Let $\{{\varphi }_j\}$ be a complete orthonormal eigenbasis of the Hamiltonian, with $H{\varphi }_j=E_j{\varphi }_j$ and $E_j\ge E_0$. Expand a nonzero admissible trial wavefunction as $\tilde{\Psi }={\sum }_j{c}_j{\varphi }_j$. ThenConsequentlyEvery numerator term is nonnegative and the denominator is positive. Equality requires the trial to lie wholly in the ground-state subspace; it need not be an individually chosen eigenvector when the ground energy is degenerate. This is the quantum variational principle, and wavefunction normalization before calculation is optional because the quotient removes the overall scale.

Linearity and the shared $l=1$ angular eigenvalue giveUsing the relative phases in part (c),Thus each real orbital remains an energy eigenstate with $E=-1/8$ hartree and has definite orbital angular momentum magnitude $\sqrt{2}\hslash$, but neither is an eigenfunction of ${\hat{l}}_z$. A measurement of $l_z$ gives $+\hslash$ or $-\hslash$, each with probability $1/2$. Its mean is zero and its variance is ${\hslash }^2$; a zero mean does not mean that a measurement gives zero.

Denote the two complex spatial atomic orbitals by ${\psi }_{+}$ and ${\psi }_{-}$, retaining the same spin state. With the phases specified by the supplied spherical harmonics, chooseThe two original atomic orbitals are orthonormal, so the squared coefficients add to one and these combinations are normalized and mutually orthogonal. Explicitly,which are real spatial wavefunctions. An overall sign on either orbital has no physical consequence.

These are the familiar $2p_x$ and $2p_y$ p orbitals. Each has two lobes of opposite wavefunction sign, with an atomic-orbital nodal plane through the atomic nucleus: $x=0$ for $2p_x$ and $y=0$ for $2p_y$. The sign is a phase convention, not a positive or negative electric charge; the probability density is nonnegative on both lobes.

Keep one normalized spin function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta ,\varphi )=\sin \theta e^{im\varphi }$, with $m=\pm 1$. Overall constants and the spin factor pass unchanged through the spatial operators.

First, in atomic units, ${\hat{l}}_z=-i{\partial }_{\varphi }$, soFor the squared orbital angular momentum, ${\partial }_{\varphi }^2{f}_m=-f_m$ andTherefore the two terms of the angular operator giveusing $1-{\cos }^2\theta ={\sin }^2\theta$. Thus ${\hat{l}}^2\Psi =2\Psi$.

Finally, the spatial hydrogen Hamiltonian isFor $R=Ar{e}^{-r/2}$,Substitute the already established angular eigenvalue $l(l+1)=2$:Hence $E=-1/8$ hartree, consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.

In atomic units, $\hslash =m_e=e=4\pi {\epsilon }_0=1$, and the hydrogen atom hasThe allowed quantum numbers areHere the fixed electron spin quantum number is $s=1/2$. The energy eigenvalues are in hartrees; restoring units gives ${\hslash }^{2}l(l+1)$ and $\hslash m$ for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.