# Rotation operator quantum mechanics

1. Apr 18, 2013

### Yoran91

Hi everyone,

I'm stuck on the concept of the rotation operator in QM.
From what I understand, one constructs a representation of SO(3) on a Hilbert space by mapping a rotation matrix $R\in SO(3)$ specified by an angle $\phi$ and a unit vector $\vec{n}$ to

$D(R) = \exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}]$.

However, I know that
$R = exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}]$,

which is just the exponential map from $\mathfrak{so}(3)$ to $SO(3)$.
This would amount to saying
$D(R)=R$,

which confuses me. What is going on here? Are we viewing $R$ as an operator on the Hilbert space?

2. Apr 18, 2013

### rubi

The $J_i$'s in the second equation are matrices of the defining representation of $\mathfrak{so}(3)$. The $J_i$'s in the first equation should better be called $\pi(J_i)$, where $\pi : \mathfrak{so}(3)\rightarrow \mathcal{L}(\mathcal{H})$ is a representation of $\mathfrak{so}(3)$ on a Hilbert space $\mathcal{H}$ (the angular momentum operators). So these are really different things and your confusion comes from the abuse of notation.

3. Apr 18, 2013

### Yoran91

I guess I don't really understand the representation of $SO(3)$ on the Hilbert space then.

What is the map $\pi$ and what is that Hilbert space $\mathcal{H}$? I've never seen a map from $\mathfrak{so}(3)$ to $\mathcal{L(H)}$ before, so I don't really now how that works.

4. Apr 18, 2013

### rubi

A representation of a Lie algebra $\mathfrak g$ is a linear map $\pi:\mathfrak g \rightarrow \mathcal L (\mathcal H)$ from the Lie algebra into the linear operators on some vector space such that $\pi([a,b]) = [\pi(a),\pi(b)]$ (the Lie bracket is mapped to the commutator). If we apply this to quantum mechanics and $\mathfrak{so}(3)$, the vector space could be $L^2(\mathbb R^3)$ for example and the map $\pi$ could be given by $\pi(J_i) = (\hat{\vec{r}}\times\hat{\vec{p}})_i$, where $\hat{\vec{r}}$ is the operator that multiplies a state by $\vec{r}$ and $\hat{\vec{p}} = -\mathrm i \hbar\nabla$. You can easily check that this representation obeys the $\mathfrak{so}(3)$ bracket structure. But you have to be careful: If you exponentiate a Lie algebra representation, you don't necessarily get a representation of the corresponding Lie group. However for the example i just gave you, you do.

But all of this is just a sophisticated way of saying that the $\vec{J}$ is really $\hat{\vec{r}}\times(-\mathrm i\hbar\nabla)$ in your first equation and a vector of 3x3 matrices $J_i$ given by $(J_i)_{jk} = \epsilon_{jik}$ (up to some factors) in your second equation.

Last edited: Apr 18, 2013
5. Apr 18, 2013

### The_Duck

Try this approach:

Presumably, for every quantum mechanical state there is a rotated version of that state. This means that, given an axis $\hat{n}$ and an angle $\phi$ there is a unitary operator $O(\phi, \hat{n})$ on the Hilbert space of your quantum mechanical system. This operator has the property that for any state $| \psi \rangle$, $O(\phi, \hat{n}) |\psi \rangle$ is the rotated version of $|\psi\rangle$, where the rotation is by an angle $\phi$ around the axis $\hat{n}$.

So we have a bunch of these operators O, one for each angle and axis. One condition they have to satisfy is that composition of rotations has to come out right. So if we rotate a state twice to get the state $O(\phi_1, \hat{n_1}) O(\phi_2, \hat{n_2}) | \psi \rangle$, this state had better be equal to $O(\phi_3, \hat{n_3}) |\psi\rangle$, where $(\phi_3, \hat{n_3})$ is the rotation that results from composing $(\phi_1, \hat{n_1})$ and $(\phi_2, \hat{n_2})$. So the operators $O$ have to have the same multiplication table as, say, the 3x3 rotation matrices. So the operators $O$ are a representation of the group SO(3).

Now consider infinitesimal rotations. The $O$'s are unitary, so we can write an infinitesimal O as

$O(\epsilon, \hat{n}) = 1 + i \epsilon J(\hat{n})$.

Where $J(\hat{n})$ is some Hermitian operator on the Hilbert space (and 1 is the identity operator on the Hilbert space). Now I think we argue that SO(3) is three-dimensional, so the $J(\hat{n})$ can be written as linear combinations of three basis operators, which we arrange in a vector $\vec{J}$. So

$O(\epsilon, \hat{n}) = 1 + i \epsilon \hat{n} \cdot \vec{J}$.

The $\vec{J}$ operators are then a set of three operators on the Hilbert space which form a representation of the Lie algebra so(3).

Finally, if we like we can use exponentials to represent finite $O$'s in terms of the $J$ operators:

$O(\phi, \hat{n}) = \exp(i \phi \hat{n} \cdot \vec{J})$.

6. Apr 19, 2013

### tom.stoer

The basic idea is rather simple. One starts with a wave function $\psi(r)$. Then one rotates 3-space with a 3*3 rotation matrix $r \to r^\prime = R r$.

On the other hand one has operators $\mathcal{O}$, e.g. the r-operator itself, operators for the momentum p represented as differential operators etc. These operators act on the wave function $\psi \to \psi^\prime = \mathcal{O}\psi$.

Note that we may chose a different representation, e.g. wave function in momentum space. We still have the same operators (in the abstract sense) but now they are represented differently. In momentum space the momentum operator simply becomes a number p, whereas now the r-operator is represented as diffential operator.

In the case of rotations in r-space defined via rotations matrices we are are looking for operators acting on the wave function. That means we want to construct a representation $D(R)$ associated with a rotation $R$

$$\psi(r) \to \psi(r^\prime) = \psi(Rr) \stackrel{!}{=} D(R)\,\psi(r) = \psi^\prime(r)$$

The last line and especially (!) is more or less the defining equation for the idea of representing an entity $R$ acting on $r$ by a new entity $D(R)$ acting on $\psi$.

Last edited: Apr 19, 2013