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In a QM text I am using (Sakurai 2nd edition 'Modern Quantum Mechanics'), he describes two rotation groups, namely the $\mathrm{SO}(3)$ rotation group and $\mathrm{SU}(2)$ rotation group (unitary unimodular group).

He defines $\mathrm{SO}(3)$ as a group with matrix multiplication on a set of orthogonal matrices (which are matrices which satisfy $R^TR = 1 = RR^T$), he then states that this group only includes rotational operators (and not also inverse operators which would be the group $\mathrm{O}(3)$). He does not ever rigorously define 'rotational operation'.

  1. How you would distinguish between rotational operators and inverse operators, would a sufficient definition be that rotational operators is a transformation with one fixed point?

He also defines the the group $\mathrm{SU}(2)$ which consists of unitary unimodular matrices, and states that the most general unitary matrix in two dimensions has four independent parameters and it is defined as $$U = e^{i \gamma} \left( {\begin{array}{cc} a & b \\ -b^* & a^* \\ \end{array} } \right) $$ where $|a|^2 + |b|^2 = 1,~~~\gamma^* = \gamma.$

  1. Am I right to assume that the $\mathrm{SO}(3)$ rotation group does not have much of application in quantum mechanics but is rather used more in classical mechanics whereas $\mathrm{SU}(2)$ is used more in quantum mechanics, particularly for $s =\frac{1}{2}$ spin systems where we work in a two dimensional Hilbert space?
  2. How does it follow that there are four independent parameters for the general unitary matrix, the way I see it there are three independent parameters, namely, $a$, $b$ and $\gamma$?
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    $\begingroup$ What do you mean by "independent parameters"--independent real or complex numbers? A $2 \times 2$ special unitary matrix can be defined by three complex numbers $a, b, \gamma$ as above, which is equivalent to six real numbers. They are subject to two constraint equations, $|a|^2 + |b|^2 = 1$ and $\gamma^* = \gamma$. Hence there are $6-2 = 4$ independent real parameters. $\endgroup$
    – jc315
    Apr 27 '17 at 16:22
  • $\begingroup$ The reason for the use of groups such as $SU(2)$ in quantum mechanics can to some extent be traced back to the fact that two Hilbert space vectors, which are complex multiples of each other, represent the same physical state. To answer the question "What is the state of my system after I rotate the whole thing some angle along some axis", one has to provide a projective unitary representation of the rotation group. Projective unitary representations of $SO(3)$ correspond to usual unitary representations of $SU(2)$ and it is more convenient to work with representations than projective ones. $\endgroup$ Apr 27 '17 at 18:36
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When classifying representations of a group in QM, it is necessary to allow for projective representations, because states are actually rays (equivalence classes) in the Hilbert space. This means that in order to study the rotational symmetry of a system, you want the projective representations of $\mathrm{SO}(3)$, which are standard representations of $\mathrm{SU}(2)$, because the latter is the universal cover of the former. This is the reason $\mathrm{SU}(2)$ is important in QM.

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The already present answers have covered the difference between $\mathrm{O}(3)$ and $\mathrm{SO}(3)$ at length, so I will not repeat that. Let me instead explain the point about the "usage" of $\mathrm{SO}(3)$ vs. the "usage" of $\mathrm{SU}(2)$, which I think has not yet been made clear:

  1. $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$, meaning there is a two-to-one group homomorphism $\mathrm{SU}(2)\overset{2:1}{\to}\mathrm{SO}(3)$, or, equivalently, $\mathrm{SO}(3)\cong\mathrm{SU}(2)/\mathbb{Z}_2$. It is also simply connected, meaning it is the universal cover. The Lie algebras of both of these Lie groups are the same, i.e. $\mathfrak{so}(3)\cong\mathfrak{su}(2)$. A representation of a Lie algebra always induces a linear representation of the simply connected Lie group associated to it, but it does not always induce a representation of the other groups. More specifically, the spin-1/2 representation is a linear representation of $\mathfrak{so}(3)$, but not of $\mathrm{SO}(3)$, only of $\mathrm{SU}(2)$.

