Appendix A Calculation of the matrix elements in the \(\vert n, k_x \rangle \)-basis
By using the explicit form of \(\Psi ^{(L_1)}_{n, k_x} (x, y)\) given by (48), it can readily be shown that
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,- \,i \,\frac{\partial }{\partial x} \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \ \nonumber \\= & {} \ k_x \,\delta (k^\prime _x - k_x) . \end{aligned}$$
(A1)
Furthermore, since \({\hat{p}}^{cons}_x ({\varvec{A}})\) reduces to the canonical momentum \({\hat{p}}^{can}_x\) in the \(L_1\)-gauge, we naturally have
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \ \nonumber \\= & {} \ k_x \,\delta (k^\prime _x - k_x) . \end{aligned}$$
(A2)
Next, the matrix element of the mechanical momentum operator becomes
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{mech}_x ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \ + \ e \,(\,- \,B \,y) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = k_x \,\delta (k^\prime _x - k_x) \, - \, e \,B \, \delta (k^\prime _x - k_x) \,\langle Y_n \,\vert \,y \,\vert \,Y_n \rangle . \end{aligned}$$
(A3)
Here, the matrix element \(\langle Y_n \,\vert \,y \,\vert \,Y_n \rangle \) can be evaluated as follows :
$$\begin{aligned} \langle Y_n \,\vert \,y \,\vert \, Y_n \rangle= & {} \ N^2_n \,\int _{- \,\infty }^\infty \,d y \,e^{\,- \,\frac{(y - y_0)^2}{l^2_B}} \,\nonumber \\\times & {} H_n \left( \frac{y - y_0}{l_B} \right) \,y \, H_n \left( \frac{y - y_0}{l_B} \right) \nonumber \\= & {} \ y_0 \ = \ \frac{k_x}{e \,B} . \end{aligned}$$
(A4)
We therefore find a remarkable relation as
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{mech}_x ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} k_x \,\delta (k^\prime _x - k_x) \ \nonumber \\- & {} \ e \,B \,\frac{k_x}{e \,B} \, \delta (k^\prime _x - k_x) \nonumber \\= & {} 0, \end{aligned}$$
(A5)
the physical significance of which is explained in the main text.
Next we turn to the matrix elements between the eigen-states \(\vert \,\Psi ^{(S)}_{n, k_x} \rangle \). With the use of the relation \(\vert \,\Psi ^{(S)}_{n, k_x} \rangle = U^\dagger \,\vert \,\Psi ^{(L_1)} \rangle \), we obtain
$$\begin{aligned} \langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \,\vert \,\Psi ^{(S)}_{n, k_x} \rangle \ = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,U \,{\hat{p}}^{can}_x \,U^\dagger \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle . \end{aligned}$$
(A6)
Noting that
$$\begin{aligned} U \,{\hat{p}}^{can}_x \,U^\dagger \ = \ {\hat{p}}^{can}_x \ - \ \frac{1}{2} \,e \,B \,y , \end{aligned}$$
(A7)
we therefore find that
$$\begin{aligned}&\langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \,\vert \,\Psi ^{(S)}_{n, k_x} \rangle \ = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{p}}^{can}_x \ - \ \frac{1}{2} \,e \,B \,y \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = k_x \,\delta (k^\prime _x - k_x) \ - \ \frac{1}{2} \,e \,B \,\,\frac{k_x}{e \,B} \, \delta (k^\prime _x - k_x) \ \nonumber \\&\quad = \ \frac{1}{2} \,k_x \,\delta (k^\prime _x - k_x) . \ \ \ \ \ \end{aligned}$$
(A8)
Similarly, we have
$$\begin{aligned}&\langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(S)}) \, \vert \,\Psi ^{(S)}_{n, k_x} \rangle \ \nonumber \\&\quad = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,U \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(S)}) \,U^\dagger \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle . \end{aligned}$$
(A9)
Taking care of the covariant gauge-transformation property of \({\hat{p}}^{cons}_x\), which means the relation \(U \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(S)}) \,U^\dagger \ = \ {\hat{p}}^{cons}_x ({\varvec{A}}^{(L_1)})\), we therefore find that
$$\begin{aligned} \langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(S)}) \, \vert \,\Psi ^{(S)}_{n, k_x} \rangle= & {} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{cons}_x ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\= & {} k_x \,\delta (k^\prime _x - k_x) . \end{aligned}$$
(A10)
For the matrix element of the mechanical momentum operator, we obtain
$$\begin{aligned}&\langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{mech}_x ({\varvec{A}}^{(S)}) \, \vert \,\Psi ^{(S)}_{n, k_x} \rangle \ \nonumber \\&\quad = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,U \,{\hat{p}}^{mech}_x ({\varvec{A}}^{(S)}) \, U^\dagger \vert \,\Psi ^{(S)}_{n, k_x} \rangle \nonumber \\&\quad = \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{mech}_x ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{p}}^{can}_x \ - \ e \,B \,y \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \ k_x \,\delta (k^\prime _x - k_x) \ - \ e \,B \,\,\frac{1}{e \,B} \,\, k_x \,\delta (k^\prime _x - k_x) \ = \ 0.\nonumber \\ \end{aligned}$$
(A11)
Here, we have used the fact that \({\hat{p}}^{mech}_x\) also transforms covariantly under a gauge transformation.
