Electric–magnetic duality as a quantum operator and more symmetries of U(1) gauge theory

Abstract

We promote the Noether charge of the electric–magnetic duality symmetry of U(1) gauge theory, “G”, to a quantum operator. We construct ladder operators, $D_{(\pm )a}^\dagger (k)$ and $D_{(\pm )a}(k)$, which create and annihilate the simultaneous quantum eigenstates of the quantum Hamiltonian (or number) and the electric–magnetic duality operators, respectively. Therefore, all the quantum states of the U(1) gauge fields can be expressed in the form of $|E,g\rangle$, where E is the energy of the state, and g is the eigenvalue of the quantum operator G, where the g is quantized in the unit of 1. We also show that ten independent bilinears comprised of the creation and the annihilation operators can form $\mathrm{SO}(2,3)$, which is as demonstrated in Dirac’s paper published in 1962. The number operator and the electric–magnetic duality operator are members of the $\mathrm{SO}(2,3)$ family of generators. We note that there are two more generators that commute with the number operator(or Hamiltonian). We prove that these generators are, indeed, symmetries of the action in U(1) gauge field theory.

Appendices

Appendix A: Quantization with the electric–magnetic duality generator and the quantum operators and states

The electric–magnetic duality generator in momentum space is given by

$$\begin{aligned} G=\frac{i}{2}\int {\mathrm{d}}^3 k \epsilon _{abc}\left[ E_a(k) \frac{k_b}{k^2} E_c(-k) + A_a(k) k_b A_c(-k) \right] , \end{aligned}$$

(37)

where $E_a(k)$ are the electric fields and $A_a(k)$ are the gauge fields. They are a canonical pair and so satisfy the following Poisson bracket relation:

$$\begin{aligned} \{ A_b(x),E_a(y)\}=\delta _{ab}\delta ^{(3)}(x-y). \end{aligned}$$

(38)

The two fields satisfy the Gauss constraint $\partial _a E^a=0$ and this ensures that they are transverse fields. Therefore, the Poisson bracket relation can be modified as

$$\begin{aligned} \left\{ A^{T}_b(x),E^{T}_a(y)\right\} =\Pi _{ab}(x)\delta ^{(3)}(x-y), \end{aligned}$$

(39)

where the $\Pi _{ab}(x)=\left( \delta _{ab}-\frac{\partial _a \partial _b}{\nabla ^2}\right)$ is the project operator.

Quantization Maxwell’s theory is mathematically a collection of two independent harmonic oscillators:

$$\begin{aligned} A_a(x)=\frac{1}{(2\pi )^{\frac{3}{2}}}\int {\mathrm{d}}^3 k \exp (ik_b x_b)\sum _{A=1,2}q_A(k)e^A_a(k), \nonumber \\ E_a(x)=\frac{1}{(2\pi )^{\frac{3}{2}}}\int {\mathrm{d}}^3 k \exp (ik_b x_b)\sum _{A=1,2}p^A(k)e^a_A(k), \end{aligned}$$

(40)

where the indices a, b... are three-dimensional spatial indices and A, B... are the polarization indices. The $e^a_A(k)$ is the polarization vector, $e^a_A(k)k_a=0$ to ensure that the fields $A_a(x)$ and $E_a(x)$ are transverse and $e^a_A(k)e^B_a(-k)=\delta ^B_A$. $p^A(k)=p^A(-k)^\star$, $q^A(k)=q^A(-k)^\star$ and $e^A_a(k)=e^A_a(-k)^\star$, because the fields are real. $e^A_b(k_1)e^a_A(-k_1)\equiv \Pi ^a_b(k_1)=\delta ^a_b-\frac{k_ak_b}{k^2}$.

