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Please excuse the lengthy question. It involves an interesting controversy which has arisen in discussions on this site:

The mass of one atom of $^1H$ is well known to be slightly less than the mass of a proton plus the mass of an electron.

Let's work in terms of rest energies rather than masses, because the numbers will be more familiar.

The rest energy of a hydrogen atom in its ground state is 938,783,066.5 eV. (Sources: Wikipedia's "Isotopes of Hydrogen" for the value in atomic mass units; Wikipedia's "Atomic mass unit" for the conversion between amu and MeV.)

The rest energy of a proton is 938,272,081.3 eV (Source: Wikipedia's "Proton").

The rest energy of an electron is 510,998.9 eV (Source: Wikipedia's "Electron")

Thus the rest energy of a hydrogen is less than the rest energy of a proton and electron by 13.7 eV.

My explanation for this, which is the very conventional one I learned at MIT and Stanford, is simple: The rest energy of a system is its total energy, consisting of the rest energy of its constituents, plus their kinetic energy, plus the potential energy of their interaction. Therefore

$$\tag{1}E_{0,H} = E_{0,p} + E_{0,e} + \langle K\rangle + \langle U\rangle$$

where $\langle K\rangle$ is the expectation value of the kinetic energy,

$$\tag{2}\langle K\rangle = \left\langle-\frac{\hbar^2}{2\mu}\nabla^2\right\rangle = \frac{1}{2}\mu c^2\alpha^2 = 13.6\;\text{eV}$$

and $\langle U\rangle$ is the expectation value of the electrostatic potential energy between the proton and electron,

$$\tag{3}\langle U\rangle = \langle -\frac{k e^2}{r}\rangle = -\mu c^2\alpha^2 = -27.6\;\text{eV}$$

For the measured precision of the rest energies, it suffices to use nonrelativistic quantum mechanics in the form of the Schrodinger equation with the potential energy $U(r)=-ke^2/r$. The expectation values are for the ground 1s state.

In the above equations, $\hbar$ is the reduced Planck constant, $c$ is the speed of light, $k$ is the Coulomb constant, $e$ is the magnitude of the charge of the proton and electron, $\alpha=ke^2/\hbar c$ is the fine structure constant, and $\mu=m_p m_e/(m_p+m_e)$ is the reduced mass of the system.

Obviously this calculation explains the observed rest mass of hydrogen to within experimental error.

However, it relies on the fact that we have assumed (as everyone conventionally does) that the potential energy is well-defined as

$$\tag{4}U = -\frac{ke^2}{r}$$

and goes to zero at infinity.

Some people in this forum have disputed this on the basis that the electrostatic potential is not gauge-invariant and assuming that it goes to 0 at infinity is merely a convention. This raises the question of what is the correct gauge-invariant calculation of the mass of hydrogen?

Some people in this forum have claimed that the invariant mass $m=\sqrt{E^2-p^2}$ (in units with $c=1$) is not a gauge invariant concept. This seems absurd to me. If it were true, why would we say that the mass of a proton, or anything else, is a particular number?

Some people in this forum have claimed that the kinetic energy contributes to the rest energy but the potential energy does not. This might be true if one moves to considering electrostatic energy as being field energy. (For example, the energy-momentum-stress tensor for a particle in an electromagnetic field separates into a "pure particle" term involving only rest energy and kinetic energy, plus a "pure field" term representing the interaction energy,) But the field energy for point particles diverges and requires renormalization, so how exactly does one get 938.783,066.5 eV for the mass of a hydrogen atom?

Some people in this forum have claimed that we cannot define mass without defining the energy of the vacuum and "boundary conditions". This seems to ignore the fact that we can measure mass simply using force and acceleration, under non-relativistic conditions.

My conventional explanation above for the mass of hydrogen has actually been downvoted in other threads multiple times as being simply "wrong". I challenge the downvoters to provide an alternate calculation.

So my main question is: Can one make a gauge-invariant calculation of the mass of a hydrogen atom, and, if so, how exactly?

