Quantum Mechanics: Does $\vec{A} \cdot \vec{p} = \frac{1}{2} \vec{B}\cdot\vec{L}$? In a quantum mechanics question context, I noticed the need to prove that for a constant magnetic field $\mathbf{B}$, The vector potential $\mathbf{A}$ and the angular momentum operator $\mathbf{L}$, satisfy:
$$\mathbf{A}\cdot\mathbf{p} = \frac{1}{2} \mathbf{B}\cdot \mathbf{L}$$
Where $\mathbf{p}$ is the momentum operator. Without loss of generality, I can take a constant $\mathbf{B} = B_0 \mathbf{\hat{z}}$, and get:
$$\mathbf{A} = \frac{1}{2} B_0 (x_2,-x_1)$$
And using a Weyl ordering for $\mathbf{A}\cdot\mathbf{p}$ I get:
$$ \frac{1}{2}(A_j p_j + p_j A_j) = \frac{1}{4} B_0 \left(x_2 p_1 - x_1 p_2 + p_1 x_2 - p_2 x_1\right) = \frac{1}{2} \mathbf{B} \cdot \mathbf{L}$$
As requested. Is it possible to prove this for a general, not necessarily constant $\mathbf{B}$?
 A: This identity doesn't hold for non constant magnetic fields. For instance with:
$$\mathbf{A} = \frac{1}{2}B_0 k_0 ( -y^2, x^2) \Rightarrow \mathbf{B} = B_0 k_0 (0,0,x+y)$$
We got on one the hand:
$$ A_j p_j + p_j A_j = \frac{1}{2} B_0 k_0 ( -y^2 p_x + x^2 p_y - p_x y^2 + p_y x^2) = B_0 k_0 (x^2 p_y-y^2 p_x)$$
Where as:
$$\frac{1}{2}\mathbf{B}\cdot \mathbf{L} = \frac{1}{2}B_0 k_0 ((x+y)L_z + L_z (x +y)) = \frac{1}{2}B_0k_0 ((x+y)(xp_y-yp_x) + (xp_y-yp_x)(x+y)) = \frac{1}{2} B_0k_0(x^2p_y-y^2p_x + yxp_y-xyp_x + xp_yx-yp_xy + xp_yy-yp_xx) = B_0k_0\left(x^2 p_y - y^2p_x+\frac{x}{2}[y,p_y]-\frac{y}{2}[x,p_x]\right) = B_0k_0 \left(x^2p_y-y^2p_x+i\frac{\hbar}{2}(x-y)\right)$$
And we can see that Proper Weyl ordering produces a difference of:
$$\mathbf{A}\cdot \mathbf{p} - \frac{1}{2}\mathbf{B}\cdot \mathbf{L} = B_0k_0 i \frac{\hbar}{2} (x-y)$$
And without weyl ordering, i.e when interpreting: $\mathbf{B}\cdot\mathbf{L}= B_j L_j = B_z (xp_y-yp_x)$ the discrepancy is even worse:
$$\mathbf{A}\cdot \mathbf{p}-\frac{1}{2}B_jL_j =  B_0k_0(yxp_y-xyp_x)$$
