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In lecture, my professor had drawn the wavefunction for a particle which encounters a potential energy barrier, whose energy it fails to exceed. The graph was highly similar to the one which appears hereafter:

enter image description here

Yet, if the wavefunction is complex (specified, for instance, in region I by $\psi(x) = Ae^{ikx}+Be^{-ikx}$ and in region III by $\psi(x) = Ce^{ikx}$), then how might we be able to thus draw the wavefunction? Are we only drawing its 'real part'? (Yet, what might this real part be, if the coefficients are themselves complex)?

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The real part being drawn is indeed misleading. Actual wavefunction, if we are talking about a particle incident on a barrier, should be complex. To draw it, you can plot separately the real and imaginary part, like in the animation of the time dependence below:

real and imaginary parts

Another option is to draw a 3D spiral where $x$ coordinate along the spiral axis is the spatial coordinate, and the $y$ and $z$ coordinates show respectively the real and imaginary part of the wavefunction:

spiral-like plot

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Yes. When drawing a wavefunction we just draw the real part. You can also write the wave function in terms of sines and cosines and avoid the exponential notation altogether.

Remember the Euler's formula

$e^{ikx}=cos{(kx)}+i\textrm{ }sin{(kx)}$

Typically waves are written with the exponential notation as it makes it easier to manipulate it mathematically. We only need to consider the real part.

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  • $\begingroup$ Thank you for your reply. I wish to inquire how I might draw this 'real part' if the coefficients which multiply $e^{ikx}$ are complex? $\endgroup$ – Vera Leighton Aug 28 '20 at 21:06
  • $\begingroup$ If the coefficients are complex, you won't be able to draw the wavefunction. You will only be able to draw the probability distribution function for that wavefunction. $\endgroup$ – Saptarshi Sarkar Aug 28 '20 at 21:09
  • $\begingroup$ Finally, might I also ask here (or need I publish a new question) why it is that the amplitude of the wavefunction is lower in region III than it is in region I? Does this arise from our need to make the wavefunction (and its derivative) continuous? $\endgroup$ – Vera Leighton Aug 28 '20 at 21:12
  • $\begingroup$ That is due to quantum tunneling. The amplitude of the wavefunction falls off exponentially in the region II. It is a consequence of boundary conditions and the solution of the wavefunction in region II. $\endgroup$ – Saptarshi Sarkar Aug 28 '20 at 21:13
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    $\begingroup$ There is no reason why we should only draw, or indeed do in general draw, the real part. The real part of the wavefunction has no physical nor mathematical meaning by itself. Refer to Ruslan's answer. $\endgroup$ – Giorgio Comitini Aug 28 '20 at 21:34

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