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Problem description

Consider quantum oscillator with $ \omega = 0 $. In other words, we have $$ \hat{H} = \hat{p}^2, $$ where $\hat{x}, \, \hat{p}$ are the usual coordinate and momentum operators with the canonical commutation relation $ [\hat{x}, \, \hat{p}] = i $.

Naive approach

We choose a vacuum $ | 0 \rangle $ satisfying the condition: $$ \hat{p} | 0 \rangle = 0. $$ Also it is normalized $\langle 0 | 0 \rangle = 1 $.

Next, we can "create" other states using the exponent of momentum operator $ \exp{i\alpha \hat{x}} $, because: $$ \hat{p} e^{i\alpha \hat{x}} |0\rangle = [\hat{p}, \, e^{i \alpha \hat{x}}]|0\rangle + e^{i \alpha \hat{x}} \hat{p} |0\rangle = \alpha e^{i \alpha \hat{x}} |0\rangle. $$ (If we choose complex $\alpha$, $\hat{p}$ will have complex eigenvalue and won't be Hermitian.) Besides, we also have the following: $$\langle 0| \hat{p} \hat{x} |0 \rangle = \langle 0 | [\hat{p}, \, \hat{x}] | 0 \rangle + \langle 0 | \hat{x} \hat{p} | 0 \rangle = - i \langle 0 | 0 \rangle = -i. $$ But that means, that $ \langle 0 | \hat{p} \neq 0 $. $\hat{p}$ isn't Hermitian. But what is $\hat{p}$ then? Is it possible to find $\hat{p}$ explicitly?

Next, even more naive perspective. If we multiply $\hat{x}$ by $i$, we'll get the commutation relation the same as the creation/annihilation operators: $[\hat{p}, \, i \hat{x}]$ = 0. Is it possible to build the usual state tower with this?

In the momentum representation the vacuum condition $ \hat{p} |0\rangle = 0 $ may be written as $\hat{p} \cdot \psi_0(p) = p\psi_0(p) = 0$. That means, that $ \psi_0(p) = \delta(p)$. In the momentum representation $\hat{x} = i d/dp$. Therefore we have $$ |n\rangle = (i\hat{x})^n |0 \rangle = i^n \delta^{(n)}(p). $$ Does it make any sense to consider them as a "basis" for the state space? Seems, like we can decompose $|\alpha \rangle = \sum_n (i\alpha)^n |n\rangle $.

More rigorous approach

I'm not very proficient in functional analysis. After some googling I think I need Gelfand triple formalism (https://math.stackexchange.com/q/1443680/), but I have no idea how it works. Maybe, there is a not super technical introduction? Nevertheless, I tried to somehow formalize the previous approach to get the answers to my questions.

By Stone-von Neumann theorem we know that any unitary representation of the canonical commutation relation algebra is isomorphic to $L^2(\mathbb{R})$ with $\hat{x} = i d/dp$ and $ \hat{p} = p$ (I choose momentum representation for an easier correspondence with the previous approach). In $L^2(\mathbb{R})$ $\hat{p}$ obviously doesn't have eigenvectors. Therefore, we need to consider a bigger space. Obvious candidate is the distribution space $\mathcal{D}'(\mathbb{R})$ (or $\mathcal{S}'(\mathbb{R})$, but I don't see any difference in the following). Our vacuum $ |0 \rangle = \delta(p) \in \mathcal{D}'(\mathbb{R})$ lies in the enlarged state space.

But $\mathcal{D}'(\mathbb{R})$ isn't a Hilbert space. It doesn't have an inner product. Thus, it doesn't make any sense to ask whether $\hat{p}$ is Hermitian or compute $ \langle 0 | \hat{p} \hat{x} | 0 \rangle $.

Is there any way to make the naive approach to work? We need simultaneously an inner product and a delta function. But they are incompatible.

Motivation

I came to consider it by studying the zero mode of a boson field on 2d cylinder. Zero mode has the angular frequency equal to zero. And we need an inner product on the state space to compute correlators. For example, the naive approach is used in Di Francesco et al, Conformal Field Theory section 6.3 or in these lectures by Lashkevich.

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    $\begingroup$ The normalization $\langle0|0\rangle=1$ isn't possible in this situation. Eigenstates of momentum are "normalized" to the delta function: $\langle p|k\rangle = \delta(p-k)$. I think you're running into this as a problem with your commutators, too. Try doing this more rigorously with $\omega > 0$ and taking the limit. Then all of your states will be normalized, etc. $\endgroup$ Commented Sep 9, 2022 at 10:20

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The spectrum of $p^2$ is fixed by the one of $p$ which, in turn, is fixed by the CCR and by the Stone-Von Neumann theorem. $p^2$ has continuous spectrum with no (proper) eigenvectors. There is no hope to define something like the ground state of the standard harmonic oscillator.

Maybe something can be said in some limit case, but you should furnish more information (especially from the physical side). As it stands, the answer is that the zero frequency oscillator is not a harmonic oscillator so that there is no ground state.

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