  2. The spin-1/2 representation, is, however, a so-called projective representation of $\mathrm{SO}(3)$. Quantum mechanics actually does not require ordinary linear representations of symmetry groups, but projective ones. For the general reason this is the case, see this Q&A of mine. In this case, it turns out that projective representations of $\mathrm{SO}(3)$ are equivalent to linear representations of $\mathfrak{so}(3)$, or equivalently linear representations of $\mathrm{SU}(2)$. This is the reason $\mathrm{SU}(2)$ appears in quantum mechanics, but not in classical mechanics, when representing the symmetry group of rotations on our space of states.

  3. The spin-1/2 representation is given by the "standard" representation of $\mathrm{SU}(2)$, i.e. just by the 2-by-2 special unitary matrices. But it is still also a representation of $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ and a projective representation of $\mathrm{SO}(3)$. The spin-1 representation is given by the "standard" representation of $\mathrm{SO}(3)$ as 3-by-3 special orthogonal matrices, but it is still also a representation of $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ and a representation of $\mathrm{SU}(2)$ via the 2-to-1 map.

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how you would distinguish between rotational operators and inverse operators, would a sufficient definition be that rotational operators is a transformation with one fixed point?

One can define a rotation as an operation that maps an arbitrary vector $\vec v$ to $\vec v'$ through an infinite sequence of infinitesimal operations which leaves the length of the vector invariant.

To exemplify, let us consider rotations in the plane. From the figure bellow enter image description here we see that the only infinitesimal operation we can do on $\vec v$ that leaves its length invariant is $$x\rightarrow x'=x-\epsilon y,\quad y\rightarrow y'=y+\epsilon x.$$ Such infinitesimal operation can be written as $$\vec v'=(I+\epsilon T)\vec v,$$ where $I$ is the identity matrix and $$T=\begin{bmatrix}0&-1\\1&0\end{bmatrix}.$$ Now do infinitely many such operations in sequence such that $n\epsilon=\theta$ where $n$ in an integer going to infinity and $\theta$ is a finite real, $$\vec v'=\left(I+\frac{\theta}{n} T\right)^n\vec v=\exp(\theta T)\vec v .$$ The last equal sign is an identity. The above equation defines the rotation by an angle $\theta$, $R(\theta)=\exp(\theta T)$. One can compute this exponential by Taylor expanding and we obtain $$R(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}.$$

We say that a matrix $M$ represents a rotation if and only if it can be written in the above form. Note that a matrix such as $$S(y)=\begin{bmatrix}-1&0\\0&1\end{bmatrix},$$ which simply maps $(x,y)$ to $(-x,y)$ is not a rotation. It is called a reflection (what you are strangely calling inverse operator).

You can easily verify that rotations matrices are orthogonal (O), $RR^T=I$, and special (S), $\det R=1$. They form the group $SO(2)$ (or $SO(3)$ in three dimensions). Reflection matrices have determinant $-1$ but are also orthogonal. Together with the rotation matrices the form the group $O(2)$ (or $O(3)$ in three dimensions).

Am I right to assume that the $\mathrm{SO}(3)$ rotation group does not have much of application in quantum mechanics but is rather used more in classical mechanics whereas $\mathrm{SU}(2)$ is used more in quantum mechanics, particularly for $s =\frac{1}{2}$ spin systems where we work in a two dimensional Hilbert space?