The Fourier transform

$$\begin{aligned} \phi (x)=\frac{1}{(2\pi )^{\frac{3}{2}}}\int {\mathrm{d}}^3 k \exp (ik_b x_b)\phi (k), \end{aligned}$$

(41)

defines the fields in momentum space as

$$\begin{aligned} A_a(k)=q_A(k)e^A_a(k)\mathrm{\ \ and\ \ }E^a(k)=p^A(k)e_A^a(k). \end{aligned}$$

(42)

Their Poisson brackets are given by

$$\begin{aligned} \{ A_b(k_1),E^a(k_2) \}= & {} \Pi _a^b(k_1)\delta ^{(3)}(k_1+k_2), \nonumber \\ \{ q_A(k_1),p^B(k_2) \}= & {} \delta _A^B\delta ^{(3)}(k_1+k_2). \end{aligned}$$

(43)

We define creation and annihilation operators as

$$\begin{aligned} a_A(k)= & {} \frac{1}{\sqrt{2}} \left( \sqrt{|k|}q_A(k)+\frac{i}{\sqrt{|k|}}p_A(k) \right) , \nonumber \\ a^\dagger _A(k)= & {} \frac{1}{\sqrt{2}} \left( \sqrt{|k|}q_A(-k)-\frac{i}{\sqrt{|k|}}p_A(-k) \right) , \end{aligned}$$

(44)

and the inverse relations are given by

$$\begin{aligned} q_A(k)=\frac{1}{\sqrt{2|k|}}(a_A(k)+a^\dagger _A(-k)), \nonumber \\ p_A(k)=-i\sqrt{\frac{|k|}{2}}(a_A(k)-a^\dagger _A(-k)). \end{aligned}$$

(45)

The final forms of the Poisson bracket are given by

$$\begin{aligned} \{ {\mathcal {A}}^\dagger _d(k),G \}= & {} \epsilon _{abc}\frac{k_b}{|k|}\Pi _{dc}(k)\mathcal A_a^\dagger (k), \nonumber \\ \{ {\mathcal {A}}_d(k),G \}= & {} \epsilon _{abc}\frac{k_b}{|k|}\Pi _{dc}(k){\mathcal {A}}_a(k), \nonumber \\ \{ {\mathcal {A}}_a(k),{\mathcal {A}}^\dagger _b(k^\prime ) \}= & {} -i\Pi _{ab}(k) \delta ^{(3)}(k-k^\prime ), \nonumber \\ \{ H,{\mathcal {A}}_a(k) \}= & {} i|k|{\mathcal {A}}_a(k), \nonumber \\ \{ H,{\mathcal {A}}^\dagger _a(k) \}= & {} -i|k|{\mathcal {A}}^\dagger _a(k), \end{aligned}$$

(46)

where ${\mathcal {A}}_a(k)=e^A_a(-k) a_A(k)$ and $\mathcal A^\dagger _a(k)=e^A_a(k) a^\dagger _A(k)$. The quantization of the fields is performed by switching the Poisson brackets to the commutators as $\{{\ \ }\} \rightarrow -i[{\ \ }]$ and promoting all fields to quantum operators. The forms of the Hamiltonian and electric–magnetic duality operators are given by

$$\begin{aligned} H=\int {\mathrm{d}}^3k |k|a^\dagger _A(k)a^A(k)\equiv \int {\mathrm{d}}^3k |k| \mathcal A^\dagger _a(k){\mathcal {A}}^a(k), \end{aligned}$$

(47)

and

$$\begin{aligned} G= & {} {-}{i}\int {\mathrm{d}}^3k \frac{\epsilon _{abc}k_b e^A_a(k)e^B_c(-k)}{|k|} a^\dagger _A(k) a_B(k)\nonumber \\\equiv & {} {-}{i}\int {\mathrm{d}}^3k \frac{\epsilon _{abc}k_b }{|k|} \mathcal A^\dagger _a(k) {\mathcal {A}}_c(k). \end{aligned}$$

(48)

The annihilation and creation operators are ladders for the Hamiltonian, but those do not increase or decrease the eigenvalues of the operator, G. To find simultaneous ladders for the H and G, we define

$$\begin{aligned} D^\dagger _{(\mp )c}(k)= & {} {\mathcal {A}}^\dagger _c(k)\pm i\epsilon _{abc}\frac{k_b}{|k|}{\mathcal {A}}^\dagger _a(k), \end{aligned}$$

