Vector Potential in Schrodinger eqn

[sinyud]

How do you get the B field from the magnetic potential?
I tried converting the curl into matrix format, but the corresponding matrix can't be inverted.

 

[Dr Transport]

The vector potential is added to the Schrodinger equation by making the substitution
$$\vec{p} = \vec{p} - e/c\vec{A}$$.

 

[R0man]

The vector potential in the Schrodinger equation is defined as A = -iħ∇ψ/2m, where ħ is the reduced Planck's constant, ∇ is the gradient operator, ψ is the wave function, and m is the mass of the particle. It is a mathematical construct that helps simplify the Schrodinger equation and allows for a more elegant and intuitive understanding of the behavior of quantum particles in a magnetic field.

To get the B field from the magnetic potential, we can use the equation B = ∇ x A, where x represents the cross product. This means that the B field is equal to the curl of the vector potential. However, as you mentioned, the matrix representation of the curl cannot be inverted. This is because the vector potential is a gauge-dependent quantity, meaning that it can vary depending on the choice of gauge. Therefore, the matrix representation of the curl will also vary and may not always be invertible.

To overcome this issue, we can use the gauge-invariant quantity known as the magnetic flux density (B-field) instead of the vector potential. The B-field is defined as B = ∇ x A + μ0J, where μ0 is the permeability of free space and J is the current density. This equation takes into account the contribution of the current density, making it independent of the choice of gauge. Therefore, we can use this equation to calculate the B-field from the magnetic potential and current density.

In summary, the B field can be obtained from the magnetic potential by using the equation B = ∇ x A, but to ensure gauge-invariance, we can also include the contribution of the current density in the form of the magnetic flux density equation B = ∇ x A + μ0J.