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In a Klein-Gordon field with Lagrangian density $$\mathcal{L} = \frac{1}{2} \left[\dot{\phi}^2 - (\nabla\phi)^2 - m^2\phi^2\right]$$ the Hamiltonian is given by \begin{align} H &= \int\operatorname{d}^{d-1}x\,\frac{1}{2}\left[\pi^2(\mathbf{x}) + (\nabla \phi(\mathbf{x}))^2 + m^2\phi^2(\mathbf{x})\right]\\ & = \int\operatorname{d}^{d-1}k\, \frac{1}{2}\left[\tilde{\pi}_+^2(\mathbf{k}) + (k^2+m^2)\tilde{\phi}_+^2(\mathbf{k}) + \tilde{\pi}_-^2(\mathbf{k}) + (k^2+m^2)\tilde{\phi}_-^2(\mathbf{k})\right] \end{align} where $\tilde{f}_{+}(\mathbf{k}) \equiv \int \frac{\operatorname{d}^{d-1}k}{(2\pi)^{(d-1)/2}} \cos(\mathbf{k}\cdot\mathbf{x})\, f(\mathbf{x})$ is the cosine transform of $f$ (the real part of the Fourier transform, symmetric under parity), and $\tilde{f}_{-}(\mathbf{k})$ is the sine transform (the imaginary part of the Fourier transform, anti-symmetric under parity). The canonical (equal time) commutation relations are then given by: \begin{align} \left[\tilde{\phi}_p(\mathbf{k}),\, \tilde{\phi}_{p'}(\mathbf{k}')\right] & = 0, \\ \left[\tilde{\pi}_p(\mathbf{k}),\, \tilde{\pi}_{p'}(\mathbf{k}')\right] & = 0,\ \mathrm{and} \\ \left[\tilde{\phi}_p(\mathbf{k}),\, \tilde{\pi}_{p'}(\mathbf{k}')\right] & = i\delta_{pp'}\,\delta^{d-1}\left(\mathbf{k}-\mathbf{k}'\right). \end{align} If we define the ladder operators in the usual way, $$ a_p(\mathbf{k}) = \tilde{\phi}_p(\mathbf{k})\frac{\sqrt[4]{k^2+m^2}}{\sqrt{2}} + i\tilde{\pi}_p(\mathbf{k}) \frac{1}{\sqrt{2}\,\sqrt[4]{k^2+m^2}},$$ The relationships become \begin{align} H-H_0 & = \sum_{p\in\{+,-\}}\int \operatorname{d}^{d-1}k\,\sqrt{k^2+m^2}\, a_p^\dagger(\mathbf{k})\, a_p(\mathbf{k}) \\ \left[a_p(\mathbf{k}),\, a_{p'}^\dagger\left(\mathbf{k}'\right)\right] & = \delta_{pp'}\,\delta^{d-1}\left(\mathbf{k}-\mathbf{k}'\right) \\ \tilde{\phi}_p(\mathbf{k}) & = \frac{a_p(\mathbf{k}) + a_p^\dagger(\mathbf{k})}{\sqrt{2}\,\sqrt[4]{k^2+m^2}} \\ \tilde{\pi}_p(\mathbf{k}) & = \frac{a_p(\mathbf{k}) - a_{p}^\dagger(\mathbf{k})}{\sqrt{2}}\sqrt[4]{k^2+m^2}, \end{align} where $H_0$ contains the divergent zero point energy.

The form of $H$ leads immediately to the total particle number operator $$ N \equiv \sum_{p\in\{+,-\}}\int \operatorname{d}^{d-1}k \, a_p^\dagger(\mathbf{k})\,a_p(\mathbf{k}). $$

In this formalism $a_p^\dagger(\mathbf{k})$ is called the creation operator because for an eigenstate of $N$ it increases number of particles with parity $p$ and momentum $\mathbf{k}$ by $1$, and $a_p(\mathbf{k})$ the annihilation for similar reasons. That's not all they do, of course. When acting on particle number density eigenstates, they spoil the normalization. If the state isn't a number density eigenstate, though, it alters the state completely because the scale is different for states of different particle number.

In the case of $d=1$, the simple harmonic oscillator, it's straightforward to construct a creation operator that preserves normalization on any state. Two alternatives are: \begin{align} A_p^\dagger(\mathbf{k}) & = \frac{1}{\sqrt{N_p}} a_p^\dagger,\ \mathrm{or} \\ A_p^\dagger(\mathbf{k}) & = a_p^\dagger \frac{1}{\sqrt{N_p+1}}. \end{align} How do I generalize the idea of a "pure" creation operator to a continuous quantum field theory? Ideally, it would satisfy something like \begin{align} |f,\psi\rangle & = \sum_{p\in\{+,-\}}\int \operatorname{d}^{d-1}k\, \tilde{f}_p(\mathbf{k})\, A_p^\dagger(\mathbf{k})\,|\psi\rangle\\ \langle f,\psi|f,\psi\rangle & = 1 \end{align} as long as $$\sum_{p\in \{+,-\}} \int \operatorname{d}^{d-1}k\ \tilde{f}_p^2(\mathbf{k}) = 1,$$ so that it could be used to construct arbitrary ensembles of multi particle states in a straightforward way.