(49)

$$\begin{aligned} D_{(\mp )c}(k)= & {} {\mathcal {A}}_c(k)\mp i\epsilon _{abc}\frac{k_b}{|k|}{\mathcal {A}}_a(k). \end{aligned}$$

(50)

Then, the commutation relations can be modified to

$$\begin{aligned} \left[ D^\dagger _{(\mp )c}(k),G\right]= & {} \pm D^\dagger _{(\mp )c}(k), {\ \ \ }[D_{(\mp )c}(k),G]={\mp } D_{(\mp )c}(k), \nonumber \\ \left[ H,D^\dagger _{(\mp )c}(k)\right]= & {} |k|D^\dagger _{(\mp )c}(k), {\ \ \ }[H,D_{(\mp )c}(k)]=-|k|D_{(\mp )c}(k), \nonumber \\ \left[ D_{(\mp )c}(k),D^\dagger _{(\mp )d}(k^\prime )\right]= & {} 2\delta ^{(3)}(k-k^\prime )\left( \Pi _{cd}(k)\pm i \epsilon _{ced}\frac{k_e}{|k|}\right) . \end{aligned}$$

(51)

Appendix B: Construction of the $[\mathrm{SO}(2,3)]$ group from bilinear operators from $D_{(\pm )a}(k)$ and $D^\dagger _{(\pm )a}(k)$

We start with new definitions of $D_{(\pm )a}(k)$ and $D^\dagger _{(\pm )a}(k)$ for further conveience, which are given by

$$\begin{aligned}&D^\dagger _{(\xi )c}(k)={\mathcal {A}}^\dagger _c(k)- i\xi \epsilon _{abc}\frac{k_b}{|k|} {\mathcal {A}}^\dagger _a(k), \end{aligned}$$

(52)

$$\begin{aligned}&D_{(\xi ^\prime )c}(k)={\mathcal {A}}_c(k)+ i\xi ^\prime \epsilon _{abc}\frac{k_b}{|k|}{\mathcal {A}}_a(k), \end{aligned}$$

(53)

where for $\xi =\pm 1$, $D^\dagger _{(\xi )c}(k)$ represents $D^\dagger _{(\pm )c}(k)$ and for $\xi ^\prime =\pm 1$, $D_{(\xi ^\prime )c}(k)$ represents $D_{(\pm )c}(k)$, respectively. The bilinear operators constructed out of the operators $D_{(\xi )c}(k)$ and $D^\dagger _{(\xi )c}(k)$ have the following forms:

$$\begin{aligned}&\int D^\dagger _{(\xi )a}(k)D^\dagger _{(\xi ')b}(k) {\mathrm{d}}^3 k, \end{aligned}$$

(54)

$$\begin{aligned}&\int D_{(\xi )a}(k)D_{(\xi ')b}(k) {\mathrm{d}}^3 k, \end{aligned}$$

(55)

$$\begin{aligned}&\int D_{(\xi )a}(k)D^\dagger _{(\xi ')b}(k){\mathrm{d}}^3k, \end{aligned}$$

(56)

$$\begin{aligned}&\int D^\dagger _{(\xi )a}(k)D_{(\xi )b}(k){\mathrm{d}}^3 k, \end{aligned}$$

(57)

where the third and the fourth operators are the same up to a constant(a c-number). Because we are interested in their commutation relations, we regard the third and the fourth operators as the same. One can also classify the operators into their trace, anti-symmetric and traceless-symmetric parts.

