# Naive Question About Wavefunctions

I was recently introduced to wavefunctions in my freshman Modern Physics class. While discussing their properties, we often drew diagrams like this: This was all fine until I was told that wavefunctions are complex functions. I take this to mean that the range is complex, but now this diagram makes little sense. How are complex numbers (with both a real and imaginary part) displayed on one axis (the vertical one in the diagram)? The problem gets worse since we then used such diagrams to look for discontinuities.

This is perhaps a silly question, but I can't find a way to make sense of this. Where am I going wrong?

## 2 Answers

More often than not, especially in freshman classes, the wavefunctions you draw are eigenstates of some time independent Hamiltonian. This means that the Schrodinger equation can be written as: $$-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + V(x)\psi(x) = E*\psi(x)$$

Where the full wavefunction would be: $$\Psi(x,t) = \psi(x) e^{iEt/\hbar}$$.

Since the complex part of these are rather trivial, all interesting properties are encoded in $$\psi(x)$$ which is a real function. In above plot, that's what you see.

Of course, in general it couldn't be done this way.

• Follow up question: in reality , the whole wavefuntion would have to be single valued, continuous, and smooth, yes ? But for simplicity, we do these checks only on a part of it at this level ? – Sal_99 Mar 27 '20 at 15:42
• Yes. When a decomposition of this sort is possible, then all these checks are only needed for the spatial part $\psi(x)$. Otherwise you need to check all of it. – Ari Mar 27 '20 at 17:43

Such plots usually have the modulus of the wavefunction plotted on the vertical axis. That's just a real number. It is important to remember that the wavefunction itself doesn't really have any physics in it, it is the modulus which represents the associated probability density (and hence the physics).

• It is important to remember that the wavefunction itself doesn't really have any physics in it That's a contentious way of putting it. The 'physics comes out' by applying the relevant quantum operator on $\psi$. – Gert Mar 25 '20 at 12:13