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  • $\begingroup$ The reason for using sin/cosine transforms instead of Fourier: this way the mode-space field operators are Hermitian. Using the regular Fourier transform would have yielded operators that have to satisfy $\tilde{\phi}^\dagger (\mathbf{k}) = \tilde{\phi}(-\mathbf{k})$, which is not as obviously Hermitian. $\endgroup$ – Sean E. Lake Nov 29 '17 at 4:50
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I've been working on this since, and the answer I've come up with is a qualified no. The qualification is that you can construct an operator that increases particle number and maintains normalization, but it doesn't maintain that property when superposed. One exception is when the state being operated on is an eigenstate of the particle number operator.

The reasoning is this. We construct a QFT as an infinite number of $d=1$ simple harmonic oscillators based on a momentum space grid $\mathbf{k}_i$. Instead of using sine and cosine transforms and separating by parity, it's easier to use a multi-demensional Hartley transform $$\tilde{\phi}(\mathbf{k}) \equiv \int \frac{\mathrm{d}^{d-1}x}{(2\pi)^{(d-1)/2}} \phi(\mathbf{x})\,\sqrt{2}\cos\left(\mathbf{k}\cdot\mathbf{x}-\frac{\pi}{4}\right),$$ which is basically saying $\tilde{\phi}\equiv\tilde{\phi}_++\tilde{\phi}_-$.

On this grid, the Hamiltonian becomes $$H = \sum_i \sqrt{m^2+k_i^2}\, a^\dagger_i\,a_i .$$ At each site we can define a pure creation operatorn as $$A_i^\dagger \equiv \sqrt{\left[a_i^\dagger a_i\right]^+}\,a_i^\dagger $$ where $\left[a_i^\dagger a_i\right]^+$ is the Moore-Penrose pseudoinverse of $\left[a_i^\dagger a_i\right]$ (basically invert the non-zero eigenvalues). These operators have the properties \begin{align} A_iA_j^\dagger &= \delta_{ij}+(1-\delta_{ij}) A_j^\dagger A_i \\ A_i^\dagger A_j &= \delta_{ij} \Pi_{\bar{0},i} + (1-\delta_{ij})A_jA_i^\dagger, \end{align} where $\Pi_{\bar{0},i}$ is the projection operator onto the complement of the ground state of the $i^{\mathrm{th}}$ site.

In order to transition to the continuum field limit, we make the substitutions $a_i\rightarrow a(\mathbf{k})\sqrt{\mathrm{d}^3k}$ and $A_i\rightarrow A(\mathbf{k})\sqrt{\mathrm{d}^3k}$, which gives \begin{align} A(\mathbf{k})\,A^\dagger\left(\mathbf{k}'\right) &= \delta\left(\mathbf{k}-\mathbf{k}'\right)+1_{\mathbf{k}\neq\mathbf{k}'} A^\dagger\left(\mathbf{k}'\right)\, A(\mathbf{k}) \\ A^\dagger(\mathbf{k})\, A\left(\mathbf{k}'\right) &= \delta\left(\mathbf{k}-\mathbf{k}'\right) \Pi_{\bar{0}}(\mathbf{k}) + 1_{\mathbf{k}\neq\mathbf{k}'} A\left(\mathbf{k}'\right)\, A^\dagger(\mathbf{k}) \end{align}

The above identities are enough to show that if \begin{align} |f,\psi\rangle &\equiv \int \mathrm{d}^{d-1}k\, \tilde{f}(\mathbf{k})\, A^\dagger(\mathbf{k})\,|\psi\rangle,\ \mathrm{then} \\ \langle f,\psi|f,\psi\rangle & =\int \mathrm{d}^{d-1}k\,\mathrm{d}^{d-1}k'\, \tilde{f}^*(\mathbf{k})\, \tilde{f}\left(\mathbf{k}'\right)\, \langle\psi|A(\mathbf{k})\,A^\dagger\left(\mathbf{k}'\right)|\psi\rangle \\ &= \int \mathrm{d}^{d-1}k\,\mathrm{d}^{d-1}k'\, \tilde{f}^*(\mathbf{k})\, \tilde{f}\left(\mathbf{k}'\right)\, \langle\psi|\delta\left(\mathbf{k}-\mathbf{k}'\right)+1_{\mathbf{k}\neq\mathbf{k}'} A^\dagger\left(\mathbf{k}'\right)\, A(\mathbf{k})|\psi\rangle \\ &= \int \mathrm{d}^{d-1}k\, f^*(\mathbf{k})\,f(\mathbf{k}) + \int_{\mathbf{k}\neq\mathbf{k}'} \mathrm{d}^{d-1}k\,\mathrm{d}^{d-1}k'\, \tilde{f}^*(\mathbf{k})\, \tilde{f}\left(\mathbf{k}'\right)\, \langle\psi| A^\dagger\left(\mathbf{k}'\right)\, A(\mathbf{k})|\psi\rangle. \tag1 \end{align} The necessary conditions for the creation operator maintaining normalization are for the second term in (1) to vanish and the first term being $1$. Ways for the second term in (1) to vanish include:

  1. $|\psi\rangle$ being an eigenstate of the total particle number operator,
  2. $f(\mathbf{k})$ being concentrated on a single value of $\mathbf{k}$, and
  3. the support of $f^*(\mathbf{k})\,f\left(\mathbf{k}'\right)$ not overlapping with the support of $\langle\psi|A^\dagger\left(\mathbf{k}'\right)\,A(\mathbf{k})|\psi\rangle$ for $\mathbf{k}\neq\mathbf{k}'$.
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