# Problem with the Landau gauge

I'm having a very simple problem which probably has an equally simple answer. I'm following the wikipedia article: https://en.wikipedia.org/wiki/Landau_quantization

We have a uniform magnetic field in $\hat{z}$-direction $$\mathbf{B}=(0,0,B).$$ The vector potential must fulfill $\nabla\times \mathbf{A}=\mathbf{B}$ and supposedly valid choice is given by the Landau gauge for which the vector potential is $$\mathbf{A}=(0,Bx,0).$$

However I find that with this choice for $\mathbf{A}$ the magnetic field becomes $$\mathbf{B}=(-Bx\partial_z,0,\partial_x(Bx))=(-Bx\partial_z,0,B).$$ Why would the first component vanish?

Or is there something which I have misunderstood completely?

EDIT: Jesus Christ, this is embarrassing. Clearly $$\nabla\times \mathbf{A}=(-\partial_z(Bx),0,\partial_x(Bx))\neq (-Bx\partial_z,0,\partial_x(Bx)).$$ This solves the problem.

You are half right. $B$ is an operator, but the curl is only acting on $A$, not on the wavefunction. In other words is should be read as $(\nabla\times A)\psi$ not $\nabla\times (A\psi)$. Mathematically this is fine because $B = \nabla\times A$ is the definition of $A$ and we are free to define it however we want.
Physically we want $B$ to be a function of position only, rather than position and momentum, in order to regain the correct classical limit, so we don't want a definition of $A$ that allows spacial derivatives to appear in $B$.
• Yes the derivatives only act on $A$. Like I said this is basically a matter of definition and saying "because this is what works". I don't know any reason this would be particularly obvious from first principles. May 10, 2018 at 13:09
The calculations were wrong and clearly, $$\nabla\times \mathbf{A}=(-\partial_z(Bx),0,\partial_x(Bx))=(0,0,\partial_x(Bx))\neq (-Bx\partial_z,0,\partial_x(Bx)).$$