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I'm studying a bit of molecular physics and having some trouble understanding the +/- symmetry in a simplest molecule like a molecular hydrogen ion (2 protons and 1 electron).

In a book, usually they solve for 2 states: one bonding and the another one is anti-bonding. One of them is the $\Sigma^+_u$ state and the other one is $\Sigma^+_g$ state. Why don't we consider the "minus" state also ($\Sigma^-_g$ and $\Sigma^-_u$) ? Because according to the symmetry of the ground sigma state, there should be 4 states: $\Sigma^-_g$, $\Sigma^+_g$, $\Sigma^+_u$ and $\Sigma^-_u$.

Also, can anyone explain to me why $\Lambda$-doubling degeneracy is not there for $\Sigma$ states? I know that we can construct simultaneous eigen-states of the energy, $L_z$ and reflection operators. But why can't they be degenerate?

Qmechanic
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thetorque
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1 Answers1

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The letter $\Sigma$ – the initial term in the sequence $\Sigma,\Pi,\Delta,\dots$ – indicates that $\Lambda=0$, the spin with respect to the axis going from one nucleus to the other.

It's a special case because in that case, the wave function is independent of $\phi$, the angle around the axis, and therefore the reflection with respect to a plane including the axis is automatically $+1$. That explains why there are no states with the minus sign at all. For $\Lambda=1,2,\dots$, there are always two signs as eigenvalues of the reflection operator, so the number of states is doubled. For $\Lambda=1,2,\dots$, the wave functions depend as $\sin \Lambda \phi$ and $\cos\Lambda \phi$ on the angle $\phi$ and on $\Lambda$. One of them is even, the other one is odd. Note that if you substitute $\Lambda=0$, the sine function is strictly zero, so it doesn't exist as a normalized state.

The $u/g$ letter refers to parity, the operation of switching all coordinates relatively to the center of the molecule. The true ground state of the electron in the field of two nuclei is one that has no zeroes anywhere, so it must be one that is an even function of the coordinates.

See Molecular term symbol and a section of Feynman's textbook for more (or complementary) comments.

Luboš Motl
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