Sub-bosonic (deformed) ladder operators

Abstract

The canonical operator ${\hat{a}}^{\dagger }$ (${\hat{a}}$) represents the ideal process of adding (subtracting) an exact amount of energy E to (from) a physical system in both elementary quantum mechanics and quantum field theory. This is a “sharp” notion in the sense that no variability around E is possible at the operator level. In this work, we present a class of deformed creation and annihilation operators that originates from a rigorous notion of fuzziness. This leads to deformed, sub-bosonic commutation relations inducing a simple algebraic structure with modified eigenenergies and Fock states. In addition, we investigate possible consequences of the introduced formalism in quantum field theories, as for instance, deviations from linearity in the dispersion relation for free quasibosons.

Appendix: Complex integration

In this appendix, we evaluate the functional

$$\begin{aligned} I_{k}[f']=\int ^{\infty }_{-\infty }\frac{x^k f'(x)}{\sqrt{x+1}} {\mathrm{d}}{x}. \end{aligned}$$

(41)

With the substitution $x+1 \rightarrow x$, we get

$$\begin{aligned} I_{k}[f']=\int ^{\infty }_{-\infty }\frac{(x-1)^k f'(x-1)}{\sqrt{x}}=\int ^{\infty }_{-\infty }\frac{g_{k}(x)}{\sqrt{x}} {\mathrm{d}}{x}. \end{aligned}$$

(42)

where $g_{k}(x)=(x-1)^k f'(x-1)$. We start by considering the following integration in the complex plane

$$\begin{aligned} I^{\gamma }_{k}[g]=\int _{\gamma }\frac{g_{k}(z)}{\sqrt{z}} {\mathrm{d}}{z}. \end{aligned}$$

(43)

Note that $\sqrt{z}$ has branch points at zero and at infinity. For the remainder of this appendix, we will work with the branch cut connecting these two branch points through the negative imaginary axis, that is, we restrict ourselves to $\theta \in [e^{\pi i/2},e^{3 \pi i/2})$. We will assume the following condition on the function $g_{k}(z)$:

• Condition (i): $g_{k}(z)$ is analytic in $\mathrm{H} \cup {\mathbb {R}}$, except for a finite number of removable singularities (poles) in $\mathrm{H}$. Here, $\mathrm{H}\equiv \left\{ z\in {\mathbb {C}} ~|~ \mathfrak {I}{(z)}> 0 \right\}$ is the upper-half of the complex plane.

Given that this condition is satisfied, let us evaluate expression (43) on the path $\gamma$ defined as

$$\begin{aligned} \gamma =\gamma _{-}+\gamma _{\varepsilon }+\gamma _{+}+\gamma _{R}, \end{aligned}$$

(44)

with the following parametrizations:

$$\begin{aligned} \gamma _{-}(r)&=re^{i\pi }, ~ r \in (R,\varepsilon ], \quad \gamma _{\varepsilon }(\theta )=\varepsilon e^{i\theta }, ~ \theta \in [0,\pi ], \\ \gamma _{+}(r)&=r,~ r \in [\varepsilon ,R),~~~~ \quad \gamma _{R}(\theta )=Re^{i \theta }, ~ \theta \in [\pi , 0], \end{aligned}$$

where we will take the limits $R \rightarrow \infty$ and $\varepsilon \rightarrow 0$. Let us now evaluate each of the integrals

$$\begin{aligned} I^{{\gamma _{-}}}_k[g]=&\,\int _{\gamma _{-}}\frac{g_{k}(z)}{\sqrt{z}} {\mathrm{d}}{z}= \lim _{\begin{array}{c} \varepsilon \rightarrow 0 \\ R \rightarrow \infty \end{array}}\int ^{\varepsilon }_{R}\frac{g_{k}(-r)}{\sqrt{re^{i\pi }}} {\mathrm{d}}{(-r)}\nonumber \\ =&\,\int ^{0}_{\infty }\frac{g_{k}(-r)}{\sqrt{-r}} {\mathrm{d}}{(-r)}\underset{(-r) \rightarrow x}{=} \int ^{0}_{-\infty }\frac{g_{k}(x)}{\sqrt{x}} {\mathrm{d}}{x} \end{aligned}$$

