# Why can we choose wavefunctions to be real?

1. May 18, 2014

### AndrewShen

There are many cases, for simplicity, we choose the wavefunctions to be real. For example, in http://en.wikipedia.org/wiki/Born–Oppenheimer_approximation, there is "The electronic wave functions \chi_k\, will be taken to be real, which is possible when there are no magnetic or spin interactions. "

I do not know why this can always be done. In fact, I think even the electron eigenstates of hydrogen cannot taken to be real. I know if the Hamiltonian has time-reversal symmetry, and the energy eigenstate is nondegenerate, then the wavefunction can taken to be real. But in most cases there is degeneracy. Therefore this assumption seems to be starnge and not reasonable?

2. May 18, 2014

### Simon Bridge

Well, in the example you show, it is used for an approximation. So it only needs to give approximate results, and the method is valid as long as the approximation holds. Welcome to real physics.

You can attempt the calculation without the approximation if you like...

3. May 19, 2014

### DrDu

I think you misunderstood here something. When the eigenstates of the Hamiltonian are degenerate, you can also chose real eigenfunctions. E.g. in the case of hydrogen, you may combine the degenerate (and complex) wavefunctions which are eigenstates of m into real eigenstates $\psi_{lm}+\psi_{l-m}$ and $i(\psi_{lm}-\psi_{l-m})$. For l=1, this are the orbitals $p_x$ and $p_y$, respectively.