Next, we evaluate the matrix elements of the three OAM operators in the two \(\vert \,n, k_x \rangle \)-basis functions, \(\Psi ^{(L_1)}_{n, k_x} (x,y)\) and \(\Psi ^{(S)}_{n, k_x} (x,y)\). Although it may sound strange, this has never been done before. The reason is probably because the symmetric gauge eigen-states \(\vert \,\Psi ^{(S)}_{n,m} \rangle \) are the most natural and convenient basis to deal with the OAM operators, and one seldom paid attention to calculating the expectation values of the OAM operators between the eigen-functions \(\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) in the 1st Landau gauge.
With the use of the definitions of \({\hat{L}}^{cons}_z ({\varvec{A}})\) and \({\hat{L}}^{mech}_z ({\varvec{A}})\), we obtain
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{cons}_z ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \ = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\qquad \ - \ \frac{1}{2} \,e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \, \nonumber \\&\qquad + \, \frac{1}{2} \,e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle , \ \ \ \ \ \end{aligned}$$
(A12)
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{mech}_z ({\varvec{A}}^{(L_1)}) \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle = \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \, \nonumber \\&\qquad + \, e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle , \end{aligned}$$
(A13)
Similarly, using the relation \(\vert \,\Psi ^{(S)}_{n, k_x} \rangle = U^\dagger \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) together with the gauge-covariant transformation properties of \({\hat{L}}^{cons}_z ({\varvec{A}})\) and \({\hat{L}}^{mech}_z ({\varvec{A}})\), we obtain
$$\begin{aligned}&\langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{cons}_z ({\varvec{A}}^{(S)}) \, \vert \,\Psi ^{(S)}_{n, k_x} \rangle \ = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad \ - \ \frac{1}{2} \,e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \, + \, \frac{1}{2} \,e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle , \nonumber \\ \end{aligned}$$
(A14)
$$\begin{aligned}&\langle \Psi ^{(S)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{mech}_z ({\varvec{A}}^{(S)}) \, \vert \,\Psi ^{(S)}_{n, k_x} \rangle \nonumber \\&\quad \ = \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \, + \, e \,B \, \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle .\nonumber \\ \end{aligned}$$
(A15)
Thus, for evaluating the matrix elements of the three OAM operators in the \(\vert \,n, k_x \rangle \)-basis, we have only to know the following three matrix elements :
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle , \ \ \ \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle , \nonumber \\&\quad \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle . \end{aligned}$$
(A16)
As one can easily verify, the two of these matrix elements, i.e. \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) and \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) can be calculated without much difficulty. On the other hand, the calculation of the second matrix element \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x^2 \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) needs some care owing to the plane-wave nature of the eigen-states \(\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) along the x-direction. (Note that this cumbersome term appears in the matrix elements of \({\hat{L}}^{can}_z\) and \({\hat{L}}^{cons}_z ({\varvec{A}})\), but it does not in those of \({\hat{L}}^{mech}_z ({\varvec{A}})\).) First, note that the eigen-functions \(\Psi ^{(L_1)}_{n, k_x} (x,y)\) are normalized as
