The question is a little unclear. I am interpreting it as follows:
Given a quantum particle in with wave function
\begin{equation} \psi_{k \ell m}(\vec{x}) = N j_{\ell}(k r) Y_{\ell m}(\theta,\phi), \end{equation}
where N is a normalization constant, how can I express the wave function in a plane wave basis? In other words what is the function $\phi_{k\ell m}(\vec{p})$ in the following
\begin{equation} \psi_{k \ell m}(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} e^{i \vec{p}\cdot \vec{x}} \phi_{k\ell m} (\vec{p}). \end{equation}
The answer is that you need to multiply by the complex conjugate of the new basis function, aka $e^{-i \vec{p}\cdot \vec{x}}$,iand integrate over $\vec{x}$. In bra ket notation this is $\phi_{k\ell m}(\vec{p})=\langle \vec{p} | k \ell m\rangle$.
This amounts to computing the Fourier tranfrom of $\psi_{k \ell m}(\vec{x})$
\begin{equation} \phi_{k\ell m}(\vec{p}) = \int d^3 x e^{-i\vec{p}\cdot \vec{x}} \psi_{k\ell m} (\vec{x}). \end{equation}
In fact this problem has a known solution. Google 'plane wave expansion in spherical Bessel functions'. You should be able to find the answer to this intergal, or at least an expansion of $e^{i \vec{p}\cdot\vec{x}}$ in terms of spherical Bessel functions and spherical harmonics (or Legendre polynomials), which will make the integral easy to do using orthogonality of the $j_{\ell}$ and $Y_{\ell m}$. (If you get stuck on this integral and you've tried looking for find a good resource, including stack exchange, this might be worth a separate question, perhaps on math.SE).
Finally some comments. You say a state can only be occupied by one boson. In fact a state can be occupied by as many bosons as you like, it is fermions that are forbidden from occupying the same state by the Pauli exclusion principle. Also you don't need the word 'oscillatory' in 'oscillatory state'. Lastly no particles get 'pulverized'--unless relativistic effects are large. What can happen though is that a single particle's wave function can be a superposition over many basis states.
Update 12/13/2016
Based on discussions below I think I have a clearer idea of the original question. Apologies for not getting it earlier.
Let's go back to the wavefunction, except choosing $N=k \sqrt{2/\pi}$ so that it is properly normalized: \begin{equation} \psi_{k \ell m}(\vec{x}) = \sqrt{\frac{2}{\pi}}\ k \ j_{\ell}(k r) Y_{\ell m}(\theta,\phi). \end{equation}
This function obeys the normalization condition \begin{equation} \int_0^\infty dk \sum_{\ell = 0}^\infty \sum_{m=-\ell}^\ell \psi^\star_{k \ell m}(\vec{x}) \psi_{k \ell m}(\vec{x}') = \delta^{(3)}(\vec{x}-\vec{x}') \end{equation}
To see this we will use the following identities: \begin{equation} \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell Y^\star_{\ell m}(\theta',\phi') Y_{\ell m}(\theta \phi) = \frac{1}{\sin \theta} \delta(\theta-\theta')\delta(\phi-\phi') \end{equation} and \begin{equation} \int_0^\infty dk k^2 j_\ell (kr) j_\ell (k r') = \frac{\pi}{2 r^2} \delta(r-r') \end{equation}
Putting everything together we have \begin{eqnarray} \int_0^\infty dk \sum_{\ell = 0}^\infty \sum_{m=-\ell}^\ell \psi^\star_{k \ell m}(\vec{x}) \psi_{k \ell m}(\vec{x}') &=& \frac{2}{\pi} \sum_{\ell =-\infty}^\infty \sum_{m=-\ell}^{\ell} Y^\star_{\ell m}(\theta',\phi') Y_{\ell m}(\theta, \phi) \left(\int_0^\infty dk k^2 j_\ell (kr) j_\ell (k r')\right) \\ &=& \frac{1}{ r^2 \sin(\theta)} \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi') \\ &=& \delta^{(3)}(\vec{x}-\vec{x}') \end{eqnarray} To see that the last line follows from the line before, note that \begin{equation} 1 = \int d^3 x \delta^{(3)}(\vec{x}-\vec{x}') = \int_0^\infty dr \int_0^\pi d\theta \int_0^{2\pi} d \phi r^2 \sin \theta \delta^{(3)} (\vec{x}-\vec{x}') \end{equation} which is consistent with the decomposition of $\delta^{(3)} (\vec{x}-\vec{x}')$ used above.
Note that there is also the 'backwards' normalization condition (which follows from a similar argument) \begin{equation} \int d^3 x\ \psi_{k \ell m}(\vec{x}) \psi_{k' \ell' m'}(\vec{x}) = \delta(k-k') \delta_{\ell \ell'}\delta_{mm'} \end{equation}\begin{equation} \int d^3 x\ \psi_{k \ell m}(\vec{x})^\star \psi_{k' \ell' m'}(\vec{x}) = \delta(k-k') \delta_{\ell \ell'}\delta_{mm'} \end{equation} (In bra ket language, what we've shown schematically is that $\sum_A |A \rangle \langle A | = \sum_B |B \rangle \langle B | = 1$, where $A$ is the $k,\ell,m$ basis and $B$ is the $\vec{x}$ basis).
One last point is how to see that the normalization condition still holds even after expressing $\psi_{k \ell m}(\vec{x})$ to the plane wave basis. In fact this is pretty easy based on what we already have:
\begin{eqnarray} \delta(k-k') \delta_{\ell \ell'}\delta_{mm'} &=& \int d^3 x\ \psi_{k \ell m}(\vec{x})^\star \psi_{k' \ell' m'}(\vec{x}) \\ &=& \int \frac{d^3 p}{(2\pi)^3} \frac{d^3 p'}{(2\pi)^3}d^3 x\ e^{i (\vec{p} - \vec{p}')\cdot \vec{x}} \phi_{k\ell m}(\vec{p}')^\star \phi_{k' \ell' m'}(\vec{p}) \\ &=& \int \frac{d^3 p}{(2\pi)^3}\phi_{k\ell m}(\vec{p})^\star \phi_{k' \ell' m'}(\vec{p}) \end{eqnarray} You could also in principle verify this by working out the $\phi_{k \ell m}(\vec{p})$ explicitly using the plane wave expansion, plugging the explicit formulas into the above expression, and show that it all works. But, that's sort of like showing $1+1=2$ by adding and then subtracting some ugly number to both sides.
Let me know if there anything is unclear.