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Matrix Elements in A v=0
<J,Lambda=0,Omega=-0.5|H|J,Lambda=0,Omega=-0.5> =
+ Origin*1
+ B*((4*J*(J+)+1)/4) + gamma*(-1/2)
<J,Lambda=0,Omega=0.5|H|J,Lambda=0,Omega=0.5> =
+ Origin*1
+ B*((4*J*(J+1)+1)/4) + gamma*(-1/2)
<J,Lambda=0,Omega=-0.5|H|J,Lambda=0,Omega=0.5> = + B*((-sqrt(4*J*(J+1)+1))/2) + gamma*(sqrt(4*J*(J+1)+1)/4)
<J,Lambda=0,Omega=0.5|H|J,Lambda=0,Omega=-0.5> = + B*((-sqrt(4*J*(J+1)+1))/2) + gamma*(sqrt(4*J*(J+1)+1)/4)
|J ± Ω=½> = 1/sqrt(2){|J,Lambda=0,Omega=+0.5> ± |J,Lambda=0,Omega=-0.5>}The 1/sqrt(2) ensures the resulting combination is normalised - all the basis states are set up to be normalised and orthogonal. There are two different ways of specifying the symmetry for linear molecules. The parity is the symmetry of the wavefunction with respect to inversion of (space fixed) co-ordinates; a properly constructed wavefunction will stay the same (+) or simply change sign (-). The parity of, say, the + combination above will alternate with J, so e and f notation is more normal; for the A state of OH the + combination has e parity always, but + for even J and - for odd J. Provided JAdjustSym is set at the molecule level, PGOPHER will display both of these, as above, and normally accept either on input. If false, e and f are not displayed. The matrix elements given above can be transformed to matrix elements between the symmetry adapted basis functions. PGOPHER will also display matrix elements between these symmetry adapted linear combinations (for linear molecules only at the moment). For the A state we obtain:
Symmetrized Matrix Elements in A v=0which a little rearranging gives the clearer form:
<J,Lambda=0,Omega=0.5 + |H| + J,Lambda=0,Omega=0.5> =
(4*Origin+(-2*B+gamma)*sqrt(4*J*(J+1)+1)+B*(4*J*(J+1)+1)-2*gamma)/4
<J,Lambda=0,Omega=0.5 - |H| - J,Lambda=0,Omega=0.5> =
(4*Origin+(2*B-gamma)*sqrt(4*J*(J+1)+1)+B*(4*J*(J+1)+1)-2*gamma)/4
<J,Ω=±½|H|J,Ω=±½> = Origin + BJ(J+1) + B/4 - ½γ ± (B-½γ)Sqrt(J(J+1)+1/4)Note that there are no matrix elements between the + and - combinations, as required by symmetry (which is of course why the combinations are used). We can compare the results of this formula with values worked out by PGOPHER - try "View, States", and ensure that the A state is selected in the box in the top line. If, say J = 4.5 and e parity is selected the display will look something like this:
Hamiltonian Matrix for A, J = 4.5, Symmetry = e (+)(This output is actually taken from the log window with the print level on the right set to "Matrices"). Note the general form of the state specification with the key given in the states window: |Name J +- Omega>. Name is the manifold and state name (here A and v=0), J is as above, +/- indicates whether the + or - combination is taken, and Omega is actually |Ω| here. (Λ is omitted in the label as it is always taken as positive in the symmetrised wavefunctions, and is thus implied by the state name.)
|A v=0 4.5 + 0.5>
<A v=0 4.5 + 0.5| 32741.6
Eigenvalues for A, J = 4.5, Symmetry = e (+)This should be read as the state name on the left has energy 32741.6 cm-1 and the overall wavefunction is equal to the (symmetrised) basis sate given. The state name key is given in the states window as Name J N Fn p. The "Name" is the manifold and state name as before, but now an additional quantum number, N, is given (here 4). (The vector J is the vector sum of N and S, so in this case the quantum number J = N ± ½.) Also given is the F number, here F1 or F2, and alternative way of specifying the spin orientation, and p indicates the parity (here e). For linear molecules, settings at the molecule level can be used to select the quantum numbers displayed - see "ShowJ", "ShowOmega", "ShowN", "ShowFNumber" and "Showef". By default these are all set on apart from "ShowOmega".