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    $\begingroup$ You should at least link to the question whose answer you are unhappy with, rather than expecting people to go looking. As it stands this reads as something of a rant along the lines of "I don't like what I was told and I'm going to keep asking until I get the answer I want.". $\endgroup$ – StephenG Dec 1 '18 at 22:23
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    $\begingroup$ I am not looking for "better answers" to some question I have asked, because I have not previously asked any questions; I've only answered them. The comments challenging some of my answers have gotten too long and we need a place to resolve the issues. The original questions I was answering are not particularly relevant, but I am happy to provide them if you would like. $\endgroup$ – G. Smith Dec 1 '18 at 23:05
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    $\begingroup$ I am trying to follow the reasoning here but having a hard time. I agree with Stephen that links to all of the previous discussions would be helpful. For example, I don't see how gauge dependence comes into this discussion. In your comments here: physics.stackexchange.com/questions/444248/… with Cham he says explicitly that rest mass is a gauge-independent property. Issues of boundary conditions are, as far as I understand, a separate thing. $\endgroup$ – Rococo Dec 1 '18 at 23:57
  • $\begingroup$ Just a comment on the rest-mass formula for a point particle : $m_0 = \sqrt{E^2 - p^2}$ ($c = 1$ here). The $E$ that appears there does NOT contain any potential energy in it. It's the relativistic momentum in the "time dimension" : $E \equiv p^0 = E_0 + K$ (where $E_0 \equiv m_0$ is the rest-energy and $K$ is the kinetic energy). You may write $m_0 = \sqrt{(\mathcal{E} - U)^2 - p^2}$ if you wish, where $\mathcal{E} = E + U$ is a gauge-dependant expression of "total energy" and $U$ is the gauge-dependant potential energy ($E = \mathcal{E} - U$ is gauge-independant but frame-dependant!) $\endgroup$ – Cham Dec 2 '18 at 1:17
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    $\begingroup$ This question arose from discussions in physics.stackexchange.com/questions/443105/… and physics.stackexchange.com/questions/444248/…. $\endgroup$ – G. Smith Dec 2 '18 at 5:11
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As I said before, energy is not a gauge invariant quantity - loosely speaking it is "energy differences" that are, as explained in the field theory textbook I linked to you in an earlier thread.

When these neat arguments about adding masses/energies are taught in undergrad, obviously the correct thing for professors to do is gloss over the subtleties of correctly defining gauge invariant quantities. The argument you've given for the mass of the hydrogen atom is fine actually, it's just that all the things there are strictly speaking energy differences.

Firstly, it is perfectly legal to shift the Hamiltonian by a constant, and therefore formally shift the ground state energy of the hydrogen atom to be whatever you want. But recall that there are both bound state solutions to the coulomb problem, and continuum (or "scattering") states. A gauge invariant statement that will stay the same no matter what you do is: "the gap between the highest energy bound state and the first continuum state is 13.6 eV."

So no matter how you define the absolute values of the energy levels themselves, it's a gauge invariant fact that you've got 13.6 eV less when you're bound than when you're free.

Regarding the masses of the proton/electrons themselves, this is a slightly more subtle point. The reason it's confusing is that "an electron" and "a proton" are actually not gauge invariant objects, because they are charged. Creating a proton/electron out of the ground state (without worrying about the "true" microscopic origins of protons, which only show up at high energies - let's just imagine they're positive test charges) requires you to create an electon-positron (or proton-antiproton) pair. The 0.5 MeV is half the energy of an electron-positron pair, and this is also a gauge invariant quantity.

So we are summing two energy differences. The first is the difference in energy between having a single electron/proton, and having no particles. The second is the difference in energy between those two particles being bound, and them being free. Both quantities are gauge invariant: the first gives us the masses of the proton and electron, the sum of them gives us the mass of the hydrogen atom.

