# Dimensionless equation schrodinger with field magnetic

1. Feb 27, 2010

### pirulito.king

$$-\frac{i\hbar^2}{2m}(\nabla-\frac{ieA}{c})^2\psi=E\psi$$
$$A=(x\^y-y\^x)\frac{H_{0}}{2}$$

I'm trying to leave it like that:
$$(\nabla-A)^2\psi=E\psi$$
I do not know if I'm posting this in the right place, sorry for mistakes in grammar

thank's

Last edited: Feb 27, 2010
2. Feb 28, 2010

### krishna mohan

Okay..let us try...

The particle you are considering has a charge e....So first, we decide that we are going to measure all charges using this charge as the basic unit..in other words, e is one unit of charge..so you can substitute 1 instead of e in the equation....

Let us also decide that m, the mass of the particle, is going to be your unit of mass...then, you can happily put m=1 in the equation...

What remains are c and $$\hbar$$....

Now you can put both of them as one and....is that it????

Not quite....you see these four parameters are not completely independent in terms of dimension..For example, for an electron, you can construct what is known as the fine structure constant as

$$\alpha =\ \frac{e^2}{(4 \pi \varepsilon_0)\hbar c}$$

This is a dimensionless quantity whose measured value is approximately 1/137. The important thing is that this value is independent of your choice of units because it is dimensionless. Whether you choose 1 kg as your unit of mass or 1 gm as the unit of mass, fine structure constant will be 1/137.

So you can put, say, e=1, $$\varepsilon_0$$=1 and c=1 but then $$\hbar$$ will correspond to $$\frac{1}{4\pi\alpha}$$ units....you are required to substitute this in place of $$\hbar$$ in the equation, if your particle is the electron.....

If you are working with some other particle , you have to use the analogous constant.....

3. Feb 28, 2010

### pirulito.king

yes the particle is a electron....its caged in square potencial with magnetic field and i try transform the equation in dimensionless for the computer

but i still confuse how i can do it