How is it possible to initialize a quantum state in a superposition of energy eigenstates?

Question

There are many experiments that prepare quantum systems in superpositions of different energy levels. For example, it is common to prepare cavities in coherent states, which are superpositions of SHO eigenstates.

One thing I know about physics is that energy is conserved. That means that if I start with a system in its ground state $E_0=0$ and want to evolve it to a superposition of states with energies $E_0$ and $E_1$, the additional energy must come from the environment, ie the rest of the universe. If before the evolution the rest of the universe had energy $E$, after the evolution I should be in a superposition of $E_0$ in the system and $E$ in the rest of the universe, and $E_1$ in the system and $E-E_1$ in the rest of the universe. Then the state of the system is entangled with the rest of the universe, and the system itself is in a mixed state rather than a superposition.

Clearly this reasoning is somehow incorrect, so what is the loophole that allows us to actually prepare superpositions instead of mixed states?

Answer 1

In most quantum optics experiments, states that are a superposition of energy states are generated using coherent states. For example the easiest way to prepare a cavity in a coherent state is to drive it with a laser, which outputs a coherent state. Coherent states have the property that removing a photon from the state does not change it, i.e., $a|\alpha\rangle \propto |\alpha\rangle$, so they can create energy superposition without creating entanglement.

But this begs the question of how to create coherent states, which are themselves superpositions of different energy states. It turns out that your reasoning is correct, and you need to bring measurement into the picture to get energy superposition. When your system is entangled with the rest of the universe like $|E_{universe}-E_0\rangle|E_0\rangle + |E_{universe}-E_1\rangle |E_1\rangle+...$, measurement of the system destroys the entanglement and can yield energy superpositions like $|E_0\rangle+e^{i\phi}|E_1\rangle+...$.

There's a very familiar example that illustrates this point nicely. Energy eigenstates of a free particle are plane waves $\psi_k(x) \propto e^{i k x}$, where the energy is just $\hbar^2 k^2/2m$. These are infinitely delocalized. We build up localized wave functions by taking coherent superpositions of them, like $\psi(x) = \int_{-\infty}^\infty \tilde{\psi}(k)\psi_k(x) dk$, where $\tilde{\psi}(k)$ is the Fourier transform of $\psi(x)$. So every position measurement generates coherence between energy eigenstates!

See this answer for more details on why realistic measurements tend to single out coherent states in systems that generate them (like lasers).

Answer 2

One intuitive argument could be that due to the size of the system, the "universe" state with energy $E$ and the universe state with energy level $E - E_1$ are more or less indistinguishable. Therefore a tensor product of $|E_0\rangle|E\rangle + |E_1\rangle|E-E_1\rangle$ is essentially the same as a pure state $(|E_0\rangle + |E_1\rangle)|universe\rangle$. Although theres probably a more rigorous argument that shows the same thing.