Why the hydrogen radial wave function is real?
Is it a coincidence?
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2Related: The book of Griffiths, Intro to QM, Problem 2.1b, p.24; and this and this Phys.SE post. – Qmechanic Apr 09 '13 at 17:01
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Thank you, but how can I conclude? – Arnaud Apr 09 '13 at 17:04
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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
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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).
zonksoft
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3It 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. – zonksoft Apr 09 '13 at 16:57
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1See http://en.wikipedia.org/wiki/Table_of_spherical_harmonics#Spherical_harmonics_with_l_.3D_1.5B1.5D . For the spherical harmonics of one angular momentum number, there is an equivalent linear combination which is real. – zonksoft Apr 09 '13 at 17:03
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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 09 '13 at 17:05
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Curiously, I can't find a proof in google right now. My guess is: Since the stationary Schrödinger equation is purely real, there is no reason for the solutions to be complex. Even the free particle SE has $\sin kx$ and $\cos kx$ as solutions which are real equivalents of $\exp\pm ikx$. – zonksoft Apr 09 '13 at 17:14
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Actually, I think so, yes. The difference is that, for a free particle, the complex solution is an eigenfunction of the momentum operator at infinity (=it has a well-defined momentum there), whereas the real solution does not. It's funny that this never occured to me before :) – zonksoft Apr 09 '13 at 17:21
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Perhaps Another way to say this is that for a matter wave the oscillations are the presence and absence of matter, mediated by the resistance above the speed of light, whereas for an electromagnetic wave the loss of electric field gives rise to the magnetic field. Here, we have no significance for the absence of matter, just empty space? That is my understanding.
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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.
joshphysics
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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 09 '13 at 16:59
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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 09 '13 at 17:02
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@Arnaud: I see what you mean. You are right, but a bound state always fulfils this property. – zonksoft Apr 09 '13 at 17:04