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I tried to give a QFT description of the hydrogen atom (at least conceptually and without going into mathematical details). The system is made by a spin$\frac{1}{2}$ fermion (electron) field $\phi_s$ coupled to the external electromagnetic field of the proton.

I can build a generic state of the electron field by acting on the vacuum with creation operators. At time $t=0$ a generic 1-electron state should be something like

$$\sum_{s} \int d^{3} \pmb{k} \, f(\pmb{k}, s) \hat{a} _{\pmb{k}, s}^{\dagger} |0\rangle$$

By choosing an appropriate $f$ it is possible to minimize the energy of the 1-electron state. I shall work with this particular state from now on and I will refer to it as $|gs\rangle$.

What if I want to predict (with a certain probability of course) where the electron is after a position measurement? Since $\pmb{x}$ is just a continuous index the ket $|\pmb{x}\rangle$ doesn't make much sense therefore I can't use it. My idea was to do the following procedure:

$|gs\rangle = \sum_{s} \int d^3 \pmb{x} \langle\phi_{s}(\pmb{x}) |gs\rangle |\phi_{s}(\pmb{x})\rangle $, where $|\phi_{s}(\pmb{x})\rangle \equiv \hat{\phi}_{s}(\pmb{x}) |0\rangle $. Since $|\phi_{s}(\pmb{x})\rangle$ describes a localized excitation of the field I thought I could interpret $\sum_{s} \lvert \langle\phi_{s}(\pmb{x}) |gs\rangle \rvert^{2} $ as the probability distribution of the electron in space.

Does what I did even make sense? Will $\langle\phi_{s}(\pmb{x}) |gs\rangle$ be related to the lowest energy solution of the Dirac equation (interpreted as the wave function of the electron)? Would lamb shift appear if I were to do the math?

Nihar Karve
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Masterme
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  • Related: Measurement of observables with continuous spectrum: State of the system afterwards – Chiral Anomaly Apr 30 '21 at 13:10
  • @ChiralAnomaly Reading the post it seems to me that my reasoning is correct. Is that so? – Masterme Apr 30 '21 at 14:39
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    You can interpret $\sum_s |\langle\phi_s(\mathbf{x})|gs\rangle|^2$ as the probability distribution of the outcomes of a position-measurement. Determining this distribution experimentally would require many measurements on many identically-prepared atoms, but you can calculate it theoretically. To get the Lamb shift, you need to use QED (nonrelativistic QED is sufficient). Presumably, the distribution would be different than it would have been without the quantum EM field. Is that what you're asking? – Chiral Anomaly Apr 30 '21 at 15:06
  • @ChiralAnomaly Yes, perfect. Basically what I did is correct and lamb shift appears with a QED calculation. Thanks a lot :) – Masterme Apr 30 '21 at 15:12
  • @ChiralAnomaly The only piece of the puzzle that's missing is: is $<\phi_s(\pmb(x)) |gs>$ the lowest energy solution of the Dirac equation? – Masterme Apr 30 '21 at 15:14
  • The question says $|gs\rangle$ is the minimum-energy 1-electron state, but I'm not sure what you mean by "lowest energy solution of the Dirac equation." If you're using nonrelativistic QED, then the Dirac equation is not involved. Even if you're using relativistic QED, the Dirac equation is only part of the picture. Energy is defined by the full Hamiltonian, which is expressed in terms of all of the quantum fields (electron and EM). The Hamiltonian generates time-evolution, so it generates the Dirac equation for the electron field operators (and Maxwell's eqs for the EM field operators). – Chiral Anomaly Apr 30 '21 at 17:14
  • By the way, if you're using relativistic QED, then $\hat\phi_s(\mathbf{x})|0\rangle$ is no longer a single-particle state, and it's not strictly localized at $\mathbf{x}$. The relationship between field operators and single-particle states is more complicated in relativistic QFT, and strictly localized single-particle states don't exist in relativistic QFT (see this post). Those technicalities might be unimportant for what you're asking, but I wanted to mention them so that I don't accidentally perpetuate any misconceptions. – Chiral Anomaly Apr 30 '21 at 17:14
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    @ChiralAnomaly I'm very new to QFT and I tried to build a probability distribution of finding an electron at a point in space in relativistic QED. I read the post you linked and it looks like one cannot do such a thing but can instead build an approximate detector that should work quite well for electrons. I was wondering if you could show (or provide some source) on how one can do such a thing. Say I have a hydrogen atom in QED: how could I predict theoretically the likelihood of the electron being in a particular region of space, say between 1 Angstrom and 1.1 Angstrom from the nucleus? – Masterme Apr 30 '21 at 18:45
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    I'd recommend starting with semiclassical QED, where the electron is quantum but the EM field is prescribed (classical Coulomb field). Then your approach is good: $|x\rangle\equiv\hat\phi_s(\mathbf{x})|0\rangle$ is a 1-particle state. In the relativistic case, it's not strictly localized, but it's still localized to a good approximation. Wou can write $\hat\phi$ as a sum of two terms, $\hat\phi_+$ and $\hat\phi_-$, involving only creation and annihilation operators, respectively. The observable $D(x)\equiv\hat\phi_+(x)\hat\phi_-(x)$ represents a detector approximately localized at $x$. (...) – Chiral Anomaly Apr 30 '21 at 19:44
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    (...) Then, insofar as the concept of a localized electron makes sense at all, the expectation value $\langle x|D(x')|x\rangle$ is the probability density of detecting the electron at $x'$, given that you prepared it in the state $|x\rangle$. You'll find that this is nonzero even if $x'\neq x$, but it's mostly concentrated near $x'=x$. Or you can calculate $\langle gs|D(x)|gs\rangle$, which is the probability density of detecting the electron at $x$ (to the same approximation), given that you prepared it in the state $|gs\rangle$. – Chiral Anomaly Apr 30 '21 at 19:44
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    (...) Semiclassical QED doesn't have the Lamb shift, though. For that, you need to treat the EM field as quantum. Then you have a model with mutually interacting quantum fields, and then constructing true $1$-particle states (or even the $0$-particle state!) becomes prohibitively difficult. This is why many QED textbooks avoid the issue altogether and just talk about things like asymptotic scattering amplitudes or energy eigenvalues, for which perturbation theory works very well. – Chiral Anomaly Apr 30 '21 at 19:54
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    Thanks a lot, that makes it all clear to me! – Masterme Apr 30 '21 at 20:15

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