$$\begin{aligned}&\int \,d x \,d y \,\Psi ^{(L_1) *}_{n, k^\prime _x} (x,y) \,\Psi ^{(L_1)}_{n, k_x} (x,y) \nonumber \\&\quad = \frac{1}{2 \,\pi } \,\int _{-\,\infty }^\infty \,d x \, e^{\,- \, i \,(k^\prime _x - k_x) \,x} \, \nonumber \\&\quad \times \int _{- \,\infty }^\infty \,d y \,[Y_n (y)]^2 \ = \ \delta (k^\prime _x - k_x) . \end{aligned}$$
(A17)
Here, we have used the fact that \(Y_n (y)\) is normalized as \(\int _{- \,\infty }^\infty \,d y \,[Y_n (y)]^2 = 1\). For later convenience, we introduce the dimensionless wave function \(\psi _n (\xi )\) by
$$\begin{aligned} Y_n (y) \ \equiv \ \frac{1}{\sqrt{l_B}} \,\,\psi _n (\xi ) , \end{aligned}$$
(A18)
with \((y - y_0) \,/\,l_B = \xi \). The normalization of \(\psi (\xi )\) is then given by
$$\begin{aligned} \int _{- \,\infty }^\infty \, d \xi \,[\psi _n (\xi )]^2 \ = \ 1. \end{aligned}$$
(A19)
We begin with the calculation of the matrix element of the canonical OAM operator :
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \,\vert \, \Psi ^{(L_1)}_{n, k_x} \rangle \ = \ - \, i \,\,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x \,\frac{\partial }{\partial y} \ - \ y \,\frac{\partial }{\partial x} \,\vert \, \Psi ^{(L_1)}_{n, k_x} \rangle . \nonumber \\ \end{aligned}$$
(A20)
Let us first consider the first part in the r.h.s. Using the variables \(\xi = (y - l^2_B \,k_x) \,/\,l_B\) and \(\xi ^\prime = (y - l^2_B \,k^\prime _x) \,/\,l_B\) together with the relation \(\frac{\partial }{\partial y} = \frac{1}{l_B} \,\frac{\partial }{\partial \xi }\), we obtain
$$\begin{aligned}&- \,i \,\,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x \,\frac{\partial }{\partial y} \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = - \,i \,\left\{ \frac{1}{2 \,\pi } \,\int _{- \,\infty }^\infty \,d x \, x \,e^{\,- \,i \,(k^\prime _x - k_x) \,x} \,\right\} \nonumber \\&\quad \times \frac{1}{l_B} \,\, \int _{- \,\infty }^\infty \,d \xi \,\psi _n (\xi ^\prime ) \, \frac{\partial }{\partial \xi } \,\psi _n (\xi ) \ \ \ \ \ \nonumber \\&\quad = \ \left\{ - \,\frac{\partial }{\partial k_x} \,\delta (k^\prime _x - k_x) \,\right\} \, \frac{1}{l_B} \,\int _{- \,\infty }^\infty \, d \xi \,\psi _n (\xi ^\prime ) \, \frac{\partial }{\partial \xi } \,\psi _n (\xi ).\nonumber \\ \end{aligned}$$
(A21)
Here, we can make use of the identity of Dirac’s delta function:
$$\begin{aligned} f (k) \,\delta ^\prime (k) \ = \ - \,f^\prime (0) \,\delta (k), \end{aligned}$$
(A22)
where f(k) is an any functions of k, which satisfies the condition \(f (0) = 0\). This gives
$$\begin{aligned}&- \,i \,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x \,\frac{\partial }{\partial y} \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \left\{ \frac{\partial }{\partial k_x} \,\frac{1}{l_B} \, \int _{- \,\infty }^\infty \,d \xi \,\psi _n (\xi ^\prime ) \,\frac{\partial }{\partial \xi } \, \psi _n (\xi ) \right\} _{k^\prime _x = k_x} \times \,\,\delta (k^\prime _x - k_x) ,\nonumber \\ \end{aligned}$$
(A23)
if the following equality holds
$$\begin{aligned} \int _{- \,\infty }^\infty \,d \xi \left. \psi _n (\xi ^\prime ) \,\frac{\partial }{\partial \xi } \,\psi _n (\xi ) \, \right| _{\xi ^\prime = \xi } \ = \ 0, \end{aligned}$$
(A24)
which can be easily verified to hold. We therefore find that
