the Schrödinger equation that describes the two-body problem is given as

\begin{equation} H = -\frac{\hbar^2}{2 m_1}\nabla^2_1 -\frac{\hbar^2}{2 m_2}\nabla^2_2 + \frac{q^2}{4 \pi \epsilon_0 | x_1 - x_2|}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \end{equation}

When we define the potential V as in (2) we describe the electric field of a charged particle as $E = q/(4 \pi \epsilon_0 r^2)$, using an expression from classical electrodynamics. This expression does not admit of an uncertainty in the position of the particle which generates the field, it implies that the particle generating the field is infinitely precisely localized at a point and that the field goes as $\sim r^{-2}$ from the infinitely precisely localized center of the particle.

So the question is:

(1) is the potential in the Hamiltonian (2) correct after all, and why?

(2) if it is not correct, then what should be the correct potential?

(3) will the Hamiltonian be exactly separable with the correct potential?

(4) in light of the above, is it even possible, at least in the static limit (which I define in the additional note), to correctly describe the two-body problem with a Schrödinger equation?

**** Note: I edited out all non-question text.

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    $\begingroup$ The third term in eq. $(2)$ is an interaction term between both particles (there still can be an external potential). There is a priori no problem with this. But of course, as always in physics, you have to compare the results of a theoretical prediction with experiments. For (most/many) non-relativistic purposes, the form of eq. $(2)$ suffices. You arrive at this interaction term by the "usual" canonical quantization: Replace $x$ by $\hat x$... $\endgroup$ Feb 18 at 11:03
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    $\begingroup$ The potential $\frac{q^2}{4\pi \epsilon_0 |X_1-X_2|}$ is actually an operator depending upon position operators of the two particles. In the position basis it reduces to an ordinary function. If the state of the two particle system is not a position eigenstate then the expectation value of potential operator in such a state will be of different form than the classical potential. $\endgroup$
    – user10001
    Feb 18 at 12:00
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    $\begingroup$ Please do not use edits to respond directly to comments; Adding information in response to comments is fine, carrying a conversation via edits is not. Also, you should always be able to post comments on your own questions, please try again. $\endgroup$
    – ACuriousMind
    Feb 18 at 12:54
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    $\begingroup$ > "This expression does not admit of an uncertainty in the position of the particle which generates the field, it implies that the particle generating the field is infinitely precisely localized at a point" This is how quantum theory works, the simplest models of atoms/molecules work with point charged particles and Hamiltonians are for point charged particles, thus $V$ is defined on configuration space of a set of point particles. The uncertainty you miss comes through $\psi$, which is not part of $V$ or $\hat{H}$, but it is part of the Schroedinger equation. $\endgroup$ Feb 18 at 19:16
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    $\begingroup$ Stop editing the question for communication purposes. $\endgroup$ Feb 18 at 21:30

2 Answers 2

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This expression does not admit of an uncertainty in the position of the particle which generates the field, it implies that the particle generating the field is infinitely precisely localized at a point [...]

The same is true of $V(x) = x^2$, the harmonic oscillator potential, if you assume that it is the potential energy of a particle at position $x$. Of course, quantum mechanical particles are not localized to a single point in space, which is reflected by the fact that the operator $\hat V(X) = X^2$ does not have any normalizable eigenfunctions.

The two body problem is precisely the same - it is an operator $\hat V(\mathbf R_1,\mathbf R_2)$ where $\mathbf R_i = (X_i,Y_i,Z_i)$. Its form is determined by analogy to the classical potential $V(\mathbf r_1,\mathbf r_2)$, in the same way as the harmonic oscillator. The only meaningful difference is that there are now more coordinates to worry about.

(1) is the potential in the Hamiltonian (2) correct after all, and why?

(2) if it is not correct, then what should be the correct potential?

(3) will the Hamiltonian be exactly separable with the correct potential?

(4) in light of the above, is it even possible, at least in the static limit (which I define in the additional note), to correctly describe the two-body problem with a Schrödinger equation?

It depends what you mean by correct. If you're asking whether it describes the $1/r$ interaction between two non-relativistic point particles, then yes. It neglects radiation, relativistic effects, spin-orbital coupling, and a host of more subtle quantum effects like vacuum polarization, but it's a perfectly reasonable idealization.

You are inserting quite a lot of metaphysics on top of the formalism which ultimately the cause of your confusion. I don't particularly understand why you're comfortable letting $V(X)$ be the potential for one particle despite quantum states necessarily having a non-zero uncertainty in the value of the observable $X$, but are not comfortable letting $V(X_1,X_2)$ be the potential for two particles because quantum states necessarily have a non-zero uncertainty in the values of $X_1$ and $X_2$.

