Finding general expression for probability current.

David Silva
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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.
 
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What about the notation confuses you?
 
vela said:
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?
 
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)##.
 
vela said:
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!
 
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