|A v=0 4.5 + 0.5>
1 A v=0 4.5 4 F1e 32741.600 1.000000
The ground state is more complicated, as the orbital angular
momentum, Λ is now not zero, so a full matrix
diagonalisation is required. To obtain the full spitting pattern
we must also consider lambda doubling. The required constants
are the spin-orbit coupling constant, A = -13.21 cm-1,
the rotational constant, B = 18.55 cm-1, two
lambda doubling constants, p = 0.235 cm-1 and
q = -0.0391 cm-1. The Hund's case (a) basis
now includes two possible values of Λ (=±1) in addition
to the electron spin S=½ and its projection
Σ=±½ as for the excited states. Given two possible
values for each of Λ and Σ there are four basis states. The
matrix elements are:
Matrix Elements in X v=0
<J,Lambda=-1,Omega=-1.5|H|J,Lambda=-1,Omega=-1.5> =
+ B*((4*J+4*J^2-7)/4)
+ A*(1/2)
<J,Lambda=-1,Omega=-1.5|H|J,Lambda=-1,Omega=-0.5> =
B*((-sqrt(4*J+4*J^2-3))/2)
<J,Lambda=-1,Omega=-0.5|H|J,Lambda=-1,Omega=-1.5> =
B*((-sqrt(4*J+4*J^2-3))/2)
<J,Lambda=-1,Omega=-0.5|H|J,Lambda=-1,Omega=-0.5> =
+ B*((4*J+4*J^2+1)/4)
+ A*(-1/2)
<J,Lambda=1,Omega=0.5|H|J,Lambda=1,Omega=0.5> =
+ B*((4*J+4*J^2+1)/4)
+ A*(-1/2)
<J,Lambda=1,Omega=0.5|H|J,Lambda=1,Omega=1.5> =
B*((-sqrt(4*J+4*J^2-3))/2)
<J,Lambda=1,Omega=1.5|H|J,Lambda=1,Omega=0.5> =
B*((-sqrt(4*J+4*J^2-3))/2)
<J,Lambda=1,Omega=1.5|H|J,Lambda=1,Omega=1.5> =
+ B*((4*J+4*J^2-7)/4)
+ A*(1/2)
<J,Lambda=-1,Omega=-0.5|H|J,Lambda=1,Omega=0.5> =
+ p*((-sqrt(4*J*(J+1)+1))/4)
+ q*((-sqrt(4*J*(J+1)+1))/2)
<J,Lambda=1,Omega=0.5|H|J,Lambda=-1,Omega=-0.5> =
+ p*((-sqrt(4*J*(J+1)+1))/4)
+ q*((-sqrt(4*J*(J+1)+1))/2)
<J,Lambda=-1,Omega=-1.5|H|J,Lambda=1,Omega=0.5> =
q*(sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1))/8)
<J,Lambda=-1,Omega=-0.5|H|J,Lambda=1,Omega=1.5> =
q*(sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1))/8)
<J,Lambda=1,Omega=0.5|H|J,Lambda=-1,Omega=-1.5> =
q*(sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1))/8)
<J,Lambda=1,Omega=1.5|H|J,Lambda=-1,Omega=-0.5> =
q*(sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1))/8)
Note that Σ = Ω - Λ, so is not specified separately in the
basis states. As before, we can simplify the matrices by
taking symmetry adapted combinations.
