I'm trying to calculate the probability current for a scattering problem. The potential is $V = V_0 > 0$ in $x>0$, with $E>V_0$

So I have in the region $x \le 0$:

$$\psi = \exp(ikx) + R \exp(-ikx)$$

And in $x>0$

$$\psi = T \exp(i \kappa x)$$

I am trying to calculate the probability current, $j = \frac{-ih}{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)$, in each region and show that it is equal.

However when I calculate the probability current in $x<0$, I get:

$$\bar{\psi} = \exp(-ikx) + \bar{R}\exp(ikx)$$ $$\psi' = ik\exp(ikx) -ikR\exp(-ikx)$$ $$\bar{\psi'} = -ik\exp(-ikx) + ik\bar{R}\exp(ikx)$$ $$\psi = \exp(ikx) + R\exp(-ikx)$$


$$\bar{\psi} \psi' = ik -ikR\exp(-2ikx) + \bar{R}ik\exp(2ikx) - ikR\bar{R}$$

$$\bar{\psi}' \psi = -ik -ikR\exp(-2ikx) +ik\bar{R}exp(2ikx) +ik R\bar{R}$$


$$j = \frac{hk}{2m} (1 + R \exp(-2ikx) - \bar{R} \exp(2ikx) - R\bar{R})$$

And in the region $x>0$:

$$j = \frac{\kappa h}{2m}(T\bar{T})$$.

I can show that (by imposing continuity conditions at the boundary):

$$k(1-|R|^2) = \kappa |T|^2$$

So I would expect the first probability current to just be: $\frac{kh}{2m}(1-|R|^2)$.

Any help with this issue is much appreciated! I'm pretty sure I'm making some stupid mistake somewhere, but it's very frustrating as I cannot find it!


  • $\begingroup$ Hm, are you sure that you calculated your probability current for $x\leq 0$ correctly? Your formula for the current has the form $-i(z-z^*)$ for a complex $z$ ($z=\bar{\psi}\psi'$), which is the imaginary part of $z$ and is thus manifestly real. However, your expression for the probability current is complex. So those two expressions are not consistent. $\endgroup$
    – Andrew
    Jun 8, 2014 at 18:47
  • $\begingroup$ Yes this is my problem! But I can't see where I am going wrong when I calculate it? I'm pretty sure it's just an annoying algebra mistake $\endgroup$
    – Wooster
    Jun 8, 2014 at 19:03
  • $\begingroup$ To me it looks like you have written down $\bar{\psi}\psi'$ and called that $j$, without subtracting off the complex conjugate. For example, $\bar{\psi}\psi'$ has a term that is $hk/2m$, but when you subtract $\psi\bar{\psi}'$ this term will cancel. $\endgroup$
    – Andrew
    Jun 8, 2014 at 19:18
  • $\begingroup$ Hm, thanks for your help. Why should $\bar{\psi} \psi'$ have those terms? I've added a bit more detail to how I am calculating $j$ to the question $\endgroup$
    – Wooster
    Jun 8, 2014 at 19:26
  • $\begingroup$ The probability current is proportional to ${\rm Im}(\bar{\psi}\psi')$ (if this part isn't clear then spend some time understanding this). So take your expressions for $\bar{\psi}$ and $\psi'$, which are correct, multiply them and work out the real and imaginary parts explicitly. Right now you are just multiplying them, without splitting into real and imaginary parts. $\endgroup$
    – Andrew
    Jun 8, 2014 at 19:33

1 Answer 1


You say $$j = \frac{hk}{2m} (1 + R \exp(-2ikx) - \bar{R} \exp(2ikx) - R\bar{R})$$ But instead as

$$j=\frac{-i\hbar}{2m}(\bar\Psi\Psi'-\bar\Psi'\Psi)=\frac{-ih}{4\pi m}(\bar\Psi\Psi'-\bar\Psi'\Psi)$$

In $x<0$, $$\bar{\psi} \psi' = ik -ikR\exp(-2ikx) + \bar{R}ik\exp(2ikx) - ikR\bar{R}$$

$$\bar{\psi}' \psi = -ik -ikR\exp(-2ikx) +ik\bar{R}exp(2ikx) +ik R\bar{R}$$


$$j = \frac{-i\hbar }{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)=\frac{-i\hbar}{2m} (2ik-2ikR\bar R)=\frac{-i\hbar}{2m} .2ik(1-1R\bar R)=\frac{hk}{2\pi m}(1-|R|^2)$$. And in $x>0$, $$\psi=T\exp(ikx)$$

$$\bar{\psi} \psi' = ik\bar TT$$

$$\bar{\psi}' \psi = -ik\bar TT$$

$$j=\frac{-i\hbar }{2m} (\bar{\psi}\psi' - \bar{\psi'}\psi)=\frac{-i\hbar }{2m} (2ik\bar TT)=\frac{-i\hbar }{2m} 2ik(|T|^2)=\frac{hk}{2\pi m}(|T|^2)$$

Your mistake was simply subtraction of two

  • $\begingroup$ Mark this as correct answer if you think it has solved your problem $\endgroup$
    – RE60K
    Jun 15, 2014 at 1:03
  • $\begingroup$ Sorry, I forgot to accept, I don't spend all that much time on this site! $\endgroup$
    – Wooster
    Jun 15, 2014 at 10:05
  • $\begingroup$ Small subtlety: when the person asking the question "accepts" the answer, that does NOT equate to it being the CORRECT answer. In this case it might well be but don't confuse the two.... $\endgroup$
    – Floris
    Jun 17, 2014 at 11:56
  • $\begingroup$ Sorry @Floris, I understand what you meant by accepted answer and correct answer, btw English is not my first language, I'm Indian $\endgroup$
    – RE60K
    Jun 18, 2014 at 4:21
  • $\begingroup$ @Aditya - no need to apologize. English is not my first language either! There are some interesting discussions on Meta about the fact that the person who asks the answer is probably not qualified to know which of a range of answers is "right" - they just mark the one they found "most helpful". When you come to the archives of Stackexchange for wisdom, you have to keep this in mind... $\endgroup$
    – Floris
    Jun 18, 2014 at 20:36

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