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I'm trying to imagine how/why the direction of a spontaneously emitted photon is random, and how that works and reconciles with the conservation of momentum and kinetic energy.

For intuition, I was trying to make a super simple 1-D scenario of one particle emitting one photon in one dimension. So the photon can either be emitted to the left or the right.

(I'm trying to treat this in a classical way for intuition's sake. And because I've no experience in quantum topics.)

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I reason that momentum and kinetic energy would be conserved, and since the system has no motion initially, initial momentum and kinetic energy will be zero.

Since it's a linear problem, conservation of angular momentum equation would just devolve into the linear momentum formula, so I ignore it.

The momentum of a photon is $p = \frac{h}{λ}$ and its kinetic energy would be $KE = \frac{hc}{λ} $ (or so I've found), so from conservation of momentum and kinetic energy, I get the system:

$0 = mv + \frac{h}{λ}$

$0 = \frac{1}{2}mv^{2} + \frac{hc}{λ}$

But this looks like it makes no sense to me! The logic of how I built the simple scenario makes sense in my mind, but when I look at this two equation system I make, it doesn't work at all. $h$ and $c$ are constants and the mass of the particle is set before the problem begins. Also, $λ$ is determined by how much energy the particle is shedding by the emission. Meaning the first equation alone solves for the velocity of the particle after the emission, giving $v = \frac{-h}{λm}$. But that implies a single and unique solution, not two possibilities. What's worse, This value of $v$ doesn't even work with the kinetic conservation formula.

So my reasoning must be going down in flames. I don't know how I should think of this scenario. Any suggestions or help?

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    $\begingroup$ Why is kinetic energy conserved? You are converting internal energy (electron in an excited state) to external energy (photon) $\endgroup$
    – Jon Custer
    Dec 3, 2020 at 21:26
  • $\begingroup$ @JonCuster that makes a lot of sense. And if I consider replacing the + in the momentum equation with a +/-, that’d produce the two possible solutions... I think $\endgroup$ Dec 3, 2020 at 23:29

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If you are worried about exact conservation of energy and momentum, you really need to do the problem relativistically. With that, the initial 4-momentum is (with $c=1$):

$$ p^{\mu}_0 = (m, 0, 0, 0) $$

If the photon has energy $\hbar\omega$ in the rest frame of the final state atom, then the atom's mass is reduced by that. With a recoil velocity of $\vec v = v\hat z$, the final state 4-momentum is:

$$ p^{mu}_{atom} = \gamma(m-\hbar\omega, 0, 0, (m-\hbar\omega)v)$$

Meanwhile, the photon has 4-momentum:

$$ p^{\mu}_{\gamma} = \sqrt{\frac{1+v}{1-v}}(\hbar\omega, 0, 0, \hbar\omega)$$

where the square root out front accounts for the Doppler shift from the final state atom's rest-frame to the center of mass.

You then solve for everything according to:

$$ p^{\mu}_0 = p^{mu}_{atom} + p^{\mu}_{\gamma} $$

But don't bother, because in an atomic transition:

$$ \hbar\omega \approx 1\,{\rm eV} $$

so the even the lightest recoiling atom is nonrelativsitic with:

$$ K.E. = \frac{p^2}{2m} = \frac 1 2 \frac{(1\,{\rm eV/c})^2}{938\,{\rm MeV/c^2}^2} \approx 5 \times 10^{-10}\,{\rm eV}$$ which is negligible: you can conserve momentum only and get the right answer.

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