(45)

$$\begin{aligned} I^{\gamma _{+}}_k[g]=&\,\int _{\gamma _{+}}\frac{g_{k}(z)}{\sqrt{z}} {\mathrm{d}}{z}= \lim _{\begin{array}{c} \varepsilon \rightarrow 0 \\ R \rightarrow \infty \end{array}}\int ^{R}_{\varepsilon }\frac{g_{k}(r)}{\sqrt{r}} {\mathrm{d}}{r}\nonumber \\ \underset{r \rightarrow x}{=}&\, \int ^{\infty }_{0}\frac{g_{k}(x)}{\sqrt{x}} {\mathrm{d}}{x} \end{aligned}$$

(46)

$$\begin{aligned} I^{\gamma _{R}}_k[g]=&\,\int _{\gamma _{R}}\frac{g_{k}(z)}{\sqrt{z}} {\mathrm{d}}{z}= \lim _{\begin{array}{c} R \rightarrow \infty \end{array}}\int _{\gamma _{R}}\frac{g_{k}(Re^{i\theta })}{\sqrt{Re^{i\theta }}} {\mathrm{d}}{(R e^{i\theta })}\nonumber \\ =&\,\lim _{\begin{array}{c} R \rightarrow \infty \end{array}}i\int ^{\pi }_{0}\frac{g_{k}(Re^{i\theta })}{\sqrt{e^{i\theta }}} R^{1/2}{\mathrm{d}}{\theta } \end{aligned}$$

(47)

$$\begin{aligned} I^{\gamma _{\varepsilon }}_k[g]=&\,\int _{\gamma _{\varepsilon }}\frac{g_{k}(z)}{\sqrt{z}} {\mathrm{d}}{z}= \lim _{\begin{array}{c} \varepsilon \rightarrow 0 \end{array}}\int ^{0}_{\pi }\frac{g_{k}(\varepsilon e^{i\theta })}{\sqrt{\varepsilon e^{i\theta }}} {\mathrm{d}}{(\varepsilon e^{i\theta })}\nonumber \\ =&\,\lim _{\begin{array}{c} \varepsilon \rightarrow 0 \end{array}}i\int ^{\pi }_{0}\frac{g_{k}(Re^{i\theta })}{\sqrt{e^{i\theta }}} \varepsilon ^{1/2}{\mathrm{d}}{\theta }. \end{aligned}$$

(48)

Now, we also need to assume the following condition on g(z):

• Condition (ii): We associate with $g_{k}(z)$ the coefficients $\alpha ,\beta$ defined as

  $$\begin{aligned} g_{k}\big (Re^{i\theta }\big ) \underset{R \rightarrow \infty }{\sim } R^\alpha , \quad \text {and} \quad g_{k}\big (\varepsilon e^{i\theta }\big ) \underset{\varepsilon \rightarrow 0}{\sim } \varepsilon ^\beta . \end{aligned}$$

  (49)

  These coefficients must be such that $\alpha <-1/2$ and $\beta >-1/2$.

If condition (ii) is satisfied, we can set

$$\begin{aligned} I^{\gamma _{R}}_k [g]\underset{R \rightarrow \infty }{\rightarrow } 0, \quad \text {and} \quad I^{\gamma _{\varepsilon }}_{k}[g]\underset{\varepsilon \rightarrow 0}{\rightarrow } 0. \end{aligned}$$

(50)

Combining these results, Eq. (43) gives

$$\begin{aligned} I^{\gamma }_{k}[g]=I^{\gamma _{-}}_k[g]+I^{\gamma _{\varepsilon }}_{k}[g]+I^{\gamma _{+}}_{k}[g]+I^{\gamma _{R}}_{k}[g]=\int ^{\infty }_{-\infty }\frac{g_{k}(x)}{\sqrt{x}} {\mathrm{d}}{(x)}. \end{aligned}$$

(51)

However, we can also apply the residue theorem to Eq. (43):

$$\begin{aligned} I^{\gamma }_{k}[g]=2 \pi i \sum _{a \in H}\underset{z=a}{{\text {Res}}}g(z). \end{aligned}$$

(52)

Hence, equating Eqs. (51) and (52), we get

$$\begin{aligned} \int ^{\infty }_{-\infty }\frac{g_{k}(x)}{\sqrt{x}} {\mathrm{d}}{x}=2 \pi i \sum _{a \in H}\underset{z=a}{{\text {Res}}}~\frac{g_{k}(z)}{\sqrt{z}}, \end{aligned}$$

(53)

which ends the Proof of Proposition 1.