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  • $\begingroup$ Okay, but (as you mention in one of the other discussions) any discussion of this issue without GR is incomplete, because (as far as I understand it) that completely changes the answer. $\endgroup$ – Rococo Dec 4 '18 at 5:10
  • $\begingroup$ Well, OP was asking about the Hydrogen atom which has nothing to do with GR, so that's the question I answered. As to whether GR "completely changes things", the answer is yes and no. If the metric is dynamical then you can measure the trace of the energy momentum tensor. But you lose energy conservation, because time translation is no longer a symmetry - it's gauged. I don't think there's any sense in which you can speak of the energy of "an electron" or "an hydrogen atom" as opposed to the trace of the energy momentum tensor. $\endgroup$ – user213887 Dec 4 '18 at 14:22
  • $\begingroup$ So you say that $m_0 = \mathcal{E} = m_p + m_e + K_e + U$ is actually made of energy differences (relative to some vacuum), and could be re-written as $m_0 = \Delta\mathcal{E} = m_p + m_e + K_e + \Delta U$ instead, is that it? This makes a lot of sense to me. $\endgroup$ – Cham Dec 4 '18 at 17:23
  • $\begingroup$ Yeah that's exactly right. The most ambiguous looking thing there is probably the "$K_e+U$" ($=H$) business, which is secretly "$\Delta H$" between the lowest energy unbound state and the highest energy bound state, ie the difference in energy between binding the electron and proton and having them float around free. (I think it's much clearer that rest mass is a gauge invariant concept and defines the energy difference between having a particle and not having one, or if you want to think about it kinematically the energy difference between the thing moving and not (they're equivalent).) $\endgroup$ – user213887 Dec 4 '18 at 17:42
  • $\begingroup$ The statement "energy is not a gauge-invariant quantity" is disproved by the gauge-invariance of the time-independent Schrodinger equation. A gauge transformation of the equation produces the identical equation, with the same energy eigenvalue $E$. For example, see quantummechanics.ucsd.edu/ph130a/130_notes/node296.html. This has also been discussed in Physics StackExchange, for example in questions such as physics.stackexchange.com/questions/394892/…. $\endgroup$ – G. Smith Dec 6 '18 at 4:05
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Yes, the rest mass of a system such as a hydrogen atom can be calculated in a fully gauge-invariant way, by integrating the $T^{00}$ component of its gauge-invariant Hilbert energy-momentum-stress tensor $T^{\mu\nu}$. This component is the energy density, and integrating it over all space gives the total gauge-invariant energy of the system.

The "canonical" energy-momentum-stress tensor' produced by Noether's Theorem is guaranteed to be conserved if the action for the system is invariant under temporal and spatial translations, but it isn't necessarily gauge-invariant or even symmetric. Thus it, and the energy found by integrating it, has no physical significance.

However, there is a well-known procedure for producing physically meaningful energy-momentum-stress tensors which are not only conserved but also manifestly gauge-invariant, manifestly Lorentz-covariant, and manifestly symmetric. These tensors do have physical significance: they represent the local energy density, momentum density, etc. of matter.

From the appearance of this kind of $T^{\mu\nu}$ on the right-hand-side of Einstein's field equations for General Relativity, it is obvious that energy density, momentum density, etc. must be measurable, because they create curvature in spacetime. Curvature is measurable, and therefore $T^{\mu\nu}$ is measurable. Thus is it not the case that only differences in energies are physically meaningful. The "absolute" value of the energy density is meaningful, because it curves spacetime.

There is a standard procedure due to Hilbert and Einstein for finding stress-tensors that are conserved, gauge-invariant, Lorentz-covariant, and symmetric, namely by functional integration of the Lagrangian density with respect to the metric:

$$T_{\mu\nu} = \frac{-2}{\sqrt{-g}} \frac{{\delta\mathcal{L}_{\text{matter}}\sqrt{-g}}}{\delta g^{\mu\nu}}.$$

When one uses such a $T^{\mu\nu}$ for a system of point particles and electromagnetic fields, the result for the hydrogen atom (in the order-$\alpha^2$ approximation that the question uses) is identical to the "naïve" calculation presented in the question. But it is completely gauge-invariant from beginning to end.

What one finds is that $U(r) = q_1 q_2/r$, previously considered the "electrostatic potential energy of the proton and electron" is in fact the gauge-invariant position-dependent part of the rest energy residing in the electrostatic field of the proton and electron. The fact that it looks like the product of one charge with the gauge-dependent electrostatic potential of the other charge is irrelevant. The calculation below does not involve any potentials at all.

There is also an infinite, position-independent, gauge-invariant, constant part of the electrostatic field energy, which simply renormalizes the masses of the two particles.