In three dimensions the infinitesimal rotations are generated by three generators, $T_1,T_2,T_3$ which play the role of $T$ above. They satisfy the commutation relations $$[T_a,T_b]=i\epsilon_{abc}T_c,$$ and form a Lie algebra namely $\mathfrak{su}(2)$. The point is that both groups $SO(3)$ and $SU(3)$ have the same Lie algebra. The infinitesimal operations are the same. Moreover in general one can represent these generators with square matrices of different size. Once we choose the size of these matrices (the choice is not arbitrary), we obtain the group associated to the Lie algebra. For example, if we start with the algebra $\mathfrak{su}(2)$ and choose to represent it by $2\times 2$ matrices, then the group obtained is $SU(2)$. On the other hand if we represent it by $3\times 3$ matrices we obtain the group $SO(3)$. This latter group is indeed important in quantum mechanics. It relates to spin $1$.

how does it follow that there are four independent parameters for the general unitary matrix, the way I see it there are three independent parameters, namely, $a$, $b$ and $\gamma$?

As already mentioned by jc315's comment, the six real parameter are subjected to two constraints which leaves four real independent parameter.

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  • $\begingroup$ Thanks this is a very good answer. Is the basic reason we choose to define rotation using infinitesimal operations first so that we can approximate $|v|\epsilon$ by a straight line? Also, do you agree with the other answer that $SO(3)$ and $SU(2)$ are isomorphic, in the text I am using it states that it is not? $\endgroup$
    – user100411
    Apr 27 '17 at 17:59
  • $\begingroup$ Yes, we approximate the arc between the two vectors by a straight line when $\epsilon$ is infinitesimal. The groups $SO(3)$ and $SU(2)$ are not isomorphic. They are distinct. However the Lie algebra associated are isomorphic. Another way to put this is to say that the groups are locally isomorphic but globally distinct. This means that infinitesimal operations are carried in the same way in both groups (remember that infinitesimal operations are related to the generators or to the algebra). However, finite operations are different. $\endgroup$
    – Diracology
    Apr 27 '17 at 18:05
  • $\begingroup$ From memory $\text{SO}(3)\cong \text {SU}(2)/\mathbb{Z}_2$. $\endgroup$ Apr 27 '17 at 23:55
  • $\begingroup$ @Diracology Could I ask one question about how the rotation matrix changes when considering the rotation of a cartesian tensor of order 2 (dyadic tensor), as I understand we have something like $$T_{ij} := U_iV_j \to \sum_{i'} \sum_{j'}R_{i' j'}U_{i'}V_{j'}$$ where $R_{i'j'}$ are elements of a $3 \times 3$ matrix, is this correct? What is the nature of the rotation matrix? Also why is the scalar product $U \cdot V$ invariant under rotation? Thanks for any assistance. $\endgroup$
    – user100411
    May 16 '17 at 20:19
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  1. One or the other is in the group, so it doesn't really matter which one to take to be the inverse of the other. Note that $SU(2)$ has $3$ (not $4$) independent parameters. $U(2)$ has an overall phase related to the determinant of its elements, in addition to the $3$ parameters in $SU(2)$.
  2. $SO(3)$ probably has more applications than $SU(2)$ since all orbital angular momentum is $SO(3)$ and not $SU(2)$. In any problem with a central potential you will label states by $SO(3)$ not $SU(2)$ irreps. The wavefunctions for rigid rotors (used to described a variety of tops and linear molecules) are $SO(3)$ group functions.
  3. The parameters are complex. If you start with the $8$ complex numbers $$ U=\left(\begin{array}{cc} A&B\\ C&D\end{array}\right)\, ,\qquad A,B,C,D\in\mathbb{C}\, , $$ then the unitarity conditions $U^\dagger U=\hat 1$ forces $4$ conditions liked to the orthogonality of rows and columns. If you enforce the condition det$=+1$ that's a fifth condition to $8$ parameters - $5$ constraints = $3$ "free" parameters.
    A general $n\times n$ unitary matrix will contain $n^2$ complex parameters, or $2n^2$ real parameters. There are $n^2$ conditions on the rows and columns leaving $n^2$ independent real parameters, from which you subtract another if you want determinant to be +1.
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$\mathrm{SO}(3)$ is the group of all $3 \times 3$ real matrices with determinant $1$. This is the definition of a proper rotation. $\mathrm{SO}(3)$ is a group in the formal mathematical sense, so

$$\forall M \in \mathrm{SO}(3)\ \exists M^{-1} \in \mathrm{SO}(3) : MM^{-1} = I.$$

