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To see how `PGOPHER` works, we will
work through the calculations involved in one fairly simple
example, but one that is complicated enough to show the full
potential complexities. The example we pick here is the A^{2}Σ^{+}
← X^{2}Π transition in the OH radical. This is a
strong transition in the UV, and is often visible in flames. We
simulate here the origin band, v'=0 - v"=0. The required steps
break down as follows:

- Calculate the ground state energies and wavefunctions
- Calculate the excited state energies and wavefunctions
- Calculate the transitions between them

We start with the excited state, as this is
simpler than the ground state in OH. The essential pattern of the
calculation is shown with just three constants - the state origin
(32402 cm^{-1}), the rotational constant, *B*=16.96
cm^{-1} and the spin rotation constant, γ=0.2 cm^{-1}.
The most recent set of constants has many more constants, but
these three will give the basic pattern of energy levels. The
first step in calculating energy levels is to choose a basis set;
for linear molecules `PGOPHER` uses a Hund's case (a)
basis, in which the basis states have well defined values of
Λ (here 0), electron spin *S*=½ and its
projection Σ=±½ on the molecular axis. A
Σ state would actually be considered classic Hund's case
(b), but a Hund's case (a) basis will give identical results,
provided the basis used is complete, which is always the case for
`PGOPHER` (unless the `OmegaSelect` option is not `All`)
. Given the choice of basis set, the next step is to work out the
matrix elements. The formulae `PGOPHER` uses can be found
by right clicking on a state and selecting "Matrix Elements",
which gives the following results in this case:

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)+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)

Any matrix elements not listed should be taken
as zero. `PGOPHER` uses a basic symbolic algebra system to
work these out for printing, so the results look a little
cumbersome. A little rearranging gives the following more
compact form:

- <J,Ω=±½|H|J,Ω=±½>
= Origin +
*BJ*(*J*+1) +*B*/4 - ½γ - <
*J*,Ω=∓½|H|J*,*Ω=±½> = (½γ-*B*)Sqrt(*J*(*J*+1)+1/4)

The first is the diagonal matrix element, and
contains the state origin and the basic *BJ*(*J*+1)
rotational dependence. The second mixes the two states in
the basis, and it is the presence of this term that means that the
state is best described as case (b), rather than case (a).

The first step in calculating the energy levels
using these matrix elements is to take symmetry adapted
combinations. Then basis states used above do not have well
defined parity, but taking plus and minus combinations does
give states with specific parity:

*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:

`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:

In this case there is only one basis symmetrised basis state for any given J and symmetry, so the wavefunctions will simply be equal to one of the basis states and the energy will be equal to one of the matrix elements above. The states window will show the resulting energy levels and wavefunctions; for J = 4.5 e parity we have:

^{-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`".

## The ground state - X^{2}Π

*J*, one for
each parity. A sample Hamiltonian matrix for *J* = 4.5, f
(-) parity shows a non zero matrix elements mixing the Ω=1/2
and Ω=3/2 states:

*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 ±1/2^{½}.
At the lowest *J=*½ the Ω=3/2 state is not
present, so the matrix is 1x1.

## Transitions

Given the upper and lower energy levels and wavefunctions the
transition frequencies and intensities can now be calculated. The
transition frequency is straightforward, just the difference
between the energies calculated above. The intensity is more
involved. The first step is the transition dipole moments between
the basis states; `PGOPHER` will print these if you right
click on the transition moment (`<v=0|T(1)|v=0>`) and
select "Matrix Elements".
The first set is:

*M*
dependence has been taken out; see Transition
Moments and Line Strengths for more details. In this case
the formula used is:

<*J'*, Ω'||T(1,±)||*J*,Ω> =
(-1)^{J'}^{-Ω'}[(2*J'*+1)(2*J*+1)]^{½}(*J'*
1 *J* -Ω' Ω'-Ω, Ω)

Note the dependence on the change in*J* and Ω. The
line strengths are also printed, which are simply the squares of
the reduced matrix elements above. 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. (Any calculation of
intensity must be done with the unsquared matrix elements.)