First of all, we discuss their trace parts, which are listed below:

$$\begin{aligned} \int {\mathrm{d}}^3 k{\ } D^\dagger _{(\xi )a}(k)D^\dagger _{(\xi ')a}(k)= & {} (1-\xi \xi ^\prime )\int {\mathrm{d}}^3 k{\ } {\mathcal {A}}^\dagger _{a}(k){\mathcal {A}}^\dagger _{a}(k) , \end{aligned}$$

(58)

$$\begin{aligned} \int {\mathrm{d}}^3 k{\ }D_{(\xi )a}(k)D_{(\xi ')a}(k)= & {} (1-\xi \xi ^\prime )\int {\mathrm{d}}^3 k{\ }{\mathcal {A}}_{a}(k)\mathcal A_{a}(k) , \end{aligned}$$

(59)

and

$$\begin{aligned} \int {\mathrm{d}}^3 k{\ } D^\dagger _{(\xi )a}(k)D_{(\xi ')a}(k)= & {} (1+\xi \xi ')\int {\mathrm{d}}^3 k{\ } {\mathcal {A}}^\dagger _{a}(k){\mathcal {A}}_{a}(k)\nonumber \\&-i(\xi +\xi ')\int {\mathrm{d}}^3 k{\ } \epsilon _{efa}\frac{k_f}{|k|}{\mathcal {A}}^\dagger _{e}(k)\mathcal A_{a}(k). \end{aligned}$$

(60)

When $\xi$ and $\xi ^\prime$ take the same sign, the bilinear operators Eqs. (58) and (59) become null identically. If they take different signs as equations $(\xi ,\xi ^\prime )=(+1,-1)$ or $(\xi ,\xi ^\prime )=(-1,+1)$, then (58) and (59)are proportional to the trace of the primitive creation and annihilation operators, respectively. Because the operators are not Hermitian, we constitute their appropriate linear combinations to be Hermitian operators. They are nothing but the $K_2$ and the $Q_2$ operators listed below.

$$\begin{aligned} K_{2}= & {} \frac{i}{8} \int {\mathrm{d}}^{3}k \big (D^{\dagger }_{(+)a}(k)D^{\dagger }_{(-)a}(k)-D_{(+)a}(k)D_{(-)a}(k)\big ) \nonumber \\= & {} \frac{i}{4} \int {\mathrm{d}}^{3}k \big ({\mathcal {A}}^{\dagger }_{a}(k){\mathcal {A}}^{\dagger }_{a}(k)-{\mathcal {A}}_{a}(k){\mathcal {A}}_{a}(k)\big ), \end{aligned}$$

(61)

$$\begin{aligned} Q_{2}= & {} -\frac{1}{8} \int {\mathrm{d}}^{3}k \big (D^{\dagger }_{(+)a}(k)D^{\dagger }_{(-)a}(k)+D_{(+)a}(k)D_{(-)a}(k)\big ) \nonumber \\= & {} -\frac{1}{4}\int {\mathrm{d}}^{3}k \big (\mathcal A^{\dagger }_{a}(k){\mathcal {A}}^{\dagger }_{a}(k)+\mathcal A_{a}(k){\mathcal {A}}_{a}(k)\big ). \end{aligned}$$

(62)

Only when $\xi$ and $\xi ^\prime$ take the same sign, does Eq. (60) become non-trivial and is a linear combination of the number operator and the electric–magnetic duality operator. Together with this, we examine the anti-symmetric parts of the bilinear operators, which are given by

$$\begin{aligned} \int {\mathrm{d}}^{3}k{\ }D^\dagger _{(\xi )[a}D^\dagger _{b](\xi ')}= & {} i(-\xi +\xi ')\int {\mathrm{d}}^{3}k{\ }\frac{k_f}{|k|}(-\epsilon _{efb}{\mathcal {A}}^\dagger _{a}\mathcal A^\dagger _{e}\nonumber \\&+\epsilon _{efa}{\mathcal {A}}^\dagger _{b}\mathcal A^\dagger _{e}), \end{aligned}$$

(63)

$$\begin{aligned} \int {\mathrm{d}}^{3}k{\ }D_{(\xi )[a}D_{b](\xi ')}= & {} i(-\xi +\xi ')\int {\mathrm{d}}^{3}k{\ }\frac{k_f}{|k|}(\epsilon _{efb}{\mathcal {A}}_{a}\mathcal A_{e}\nonumber \\&-\epsilon _{efa}{\mathcal {A}}_{b}{\mathcal {A}}_{e}), \end{aligned}$$

