I am reading about the solution of the Schrödinger equation for the hydrogen atom and have a question about the choice of the $z$-axis. Most websites say that the $z$-axis is arbitrarily chosen. If so, why is the choice of the $z$-axis not part of the quantum numbers? Wouldn't two electrons with the same quantum numbers but different $z$-axes have different wavefunctions?

Say, I have a wavefunction of an electron with $n$=2, $\ell$=1, $m_\ell$=0, $m_s$ = 1/2 and the z-axis pointing in some particular direction. I can create a new wavefunction with the same quantum numbers by, say, rotating the z-axis around the x-axis by 1 degree. In fact, I can create an infinite number of different wavefunctions with the same quantum numbers by just rotating the z-axis. Since they are different wavefunctions, they should be able to fit into the same atom without violating the Pauli exclusion principle. Therefore, there should have been many more electrons in the hydrogen atom at $n$=2. But this is not the case. Could you please tell me what I am missing?

  • $\begingroup$ Wave-functions must be orthogonal, and only 4 fit at the n=2 level. More concretely, of course the 'z' axis can rotate - we expect the atoms are all in arbitrary orientations, so unless we apply a field that 'squeezes' the atoms, we can't tell which state of the n=2 electrons we're looking at anyway. $\endgroup$
    – user121330
    Nov 5, 2014 at 2:22

1 Answer 1


Say we have two different coordinate systems $F=(x,y,z)$ and $F'=(x',y',z')$.

Consider the basis spanned by the eigenvectors of the $L_z$ operator for a given $l$, $\{|l \; m\rangle\; ; \;m=-l,\dots,l\}$. Now, one can find the eigenvectors of the $L_{z'}$ on this basis. In general, the eigenvectors of $L_{z'}$ will be linear combinations of the eigenvectors of $L_z$.

This means that if $|\psi'\rangle$ is an eigenvector of $L_{z'}$, then, there are values of $m$ for which $$ \langle l \; m | \psi'\rangle \neq 0. $$ So stating that you can create different wavefunctions is not true.


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