We are learning about step potential in class. I have completely understood that the behavior of the wave function representing the particle, can have different responses depending on the energy of the particle as compared to the energy of the potential step.

Our teacher instructed us to divide it into a two region problem, one where the potential $V=0$ and the second where $$V=V_0$$

The solution of the Schrodinger wave equation in region 1, is $$Ae^{ik_1x}+Be^{-ik_2x}$$ While finding the reflection and transmission coefficient, our teacher asked us to take $A^2$ and $B^2$. Does this mean that $A$ and $B$ are real in nature? Because if they are complex, then we need to have a modulus around them, right?

Also, if they are complex in nature, isn't there a possibility that the solution will turn out to be real?

I am a beginner to this study, so kindly help me accordingly.

  • 2
    $\begingroup$ The wave function is always complex in general, so yes, you should always have $|~|$ around them. $\endgroup$
    – kaylimekay
    Jan 12, 2021 at 5:38
  • $\begingroup$ Can you please tell me why exactly we are "multiplying" two wavefunctions? $\endgroup$ Jan 12, 2021 at 6:55
  • $\begingroup$ You are trying to find the probabilities of reflection/transmission, right? $\endgroup$
    – kaylimekay
    Jan 12, 2021 at 7:16
  • $\begingroup$ Yes, and the above equation is the solution when v=0...we have similar solutions for V>E and V<E... $\endgroup$ Jan 12, 2021 at 7:24
  • $\begingroup$ I might suggest reviewing your notes or textbook regarding how one gets probabilities from the wave function. $\endgroup$
    – kaylimekay
    Jan 12, 2021 at 7:27

2 Answers 2


Either your teacher is wrong, or you misunderstood what was meant: the constants $A$ and $B$ are in general complex numbers, and so what you should be using (when you're finding the transmission and reflection probabilities, for example) are $|A|^2$ and $|B|^2$, as @kaylimekay points out.

It turns out that you could choose one of the numbers $A$ or $B$ to be real, but there's no contradiction there: real numbers are a subset of complex numbers. In that case (if, say, you've chosen $A$ to be real) $|A|^2 \equiv A^2$.


Like others have said, A and B are complex numbers in general, so you should have the square modulus signs.

Overall, the solution to the wavefunction is in general a complex number. A basic example to illustrate:

We can express any complex number like this: $x + iy = re^{i\phi}$. In our case $\phi = kx$, and we can ignore the scaling factor and let $r=1$.

Let's allow $A$ to be a general complex number: $A = a + ib$, and multiply it with our exponential:

$Ae^{ikx} = (a+ib)(x+iy) = ax + ibx + iay - by = ax -by + i(bx + ay)$

Assuming I made no mistake, we can see that, in general , the product of two complex numbers is itself complex. However, it could be real, if the numbers are such that $bx = - ay$ (making the imaginary component zero).

There is no reason why the solution to the wavefunction cannot be real, just keep in mind that in general it is a complex number. Real numbers are just a subset of complex numbers.

All that you need to worry about is that the modulus squared of the wavefunction, when correctly normalised, yields a probability of observing something.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.