Why two s=1/2 states can be linked by a rotation, but not two s=1 states?

[DrDu]

Here is another example: Consider a molecule like NH3. It has symmetry C3v (6 elements). There exists a two dimensional representation spanned by the orbitals px and py. It is clear that with only 6 operations, I cannot span a whole vector space. In fact, I cannot even convert px into py.

 

[tom.stoer]

@DrDu: You are saying that the j-rep. of the rotation group does not act transitively on its defining vector space (at least not for j=1; for other's we did not discuss).

Is this a property of the group SO(3)? Or of specific representations only? What about SO(N) and SU(N) in general? How many orbits are there?

Do you have any references?

 

[wdlang]

i am so happy that my post induced so much discussion

very stimulating!

 

[DrDu]

For a space of dimension n, the set of all normalized vectors in this vector space forms a sphere C^(n-1) (or however mathematicians may abbreviate this object). The group of all possible unitary transformations of this set is U(n-1).

 

[peteratcam]

I wish people had agreed on what they mean by rotation early on!

SU(2) is smaller (fewer parameters) than U(3). A 3-dimensional rep of SU(2) obviously can't do everything that U(3) can do. There should be no surprise here.

U(3) will 'rotate' any two normalized vectors in a 3-dim complex vector space into each other. All vectors with the same norm look identical from the point of view of U(3) which is why this will work.

Given two arbitrary vectors (from our 3-dim complex vector space), SU(2) (ie the j=1 rep of SU(2) ) will rotate them into each other only if the two vectors 'look the same' from the point of view of SU(2). What does it mean to look the same? Well states which are highest-weight states (|j,j> ) all look the same (so a state which is physically polarised along any direction can be rotated to a state polarised along any other direction). But a non-highest weight state doesn't 'look like' a highest-weight state, from the point of view of SU(2), so the rep won't rotate them into each other. (This is the cone/plane idea from earlier in the thread)

The spin coherent states are constructed precisely in this way, by acting on highest-weight states with the euler-angle parametrised SU(2) matrices in whichever dimension rep we are working in. Perhaps another way to answer the question is that m=0 is not a coherent state of spin.

 

[Dickfore]

For some reason, the array command does not work anymore

$$(1, 0, 0)^{\top} \equiv |j = 1, m = 1 \rangle$$

$$(0, 1, 0)^{\top} \equiv |j = 1, m = 0 \rangle$$

$$(0, 0, 1)^{\top} \equiv |j = 1, m = -1 \rangle$$

are the standard notations for the basis column vectors in the $|j, m \rangle$ representation.

 

[tom.stoer]

U(3) is larger than SO(3); therefore 6 out of 9 angles cannot be determined. OK, by looking at U(3) which can do the job it may become clear that SU(2) is too small and therefore can't do the job. Was that the original idea?

Anyway - thanks a lot for the clarification!

 

[peteratcam]

U(3) is larger than SO(3); therefore 6 out of 9 angles cannot be determined. OK, by looking at U(3) which can do the job it may become clear that SU(2) is too small and therefore can't do the job. Was that the original idea?

Well the original question makes out it is to be expected that the states should be linked by a rotation (i.e R(g) for g in SU(2) ), but there is no reason to expect that.

Maybe its better to answer the question by explaining away why j=1/2 is special: all 2-dim complex vectors can be obtained by applying an SU(2) matrix to (1,0) (up to a phase). In any higher dim representation space, this is simply not true, nor would one expect it to be true, so there is no explaining necessary for j=1 and higher.

Part of the trouble in this thread has been the choice of basis.

It is obvious there exists a matrix in SO(3) which rotates (0,1,0) to (1,0,0).

But there is no set of euler angles such that
$$e^{i\alpha S_x}e^{i\beta S_y}e^{i\gamma S_z}|1,1\rangle = |1,0\rangle$$

Distinguishing between these two statements seems to have taken most of the first page of this thread.

 

[tom.stoer]

Unfortunately this was clear for me from the very beginning, nevertheless it seemed obvious that because one can rotate two arbitrary verctor in R³ into each other, one can also rotate two arbitarry states in the vector representation j=1 into each other. I even hesitated to start with the calculation simply because "it had to be true". After some exercise I was shocked and I double checked my calculations. Then I started think about being on the wrong track ...

 

[Dickfore]

The transformation of states in j = 1 representation, is not in $\mathbb{R}^{3}$, but in $\mathbb{C}^{3}$. Admittedly, there is the requirement of the invariance of the scalar product $\langle x | y \rangle$, which imposes the condition of unitarity and the requirement that the identical transformation is in the class of transformations, which gives the condition of speciality, but then the matrices which transform these states are in SU(3), and this group has 8 generators (corresponding to the Gell-Mann matrices). You cannot parametrize this group completely with only 3 parameters (Euler angles).

 

[DrDu]

I want to continue where I stopped yesterday. As others have said already, for a representation with given j, the group U(n) with n=2j+1 acts transitively on that vector space.
for j=1/2, the group is U(2) and the quotient U(2)\SU(2)=U(1), that is, the group SU(2) acts transitively on the physical states which are defined up to a phase factor from U(1).
For the vector representation the different vectors which cannot be transformed into each other with a rotation are parameterized by the quotient U(3)\SU(2) which is parameterized by 6 parameters. Even if one is the phase of the wavefunction, there are still 5 left.