$$\begin{aligned}&- \,i \,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x \,\frac{\partial }{\partial y} \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \delta (k^\prime _x - k_x) \,\left( - \,l_B \,\frac{\partial }{\partial \xi } \right) \, \frac{1}{l_B} \,\int _{- \,\infty }^\infty \,d \xi \,\,\left. \psi _n (\xi ^\prime ) \,\frac{\partial }{\partial \xi } \,\psi (\xi ) \, \right| _{\xi ^\prime = \xi } \nonumber \\&\quad = - \,\delta (k^\prime _x - k_x) \,\int _{- \,\infty }^\infty \, \psi (\xi ) \,\frac{\partial ^2}{\partial \xi ^2} \,\psi _n (\xi )\nonumber \\&\quad = \frac{1}{2} \,( 2 \,n + 1) \,\delta (k^\prime _x - k_x) . \end{aligned}$$
(A25)
The second half part of (A20) can be evaluated as
$$\begin{aligned}&i \,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,y \,\frac{\partial }{\partial x} \, \vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \ = \ - \,k_x \,\delta (k^\prime _x - k_x) \nonumber \\&\quad \times \,\int _{- \,\infty }^\infty \, d y \,Y_n (y) \,y \,Y_n (y) \nonumber \\&\quad = \ - \,k_x \,\delta (k^\prime _x - k_x) \,y_0 \ = \ - \,\frac{k_x^2}{e \,B} \,\delta (k^\prime _x - k_x) . \end{aligned}$$
(A26)
Combining the two terms, we therefore find that
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, {\hat{L}}^{can}_z \,\vert \, \Psi ^{(L_1)}_{n, k_x} \rangle \ = \ \left\{ \, n \, + \, \frac{1}{2} \ - \ \frac{k^2_x}{e \,B} \,\right\} \, \delta (k^\prime _x - k_x) .\nonumber \\ \end{aligned}$$
(A27)
Since the calculation of the matrix element \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) is straightforward, here we show only the answer given as
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, y^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle = \frac{1}{e \,B} \,\left\{ \, n + \frac{1}{2} + \frac{k^2_x}{e \,B} \right\} \, \delta (k^\prime _x - k_x) .\nonumber \\ \end{aligned}$$
(A28)
Finally, as already pointed out, the calculation of the matrix element \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \, x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \) is a little technical, so that here we show only the final answer by leaving the explicit derivation to another Appendix B. The answer reads as
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \frac{1}{e \,B} \,\left( n + \frac{1}{2} \right) \,\delta (k^\prime _x - k_x) \ \nonumber \\&\quad - \ \delta ^{\prime \prime } (k^\prime _x - k_x) , \end{aligned}$$
(A29)
where \(\delta ^{\prime \prime } (k)\) represents the second derivative of \(\delta (k)\). In this way, we now have all the necessary matrix elements as follows :
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,{\hat{L}}^{can}_z \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \left\{ \, n \, + \,\frac{1}{2} \,- \,\frac{k^2_x}{e \,B} \right\} \, \delta (k^\prime _x - k_x),\nonumber \\ \end{aligned}$$
(A30)
$$\begin{aligned} e \,B \,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,y^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \left\{ \, n \, + \, \frac{1}{2} \,+ \,\frac{k^2_x}{e \,B} \right\} \, \delta (k^\prime _x - k_x),\nonumber \\ \end{aligned}$$
(A31)
$$\begin{aligned} e \,B \,\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle= & {} \ \left( n \, + \,\frac{1}{2} \right) \,\delta (k^\prime _x - k_x) \ \nonumber \\&\quad - \ e \,B \,\delta ^{\prime \prime } (k^\prime _x - k_x) . \end{aligned}$$
(A32)
Using these answers, we can now readily write down the final answers for the matrix elements of the three OAM operators in the \(\vert \,n, k_x \rangle \)-basis. The answers are summarized in Table 1 in the main text together with the matrix elements of the three momentum operators.