Thus electrostatic "potential energy" can be understood as the position-dependent part of the gauge-invariant rest energy residing in the electrostatic field. It really goes to zero at infinity, because a gauge-invariant calculation tells us so. Going to zero at infinity is not simply a convention.

If you want to say that the mass of a proton and an electron are their standard measured values, then you must take the "potential energy" between them to be exactly $-e^2/r$ and not add a constant or any other gauge-dependent term.

Here are the mathematical details:

Consider a collection of point particles with masses $m_i$ and charges $q_i$, moving under the influence of an electromagnetic field, which could be a combination of the fields of the particles themselves and some external field.

The conserved, symmetric, manifestly-covariant, and gauge-invariant energy-momentum-stress tensor for a system of point particles and electromagnetic fields divides cleanly into two terms,

$$T^{\mu\nu} = T_{(p)}^{\mu\nu} + T_{(f)}^{\mu\nu},$$

where the particles-only term is

$$T_{(p)}^{\mu\nu} = \sum_i m_i \int \dot{X}_i^\mu (\tau) \dot{X}_i^\nu (\tau) \; \delta^{(4)}(x-X_i(\tau)) \; d\tau$$

and the fields-only term is

$$T_{(f)}^{\mu\nu}=\frac{1}{4\pi} \left( F^{\mu\alpha} {F^\nu}_\alpha - \frac{1}{4} \eta^{\mu\nu} F^{\alpha\beta} F_{\alpha\beta} \right).$$

Here $m_i$ is the mass of the $i^\text{th}$ particle, $X_i(\tau)$ is its world-line as a function of the proper time $\tau$ along the worldline, and $F^{\mu\nu}$ is the gauge-invariant electromagnetic field tensor. Note that the gauge-dependent electromagnetic potential appears nowhere in this energy-momentum-stress tensor.

The $T^{00}$ component for the particles can be written, after doing the integration over $\tau$, as

$$T_{(p)}^{00} = \sum_i m_i( 1-\mathbf{v}_i^2)^{-1/2} \; \delta^{(3)}(\mathbf{x}-\mathbf{X}_i(t)),$$

and integrating this over space gives the energy of the particles,

$$E_{(p)} = \sum_i m_i( 1-\mathbf{v}_i^2)^{-1/2} = \sum_i m_i + \sum_i \frac{1}{2}m_i \mathbf{v}^2 + \dots \; .$$

Obviously this is the usual expansion into the energy of the rest masses plus the kinetic energy. We're doing a non-relativistic approximation and don't need the higher terms. In the case of the hydrogen atom, the energy of the particles in the center-of-mass frame takes the form

$$E_{(p)} = m_p + m_e + \frac{\mathbf{p}^2}{2\mu}$$

where $\mu = m_p m_e/(m_p + m_e)$ is the reduced mass and $\mathbf{p}$ is the relative momentum. The expectation value of this third term is what was denoted $\langle K \rangle$ in the question. So we have reproduced the first three terms for the rest energy of hydrogen.

The $T^{00}$ component for the fields can be written in terms of the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$,

$$T_{(f)}^{00} = \frac{\mathbf{E}^2 + \mathbf{B}^2}{8 \pi},$$

and integrating this over space gives the energy of the fields,

$$E_{(f)} = \int \frac{\mathbf{E}^2 + \mathbf{B}^2}{8 \pi} \, dV.$$

For the order-$\alpha^2$ approximation of the hydrogen rest energy that we care about, we can calculate this field energy by ignoring the motion of the charges. The electric field is just the usual Coulomb one for a static charge, and there is no magnetic field. There is no external field to consider.

The fields of $q_1$ are

$$\mathbf{E}_1 = q_1 \frac{\mathbf{r}-\mathbf{r_1}}{|\mathbf{r}-\mathbf{r_1}|^3}, \quad \mathbf{B}_1 = 0,$$

and the fields of $q_2$ are

$$\mathbf{E}_2 = q_2 \frac{\mathbf{r}-\mathbf{r_2}}{|\mathbf{r}-\mathbf{r_2}|^3}, \quad \mathbf{B}_2 = 0.$$

