Do loop so that I do not get double terms?

Question

I have a Do loop where I generate quantum state equations <111||111>,<111||112>...<232||232>. This is 144 equations in total but I want both bra and kets to be seen as the same so that for example <121||111> is the same as <111||121> and only one of these is generated namely the <111||121> state. This would mean that when we move from <111| to the <112| bra we go straight to <112||112> rather than <112||111> so this would be 78 equations in total.

Here is the code for it is using the Quantum Notation Package I will also attach a screenshot of the code and some of its output for further clarity.

Do[Print[\!\(\*
TagBox[
RowBox[{"\[LeftAngleBracket]", 
TagBox[
RowBox[{
SubscriptBox["x", 
OverscriptBox["L", "^"]], ",", 
SubscriptBox["y", 
OverscriptBox["M", "^"]], ",", 
SubscriptBox["z", 
OverscriptBox["R", "^"]]}],
Quantum`Notation`zz080BraArgs,
BaseStyle->{ShowSyntaxStyles -> True},
Editable->True,
Selectable->True], "\[VerticalSeparator]"}],
Quantum`Notation`zz080Bra,
BaseStyle->{ShowSyntaxStyles -> False},
Editable->False,
Selectable->False]\) \!\(\*
TagBox[
RowBox[{"\[VerticalSeparator]", 
TagBox[
RowBox[{
SubscriptBox["i", 
OverscriptBox["L", "^"]], ",", 
SubscriptBox["j", 
OverscriptBox["M", "^"]], ",", 
SubscriptBox["k", 
OverscriptBox["R", "^"]]}],
Quantum`Notation`zz080KetArgs,
BaseStyle->{ShowSyntaxStyles -> True},
Editable->True,
Selectable->True], "\[RightAngleBracket]"}],
Quantum`Notation`zz080Ket,
BaseStyle->{ShowSyntaxStyles -> False},
Editable->False,
Selectable->False]\), {i, 1, 2}, {j, 1, 3}, {k, 1, 2}, {x, 1, 2}, {y, 
   1, 3}, {z, 1, 2}], {i, 2}, {j, 3}, {k, 2}, {x, 2}, {y, 3}, {z, 2}]

Answer 1

Based on the suggestion of Henrik Schumacher, and the answers of Hausdorff and Daniel Huber, you can do it as follows:

 Format[braket[{{a_, b_, c_}, {d_, e_, f_}}]] := 
 Row[{"\[LeftAngleBracket]", 
   StringRiffle[{ToString[a], ToString[b], ToString[c]}], "|", "|", 
   StringRiffle[{ToString[d], ToString[e], ToString[f]}], 
   "\[RightAngleBracket]"}]

When we apply the function to all pairs of states, we obtain the following:

MapAt[braket, pairs, All]

Quantum Notation Package works well:

It's just the following:

Needs["Quantum`Notation`"]
Table[braket /. Thread[{a, b, c, d, e, f} -> Flatten@pairs[[i]]], {i, 1, Length[pairs]}]

Answer 2

A more "Mathematica" way of generating your base states is via Tuples

states = Tuples[Range/@{2,3,2}]

To then get the independent pairings of states, you can use

pairs=Flatten[Table[states[[{i,j}]],{i,1,Length@states},{j,i,Length@states}],1];

{{{1, 1, 1}, {1, 1, 1}}, {{1, 1, 1}, {1, 1, 2}}, 
 {{1, 1, 1}, {1, 2, 1}}, {{1, 1, 1}, {1, 2, 2}},
 {{1, 1, 1}, {1, 3, 1}}, {{1, 1, 1}, {1, 3, 2}},
 {{1, 1, 1}, {2, 1, 1}}, {{1, 1, 1}, {2, 1, 2}},
  ...}

Length@pairs
(* 78 *)

Edit As suggested by Henrik Schumacher in the comments, you can also use Subsets here, for example via

pairs=Join[Subsets[states,{2}],Thread[{states,states}]]

Edit 2 Another fun method which generalizes more nicely to similar problems is interpreting the problem as obtaining all independent components of a symmetric two-tensor

pairs = states[[#]] & /@ 
  SymmetrizedIndependentComponents[ConstantArray[Length@states,2],Symmetric[{1,2}]]

Length @ pairs
(* 78 *)

Answer 3

Often a problem is easier, seen from another perspective.

You have different quantum states, specified by 3 integers. We may e.g. represent these as lists:

states=Flatten[Array[{#1, #2, #3} &, {3, 3, 3}], 2]

Now, you want bra-kets that are symmetrical, that is <a||b> and <b||a> are identified. Instetad of thinking in "bra-kets", think about it as tuples. You are looking for all tuples than can be assembled from our states. Here the command "Tuples" comes in handily:

tup=Tuples[states, 2];

Finally, we may change the representation back into the bra-ket form by::

tup /. {x1 : {__Integer}, x2 : {__Integer}} :> 
   Print["<" <> ToString /@ x1 <> "||" <> ToString /@ x2 <> ">"];

.......