Appendix B Calculation of the matrix elements \(\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \)
The calculation starts with
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \ \left\{ \frac{1}{2 \,\pi } \,\int _{- \,\infty }^\infty \,d x \, e^{\,- \,i \,(k^\prime _x - k_x) \,x} \,x^2 \right\} \, \nonumber \\&\qquad \times \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \,Y_n (y - y_0) \nonumber \\&\quad = \ \left\{ - \,\frac{\partial ^2}{\partial k^2_x} \, \delta (k^\prime _x - k_x) \right\} \, \nonumber \\&\qquad \times \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \,Y_n (y - y_0) . \end{aligned}$$
(B33)
With use of the following identity of Dirac’s delta function,
$$\begin{aligned} \delta ^{\prime \prime } (k) \,f (k) \ = \ f^{\prime \prime } (0) \,\delta (k) - 2 \,f^\prime (0) \,\delta ^\prime (k) + f (0) \,\delta ^{\prime \prime } (k) ,\nonumber \\ \end{aligned}$$
(B34)
we obtain
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \nonumber \\&\quad = \ - \,\left\{ \frac{\partial ^2}{\partial k^2_x} \, \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \nonumber \\&\qquad \times \,\,\delta (k^\prime _x - k_x) \nonumber \\&\qquad + \ 2 \, \left\{ \frac{\partial }{\partial k_x} \, \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \nonumber \\&\qquad \times \delta ^\prime (k^\prime _x - k_x) \nonumber \\&\qquad - \ \left\{ \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \nonumber \\&\qquad \times \,\,\delta ^{\prime \prime } (k^\prime _x - k_x) \end{aligned}$$
(B35)
It is not so difficult to verify the equalities
$$\begin{aligned} \left\{ \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \ = \ 1, \end{aligned}$$
(B36)
and
$$\begin{aligned} \left\{ \frac{\partial }{\partial k_x} \, \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \ = \ 0 , \end{aligned}$$
(B37)
so that we show below how to evaluate
$$\begin{aligned}&\left\{ \frac{\partial ^2}{\partial k^2_x} \!\! \int _{- \,\infty }^\infty \!\! d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \nonumber \\&\quad = l^2_B \,\int _{- \,\infty }^\infty d \xi \,\psi _n (\xi ) \, \frac{\partial ^2}{\partial \xi ^2} \,\psi _n (\xi ). \end{aligned}$$
(B38)
Using the familiar recursion formulas for the harmonic oscillator wave functions
$$\begin{aligned} \frac{\partial }{\partial \xi } \,\psi _n (\xi )= & {} \ \sqrt{\frac{n}{2}} \,\psi _{n-1} (\xi ) \ - \ \sqrt{\frac{n+1}{2}} \,\psi _{n+1} (\xi ), \end{aligned}$$
(B39)
$$\begin{aligned} \frac{\partial ^2}{\partial \xi ^2} \,\psi _n (\xi )= & {} \ - \,\frac{2 \,n + 1}{2} \,\psi _n (\xi ) \nonumber \\&+ \ \sqrt{\frac{n \,(n - 1)}{2}} \,\psi _{n-2} (\xi ) \nonumber \\&+ \ \sqrt{\frac{(n+1) \,(n+2)}{2}} \,\psi _{n+2} (\xi ) , \ \ \ \ \ \end{aligned}$$
(B40)
we find that
$$\begin{aligned} \int _{- \,\infty }^\infty \,d \xi \,\psi _n (\xi ) \,\frac{\partial ^2}{\partial \xi ^2} \, \psi _n (\xi ) \ = \ - \,\frac{1}{2} \,(2 \,n + 1) , \end{aligned}$$
(B41)
which in turn gives
$$\begin{aligned}&\left\{ \frac{\partial ^2}{\partial k^2_x} \, \int _{- \,\infty }^\infty \,d y \,Y_n (y - y^\prime _0) \, Y_n (y - y_0) \right\} _{k^\prime _x = k_x} \nonumber \\&\quad \ = \ - \,l^2_B \,\,\frac{1}{2} \,(2 \,n + 1) . \end{aligned}$$
(B42)
Collecting the above formulas, we finally get
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n, k^\prime _x} \,\vert \,x^2 \,\vert \,\Psi ^{(L_1)}_{n, k_x} \rangle \ = \ \frac{1}{e \,B} \,\left( n + \frac{1}{2} \right) \,\delta (k^\prime _x - k_x)\nonumber \\&\quad \ - \ \delta ^{\prime \prime } (k^\prime _x - k_x) . \end{aligned}$$
(B43)
which reproduces (A29) in the Appendix A.
