Does spontaneous emission actually emit in a random direction, or is it measured in a random direction?

When an excited state couples to the vacuum, it has an infinite number of directions of the quantized electromagnetic field to couple to. Does it evolve into a superposition of all those directions at the same time and only collapses once the photon is measured, or does it couple to only one? (Or, of course, is there no experimental way to tell?)

There isn't really an answer to this. If your initial conditions are spherically symmetric then the system remain spherically symmetric and the emitted light will be in a superposition of all directions. The superposition will collapse, and the symmetry be broken, only when something interacts with your system e.g. a CCD detector.

On the other hand, if the emission was triggered by an incoming photon then your system isn't sperically symmetric, and the outgoing photon will be emitted in the same direction of the incoming one.

So how your system evolves with time depends on how you set it up to begin with. Your question implies that you're thinking of a spherically symmetric initial state, and in this case yes the emission will be in a superposition of all directions.

• " if the emission was triggered by an incoming photon then your system isn't spherically symmetric, and the outgoing photon will be emitted in the same direction of the incoming one" -- this is true for an ensemble of atoms that get excited coherently by the same pulse, but it's false for a single emitter. Atoms are too small to see the direction of the beam (they see only a uniform field within the dipole approximation), and so the excited state is spherically symmetric (modulo things like the dipole pattern in the radiation, along the direction of polarization of the exciting beam. Mar 25 at 11:06
• Ensembles of atoms are able to emit directionally because they act as a phased array. But each microscopic emitter is as likely to emit forward as it is backwards. Mar 25 at 11:06
• @EmilioPisanty The excited state is spherically symmetric, if we treat the emission classically. In quantum picture however the probability amplitude of an emission into a mode is proportional to the square root of the number of photons in the mode, as per bosonic relation $a_{\mathbf{k},\nu}^\dagger |n_{\mathbf{k},\nu}\rangle=\sqrt{n_{\mathbf{k},\nu}+1}|n_{\mathbf{k},\nu}+1\rangle$. Mar 26 at 7:59

There are two principle factors that affect the polarization and direction of the spontaneously emitted radiation:

• The mode structure of the field
• The properties of the transition (i.E., the properties of the two levels between which the transition occurs).

The mode structure of the field
The question suggests that the emission occurs in vacuum, but this is not necessarily the case in practice - certainly not in the human-made lasers and masers, where the emission occurs inside of a resonator/cavity. (The quantization of field in a cavity implies expansion in the cavity modes, rather than in the plane waves - which can be thought of as a special case of a cavity.) This cavity modes may (and do have) preferred directions of propagation and polarization.

The properties of the transition
These are known as the selection rules: e.g., if the dipole moment between the two states is polarized in $$z$$-direction, this is the direction of the polarization of the emitted waves (in practice it may a be a superposition of the waves with the net moment polarized along the $$z$$-direction). Similarly, the direction of propagation of the emitted wave depends on the moment transmitted to this wave during the emission - this latter fact is a rather fine point, since this momentum is small and omitted in many discussions (the wave length is large compared to the size of a molecule/atom), but it is actually not zero.

Remark Of course, if the emission is triggered by presence of another photon, the second moment will be identical to the first one - but in this case we talk about stimulated rather than the spontaneous emission.

Update
Although the above technically answers the question as its posed (the direction of the photon), it does not address the premise of the question; indeed, even though the direction of the emission is not completely arbitrary, there are still many modes in which photon may find itself.

Generic emission process can be described by Jaynes-Cummings-like Hamiltonian $$H=\frac{\hbar\omega_a}{2}\sigma_z + \sum_{k}\hbar\omega_ka_k^\dagger a_k + \sum_k\left(\lambda_k \sigma_+a_k + \lambda_k^*a^\dagger\sigma_-\right),\\ \sigma_z = |e\rangle\langle e| - |g\rangle\langle g|, \sigma_+=|e\rangle\langle g|, \sigma_-=|g\rangle\langle e|$$ (where $$k$$ includes all the quantum numbers of a mode). We start with an atom in the excited state and no photons $$|\psi(0)\rangle=|e,0\rangle$$ and follow the unitary evolution of this state due to the Hamiltonian. After time $$t$$ it becomes $$|\psi(t)\rangle = c_e|e,0\rangle + \sum_kc_k|g, 1_k\rangle,$$ which is a superposition of the states in all available photon modes!

This wave function is a usual superposition in quantum mechanics - i.e., the system collapses to a particular state only when we measure it. It is also worth noting that measurement in this context does not literally imply a human observer, but rather interaction with the rest of the world, which can be here treated as a macroscopic object (i.e., the observer).

Moreover, there is a subtle point with taking the limit with the infinite number of modes. If the number of modes is finite, we would eventually see (after a long time) the revival of the wave function - i.e., the return of the atom to the excited state. The infinite number of modes means that this never happens - the emitted photon remains "spread" over an infinite number of modes, till it is localized in a particular state via an interaction with an observer/bath.