You seem to be getting turned around thinking about the potential as being generated by one particle, and thinking that somehow there must be some uncertainty which isn't being taken into account. I would suggest thinking of the potential as merely a function of both positions, at which point the uncertainty in those positions has the same character as the uncertainty in a single position for a single-body problem.


As far as separability goes, you are talking about re-factoring the Hilbert space into center-of-mass coordinates. When you do this, the Hamiltonian can indeed be separated into the form $H = H_{cm} \otimes \mathbf 1 + \mathbf 1 \otimes h$, where $h$ is the standard single-particle Hamiltonian for the $1/r$ potential centered at the origin.

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  • $\begingroup$ The fact that coordinate operators do not have normalizable eigenfunctions is due to assuming the domain of coordinates to be a continuous space. If the domain of coordinates was a discrete lattice, coordinate operators would have normalizable eigenfunctions, but this would not imply the particles were localized to a single point of space. These are two different things. $\endgroup$ Feb 18 at 19:12
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    $\begingroup$ @JánLalinský It would imply that they could be localized to a single point in "space" (or rather, a single point on the lattice), and therefore there exist states with no positional uncertainty. The fact that particles on a continuous domain cannot be localized in this way implies the existence of an inherent uncertainty in position for all states, which is the point I was trying to make. $\endgroup$
    – J. Murray
    Feb 19 at 2:12
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When we solve the single coordinate Schrödinger equation...

we imply the potential $V(x)$ is exactly defined everywhere on $x$ with no uncertainties...

The potential is taken to be a well-defined function of space. This means that for every point $x$ there is a single value $V(x)$. This is just the definition of a function.

When the function "acts on" a quantum particle, the function is evaluated "at the position" of the particle, but in quantum mechanics "at the position" really means that we evaluate the function by inserting the position operator: $$ V(x) \to V(\hat X)\;, $$ where $\hat X$ is the position operator of interest. These two things are not the same. The first is a real number, whereas the second is an operator on a Hilbert space.


Update (to briefly address the problem in terms of convolution with the wavefunction rather than operators): In order to return to functions, rather than dealing with operators on a Hilbert space, one can act on Hilbert space vector $|\psi\rangle$ and project onto the position space basis $|x\rangle$ in the Hilbert space like: $$ \langle x|V(\hat X)|\psi\rangle = V(x)\langle x|\psi\rangle = V(x)\psi(x)\;, $$ where now we have returned to a real valued function $V(x)$, but it is also multiplied by another complex-valued function $\psi(x)$.

In order to understand certain aspects of the uncertainty associated with $V(\hat X)$ one could look at the expectation value of $V(\hat X)$: $$ \langle \psi |V(\hat X)|\psi\rangle =\int dx |\psi(x)|^2 V(x)\;, $$ where again we have returned to the problem of dealing with functions to and from real numbers (i.e., $V(x)$ and $|\psi(x)|^2$), but we have to convolve those functions to understand the uncertainty: $$ \delta V^2 \equiv \langle \hat V^2\rangle - \langle \hat V\rangle^2 $$ $$ = \int dx |\psi(x)|^2V(x)^2 - \left(\int dx |\psi(x)|^2 V(x)\ \right)^2$$


But when we turn to the two-body problem, $V(x)$ is generated by a particle...

First, be careful, you seem to be potentially confusing an external potential field with an interaction energy between two quantum particles.

In both cases we use the correspondence principle, but in the latter case we substitute both positions arguments with two different position operators, whereas in the former case we substitute only one.

An example may be helpful. Classical the interaction between an electron and a nucleus is: $$ V_{ei}(\vec x,\vec y) = \frac{-Ze^2}{|\vec x - \vec y|}\;.\tag{1} $$ Often, we only treat the electron as a dynamical quantum particle (this is basically the Born-Oppenheimer approximation). This means that the location of the nucleus is taken to be fixed and equal to a set of real numbers (and usually equal to zeros if there is only one nucleus, i.e., the origin of the coordinate system), but we replace the position of the electron with the electron's position operator: $$ V_{ei}(\vec x, \vec y)\to V_{ei}(\vec{ \hat{X}}, 0) = \frac{-Ze^2}{|\vec{ \hat{X}}|} $$

OTOH, when we look at the interaction energy between two electrons, we can not use the Born-Oppenheimer approximation. We still start from basically the same equation as Eq. (1), except that the charge of both particles is $-e$ and so we have: $$ V_{ee}(\vec x,\vec y) = \frac{e^2}{|\vec x - \vec y|}\;.\tag{2} $$ Since both the electrons have to be treated dynamically we substitute both position arguments with the electron position operators: $$ V_{ee}(\vec x,\vec y) \to V_{ee}(\vec{\hat X}, \vec{\hat Y}) = \frac{e^2}{|\vec{ \hat{X}} - \vec{ \hat{Y}}|} $$

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