# Where is spin in the Schroedinger equation of an electron in the hydrogen atom?

In my current quantum mechanics, course, we have derived in full (I believe?) the wave equations for the time-independent stationary states of the hydrogen atom.

We are told that the Pauli Exclusion principle is a consequence of two electrons not being able to share the same wave equation.

However, in our derived equation, we did not have anything including the spin. We defined $\psi (r,\theta,\phi)$ as $\psi_{n,l,m} (r,\theta,\phi) = R_{n,l}(r) Y_{l,m}(\theta,\phi)$, where $Y_{l,m}(\theta,\phi) = f_{l,m}(\theta) e^{i m \phi}$. We then were given well-defined $R_{n,l}$ and $f_{l,m}$ that satisfied the partial differential equations in the Schroedinger Equation.

Nowhere in our final $\psi$ do we find anything that varies depending on a fourth degree of freedom, not to mention one that behaved as $m_s$ should.

Am I missing the point of the Pauli Exclusion Principle? Is there a part of the solutions for $\psi$ that I am not understanding?

EDIT: I am referring to an $H^-$ ion, where there are two electrons each with their own wave equation. If we imagine the case where both have the same quantum numbers n,l,m, but different spin $m_s$, would not their wave equations be exactly the same, and therefore not allowed?

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I think that the solutions that you exhibit are only good for "hydrogen-like" atoms meaning those with only one electron. In the two electron cases there is a interaction between the individual electrons, and the solution must be written $\psi(r_1,\theta_1,\phi_1,r_2,\theta_2,\phi_2)$. – dmckee Jun 6 '11 at 22:55
@dmckee In that case, does spin show up? – Justin L. Jun 7 '11 at 1:17

What you've written down is the spatial part of the electron wavefunction. The spin state is not included. The full wavefunction of the electron involves both the spatial part and the spin part. Sometimes in quantum mechanics books the full electron wavefunction is written as the tensor product of the spatial and spinor parts, sometimes you'll just see it as one written after the other. Sometimes you'll just see it written as $|n,l,m_l, s,m_s \rangle$. I'm not sure what book you're using.

When you derived the solutions of the TISE for the hydrogen atom Hamiltonian, you probably neglected the relativistic corrections, fine-structure interactions, and hyperfine interactions, leaving you with just the Coulomb potential. Since the Coulomb potential doesn't depend on electron spin, you could ignore that part of the wavefunction. (In other words, since the spin operators all commute with the $1/r$ potential, the spin eigenstates are all eigenstates of the $1/r$ Hamiltonian, so it's Kosher to just throw in the $|s, m_s \rangle$ stuff afterwards).

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So basically, Schroedinger's Equations don't tell the complete story? – Justin L. Jun 7 '11 at 1:19
@Justin: Schroedinger's equation can do a fine job, but you have to include the spin dependent potentials as well as the Coulomb potential. – dmckee Jun 7 '11 at 2:17
I'll echo what dmckee said: the Schroedinger Equation does tell the complete (nonrelativistic) story, but it's up to the user to put everything in the Hamiltonian, and make sure the state vector space includes everything about the system. In this case, you approximated Hamiltonian (as explained above) to include only the Coulomb term, and you used a simplified vector space that included the spatial wavefunction but not the spinor wavefunction. That's A-OK for the Hamiltonian you used, since it has no dependence on spin. – Anonymous Coward Jun 7 '11 at 2:28
So basically, the $V(\mathbf{r})$ term of the schroedinger equation we solved was only an approximation? – Justin L. Jun 7 '11 at 5:00
I dunno, Justin L. Why don't you write down the Hamiltonian (or the Schroedinger equation) you used? – Anonymous Coward Jun 7 '11 at 16:51

A description of electron spin and the Pauli exclusion principle needs to go beyond the Schrödinger-equation to the spinor-valued Dirac-equation.

I don't remember my atomic physics course very well but at the level of your analysis I think you just add the rule of the two parallel spin orbitals explicitly.

For a discussion on this topic the wikipedia-page is helpful.

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You have probably solved the non-relativistic Schroendinger's equation which gives rise to only three of quantum numbers i.e. n, l , ml . The property of spin comes as a result of the relativistic effects. The equation was worked out by Dirac.

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For the Pauli principle to be explicitly seen/implemented, one has to have at least two electrons. It is not the case in H.

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I am referring to an $H^-$ ion, where there are two electrons each with their own wave equation. If we imagine the case where both have the same quantum numbers n,l,m, but different spin $m_s$, would not their wave equations be exactly the same? – Justin L. Jun 6 '11 at 22:35
No, we have one equation for one wave function that depends on two arguments $\psi(x_1,x_2)$. The total wave function (including spin variables) must then be antisymmetric on these arguments. – Vladimir Kalitvianski Jun 7 '11 at 9:01

## protected by Qmechanic♦Mar 14 '14 at 14:21

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