Appendix C Calculation of the matrix elements in the \(\vert n, m \rangle \)-basis
Let us start with the calculation of the momentum operators. Using (107), we obtain
$$\begin{aligned}&\langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,p^{cons}_x ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\sqrt{\frac{e \,B}{2}} \,\,{}^A \langle n \,\vert \,{}^B \langle n - m^\prime \,\vert \, b - b^\dagger \,\vert \,n \rangle ^A \,\vert \,n - m \rangle ^B \nonumber \\&\quad = - \,i \,\sqrt{\frac{e \,B}{2}} \left\{ \sqrt{n - m} \,\delta _{m^\prime , m+1} \, - \, \sqrt{n - m + 1} \,\delta _{m^\prime , m-1} \right\} .\nonumber \\ \end{aligned}$$
(C44)
On the other hand, we get
$$\begin{aligned}&\langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,p^{mech}_x ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\sqrt{\frac{e \,B}{2}} \,\,{}^A \langle n \,\vert \,{}^B \langle n - m^\prime \,\vert \, a - a^\dagger \,\vert \,n \rangle ^A \,\vert \,n - m \rangle ^B \ = \ 0.\nonumber \\ \end{aligned}$$
(C45)
The matrix element of the canonical momentum operator can be calculated by using the relation \(p^{can}_x = \frac{1}{2} \,\left( p^{mech}_x ({\varvec{A}}^{(S)}) + p^{cons}_x ({\varvec{A}}^{(S)}) \right) \), which gives
$$\begin{aligned}&\langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,p^{can}_x \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\,\frac{1}{2} \,\sqrt{\frac{e \,B}{2}} \,\left\{ \sqrt{n - m} \,\delta _{m^\prime , m+1} \ - \ \sqrt{n - m + 1} \,\delta _{m^\prime , m-1} \right\} .\nonumber \\ \end{aligned}$$
(C46)
The matrix elements of the momentum operators between the eigen-states \(\vert \,\Psi ^{(L_1)}_{n,m} \rangle \) are obtained by using the relation \(\vert \,\Psi ^{(L_1)}_{n,m} \rangle = U \,\vert \,\Psi ^{(S)}_{n,m} \rangle \). For the canonical momentum operator, we get
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \, p^{can}_x \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle= & {} \ \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \, U^\dagger \,p^{can}_x \,U \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\= & {} \ \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \, p^{can}_x \ + \ \frac{1}{2} \,e \,B \,y \,\vert \,\Psi ^{(S)}_{n,m} \rangle . \nonumber \\ \end{aligned}$$
(C47)
The second part can be evaluated by using
$$\begin{aligned} \frac{1}{2} \,e \,B \,y \ = \ i \,\,\frac{1}{2} \,\sqrt{\frac{e \,B}{2}} \,\, ( a \ - \ b \ - \ a^\dagger \ + \ b^\dagger ) , \end{aligned}$$
(C48)
which gives
$$\begin{aligned}&\frac{1}{2} \,e \,B \,\langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,y \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\,\frac{1}{2} \,\sqrt{\frac{e \,B}{2}} \,\left\{ \sqrt{n - m} \,\delta _{m^\prime , m+1} \, - \, \sqrt{n - m + 1} \,\delta _{m^\prime , m-1} \right\} .\nonumber \\ \end{aligned}$$
(C49)
Combining the two terms, we obtain
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \, p^{can}_x \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\sqrt{\frac{e \,B}{2}} \,\left\{ \sqrt{n - m} \,\delta _{m^\prime , m+1} \ - \ \sqrt{n - m + 1} \,\delta _{m^\prime , m-1} \right\} .\nonumber \\ \end{aligned}$$
(C50)
Similarly, we can readily show that
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \, p^{cons}_x ({\varvec{A}}^{(L_1)}) \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle \nonumber \\&\quad = - \,i \,\sqrt{\frac{e \,B}{2}} \,\left\{ \sqrt{n - m} \,\delta _{m^\prime , m+1} \, - \, \sqrt{n - m + 1} \,\delta _{m^\prime , m-1} \right\} ,\nonumber \\ \end{aligned}$$
(C51)
and that
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \, p^{mech}_x ({\varvec{A}}^{(L_1)}) \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle \ = \ 0 . \end{aligned}$$
(C52)
Next, we evaluate the matrix elements of the three OAM operators. First, for the canonical OAM operator, we obtain
