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Why the hydrogen radial wave function is real?

Is it a coincidence?

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    $\begingroup$ Related: The book of Griffiths, Intro to QM, Problem 2.1b, p.24; and this and this Phys.SE post. $\endgroup$ – Qmechanic Apr 9 '13 at 17:01
  • $\begingroup$ Thank you, but how can I conclude? $\endgroup$ – Arnaud Apr 9 '13 at 17:04
  • $\begingroup$ The main point is that a wave function solution to the TISE is not necessarily real, but it can be chosen so. $\endgroup$ – Qmechanic Apr 13 '13 at 15:09
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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).

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  • $\begingroup$ The angular part is not real ! $\endgroup$ – Arnaud Apr 9 '13 at 16:55
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    $\begingroup$ 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. $\endgroup$ – Rafael Reiter Apr 9 '13 at 16:57
  • $\begingroup$ What is the "$p_x$ wave function" ? $\endgroup$ – Arnaud Apr 9 '13 at 17:01
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    $\begingroup$ See en.wikipedia.org/wiki/… . For the spherical harmonics of one angular momentum number, there is an equivalent linear combination which is real. $\endgroup$ – Rafael Reiter Apr 9 '13 at 17:03
  • $\begingroup$ 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 ? $\endgroup$ – Arnaud Apr 9 '13 at 17:05
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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.

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

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