# Determining wave function for term symbol 1D

I am trying to follow a book (Introduction to Ligand Field Theory by Ballhausen in 1962 on pg 15), but it isn't clear how they make a particular leap.

Background I want to find the wave function for the term $^1 D$. We know that $\psi(L,M_L,S,M_S) = \psi (2,2,0,0)$ is made up of three micro states: $(2^+,0^-)$, $(2^-,0^+)$, $(1^+,1^-)$. Thus, a linear combination must be taken of these micro states. Now in the book they say they must be orthogonal to $\psi(3,2,1,0)$ and $\psi(4,2,0,0)$. Presumably these are picked because we just determined those via lowering operators in the previous section. I don't take it that there is another reason those two wave functions are mentioned.

$(2^+,0^−)$ means that electron 1 has $m_s=+1/2$ and $m_l=2$ and that electron 2 has $m_s=−1/2$ and $m_l=0$. We write that $\Psi=|(\psi^+_1)(\psi^−_2)|$ where we have written the short form of the determinantal antisymmetrized normalized wave function, which comes from the diagonal element in the Slater determinant AND separated the orbital- and spin-dependent parts of the wave function (spin is denoted by super plus or minus sign).

Problem Anyway, I understand that $\psi(2,2,0,0) = a (2^+,0^-) + b (2^-,0^+) + c(1^+,1^-)$. Then they write "and we get $a \sqrt{3} - b \sqrt{3} + c \sqrt{8} = 0$ and $a + b = 0$." I'm totally lost how they made this leap and why these must be equal to zero. Does this have to do with the orthogonal wave functions just mentioned? Where did the numbers in the sqrt come from? And the minus sign! Greatly appreciated.

• Which book is it? Your question seems to be missing something as is, so it would help if we could check the book to fill in any gaps. – David Z Jun 20 '11 at 6:13
• it's "Introduction to Ligand Field Theory" by Ballhausen (1962) on page 14. The point of this is that we "must develop a technique for picking out suitable linear combinations of these micro states in order to form eigenfunctions for $\hat{L}$ and $\hat{S}$." – Chris Jun 20 '11 at 6:38
• I presume this problem is about expressing one state in some other basis. If so, we can't really help you until you tell us what that basis is. I can only guess that this is a system of two spin-$1/2$ particles and you are trying to express the common wavefunction as the wavefunction of individual particles. But if so, I don't understand why there are only three microstates and not $(0^+,2^-)$ etc. Please, fill in the details. – Marek Jun 20 '11 at 9:39
• Or is $(2^+,0^-)$ supposed to represent already antisymmetrized product? – Marek Jun 20 '11 at 9:40
• $(2^+,0^-)$ means that electron 1 has $m_s=+1/2$ and $m_l=2$ and that electron 2 has $m_s = -1/2$ and $m_l=1$. We write that $\Psi = | (\psi_1^+) (\psi_2^-) |$ where we have written the short form of the determinantal antisymmetrized normalized wave function, which comes from the diagonal element in the Slater determinant AND separated the orbital- and spin-dependent parts of the wave function. In the complete expanded wave function, we have all possible permutations of electrons. – Chris Jun 20 '11 at 17:50

Previously on pg. 14 we determined using ladder operators that $\psi(4,2,0,0) = \sqrt{3/14} (2^+, 0^-) + \sqrt{8/14} (1^+,1^-) - \sqrt{3/14}(2^-,0^+)$ and $\psi(3,2,1,0) = sqrt{1/2} [(2^+,0^-) + (2^-,0^+)]$.
For the singlet D term, we have $\psi(2,2,0,0) = a (2^+,0^-) + b(2^-,0^+) + c(1^+,1^-)$.
And if we take $\psi(2,2,0,0) \cdot \psi(4,2,0,0) = \sqrt{3/14}a + \sqrt{8/14}c - \sqrt{3/14}b = 0$ because he has said that these micro states must be orthogonal. Multiply by $\sqrt{14}$ to obtain his result. It also follows that $\psi(2,2,0,0) \cdot \psi(3,2,1,0) = \sqrt{1/2} a + \sqrt{1/2} b = 0$. Multiplication by $\sqrt{2}$ will give his result of $a + b = 0$. Also realize that $a^2 + b^2 + c^2 = 1$ must be true as a normalizing condition. Now we have three equations and three unknowns, so solving it is simple.
$\psi(3,2,1,0)$ and $\psi(4,2,0,0)$ were picked in the first place because we had three unknowns and needed three equations to solve it. These wave functions must be orthogonal in the first place, therefore it was a matter of convenience AND that they were made up of the corresponding micro states in the first place.