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In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{1}{2} \, m \omega^2 (x^2 + y^2) + i \hbar \omega \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}) \equiv i \hbar \omega \, (\vec{\mathrm{r}} \cdot \vec{\mathrm{\pi}} + \vec{\mathrm{\pi}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators? I have the impression that they may be related to the arbitrary choice of axes origin, while the particle may evolve around any point in the $(x, y)$ plane.

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the symmetric gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{1}{2} \, m \omega^2 (x^2 + y^2) + i \hbar \omega \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}) \equiv i \hbar \omega \, (\vec{\mathrm{r}} \cdot \vec{\mathrm{\pi}} + \vec{\mathrm{\pi}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators? I have the impression that they may be related to the arbitrary choice of axes origin, while the particle may evolve around any point in the $(x, y)$ plane.

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{q^2 B^2}{8 m} (x^2 + y^2) + i \hbar \frac{q B}{2 m} \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators?

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{1}{2} \, m \omega^2 (x^2 + y^2) + i \hbar \omega \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}) \equiv i \hbar \omega \, (\vec{\mathrm{r}} \cdot \vec{\mathrm{\pi}} + \vec{\mathrm{\pi}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators? I have the impression that they may be related to the arbitrary choice of axes origin, while the particle may evolve around any point in the $(x, y)$ plane.

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ So I need to confirm this. Is (6) correct?

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ Is this right and how should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field, I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$) $$\tag{9} H = \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} = \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{q^2 B^2}{8 m} (x^2 + y^2) + i \hbar \frac{q B}{2 m} \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), $$ I get $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}). $$ This thing is weird to me. Is (10) correct too? I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?), if these relations are correct (I believe they are).

In non-relativistic quantum mechanics, the canonical momentum of a particle is defined as $$\tag{1} p_i = - i \hbar \: \partial_i. $$ When there's an external magnetic field (suppose for simplicity that it's constant and homogeneous), the observable momentum (or dynamical momentum) isn't $p_i$, but $$\tag{2} \pi_i = p_i - q A_i, $$ where $A_i$ is the gauge dependent potential-vector associated to the external magnetic field. While $[ p_i, p_j] = 0$, it's easy to get the following commutator: $$\tag{3} [ \pi_i, \pi_j] = i \hbar \, \varepsilon_{ijk} \, q B_k. $$ So, in the presence of a magnetic field, the observable orbital angular momentum isn't anymore $$\tag{4} L_i = \varepsilon_{ijk} \, x_j p_k, $$ and should be replaced by $$\tag{5} M_i = \varepsilon_{ijk} \, x_j \, \pi_k. $$ The commutator of this thing isn't simple. Unless I made a mistake, I get this: $$\tag{6} [ M_i, M_j] = i \hbar \, \varepsilon_{ijk} \, M_k + i \hbar \, \varepsilon_{ijk} \, x_k \, (q \, \vec{\mathrm{r}} \cdot \vec{\mathrm{B}}). $$ The last term is weird to my eyes!

Also, for a simple charged particle in the homogeneous magnetic field $\vec{\mathrm{B}} = B \, \hat{\mathrm{z}}$, the component $M_z$ doesn't commute with the Hamiltonian, while $L_z$ do commute: $$\tag{7} [L_z, H] = 0, \qquad [M_z, H] \ne 0. $$ How should I interpret this? I guess this should be right, since the system isn't closed (there's an external magnetic field!), but I feel a bit hesitant with having $[ M_z, H ] \ne 0$. For the simple particle in a constant magnetic field (I'm using the gauge $\vec{\mathrm{A}} = \tfrac{1}{2} \, B \, (- y \, \hat{\mathrm{x}} + x \, \hat{\mathrm{y}})$), I get $$\tag{8} M_z = L_z - \frac{1}{2} \, q B \, (x^2 + y^2). $$ So, for the Hamiltonian \begin{align} H &= \frac{(\vec{\mathrm{p}} - q \vec{\mathrm{A}})^2}{2 m} \\[2ex] &= \frac{p_x^2}{2 m} + \frac{p_y^2}{2m} + \frac{q^2 B^2}{8 m} (x^2 + y^2) + i \hbar \frac{q B}{2 m} \Bigl( x \, \frac{\partial}{\partial y} - y \, \frac{\partial}{\partial x} \Bigr), \tag{9} \end{align} I get this commutator: $$\tag{10} [M_z, H]= - i \hbar \, \frac{q B}{2m} (\vec{\mathrm{r}} \cdot \vec{\mathrm{p}} + \vec{\mathrm{p}} \cdot \vec{\mathrm{r}}). $$ This thing is weird! I feel uneasy with the usual angular momentun (4) not observable anymore, since it is implicitly gauge-dependent, while (5) is gauge-independent.

So I'm looking for an interpretation of the extra-term in (6) and the non-commutativity (10) (the physical angular momentum isn't conserved?). How should I consider these commutators?