$$\begin{aligned}&\quad \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,L^{can}_z \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&= {}^A \langle n \,\vert \,{}^B \langle n - m^\prime \,\vert \, a^\dagger \,a - b^\dagger \,b \, \vert \,n \rangle ^A \,\vert \,n - m \rangle ^B \nonumber \\&= n \,\delta _{m^\prime , m} \ - \ ( n - m) \,\delta _{m^\prime , m} \ = \ m \,\delta _{m^\prime , m} .\nonumber \\ \end{aligned}$$
(C53)
Since \(L^{cons}_z ({\varvec{A}})\) reduces to \(L^{can}_z\) in the symmetric gauge, we naturally get
$$\begin{aligned} \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,L^{cons}_z ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \ = \ m \,\delta _{m^\prime , m} , \end{aligned}$$
(C54)
Finally, for the matrix element of the mechanical OAM operator, we find that
$$\begin{aligned} \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,L^{mech}_z ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \ = \ ( 2 \,n + 1) \,\delta _{m^\prime , m} , \end{aligned}$$
(C55)
Next, we evaluate the matrix elements of the OAM operators between the eigen-states \(\vert \,\Psi ^{(L_1)}_{n,m} \rangle \), which also belong to the \(\vert n,m \rangle \)-basis class. Here we start with the matrix elements of the canonical OAM operator given by
$$\begin{aligned}&\quad \langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \,L^{can}_z \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle = \ \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,U^\dagger \,L^{can}_z \,U \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&= \ \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \, L^{can}_z \ + \frac{1}{2} \,e \,B \,(x^2 - y^2) \,\vert \,\Psi ^{(S)}_{n,m} \rangle . \end{aligned}$$
(C56)
The second part can be calculated by using the relations
$$\begin{aligned} x^2= & {} \ l^4_B \,(\,\Pi _y - {\tilde{\Pi }}_y )^2 = \frac{1}{2} \,\,l^2 _B \,(\,a + a^\dagger + b + b^\dagger )^2,\quad \end{aligned}$$
(C57)
and
$$\begin{aligned} y^2= & {} \ l^4_B \,(\,\Pi _x - {\tilde{\Pi }}_x )^2 = - \,\frac{1}{2} \,l^2_B \,(\, a - a^\dagger - b + b^\dagger )^2,\nonumber \\ \end{aligned}$$
(C58)
which gives
$$\begin{aligned}&x^2 \ - \ y^2 \nonumber \\&= \ l^2_B \,\, \left\{ \,a^2 \,+ \,b^2 \,+ \,(a^\dagger )^2 \,+ \,(b^\dagger )^2 \,+ \,a \,b^\dagger \,\right. \nonumber \\&\quad \left. + \,b \,a^\dagger \,+ \,b^\dagger \,a \,+ \,a^\dagger \,b \, \right\} . \end{aligned}$$
(C59)
After some tedious but straightforward algebra, we find that
$$\begin{aligned}&\frac{1}{2} \,e \,B \,\langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,x^2 - y^2 \, \vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\&\quad = \frac{1}{2} \,\left\{ \sqrt{(n-m) \,(n-m+1)} \,\delta _{m^\prime , m+2} \right. \nonumber \\&\quad + \left. \sqrt{(n-m+1) \,(n-m+2)} \,\delta _{m^\prime ,m-2} \right\} , \end{aligned}$$
(C60)
which in turn gives
$$\begin{aligned}&\langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \,L^{can}_z \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle \nonumber \\&\quad = m \,\delta _{m^\prime ,m} \nonumber \\&\quad + \frac{1}{2} \,\left\{ \sqrt{(n-m) \,(n-m+1)} \,\delta _{m^\prime , m+2} \right. \nonumber \\&\quad + \left. \sqrt{(n-m+1) \,(n-m+2)} \,\delta _{m^\prime ,m-2} \right\} . \end{aligned}$$
(C61)
Since both of \(L^{cons}_z ({\varvec{A}})\) and \(L^{mech}_z ({\varvec{A}})\) transform covariantly under a gauge transformation, their matrix elements between the eigen-state \(\vert \,\Psi ^{(L_1)}_{n,m} \rangle \) can easily be obtained from those between the eigen-states \(\vert \,\Psi ^{(S)}_{n,m} \rangle \) as
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \,L^{cons}_z ({\varvec{A}}^{(L_1)}) \,\vert \,\Psi ^{(L_1)}_{n,m} \rangle= & {} \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,L^{cons}_z ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\= & {} m \,\delta _{m^\prime , m} , \end{aligned}$$
(C62)
and
$$\begin{aligned} \langle \Psi ^{(L_1)}_{n,m^\prime } \,\vert \,L^{mech}_z ({\varvec{A}}^{(L_1)}) \,\vert \,\Psi ^{(L_1)} \rangle= & {} \langle \Psi ^{(S)}_{n,m^\prime } \,\vert \,L^{mech}_z ({\varvec{A}}^{(S)}) \,\vert \,\Psi ^{(S)}_{n,m} \rangle \nonumber \\= & {} ( 2 \,n + 1 ) \,\delta _{m^\prime , m} . \end{aligned}$$
(C63)
