# Probability current calculations

I have a question about the probability current density. Because I cant really understand the meaning of that (how can we relate something real like a current to something abstract such as probability), I dont really have a good intuition for that.

The question is a part of a problem of finite potential barrier $$V\left(x\right)=\begin{cases} 0 & x<0\\ V_{0} & x>0 \end{cases}$$

An incident particel comes from the negative part with energy given energy $$E$$.

What Im trying to do is to find the probability current density for $$x<0$$ and $$x>0$$ for both cases $$E and $$E>V_0$$.

The wave function for the case $$E>V_0$$ is given by:

$$\psi\left(x\right)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x<0\\ Ce^{iqx} & x>0 \end{cases},\thinspace\thinspace\thinspace\thinspace k=\sqrt{\frac{2mE}{\hbar^{2}}},\,\thinspace\thinspace q=\sqrt{\frac{2m\left(E-V_{0}\right)}{\hbar^{2}}}$$

And the wave function for the case $$E is given by

$$\psi\left(x\right)=\begin{cases} Ae^{ikx}+Be^{-ikx} & x<0\\ Ce^{-\alpha x} & x>0 \end{cases},\thinspace\thinspace\thinspace\thinspace k=\sqrt{\frac{2mE}{\hbar^{2}}},\,\thinspace\thinspace\alpha=\sqrt{\frac{2m\left(V_{0}-E\right)}{\hbar^{2}}}$$

Now, in order to calculate the probability current, I want to use $$J\left(x\right)=\frac{\hbar}{m}|\psi\left(x\right)|^{2}\overrightarrow{k}$$.

Define:

$$r=\frac{B}{A},\thinspace\thinspace\thinspace t=\frac{C}{A}$$.

Thus, for the case $$E>V_0$$ and for $$x>0$$ we have: $$J\left(x\right)=\frac{\hbar}{m}|C|^{2}q=\frac{\hbar}{m}|A^{2}||t|^{2}q$$

But for $$x<0$$ Im not sure how it should be, because the result that I got is different from my book result. Here's what I have tried:

$$J\left(x\right)=\frac{\hbar}{m}|\psi\left(x\right)|^{2}k=\frac{\hbar}{m}|Ae^{ikx}+Are^{-ikx}|^{2}k=\frac{\hbar}{m}\left(A^{*}e^{-ikx}+A^{*}r^{*}e^{ikx}\right)\left(Ae^{ikx}+Are^{ikx}\right)k=\frac{\hbar}{m}|A|^{2}|r+e^{2ikx}|^{2}k$$

Which is different from my book's result. The book result for $$x<0$$ is given by:

$$J\left(x\right)=\frac{\hbar}{m}|A|^{2}\left(1-|r|^{2}\right)k$$

And Im not sure how to get to this expression.

That was my questions for the case $$E>V_0$$. Now for the case $$E, I know that the probability of transition is 0, that is to say, the probability of reflection is $$1$$, and in this case the answer in the book is that the probability current is just $$0$$, which I cannot understand why (Can we see it by the equation? or is it just an assumption we have to make by physics reasons?)

Thanks in advance, it is highly appreciated.

• If you use the general definition of the probability current, i.e. $J(x) \propto \Im (\Psi^*(x)\, \partial_x \Psi(x))$, you should be able to derive the correct expression. Commented Mar 4, 2021 at 13:40
• @Jacob The expression that I got is not correct? Commented Mar 4, 2021 at 13:40
• Which expression do you mean? Commented Mar 4, 2021 at 13:43
• @Jakob For the case $E>V_0$ and $x<0$: $J\left(x\right)=\frac{\hbar}{m}|A|^{2}|r+e^{2ikx}|^{2}k$ Commented Mar 4, 2021 at 13:55
• I just followed the formula, cant see why this is not correct Commented Mar 4, 2021 at 13:56

Now, in order to calculate the probability current, I want to use $$J(x)= \frac{\hbar}{m}|\psi(x)|^2\vec{k}$$.

I don't know why you think this formula is applicable here. As @Jakob already stated in his comments, this formula is not correct in general. The correct general formula can also be found e.g. in Wikipedia - Probability current: $$J(x) = \frac{\hbar}{2mi}\left( \psi^*(x)\frac{\partial\psi(x)}{\partial x} -\psi(x)\frac{\partial\psi^*(x)}{\partial x} \right) \tag{1}$$

This formula (1) reduces to $$J(x)=\frac{\hbar k}{m}|\psi(x)|^2 \tag{2}$$ only in case of $$\psi(x)=Ae^{ikx}$$ (i.e. a single wave travelling only in one direction). For other types of wave functions $$\psi(x)$$ (especially for the superposition of a left-to-right and a right-to-left travelling wave) the simple formula (2) is no longer true.

Now you should be able to apply formula (1) to the wave functions from your question and do the calculations by yourself. Since your question is tagged as homework-and-exercises, I will not walk through the details here. You should arrive at these final results (and you know these results from your text-book anyway):

If $$E>V_0$$ $$J(x)=\begin{cases} \frac{\hbar k}{m}(|A|^2-|B|^2) &, x<0 \\ \frac{\hbar q}{m}|C|^2 &, x>0 \end{cases} \tag{3a}$$ and if $$E $$J(x)=\begin{cases} \frac{\hbar k}{m}(|A|^2-|B|^2) &, x<0 \\ 0 &, x>0 \end{cases} \tag{3b}$$

Because I cant really understand the meaning of that (how can we relate something real like a current to something abstract such as probability), I dont really have a good intuition for that.

It is often hard to give intuitive explanations, because what it is intuitive to one person, may be non-intuitive to another person. You will develop intuition only after much experience and practice. Having said that, may be you find these aspects intuitive:

• The incident wave function part, $$\psi(x)=Ae^{ikx}$$,
contributes a probability current $$J(x)=\frac{\hbar k}{m}|A|^2$$, i.e. pointing to right.
• The reflected wave function part, $$\psi(x)=Be^{-ikx}$$,
contributes a probability current $$J(x)=-\frac{\hbar k}{m}|B|^2$$, i.e. pointing to left.
• Therefore the superposition of these both parts, $$\psi(x)=Ae^{ikx}+Be^{-ikx}$$,
makes a probability current $$J(x)=\frac{\hbar k}{m}(|A|^2-|B|^2)$$.