Understanding the Probability Current J

[Rahmuss]

I actually have a homework problem on this; but I don't want help specifically with that. What I'd like to know is, what does the probability current represent?

Does it represent how $$\left\langle x\right\rangle$$ changes? And if so, is it how it changes over time, or over a change in the range used to evaluate $$\left\langle x\right\rangle$$. Or does it have to do with $$\left\langle p\right\rangle$$ instead?

To me it seems like it would be how the value of $$\sigma_{x}$$ and / or $$\sigma_{p}$$ change based off of the size of the well, or the range which the integral is being evaluated over to find $$\left\langle x\right\rangle$$ and $$\left\langle p\right\rangle$$ (and $$\left\langle x^{2}\right\rangle$$ and $$\left\langle p^{2}\right\rangle$$).

 

[Demystifier]

In the Bohmian interpretation, the current (divided by the probability density) at a given point determines the particle velocity at that point.

 

[Rahmuss]

Demystifier - Ok, that makes sense. I've been looking more into this and from what I've learned it sounds like you measure the probability at some point (A) and at some time (t) and then measure the probability at some point (B) and at the same time (t), and take the difference and you get the 'current'. That seems absurd, so I'm sure I'm missing something.

How could measuring the probability at two different points with the same time t tell you anything about how the probability is changing between those two points? It seems like you would need to find the total probability over that region between A and B at time t and then increment time by $$\Delta t$$ and measure it again, and then take the difference and see if it's increasing or decreasing in that region.

I'm working on a problem and need to know what the difference between $$\Psi \frac{\partial \Psi^{*}}{\partial x}$$ and $$\Psi^{*} \frac{\partial \Psi}{\partial x}$$.
It seems to me that $$\Psi^{*}$$ is the complex part of the wave function, which in this case is $$\Psi = Ae^{i(kx - \frac{\hbar k^{2}t}{2m})}$$; but if that's the case, then the real part would be $$\Psi = 0$$. Or if $$\Psi$$ and $$\Psi^{*}$$ are the same, then we would have $$\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x} = 0$$

 

[monish]

I'm working on a problem and need to know what the difference between $$\Psi \frac{\partial \Psi^{*}}{\partial x}$$ and $$\Psi^{*} \frac{\partial \Psi}{\partial x}$$.
It seems to me that $$\Psi^{*}$$ is the complex part of the wave function, which in this case is $$\Psi = Ae^{i(kx - \frac{\hbar k^{2}t}{2m})}$$; but if that's the case, then the real part would be $$\Psi = 0$$. Or if $$\Psi$$ and $$\Psi^{*}$$ are the same, then we would have $$\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x} = 0$$

I quite sure that $$\Psi^{*}$$ is not the complex part of the wave function as you say, but the complex conjugate of the wave function. As for the two terms whose difference you are seeking, I think one must be simply the negative of the other.

What you ultimately do with these products is integrate them over all space. The products are always basically the wave function times itself...with an operator stuck in between. If the operator is the unity operator (=1), then you just integrate the wave function against itself and get the total charge. If the operator is the position operator (=x) then you are finding the average position of the charge. If the operator is the momentum operator (= differentiation), as it is in your case, then you get the average momentum.

I'm not sure why you have to take the difference term. It's something to do with the fact that the integral comes out differently depending if you operate to the left or to the right with the differentiation operator.

I hope some of this makes sense to you.

 

[Rahmuss]

monish - Yes, thanks for the info. The book describes the probability current $$J(x,t)$$ as :

$$J(x,t) \equiv \frac{i\hbar}{2m}(\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x})$$

See, I thought the same thing as you. I figured I'd want to integrate over a region, and then see how it changes with time; but they say the "probability current... tells you the rate at which probability is 'flowing' past the point x." So it sounds like it doesn't care what the probability is over a region, it just wants to know at that point if it is going up or down. But even with that I figure you'd need to know how it's changing over time rather than taking a completely different point (B) to figure it out. With the exception being if that second point is simply $$A + \Delta x$$. Then you could measure the change; but you still wouldn't know which way it was flowing unless you took into account a change in time. I guess that's why I'm lost. I don't understand the probability current.

 

[monish]

monish - Yes, thanks for the info. The book describes the probability current $$J(x,t)$$ as :

$$J(x,t) \equiv \frac{i\hbar}{2m}(\Psi \frac{\partial \Psi^{*}}{\partial x} - \Psi^{*} \frac{\partial \Psi}{\partial x})$$


OK, then I was wrong when I said the two terms in the brackets are simply the negative of one another. They must be complex conjugates of each other. That's why when you subtract them you get a pure complex term; and then when you multiply them up front by i, you get a pure real term. Which is what you need for the probability flow. Which is equivalent to the current density at that point.

I think the point to notice hear is you don't get current flow by taking the time derivative of the charge density, but by taking the space derivative of the wave function. I don't know how to justify this by using intuitive pictures. Except that if you look at it in momentum space (the Fourier transform of position space) the differention operator simply corresponds to multiplication by p, which is the analog of what you do with x operator in position space.

But I don't have an intuitive explanation for why momentum space should be the Fourier transform of position space. If I knew a good explanation for that, a lot of these operator things would make more sense to me.

 

[Rahmuss]

monish - Thanks again for the help. So then maybe I'm just misunderstanding what they mean by "flow". I figured that you'd have to know something about it over time or else you wouldn't know which direction it is flowing. But if they don't care about direction, only change, then I can see how to calculate it that way.

 

[monish]

Well, the problem you've been given is one-dimensional, so the answer, either positive or negative, will give a direction of current flow to the left or to the right. But if you do the analogous calculation in three dimensions, you definitely get a vector direction. Instead of differention you use the del operator and that turns your scalar (complex-valued) wave function into a vector.