*J* does not
change, and + to + and - to - if *J* changes. `PGOPHER`
also gives the line strengths, which are simply the squares of the
matrix elements. As before the

The numerical values of the above can be displayed with "View, Transitions" and selecting the required upper and lower state*J*
and parity, or looking in the log window with log window with the
print level on the right set to "Matrices". The matrix elements
between the two *J*= 4.5 states shown above are:

*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:

|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 withJ± Ω=½> = 1/sqrt(2){|J,Lambda=0,Omega=+0.5> ± |J,Lambda=0,Omega=-0.5>}

Symmetrized Matrix Elements in A v=0which with a little rearranging gives the clearer form:

<J,Lambda=0,Omega=0.5 + |H| + J,Lambda=0,Omega=0.5> =

+ Origin*1

+ B*((4*J*(J+1)-2*sqrt(4*J*(J+1)+1)+1)/4)

+ gamma*((sqrt(4*J*(J+1)+1)-2)/4)

<J,Lambda=0,Omega=0.5 - |H| - J,Lambda=0,Omega=0.5> =

+ Origin*1

+ B*((4*J*(J+1)+2*sqrt(4*J*(J+1)+1)+1)/4)

+ gamma*((-sqrt(4*J*(J+1)+1)-2)/4)

<J,Ω=±½|H|J,Ω=±½> = Origin +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 byBJ(J+1) +B/4 - ½γ ± (B-½γ)Sqrt(J(J+1)+1/4)

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

In this case there is only one basis symmetrised basis state for any given J and symmetry, so the wavefunctions will simply be equal to one of the basis states and the energy will be equal to one of the matrix elements above. The states window will show the resulting energy levels and wavefunctions; for J = 4.5 e parity we have:

Eigenvalues for A, J = 4.5, Symmetry = e (+)This should be read as the state name on the left has energy 32741.6 cm

|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* = -139.21 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*(J+1)-3)/4)

+ A*(1/2)

<J,Lambda=-1,Omega=-1.5|H|J,Lambda=-1,Omega=-0.5> =

B*((-sqrt(4*J*(J+1)-3))/2)

<J,Lambda=-1,Omega=-0.5|H|J,Lambda=-1,Omega=-1.5> =

B*((-sqrt(4*J*(J+1)-3))/2)

<J,Lambda=-1,Omega=-0.5|H|J,Lambda=-1,Omega=-0.5> =

+ B*((4*J*(J+1)+5)/4)

+ A*(-1/2)

<J,Lambda=1,Omega=0.5|H|J,Lambda=1,Omega=0.5> =

+ B*((4*J*(J+1)+5)/4)

+ A*(-1/2)

<J,Lambda=1,Omega=0.5|H|J,Lambda=1,Omega=1.5> =

B*((-sqrt(4*J*(J+1)-3))/2)

<J,Lambda=1,Omega=1.5|H|J,Lambda=1,Omega=0.5> =

B*((-sqrt(4*J*(J+1)-3))/2)

<J,Lambda=1,Omega=1.5|H|J,Lambda=1,Omega=1.5> =

+ B*((4*J*(J+1)-3)/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)-3)/4)

+ A*(1/2)

<J,Lambda=1,Omega=1.5 - |H| - J,Lambda=1,Omega=1.5> =

+ B*((4*J*(J+1)-3)/4)

+ A*(1/2)

<J,Lambda=1,Omega=1.5 + |H| + J,Lambda=1,Omega=0.5> =

+ B*((-sqrt(4*J*(J+1)-3))/2)

+ 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> =

+ B*((-sqrt(4*J*(J+1)-3))/2)

+ 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> =

+ B*((-sqrt(4*J*(J+1)-3))/2)

+ 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> =

+ B*((-sqrt(4*J*(J+1)-3))/2)

+ 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)+5)/4)

+ A*(-1/2)

+ 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> =

+ B*((4*J*(J+1)+5)/4)

+ A*(-1/2)

+ p*(sqrt(4*J*(J+1)+1)/4)

+ q*(sqrt(4*J*(J+1)+1)/2)

<J,Ω=3/2,±|H|J,Ω=3/2,±> =The symmetry adapted combinations mean that the four basis states split into two pairs of states for any givenA/2 +BJ(J+1) - 3B/4

<J,Ω=1/2,±|H|J,Ω=1/2,±> = -A/2 +BJ(J+1) + 5B/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| 375.595 -90.397194212542

<X v=0 4.5 - 0.5| -90.397194212542 552.297

Eigenvalues for X, J = 4.5, Symmetry = f (-)This should be read 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

|X v=0 4.5 - 1.5>|X v=0 4.5 - 0.5>

1 X v=0 4.5 4 F1f 337.544 -0.921674 -0.387965

2 X v=0 4.5 5 F2f 590.348 -0.387965 0.921674

Reduced Matrix Elements of <v=0|T(1)|v=0>These are Reduced transition matrix elements as the

