Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why the hydrogen radial wave function is real?

Is it a coincidence?

share|cite|improve this question
Related: The book of Griffiths, Intro to QM, Problem 2.1b, p.24; and this and this Phys.SE post. – Qmechanic Apr 9 '13 at 17:01
Thank you, but how can I conclude? – Arnaud Apr 9 '13 at 17:04
The main point is that a wave function solution to the TISE is not necessarily real, but it can be chosen so. – Qmechanic Apr 13 '13 at 15:09
up vote 3 down vote accepted

Wave functions which are Eigenfunctions of the stationary Schrödinger equation can always be chosen to be real. That's because the equation itself is real. Depending on the boundary conditions, the solution can also be complex (e.g. for scattering BC they are complex).

share|cite|improve this answer
The angular part is not real ! – Arnaud Apr 9 '13 at 16:55
It can be chosen to be real. See e.g. $p_x$, $p_y$ and $p_z$ wave function which are equivalent to the complex counterparts for l=1. – Rafael Reiter Apr 9 '13 at 16:57
What is the "$p_x$ wave function" ? – Arnaud Apr 9 '13 at 17:01
See… . For the spherical harmonics of one angular momentum number, there is an equivalent linear combination which is real. – Rafael Reiter Apr 9 '13 at 17:03
Oh, thank you very much ! Have you a link that prove the fact that wave functions of bound systems can be chosen to be real ? – Arnaud Apr 9 '13 at 17:05

Since quantum states that differ by multiplication by a complex number of length $1$ are all equivalent, you can multiply any wavefunction of the Hydrogen atom by such a complex number, and you'll get a vector in Hilbert space that is an equivalently valid description of the corresponding physical state.

share|cite|improve this answer
Yes, but it doesn't answer my question. $e^{ix}$ is never a real function, even when you multiply it by a constant. – Arnaud Apr 9 '13 at 16:59
Yes, because it is not bound! – Rafael Reiter Apr 9 '13 at 17:00
And $e^{ix}/(x^2+1)$ ? – Arnaud Apr 9 '13 at 17:00
@Arnaud: That's not a constant. – Rafael Reiter Apr 9 '13 at 17:01
Yes, but it doesn't answer my question. $e^{ix}/(x^2+1)$ is never a real function, even when you multiply it by a constant. – Arnaud Apr 9 '13 at 17:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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