#### [SOLVED] Can Quantum Mechanical Potential have a Probability Distribution

I am currently in my second semester of undergraduate quantum mechanics. We have recently starting discussing two particle systems, usually in relation to spin interactions. In all of our calculations, we have assumed entirely non-interacting particles. This does not seem physical, especially if we are considering charged particles as most particles are.

Say we have two electrons in an infinite square potential well. Classically, there should be potential between the charged electrons. But, since the electrons are defined by wave functions rather than positions, we cannot use the standard $$kqq/r$$ formula for potential.

A quick Wikipedia search told me that an electron's charge density is proportional to its wave function. Can you use this to define some sort of probability distribution for the potential due to the uncertainty in the position of the electron? How would an uncertainty in potential manifest itself in the Schrodinger Equation? What would the probability density of two electrons in an infinite square potential well look like assuming the two electrons have some sort of potential between them? #### @flaudemus 2019-02-05 07:33:59

Can you use this to define some sort of probability distribution for the potential due to the uncertainty in the position of the electron?

A formal answer to your question: in quantum mechanics, the interaction potential between two particles $$V({\mathbf x}_1,{\mathbf x}_2)$$ is regarded as an operator. You can obtain a matrix representation of this operator by calculating the matrix elements $$V_{mn} = \int d^3x_1\int d^3x_2 \psi_m^\star({\mathbf x}_1,{\mathbf x}_2)V({\mathbf x}_1,{\mathbf x}_2)\psi_n({\mathbf x}_1,{\mathbf x}_2).$$ Here, the wave functions $$\psi_n({\mathbf x}_1,{\mathbf x}_2)$$ are a complete set of basis functions of your problem. You can now imagine that this operator of the potential (i.e., this matrix) has eigenvalues $$v_k$$ and eigenstates $$\psi_k({\mathbf x}_1,{\mathbf x}_2)$$. The eigenvalues $$v_k$$ are the possible values of the interaction potentials that could be measured. Each eigenvalue has a probability to be measured that is given by the overlap integral between the corresponding eigenstate of $$V$$ and the actual wave function $$\psi_\mathrm{el}({\mathbf x}_1,{\mathbf x}_2)$$ describing the state of the two electrons. Quantitatively, the probability $$p_k$$ to find the eigenvalue $$v_k$$ of the interaction potential is $$p_k = \left|\int d^3r_1 \int d^3r_2 \psi_k^\star({\mathbf x}_1,{\mathbf x}_2)\psi_\mathrm{el}({\mathbf x}_1,{\mathbf x}_2)\right|^2.$$ This is the 'probability distribution for the potential' that you were looking for. You can use it, to calculate the expectation value of the interaction potential, or the variance (the fluctuations) of the interaction potential like you do with any other statistical quantity. #### @anna v 2019-02-04 11:05:06

electron's charge density is proportional to its wave function.

This is an image of the electron orbitals, calculated from the wavefunction of the Hydrogen atom As a point particle, the electron is identified with the probability distribution ( which is what the orbitals are,) and thus one can say that it is also the probable charge density. BUT the wavefunction has been derived by using the coulomb potential in the Schrodinger equation. In the quantum mechanical frame , where nuclei and particles have to be described, the potential between the proton and the electron is already described, it makes no sense to go into this circular logic, mixing up the quantum frame with classical thinking of attraction.

Attraction in quantum mechanics is described in quantum field theory as exchanges of virtual photons, which in the limit of classical theory end up in building the coulomb potential.

Can you use this to define some sort of probability distribution for the potential due to the uncertainty in the position of the electron? How would an uncertainty in potential manifest itself in the Schrodinger Equation? What would the probability density of two electrons in an infinite square potential well look like assuming the two electrons have some sort of potential between them? #### @Chiral Anomaly 2019-02-04 04:55:35

A system of two interacting electrons (neglecting their spins for simplicity) can be described by the Schrödinger equation $$i\frac{d}{dt}\psi(t,\mathbf{x}_1,\mathbf{x}_2) =H\psi(t,\mathbf{x}_1,\mathbf{x}_2) \tag{1}$$ with the Hamiltonian $$H$$ $$H = \frac{\mathbf{P}_1^2}{2m_1} + \frac{\mathbf{P}_2^2}{2m_1} +V(\mathbf{x}_1-\mathbf{x}_2) \tag{2}$$ with $$\mathbf{P}_n=-i\hbar\nabla_n$$ being the momentum operator for the $$n$$-th electron, and $$\nabla_n$$ is the gradient with respect to $$\mathbf{x}_n$$. Here, $$V$$ is an ordinary function of its argument $$\mathbf{x}_1-\mathbf{x}_2$$. For a Coulomb interaction we can take this function to be $$V(\mathbf{x}_1-\mathbf{x}_2)\propto\frac{1}{|\mathbf{x}_1-\mathbf{x}_2|}. \tag{3}$$ Classically, we would interpret $$V$$ as describing the interaction between two point particles with locations $$\mathbf{x}_1$$ and $$\mathbf{x}_2$$, respectively. Quantum mechanically, that doesn't make sense, because the particles don't have definite locations; their state is described by the wavefunction $$\psi$$ instead. However, the offending part of the interpretation is the part that says "with locations $$\mathbf{x}_1$$ and $$\mathbf{x}_2$$." That's the part that doesn't make sense quantum mechanically. The function $$V$$ is still a perfectly well-defined term in the Hamiltonian, and we can still say that the function $$V$$ describes the interaction between the particles, because the presence of the function $$V$$ in the Schròdinger equation still implies that neither particle's time-evolution is independent of the other particle.

If we consider an initial wavefunction in which the two electrons do have relatively sharply-defined locations, then we can interpret $$V$$ approximately as the interaction between two point particles with locations. The approximation is because the locations are only approximately-defined, not because there is any approximation involved in the definition of $$V$$ itself. Over time, though, the particles' locations will spread, and this approximation will break down.