Finding general expression for probability current.

[David Silva]

The conservation of probability says:

$$\partial_t J^{0} + \partial{i}J^{i} = 0$$

Use the Schrodinger equation to obtain$$ J^{i} (\vec r)$$.

I have no idea where to start this kind of problem because the notation makes no sense to me. I would appreciate a hint or nudge in the correct direction.

 

[vela]

What about the notation confuses you?

 

[David Silva]

What about the notation confuses you?

What do those partials mean? I am not used to seeing them this way? Also would I be using the general SE eq or the radial one since we are in r?

 

[vela]

Typically $x^0 = t$, $x^1 = x$, $x^2 = y$, and $x^3 = z$, so $\partial_1$ would mean $\partial_x$. The repeated index $i$ implies a summation from $i=1$ to $i=3$.

You have $J^\mu = (\rho,J_x, J_y, J_z)$, so in more traditional notation, conservation of probability is
$$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$$ where $\vec{j} = (J_x, J_y, J_z)$.

 

[David Silva]

Typically $x^0 = t$, $x^1 = x$, $x^2 = y$, and $x^3 = z$, so $\partial_1$ would mean $\partial_x$. The repeated index $i$ implies a summation from $i=1$ to $i=3$.

You have $J^\mu = (\rho,J_x, J_y, J_z)$, so in more traditional notation, conservation of probability is
$$\frac{\partial \rho}{\partial t} + \nabla \cdot \vec{j} = 0$$ where $\vec{j} = (J_x, J_y, J_z)$.

Thanks, I will look at this later today and see if it helps!