<J-1,Lambda=0,Omega=-0.5|T(1)|J,Lambda=-1,Omega=-1.5> = sqrt(((2*J+1)*(2*J+3))/(2*J))/2

<J-1,Lambda=0,Omega=-0.5|T(1)|J,Lambda=1,Omega=0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2

<J-1,Lambda=0,Omega=0.5|T(1)|J,Lambda=-1,Omega=-0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2

<J-1,Lambda=0,Omega=0.5|T(1)|J,Lambda=1,Omega=1.5> = sqrt(((2*J+1)*(2*J+3))/(2*J))/2

<J,Lambda=0,Omega=-0.5|T(1)|J,Lambda=-1,Omega=-1.5> = ((-2*J-1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2

<J,Lambda=0,Omega=-0.5|T(1)|J,Lambda=1,Omega=0.5> = (2*J+1)^2/(2*sqrt(2*J*(2*J+1)*(J+1)))

<J,Lambda=0,Omega=0.5|T(1)|J,Lambda=-1,Omega=-0.5> = ((-2*J-1)*(2*J+1))/(2*sqrt(2*J*(2*J+1)*(J+1)))

<J,Lambda=0,Omega=0.5|T(1)|J,Lambda=1,Omega=1.5> = ((2*J+1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2

<J+1,Lambda=0,Omega=-0.5|T(1)|J,Lambda=-1,Omega=-1.5> = sqrt(((2*J-1)*(2*J+1))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=-0.5|T(1)|J,Lambda=1,Omega=0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=0.5|T(1)|J,Lambda=-1,Omega=-0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=0.5|T(1)|J,Lambda=1,Omega=1.5> = sqrt(((2*J-1)*(2*J+1))/(2*(J+1)))/2

<

Note the dependence on the change in

The transition matrix elements between the symmetry adapted states are also given:

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-1,Lambda=0,Omega=0.5 + |T(1)| + J,Lambda=1,Omega=1.5> = sqrt(((2*J+1)*(2*J+3))/(2*J))/2

<J-1,Lambda=0,Omega=0.5 - |T(1)| - J,Lambda=1,Omega=1.5> = sqrt(((2*J+1)*(2*J+3))/(2*J))/2

<J-1,Lambda=0,Omega=0.5 + |T(1)| + J,Lambda=1,Omega=0.5> = sqrt(((2*J-1)*(2*J+1))/(2*J))/2

<J-1,Lambda=0,Omega=0.5 - |T(1)| - J,Lambda=1,Omega=0.5> = (-sqrt(((2*J-1)*(2*J+1))/(2*J)))/2

<J,Lambda=0,Omega=0.5 + |T(1)| - J,Lambda=1,Omega=1.5> = ((2*J+1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2

<J,Lambda=0,Omega=0.5 - |T(1)| + J,Lambda=1,Omega=1.5> = ((2*J+1)*sqrt(((2*J-1)*(2*J+3))/(2*J*(2*J+1)*(J+1))))/2

<J,Lambda=0,Omega=0.5 + |T(1)| - J,Lambda=1,Omega=0.5> = (2*J+1)^2/(2*sqrt(2*J*(2*J+1)*(J+1)))

<J,Lambda=0,Omega=0.5 - |T(1)| + J,Lambda=1,Omega=0.5> = ((-2*J-1)*(2*J+1))/(2*sqrt(2*J*(2*J+1)*(J+1)))

<J+1,Lambda=0,Omega=0.5 + |T(1)| + J,Lambda=1,Omega=1.5> = sqrt(((2*J-1)*(2*J+1))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=0.5 - |T(1)| - J,Lambda=1,Omega=1.5> = sqrt(((2*J-1)*(2*J+1))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=0.5 + |T(1)| + J,Lambda=1,Omega=0.5> = sqrt(((2*J+1)*(2*J+3))/(2*(J+1)))/2

<J+1,Lambda=0,Omega=0.5 - |T(1)| - J,Lambda=1,Omega=0.5> = (-sqrt(((2*J+1)*(2*J+3))/(2*(J+1))))/2

The numerical values of the above can be displayed with "View, Transitions" and selecting the required upper and lower state

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| 0.702657240207629Note the switch it notation - the first matrix was between basis states (labelled with

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 32404.0563 .0124568 32741.6000 337.5437 2.805936 8982545 0.000000 : 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 32151.2517 .0006520 32741.6000 590.3483 .4937272 1543847 0.000000 : pQ12(4.5) : A v=0 4.5 4 F1e - X v=0 4.5 5 F2f