The $\mathrm{O}$ in $\mathrm{SO}(3)$ stands for 'orthogonal', which means that

$$M^{-1} = M^{T} \iff \mathrm{det}(M) = \pm1,$$

and the $\mathrm{S}$ standard for 'special', which limits this to positive determinants only. I don't know where you got the idea that the inverses only exist in $\mathrm{O}(3)$ (all orthogonal matrices with positive or negative determinant), but it's not true. The elements of $\mathrm{O}(3)$ not in $\mathrm{SO}(3)$ (i.e., elements of $\mathrm{O}(3) \backslash \mathrm{SO}(3)$) are the matrices with strictly negative determinant, and these are called improper rotations. They invert the coordinate axes as well as rotating, which may be where the confusion arose. (To be clear: proper rotation $\iff\mathrm{det}(M) = 1$, improper rotation $\iff\mathrm{det}(M) = -1$.)

$\mathrm{SU}(2)$ is the group of all $2 \times 2$ complex matrices with determinant $1$. The $\mathrm{U}$ stands for unitary, which is the complex version of orthogonal:

$$U^{-1} = U^{\dagger} \iff \mathrm{det}(U) = \pm1,$$

where the dagger is the Hermitian conjugate, but everything else is the same (other than the entries being complex). There are four free real parameters because there are six (not-free) real parameters and two conditions, $6-2=4$. Specifically, these are the complex phases of $a$ and $b$ (two real numbers), the relative magnitude of $a$ and $b$, and the value of $\gamma$, which is a single free real parameter (or two reals and a complex condition ensuring it is real).

The important thing about these two groups is that $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$. This is why you need four parameters to specify a rotation in 3D space, rather than just three. The Bloch sphere in quantum mechanics is a manifestation of this relationship. The double cover is why the angle is halved when going to the Bloch representation of a qubit.

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  • $\begingroup$ Thanks for your answer. Your use of $\Longleftrightarrow$ is confusing me a bit. Are you saying any 3 x 3 matrix with $det(M) = \pm 1$ is orthogonal (defined by $M^{-1} = M^{T}$) and vice versa? And any 2x2 matrix with $det(U) = \pm 1$ is unitary and vice versa. Is this what you are saying? Secondly, it states explicilty in Sakurai that $SO(3)$ and $SU(2)$ are not isomorphic, since for example a rotation by $2 \pi$ and $4 \pi$ in $SO(3)$ is both the identity matrix where as in $SU(2)$ they are the -1 times the identity matrix and the identity matrix respectively. What do you think? $\endgroup$
    – user100411
    Apr 27 '17 at 17:46
  • $\begingroup$ Re $\iff$, yes. It means 'if and only if' and is a two way implies. Sorry, yes, I made a mistake. $\mathrm{SU}(2)$ is a double cover of $\mathrm{SO}(3)$. $\mathrm{SU}(2)$ is isomorphic to the quaternions of unit norm. $\endgroup$
    – gautampk
    Apr 27 '17 at 21:56
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Firstly, the S means "special", which means that the matrices have determinant=1. Orthogonal matrices satisfying $O^T O=I$ with determinant -1 are rotations combined with parity transformations- reflection in a mirror. For rotations $O^T$ is of course the matrix inverse of $O$; but parity transformations are sometimes called "inversion".

Both groups are used in quantum mechanics to describe the properties under rotation of different physical systems, depending on their angular momentum. $SU(2)$ describes spin-1/2 particles- fermions, such as the electron. $SO(3)$ describes spin-1 systems, such as the p-orbital of a hydrogen atom or the polarisation of a massive vector boson.

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