Symmetrized Matrix Elements in X v=0which can be rearranged to give the more compact form:
<J,Lambda=1,Omega=1.5 + |H| + J,Lambda=1,Omega=1.5> =
(B*(4*J*(J+1)-7)+2*A)/4
<J,Lambda=1,Omega=1.5 - |H| - J,Lambda=1,Omega=1.5> =
(B*(4*J*(J+1)-7)+2*A)/4
<J,Lambda=1,Omega=1.5 + |H| + J,Lambda=1,Omega=0.5> =
(-4*B*sqrt(4*J*(J+1)-3)+q*sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1)))/8
<J,Lambda=1,Omega=1.5 - |H| - J,Lambda=1,Omega=0.5> =
(-4*B*sqrt(4*J*(J+1)-3)-q*sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1)))/8
<J,Lambda=1,Omega=0.5 + |H| + J,Lambda=1,Omega=1.5> =
(-4*B*sqrt(4*J*(J+1)-3)+q*sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1)))/8
<J,Lambda=1,Omega=0.5 - |H| - J,Lambda=1,Omega=1.5> =
(-4*B*sqrt(4*J*(J+1)-3)-q*sqrt((4*J*(J+1)-3)*(4*J*(J+1)+1)))/8
<J,Lambda=1,Omega=0.5 + |H| + J,Lambda=1,Omega=0.5> =
(B*(4*J*(J+1)+1)-2*A+(-p-2*q)*sqrt(4*J*(J+1)+1))/4
<J,Lambda=1,Omega=0.5 - |H| - J,Lambda=1,Omega=0.5> =
(B*(4*J*(J+1)+1)-2*A+(p+2*q)*sqrt(4*J*(J+1)+1))/4
<J,Ω=3/2,±|H|J,Ω=3/2,±> = A/2 + BJ(J+1) + 7B/4The symmetry adapted combinations mean that the four basis states split into two pairs of states for any given J, one for each parity. A sample Hamiltonian matrix for J = 4.5, f (-) parity shows a non zero matriox elements mixing the Ω=1/2 and Ω=3/2 states:
<J,Ω=1/2,±|H|J,Ω=1/2,±> = -A/2 + BJ(J+1) + B/4 - ±½(p+2q)(J+1/2)
<J,Ω=3/2,±|H|J,Ω=1/2,±> = -B(J(J+1)-¾)½ ± ½q{(J(J+1)-¾)(J(J+1)+¼)}½
Hamiltonian Matrix for X, J = 4.5, Symmetry = f (-)The resulting energy levels are thus mixtures of these two states:
|X v=0 4.5 - 1.5>|X v=0 4.5 - 0.5>
<X v=0 4.5 - 1.5| 357.045 -90.397194212542
<X v=0 4.5 - 0.5| -90.397194212542 533.747
Eigenvalues for X, J = 4.5, Symmetry = f (-)This should be rad\as the lowest state, labelled 1, has energy 318.994 and is mainly Ω=3/2 - the coefficient in the wavefunction is -0.92. The fact that it is not 1 reflects partial Hund's case (b) character; for higher values of J the matrix elements mixing the two values of Ω increase, and the magnitudes of the coefficients become more nearly equal. In the case (b) limit at high J the coefficients are ±1/2½. At the lowest J=½ the Ω=3/2 state is not present, so the matrix is 1x1.
|X v=0 4.5 - 1.5>|X v=0 4.5 - 0.5>
1 X v=0 4.5 4 F1f 318.994 -0.921674 -0.387965
2 X v=0 4.5 5 F2f 571.798 -0.387965 0.921674
Reduced Matrix Elements of <v=0|T(1)|v=0>These are Reduced transition matrix elements as the M dependence has been taken out; see Transition Moments and Line Strengths for more details. In this case the formua used is:
<J-1,Lambda=-1,Omega=-1.5|T(1)|J,Lambda=0,Omega=-0.5> = sqrt(((2*J-1)*(2*J-3))/(2*J))/2
<J,Lambda=-1,Omega=-1.5|T(1)|J,Lambda=0,Omega=-0.5> = ((2*J+1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2
<J+1,Lambda=-1,Omega=-1.5|T(1)|J,Lambda=0,Omega=-0.5> = sqrt(((2*J+3)*(2*J+5))/(2*(J+1)))/2
<J-1,Lambda=-1,Omega=-0.5|T(1)|J,Lambda=0,Omega=0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2
<J,Lambda=-1,Omega=-0.5|T(1)|J,Lambda=0,Omega=0.5> = (2*J+1)^2/(2*sqrt(2*J*(2*J+1)*(J+1)))
<J+1,Lambda=-1,Omega=-0.5|T(1)|J,Lambda=0,Omega=0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2
<J-1,Lambda=1,Omega=0.5|T(1)|J,Lambda=0,Omega=-0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2
<J,Lambda=1,Omega=0.5|T(1)|J,Lambda=0,Omega=-0.5> = ((-2*J-1)*(2*J+1))/(2*sqrt(2*J*(2*J+1)*(J+1)))
<J+1,Lambda=1,Omega=0.5|T(1)|J,Lambda=0,Omega=-0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2
<J-1,Lambda=1,Omega=1.5|T(1)|J,Lambda=0,Omega=0.5> = sqrt(((2*J-1)*(2*J-3))/(2*J))/2