The field energy obviously breaks into three integrals:

$$E_f = \int \frac{(\mathbf{E}_1 + \mathbf{E}_2)^2}{8 \pi} \, dV = \int \frac{\mathbf{E}_1^2}{8 \pi} \, dV + \int \frac{\mathbf{E}_1 \cdot \mathbf{E}_2}{4 \pi} \, dV + \int \frac{\mathbf{E}_2^2}{8 \pi} \, dV.$$

The first integral,

$$E_{f,1} = \int \frac{\mathbf{E}_1^2}{8 \pi} \, dV = \frac{q_1^2}{8\pi} \int \frac{1}{|\mathbf{r}-\mathbf{r_1}|^4} \, dV = \frac{q_1^2}{2} \int_0^\infty \frac{dr}{r^2},$$

diverges. It is the classical electrostatic self-energy of point charge $q_1$ interacting with its own field. It has nothing to do with $q_2$ and does not depend on the distance between the charges or the position of $q_1$. It is rest energy that is just as intrinsic to $q_1$ as its mass-energy is, and this energy simply renormalizes the mass $m_1$, in the same way as happens in QED.

The third integral,

$$E_{f,2} = \int \frac{\mathbf{E}_2^2}{8 \pi} \, dV = \frac{q_2^2}{8\pi} \int \frac{1}{|\mathbf{r}-\mathbf{r_2}|^4} \, dV = \frac{q_2^2}{2} \int_0^\infty \frac{dr}{r^2},$$

similarly diverges and simply renormalizes $m_2$.

The second integral,

$$E_{f,12} = \int \frac{\mathbf{E}_1 \cdot \mathbf{E}_2}{4 \pi} \, dV = \frac{q_1 q_2}{4\pi} \int \frac{\mathbf{r}-\mathbf{r_1}}{|\mathbf{r}-\mathbf{r_1}|^3} \cdot \frac{\mathbf{r}-\mathbf{r_2}}{|\mathbf{r}-\mathbf{r_2}|^3} \; d^3\mathbf{r} = \;?$$

is the interesting one. It involves both gauge-invariant fields $\mathbf{E}_1$ and $\mathbf{E}_2$ and can be described as a gauge-invariant position-dependent interaction energy. It looks like it might be divergent, but it turns out that this integral can be performed (see below) and the result is finite! In fact, it is just the usual "potential energy" of two point charges:

$$E_{f,12} = \frac{q_1 q_2}{|\mathbf{r_1}-\mathbf{r_2}|}$$

For the hydrogen atom, this means that the "potential energy" term $\langle -e^2/r \rangle$ is completely legitimate; it is the $r$-dependent part of the gauge-invariant field energy. The $r$-independent part of the gauge-invariant field energy is divergent and renormalizes the masses.

This term coming from the fields produces the fourth term in the hydrogen rest energy, which was previously denoted $\langle U\rangle$.

To do the integral, introduce Cartesian coordinates with $q_1$ is at $\mathbf{r}_1=a\hat{\mathbf{z}}$ and $q_2$ is at $\mathbf{r}_2=-a\hat{\mathbf{z}}$.

The electric field of $q_1$ is

$$\mathbf{E}_1 = q_1 \frac{x\hat{\mathbf{x}} + y\hat{\mathbf{y}} + (z-a)\hat{\mathbf{z}}}{[x^2 + y^2 + (z-a)^2]^{3/2}}$$

and the electric field of $q_2$ is

$$\mathbf{E}_2 = q_2 \frac{x\hat{\mathbf{x}} + y\hat{\mathbf{y}} + (z+a)\hat{\mathbf{z}}}{[x^2 + y^2 + (z+a)^2]^{3/2}},$$

so the field interaction energy is

$$E_{f,12} = \frac{q_1 q_2}{4 \pi} \int_{-\infty}^\infty dx \int_{-\infty}^\infty dy \int_{-\infty}^\infty dz \; \frac{x^2 + y^2 + z^2 - a^2}{[(x^2 + y^2 + z^2 + a^2)^2 - 4 a^2 z^2]^{3/2}}.$$

Converting to spherical polar coordinates, the integral becomes

$$E_{f,12} = \frac{q_1 q_2}{4 \pi} \int_0^\infty r^2 dr \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\phi \; \frac{r^2 - a^2}{[(r^2 + a^2)^2 - 4 a^2 r^2 \cos^2\theta]^{3/2}}.$$

