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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.

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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 $B$.

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  • $\begingroup$ I get that the cross product only acts on A. But are you saying that the partial derivative therefore also only acts on A? Perhaps it is trivially so but it just isnt lear to me. $\endgroup$ – Jens Roderus May 10 '18 at 12:59
  • $\begingroup$ 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. $\endgroup$ – By Symmetry May 10 '18 at 13:09
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The calculations were wrong and clearly, $$\nabla\times \mathbf{A}=(-\partial_z(Bx),0,\partial_x(Bx))=(0,0,\partial_x(Bx))\neq (-Bx\partial_z,0,\partial_x(Bx)).$$

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