(64)

$$\begin{aligned} \int {\mathrm{d}}^{3}k{\ }D^\dagger _{(\xi )[a}D_{b](\xi ')}= & {} \int {\mathrm{d}}^{3}k{\ }\left[\vphantom {\left. -\epsilon _{efa}\frac{k_f}{|k|}(i\xi '\mathcal A^\dagger _{b}{\mathcal {A}}_{e}+i\xi {\mathcal {A}}^\dagger _{e}\mathcal A_{b})\right] }{\mathcal {A}}^\dagger _{a}{\mathcal {A}}_{b}-\mathcal A^\dagger _{b}{\mathcal {A}}_{a}\right. \nonumber \\&\left. +\xi \xi '\frac{k_dk_f}{|k|}\mathcal A^\dagger _{c}\mathcal A_{e}(\epsilon _{cda}\epsilon _{efb}-\epsilon _{cdb}\epsilon _{efa})\right. \nonumber \\&+\epsilon _{efb}\frac{k_f}{|k|}(i\xi '\mathcal A^\dagger _{a}{\mathcal {A}}_{e}+i\xi {\mathcal {A}}^\dagger _{e}\mathcal A_{a})\nonumber \\&\left. -\epsilon _{efa}\frac{k_f}{|k|}(i\xi '\mathcal A^\dagger _{b}{\mathcal {A}}_{e}+i\xi {\mathcal {A}}^\dagger _{e}\mathcal A_{b})\right] . \end{aligned}$$

(65)

Equations (63) and (64) turns out not to be independent operators. Once we contract the operators with a tensor $\frac{k_c}{|k|}\epsilon _{abc}$, they become proportional to Eqs. (58) and (59), respectively. The same contractions acting on Eq. (65) leads another linear combination of the number and electric–magnetic duality operators. Therefore, appropriate combinations of Eqs. (65) and (60) provide the operators:

$$\begin{aligned} S_{3}= & {} \frac{1}{16} \int {\mathrm{d}}^{3}k (D^{\dagger }_{(+)a}(k)D_{(+)a}(k)+D^{\dagger }_{(-)a}(k)D_{(-)a}(k)\\&+D_{(-)a}(k)D^{\dagger }_{(-)a}(k) +D_{(+)a}(k)D^{\dagger }_{(+)a}(k))\\= & {} \frac{1}{4} \int {\mathrm{d}}^{3}k (\mathcal A^{\dagger }_{a}(k){\mathcal {A}}_{a}(k)+{\mathcal {A}}_{a}(k)\mathcal A^{\dagger }_{a}(k))=\frac{1}{2}{\hat{N}}, \\ L_{2}= & {} -\frac{1}{8} \int {\mathrm{d}}^{3}k (D^{\dagger }_{(+)a}(k)D_{(+)a}(k)-D^{\dagger }_{(-)a}(k)D_{(-)a}(k)) \\= & {} -\frac{1}{2i}\int {\mathrm{d}}^{3}k \epsilon _{efa}\frac{k_f}{|k|}{\mathcal {A}}^{\dagger }_{e}(k) \mathcal A_{a}(k)=-\frac{1}{2} \hat{G}. \end{aligned}$$

To construct the symmetric parts of the bilinear operators, we utilize the following identities:

