Talk:Probability current

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Derivation

I added another line to the continuity equation derivation to make it a bit easier to follow.

\int _{V}\left({\frac {\partial |\Psi |^{2}}{\partial t}}\right)dV=-\int _{V}\left({\vec {\nabla }}\cdot {\vec {j}}\right)dV

--146.232.75.208 (talk) 18:56, 26 February 2008 (UTC)[reply]

Current and stationary states

The article states

Note that the probability current is nonzero despite the fact that plane waves are stationary states and hence
${\frac {d|\Psi |^{2}}{dt}}=0\,$
everywhere. This demonstrates that a particle may be in motion even if its spatial probability density has no explicit time dependence.

But doesn't this only work because the plane wave is unnormalizable? Any normalizable wavefunction would have to be localized. If it is localized and has non-zero probability current, then the "center of gravity" (i.e. the position expectation value) of the wavefunction must be moving. So the wavefunction will have time dependence. This is like saying that an infinite stream has flowing water but no movement.

However, I can imagine a particle traveling in a circle. Then I suppose it could have a non-zero flux without time dependence of the probability distribution.128.112.50.18 01:26, 12 March 2007 (UTC)[reply]

ψ* What does this mean?

I can't understand what this symbol means. I understand that, ρ=ψ*ψ, but I can't find anywhere that actually says what ψ* is or how to do calculations with it. The page in wikibooks for probability flux uses this symbol and it changes how Schrödinger's Equation is written. Can anyone help? Thanks, Brian.

ψ* stands for the Complex conjugate of the wavefunction ψ.

Last equation

I can't see how the second equality

{\frac {1}{2m}}\left(\Psi ^{*}{\vec {P}}\Psi -\Psi {\vec {P}}\Psi ^{*}\right)={\frac {\hbar }{2mi}}\left(\Psi ^{*}{\vec {\nabla }}\Psi -\Psi {\vec {\nabla }}\Psi ^{*}\right)+(1-\beta ){\frac {q}{m}}{\vec {A}}(t)|\Psi |^{2}

in the last equation in the article follows. I would have thought that the vector potential part of one term would cancel the other due to the minus sign, rather than adding as they seem to here. This is probably an embarrassing algebraic mistake on my part, but I thought I should bring it up in case it is actually a mistake. 163.1.176.253 (talk) 14:45, 13 June 2009 (UTC)[reply]

The idea is more general...

...than the article suggests. Probability current is not specific to quantum mechanics -- it's relevant to the Fokker-Planck equation for instance, and many related topics like the forwards and backwards equations. 128.97.41.120 (talk) 00:57, 22 April 2010 (UTC)[reply]