# Ground eigenstate of the quantum harmonic oscillator with the interacting vacuum $| \Omega \rangle$

According to this video (at the timestamp), the professor writes down the derivation of the ground state of the 1D quantum harmonic oscillator.

Here is the screenshot from the slides of where she does it:

Now that is well and good. However, I am wondering what if we tried to derivate the ground state wavefunction for the interacting vacuum $$| \Omega \rangle$$ in the same way. Could we?

I know that in general

$$| \Omega \rangle \neq |0\rangle$$

And maybe we could do

$$\hat{a}| \Omega \rangle = 0$$

Just like in the screenshot and video to attain the ground state wavefunction, but this should not work since

$$\hat{a}| \Omega \rangle \neq 0$$

Because the annihilation operator $$\hat{a}$$ is from the free theory and we need the interacting annihilation operator instead because we are working with the interacting vacuum. I have seen this discussed here. Also, from the QFT wiki (the previous section from the one I linked, just before the Lagrangian), it says

However, creation and annihilation operators are only well defined in the simplest theories that contain no interactions (so-called free theory)

So I am not sure if these exist.

Now, I have never seen this done anywhere (from textbooks or videos) so could someone please provide an example of this or a link to some resources that do this?

Is there such a thing as the QHO in the interacting theory with the ground state wavefunction being

$$\Psi_\Omega (x) = \langle x|\Omega\rangle$$

• The quantum harmonic oscillator has a specific Hamiltonian. If you are talking of an interacting theory then it implies some interaction terms that you are adding to the Hamiltonian, which as a result would not be the harmonic oscillator anymore. Perhaps you can start by defining precisely the Hamiltonian that you want to study and then ask what the ground state of that theory would be. Commented May 22, 2023 at 2:58

The basic problem is that the state of the QHO system that we represent by $$|0\rangle$$, which is what you mean when you say $$a|0\rangle = 0$$, is not the same as the vacuum of a free QFT, which is the one that you're referring to in $$|\Omega\rangle \neq |0\rangle$$. One reason that we use the same symbol for both is that it's sometimes useful to think of the vacuum of a free QFT, e.g. that of the free scalar theory, as a state that's annihilated by $$\hat{a}_\mathbf{k}$$ for any four-vector $$\mathbf{k}$$–the physical picture here is that the Hilbert space is spanned by states $$|\mathbf{k}_1, \mathbf{k}_2, \ldots\rangle = a^\dagger_{\mathbf{k}_1} a^\dagger_{\mathbf{k}_1} \ldots |0\rangle,$$ and we can conceive of a Hamiltonian that looks something like $$H \sim \int d^4 \mathbf{k}\, \left(a_\mathbf{k}^\dagger a_\mathbf{k} + \frac{1}{2}\right),$$ which is diagonalized by this basis of states.
There are a few conceptual problems with the picture that I just presented. One of them is that the value of this Hamiltonian will diverge for all states, which is no good, but we can sort of fix this by just "changing our energy zero" and getting rid of the $$1/2$$–this is the first rumbling of troublesome infinities baked into QFT. The other problem is that when you turn on interactions, those states are no longer eigenstates of our interaction Hamiltonian, and not really clear what should take their place. One way to think about this is that in QED, for instance, a lone photon with a plane-wave wavefunction can't be a stationary state, because QED allows it to spontaneously turn into an electron-positron pair. You can still define the interacting vacuum $$|\Omega\rangle$$, though, but now we define it as "the unique state which is invariant under all translations and rotations and boosts," which is not very constructive but is still a helpful notion, and we can consider electrons and photons and such to be perturbations to this vacuum.