This is also not the entirely correct way to view photons, I'm sorry to say. First of all as massless quanta with spin ##1## there's no position operator. So the naive uncertainty relation, valid for massive particles, doesn't make sense to begin with. For a review, see
http://arnold-neumaier.at/physfaq/topics/position.html
The most easy way to introduce photons in a correct way is to look at the free classical electromagnetic field and fix the gauge completely, i.e., you choose the radiation gauge for the four potential ##A^{\mu}##, which consists in the two constraints
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
and then quantize the remaining two physical transverse components canonically. This leads to the mode decomposition
$$\hat{\vec{A}}(x)=\sum_{\lambda= \pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 |\vec{p}|}} [\vec{\epsilon}(\vec{k},\lambda) \hat{a}(\vec{p},\lambda) \exp(-\mathrm{i} x \cdot p) + \text{h.c.}]_{p^0=|\vec{p}|}.$$
The total energy and momentum, i.e., the Hamiltonian and the momentum of the em. field are defined as the normal ordered expressions
$$\hat{P}^{\mu} = \sum_{\lambda= \pm 1} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} p^{\mu} \hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda), \quad p^0=|\vec{p}|.$$
the annihilation and creation operators fulfill the commutator relations
$$[\hat{a}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')]= \delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'}.$$
Each single mode indeed fulfills the commutator relations of the harmonic oscillator, i.e., the electromagnetic field is equivalent to an infinite number of uncoupled harmonic oscillators. Thus the Fock states, i.e., the common occupation-number eigenstates of the number operators ##\hat{N}(\vec{p},\lambda)=\hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda)## (where only a finite number of occupation numbers is different from 0, ##N(\vec{p},\lambda)\in \{0,1,2,\ldots \}##.
Now we can unanimously define what a photon is: It's a single-photon state, i.e.,
$$|\psi \rangle=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \hat{a}^{\dagger}(\vec{p},\lambda) \phi(\vec{p},\lambda) |\Omega \rangle,$$
where ##|\Omega \rangle## is the vacuum state, for which all ##N(\vec{p},\lambda)=0## (ground state of the system).