<J,Lambda=1,Omega=1.5|T(1)|J,Lambda=0,Omega=0.5> = ((-2*J-1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2
<J+1,Lambda=1,Omega=1.5|T(1)|J,Lambda=0,Omega=0.5> = sqrt(((2*J+3)*(2*J+5))/(2*(J+1)))/2
Symmetrized Reduced Matrix Elements of <v=0|T(1)|v=0>As before,the + and - in the bra and ket indicate the sign of the combination. The selection rules are + to - if J does not change, and + to + and - to - if J changes. PGOPHER also gives the line strengths, which are essentially the squares of the matrix elements. In the absence of state mixing these are Honl-London factors, but in the current case there is significant mixing so they are not useful. (The calculation must be done with the unsquared matrix elements.) TODO: factor of 2k+1???b
<J-1,Lambda=1,Omega=1.5 + |T(1)| + J,Lambda=0,Omega=0.5> = sqrt(((2*J-1)*(2*J-3))/(2*J))/2
<J-1,Lambda=1,Omega=1.5 - |T(1)| - J,Lambda=0,Omega=0.5> = sqrt(((2*J-1)*(2*J-3))/(2*J))/2
<J,Lambda=1,Omega=1.5 + |T(1)| - J,Lambda=0,Omega=0.5> = ((-2*J-1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2
<J,Lambda=1,Omega=1.5 - |T(1)| + J,Lambda=0,Omega=0.5> = ((-2*J-1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2
<J+1,Lambda=1,Omega=1.5 + |T(1)| + J,Lambda=0,Omega=0.5> = sqrt(((2*J+3)*(2*J+5))/(2*(J+1)))/2
<J+1,Lambda=1,Omega=1.5 - |T(1)| - J,Lambda=0,Omega=0.5> = sqrt(((2*J+3)*(2*J+5))/(2*(J+1)))/2
<J-1,Lambda=1,Omega=0.5 + |T(1)| + J,Lambda=0,Omega=0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2
<J-1,Lambda=1,Omega=0.5 - |T(1)| - J,Lambda=0,Omega=0.5> = (-sqrt(((2*J-1)*(2*J+1))/(2*J)))/2
<J,Lambda=1,Omega=0.5 + |T(1)| - J,Lambda=0,Omega=0.5> = (2*J+1)^2/(2*sqrt(2*J*(2*J+1)*(J+1)))
<J,Lambda=1,Omega=0.5 - |T(1)| + J,Lambda=0,Omega=0.5> = ((-2*J-1)*(2*J+1))/(2*sqrt(2*J*(2*J+1)*(J+1)))
<J+1,Lambda=1,Omega=0.5 + |T(1)| + J,Lambda=0,Omega=0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2
<J+1,Lambda=1,Omega=0.5 - |T(1)| - J,Lambda=0,Omega=0.5> = (-sqrt(((2*J+1)*(2*J+3))/(2*(J+1))))/2
Rank 1 Transition Matrix for <X, J = 4.5, Symmetry = f (-)| |A, J = 4.5, Symmetry = e (+)>These are between the symmetry adapted basis states, but as we saw above the X state energy levels are mixtures of these basis states, so the above matrix need to be transformed with these. The above matrix becomes:
|A v=0 4.5 + 0.5>
<X v=0 4.5 - 1.5| -1.2712834523275
<X v=0 4.5 - 0.5| -1.2974982402692
Transformed Rank 1 Transition Matrix for <X, J = 4.5, Symmetry = f (-)| |A, J = 4.5, Symmetry = e (+)>
|A v=0 4.5 4 F1e> <X v=0 4.5 4 F1f| 1.67509286382788 <X v=0 4.5 5 F2f| -.702657240207629Note the switch it notation - the first matrix was between basis states (labelled with J, phase and Ω) and the second is between the final energy levels, labelled with J, N and F1/F2. The squares of the numbers above are n ow proportional to transition intensity; to calculate absolute values the population and some units conversion must be done as described in Intensity Formulae. For the states above the following linelist entries are generated:
Mo M' J' S'#'M" J" S"#" Position I Eupper Elower S A WidthNote #' labels the upper state eigenvalue number , here always 1 as a 1x1 matrix was used for the A state. #" is the lower state eigenvalue number, here 1 or 2 to indicate the which of the two states in the matrix was used.
OH A 4.5 e 1 X 4.5 f 1 308.3372 .0068728 32741.6000 318.9937 2.805936 7.738899 0 : qQ1(4.5) : A v=0 4.5 4 F1e - X v=0 4.5 4 F1f OH A 4.5 e 1 X 4.5 f 2 310.7604 .001025 32741.6000 571.7983 .4937272 1.394079 0 : pQ12(4.5) : A v=0 4.5 4 F1e - X v=0 4.5 5 F2f