The integral over $\phi$ just gives $2\pi$, and the integral over $\theta$ becomes elementary with the subsitution $u=\cos\theta$:

\begin{align*} E_{f,12} &= \frac{q_1 q_2}{2} \int_0^\infty r^2 (r^2 - a^2) \; dr \int_{-1}^1 \frac{du}{[(r^2 + a^2)^2 - 4 a^2 r^2 u^2]^{3/2}} \\ &= \frac{q_1 q_2}{2} \int_0^\infty r^2 (r^2 - a^2) \; dr \left[ \frac{u}{(r^2 + a^2)^2 [(r^2 + a^2)^2 - 4 a^2 r^2 u^2]^{1/2}} \right]_{-1}^1 \\ &= q_1 q_2 \int_0^\infty \frac{r^2 dr}{(r^2 + a^2)^2} \frac{r^2 - a^2}{|r^2 - a^2|}. \end{align*}

The integral over $r$ must be split into two parts, one from 0 to $a$, and one from $a$ to $\infty$.

$$E_{f,12} = q_1 q_2 \left( \int_0^a \frac{r^2 dr}{(r^2 + a^2)^2}(-1) + \int_a^\infty \frac{r^2 dr}{(r^2 + a^2)^2} (+1) \right).$$

The resulting integrals can be done with the substitution $r=a\tan{u}$ and give

\begin{align*} E_{f,12} &= q_1 q_2 \left( \left[ \frac{r}{2(r^2 + a^2)} - \frac{\tan^{-1}(r/a)}{2 a} \right]_0^a + \left[ \frac{\tan^{-1}(r/a)}{2 a} - \frac{r}{2(r^2 + a^2)} \right ]_a^\infty \right) \\ &= q_1 q_2 \left[ \left( \frac{1}{4a} - \frac{\pi}{8a} \right) - (0) + \left( \frac{\pi}{4a} \right) - \left( \frac{\pi}{8a} - \frac{1}{4a} \right) \right] \\ &= \frac{q_1 q_2}{2a} \end{align*}

Thus the final result is

$$E_{f,12} = \frac{q_1 q_2}{d}$$

where $d=2a$ is the separation between the charges.

The claimed result,

$$E_{f,12} = \int \frac{\mathbf{E}_1 \cdot \mathbf{E}_2}{4 \pi} \, dV = \frac{q_1 q_2}{4\pi} \int \frac{\mathbf{r}-\mathbf{r_1}}{|\mathbf{r}-\mathbf{r_1}|^3} \cdot \frac{\mathbf{r}-\mathbf{r_2}}{|\mathbf{r}-\mathbf{r_2}|^3} \; d^3\mathbf{r} = \frac{q_1 q_2}{|\mathbf{r_1}-\mathbf{r_2}|}$$

follows from the fact that this is a rotationally-invariant equation which we've verified with one particular choice of coordinates.

Addendum:

A much simpler approach is simply to realize that the Schrodinger equation is gauge-invariant. For an electron in an electromagnetic field described by a scalar potential $\phi(\mathbf{r})$ and a vector potential $\mathbf{A}(\mathbf{r})$, the Schrodinger equation is

$$\left[ \frac{1}{2m} \left( \frac{\hbar}{i}\nabla + \frac{e}{c} \mathbf{A}(\mathbf{r}) \right)^2 - e \phi(\mathbf{r}) \right] \psi(\mathbf{r}, t)=E \psi(\mathbf{r}, t).$$

When one makes the gauge transformations

$$\psi(\mathbf{r}, t) \rightarrow \exp({i \lambda(\mathbf{r}, t)}) \psi(\mathbf{r}, t)$$

$$\phi(\mathbf{r}) \rightarrow \phi(\mathbf{r}) + \frac{\hbar}{e} \frac{\partial\lambda(\mathbf{r}, t)}{\partial t}$$

$$\mathbf{A}(\mathbf{r}) \rightarrow \mathbf{A}(\mathbf{r}) - \frac{\hbar c}{e} \nabla \lambda(\mathbf{r}, t)$$

and uses the relationship

$$i \hbar \frac{\partial \psi(\mathbf{r}, t)}{\partial t} = E \psi(\mathbf{r}, t),$$

one finds that the equation is unchanged, showing that the energy $E$ is gauge-invariant. This is a standard part of most undergraduate courses in Quantum Mechanics.

