Undergrad The imaginary part of the wave function

[LarryS]

TL;DR

How is the imaginary part of the wave function related to the real part?

In non-relativistic QM, the wave function is complex-valued. But in my experience, most of the books and lectures, when solving the Schrodinger Equation, focus only on the real part of the solution. Is that because the imaginary part of the solution can always be determined from the real part by applying a 90 phase shift?

Thanks in advance.

 

[renormalize]

In non-relativistic QM, the wave function is complex-valued. But in my experience, most of the books and lectures, when solving the Schrodinger Equation, focus only on the real part of the solution.

I don't see that at all.
Consider, e.g., Griffiths & Schroeter, Introduction to Quantum Mechanics (3rd ed.) in which the authors solve for the wave functions of the hydrogen atom:

Here the spherical harmonics $Y^m_l\left(\theta,\phi\right)$ are given by:

The $\psi_{nlm}$ are clearly complex for all $m\neq 0$.
What books are you finding that "focus only on the real part of the solution"?

 

[WernerQH]

We had a related discussion a while ago:

In chemistry, often real-valued orbitals are preferred, while in physics often complex-valued ones are used.

One can use different linear combinations of the spherical harmonics, combining the $e^{+im\phi}$ and $e^{-im\phi}$ terms into $\sin m\phi$ and $\cos m\phi$ terms. Chemists are rarely interested in systems with exact spherical symmetry.

The time-dependent Schrödinger equation has a companion that has $i$ replaced by $-i$, describing a wave function evolving backwards in time. For systems invariant under time reversal the functions are identical, and the stationary states can be represented by real functions ($\Phi^* = \Phi$). But if you wish you can also multiply by $i$ and choose purely imaginary eigenfunctions. They represent exactly the same states.

This changes when magnetic fields are involved, because they destroy time reversal symmetry.