$$\begin{aligned} D^\dagger _{(\xi )(a}D^\dagger _{b)(\xi ')}= & {} 2\mathcal A^\dagger _{a}{\mathcal {A}}^\dagger _{b}-2\xi \xi ' \epsilon _{cdb}\epsilon _{efa}\frac{k_dk_f}{|k|^2}\mathcal A^\dagger _{c}\mathcal A^\dagger _{e}\\&-i(\xi +\xi ')\frac{k_f}{|k|}\big (\epsilon _{efb}\mathcal A^\dagger _{a}{\mathcal {A}}^\dagger _{e}+\epsilon _{efa}\mathcal A^\dagger _{b}{\mathcal {A}}^\dagger _{e}\big ),\\ D_{(\xi )(a}D_{b)(\xi ')}= & {} 2{\mathcal {A}}_{a}{\mathcal {A}}_{b}-2\xi \xi ' \epsilon _{cdb}\epsilon _{efa}\frac{k_dk_f}{|k|^2}\mathcal A_{c}\mathcal A_{e}\\&+i(\xi +\xi ')\frac{k_f}{|k|}\big (\epsilon _{efb}\mathcal A_{a}{\mathcal {A}}_{e}+\epsilon _{efa}{\mathcal {A}}_{b}{\mathcal {A}}_{e}\big ),\\ D^\dagger _{(\xi )(a}D_{b)(\xi ')}= & {} \mathcal A^\dagger _{a}{\mathcal {A}}_{b}+{\mathcal {A}}^\dagger _{b}\mathcal A_{a}+\xi \xi '\frac{k_dk_f}{|k|}{\mathcal {A}}^\dagger _{c}\mathcal A_{e}\big (\epsilon _{cda}\epsilon _{efb}+\epsilon _{cdb}\epsilon _{efa}\big )\\&+\epsilon _{efb}\frac{k_f}{|k|}\big (i\xi '\mathcal A^\dagger _{a}{\mathcal {A}}_{e}-i\xi {\mathcal {A}}^\dagger _{e}\mathcal A_{a}\big )\\&+\epsilon _{efa}\frac{k_f}{|k|}\big (i\xi '\mathcal A^\dagger _{b}{\mathcal {A}}_{e}-i\xi {\mathcal {A}}^\dagger _{e}\mathcal A_{b}\big ). \end{aligned}$$

Because they are tensor components, they change under $\mathrm{SO}(3)$ spatial rotation. Therefore, we choose our frame as $\vec {k}=(0,0,k)$. then, the spatial index a in $D_{a(\xi )}(k)$ and $D^\dagger _{a(\xi )}(k)$ can take 1 or 2. All possible (linearly independent of one another) choices are listed below:

$$\begin{aligned} L_{1}= & {} \frac{1}{8}\int {\mathrm{d}}^{3}k \big [D^{\dagger }_{+(1}(k)D_{2)-}(k)+D^{\dagger }_{-(1}(k)D_{2)+}(k)],\\ L_{3}= & {} -\frac{i}{8}\int {\mathrm{d}}^{3}k [D^{\dagger }_{+(1}(k)D_{2)-}(k)-D^{\dagger }_{-(1}(k)D_{2)+}(k)\big ],\\ K_{1}= & {} \frac{i}{16}\int {\mathrm{d}}^{3}k \big [D^{\dagger }_{+(1}(k)D^{\dagger }_{2)+}(k)-D^{\dagger }_{-(1}(k)D^{\dagger }_{2)-}(k)\\&+D_{-(1}(k)D_{2)-}(k)-D_{+(1}(k)D_{2)+}(k)\big ],\\ K_{3}= & {} \frac{1}{8}\int {\mathrm{d}}^{3}k \big [D^{\dagger }_{+(1}(k)D^{\dagger }_{2)+}(k)+D^{\dagger }_{-(1}(k)D^{\dagger }_{2)-}(k)\\&+D_{-(1}(k)D_{2)-}(k)+D_{+(1}(k)D_{2)+}(k)\big ],\\ Q_{1}= & {} -\frac{1}{16}\int {\mathrm{d}}^{3}k \big [D^{\dagger }_{+(1}(k)D^{\dagger }_{2)+}(k)-D^{\dagger }_{-(1}(k)D^{\dagger }_{2)-}(k)\\&-D_{-(1}(k)D_{2)-}(k)+D_{+(1}(k)D_{2)+}(k)\big ],\\ Q_{3}= & {} \frac{i}{8}\int {\mathrm{d}}^{3}k \big [D^{\dagger }_{+(1}(k)D^{\dagger }_{2)+}(k)+D^{\dagger }_{-(1}(k)D^{\dagger }_{2)-}(k)\\&-D_{-(1}(k)D_{2)-}(k)-D_{+(1}(k)D_{2)+}(k)\big ].\\ \end{aligned}$$