Thus the original calculation in the question is completely gauge-invariant (although not manifestly, like the one in my answer), because the gauge-invariance of the Schrodinger equation guarantees that the energy is the same in any gauge. Thus it is fine to calculate the energy in a gauge where the proton's potential is $\phi=e/r$.

A dissenting commenter seems not to understand that the Schrodinger equation is gauge-invariant. He has argued (in the "Energy/mass of Quantum Vacuum" thread) that rephasing the wavefunction as $\psi \rightarrow e^{i \alpha t} \psi$, where $\alpha$ is constant, shifts the spectrum by $\alpha$. He seems to have forgotten that rephasing the wavefunction is only one part of a gauge transformation. The other part is changing the electromagnetic potentials. When one does both, the energy does not change. This is the entire point of having gauge fields... they are there to make the equations gauge-invariant. His contention in the other thread that "The energy of the ground state is not an observable quantity", apparently because he thinks that it is gauge-dependent, is false.

One potential confusion is that, although the Schrodinger equation for a charge in an electromagnetic field is gauge-invariant, the Hamiltonian is not, in general. But this is not inconsistent with the energy being gauge-invariant, because the Hamiltonian is not always the energy operator. The problem is that a gauge transformation can turn a non-time-dependent Hamiltonian into a time-dependent one, in which case it is no longer true that $\hat{H}\psi=E\psi$.

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    $\begingroup$ Okay firstly, the expression $\int dV \vec{E}^2/4\pi$ is the energy stored in the electric field. But that's not what we're interested in here. When we calculate the binding energy of the hydrogen atom, which is the $\langle U \rangle$ in your expression, i.e. the potential energy of the electron/proton, we're interested in the energy of the electron/proton which is not the energy stored in the electric field. The energy stored in the electric field is totally gauge invariant - this is trivially obvious, as it is expressed in terms of the gauge invariant quantities $ \vec{E}^2$ and $\vec{B}^2$ $\endgroup$ – user213887 Dec 5 '18 at 15:05
  • $\begingroup$ So $\tfrac{e^2}{r}$ is not a gauge invariant quantity, in classical or quantum mechanics. But secondly, the expression $m_e+m_p+\tfrac{\vec{p}^2}{2\mu}$ is not gauge invariant due to a different gauge redunancy: the phase redundancy of quantum mechanics - the problem here is you've basically treated the electron and proton as classical point particles, but the 13.6 eV number comes from solving the Schrodinger equation. $\endgroup$ – user213887 Dec 5 '18 at 15:07
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    $\begingroup$ What you've done is equivalent to saying what you said in an earlier thread: "ah, the Minkowski metric tells us the dispersion is $p^2=m^2$." This is indeed the energy of a single electron classically. But in quantum mechanics there's a new gauge redundancy: the energy of the electron is given the eigenvalues of the Hamiltonian, and we are free to shift these by rephasing the wavefunction. This allows us to shift the energy of the Coulomb bound state to whatever we want. $\endgroup$ – user213887 Dec 5 '18 at 15:07
  • $\begingroup$ Thirdly, I'd also point out (though it's a quibble and not relevant to the essence of the issue here) that the mass/charge renormalisation of the electron/proton is not really playing a role here. The masses in your equation are the physical masses, so the running under RG isn't visible here - the only quantum mechanical object that will play a role here is the vacuum energy. $\endgroup$ – user213887 Dec 5 '18 at 15:07
  • $\begingroup$ So, this is a nice calculation, but does not solve the problem you're trying to solve. You've not shown that the electrostatic energy of the electron is gauge-invariantly $\tfrac{e^2}{r}$ - good luck trying, $qV$ is never going to be gauge invariant no matter how hard you try to spin it - and you've not shown the kinetic energy of the electron is $m+\tfrac{\vec{p}^2}{2\mu}$ (though there are different gauge redundancies in both cases). $\endgroup$ – user213887 Dec 5 '18 at 15:09

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