# Angular momentum operators

1. Mar 30, 2014

### Matterwave

Hi guys,

This is a problem which is bothering me right now. The angular momentum operators (Lx, Ly, Lz), when expressed in spatial rotations consists of derivatives in $$\theta$$ and $$\phi$$. This would suggest that there are, at any point in space, only two linearly independent operators (since there are derivatives in only two directions).

When we talk about spin 1/2, which is still a type of angular momentum, we often represent the spin angular momentum operators (Sx, Sy, Sz) in terms of Pauli matrices which are three linearly independent matrices.

Maybe I'm just being dense, but why the disparity in the dimensionality of angular momenta?

When I think about it, it's often stated "angular momenta are the 'generators' of rotations". In which case, you'd think there would be 3 operators since the space of rotations in 3-dimensional space is of dimension 3 (e.g. given by the three Euler angles).

Also just thinking about describing angular momenta, you'd think there would be 3 numbers necessary (e.g. 2 numbers specifying the axis of rotation, and 1 number specifying the rate of rotation).

I'm getting really confused. =[

2. Mar 30, 2014

### The_Duck

This conclusion is incorrect; there are three linearly independent angular momentum operators. It sounds like you have the spherical representations of the angular momentum operators in front of you; you should convince yourself that $L_x$, $L_y$, and $L_z$ are indeed linearly independent, contrary to your intuition here.

Right, this is the correct way of thinking about it. Consider the operator $\partial/\partial\phi$. This operator is associated with rotations around the z axis. You can think of this derivative as "the rate of change of a function as I rotate around the z axis." There are similar operators for rotations around any axis. In principle, all of these rotation operators can be expressed in terms of $\theta$ and $\phi$, but rotation operators around axes other than z look really messy in these coordinates. It can be more convenient to write the angular momentum operators as follows:
$$L_x = i \hbar \left(y \frac{\partial}{\partial z} - z \frac{\partial}{\partial y}\right)$$
$$L_y = i \hbar \left(z \frac{\partial}{\partial x} - x \frac{\partial}{\partial z}\right)$$
$$L_z = i \hbar \left(x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}\right)$$
(I might be off by an overall sign from the usual conventions). Note that these three operators are clearly linearly independent: none can be expressed as a linear combination of the other two. A rotation operator around any axis can be written as a linear combination of these three basis elements.

Yes. Often an easier set of numbers to work with is the components of the angular momentum around the x, y, and z axes.

3. Mar 30, 2014

### Matterwave

The representations of the angular momentum operators I have in front of me right now are:

$$L_x=i\hbar(\sin\phi\frac{\partial}{\partial\theta}+\cot\theta\cos\phi \frac{\partial}{\partial\phi})$$

$$L_y=i\hbar(-\cos\phi\frac{\partial}{\partial\theta}+\cot\theta\sin\phi \frac{\partial}{\partial\phi})$$

$$L_z=-i\hbar\frac{\partial}{\partial\phi}$$

Since only $\frac{\partial}{\partial\theta}$ and $\frac{\partial}{\partial\phi}$ appear in these 3 operators, can I not express one of them as a sum of the other two?

Specifically it seems that:

$$L_z=\frac{1}{\cot\theta}(\cos\phi L_x+\sin\phi L_y)$$

4. Mar 31, 2014

### The_Duck

Sure, this equality holds (except maybe it is missing a minus sign if the expressions you gave are right?). You've used operator-valued coefficients on the right-hand side. Like x, y, and z, $\theta$ and $\phi$ should be thought of as operators.

If you want to express all rotation operators in terms of linear combinations of a set of basis operators using *real-valued coefficients*, then you need a basis of three operators, and two won't do. This is the sense in which there are three linearly independent rotation operators. This is related to the fact that you can pick three axes and then build up any rotation as a sequence of three rotations, one around each axis; but you can't do this with only two axes.

5. Mar 31, 2014

### Matterwave

Yes you are right, I'm missing a minus sign. Can you elaborate on the requirement that I need to build linear combinations using "real-valued coefficients" rather than operator-valued coefficients?

Certainly at some point P in space the coefficients in my construction of Lz are just real numbers. But even the number "1" itself can be thought of as the identity operator on my Hilbert space of functions right? So how can I make something not an operator?

6. Mar 31, 2014

### Matterwave

By the way, I am not fighting the fact that there should be a 3-dimensional space of rotations of 3-space. That fact seems obvious. What I can't understand is why the angular momentum operators, as expressed in the way I expressed them, do not at first glance appear to be linearly independent.

7. Mar 31, 2014

### PhilDSP

It's good to remember that the sine and cosine functions are orthogonal to each other when their arguments are equal (as there are in your expressions).

8. Mar 31, 2014

### The_Duck

Well, you can do anything you want! But here is why linear combinations with real coefficients are interesting, and why I implicitly assumed real coefficients when I originally asserted that $L_x$, $L_y$, and $L_z$ are linearly independent.

Let $\alpha$ be an infinitesimal angle. Then if you act with the operator $1 + \frac{i\alpha}{\hbar} L_z$ on a wave function, you rotate the whole wave function by an angle $\alpha$ around the z axis. Convince yourself of this! Similarly $1 + \frac{i\beta}{\hbar}L_x$ will rotate a wave function by an infinitesimal angle $\beta$ around the x axis.

Suppose you apply two successive infinitesimal rotations, for example by acting with the operator

$(1 + \frac{i\alpha}{\hbar} L_z)(1 + \frac{i\beta}{\hbar}L_x)$

This operator rotates by a tiny angle $\beta$ around x and then a tiny angle $\alpha$ around z. The net effect is equivalent to a rotation by some tiny angle around the axis $\alpha \hat z + \beta \hat x$. In fact if we multiply out the above expression and ignore the extremely tiny number $\alpha \beta$, we get

$1 + \frac{i}{\hbar}(\alpha L_z + \beta L_x)$

which corresponds nicely to the fact that this operator performs a rotation around the $\alpha \hat z + \beta \hat x$ axis.

In fact any infinitesimal rotation operator can be written in the form

$1 + \frac{i}{\hbar}(\alpha L_z + \beta L_x + \gamma L_y)$

where $\alpha, \beta, \gamma$ are infinitesimal angles. So the linear combinations, with real coefficients, of this basis of three rotation operators correspond to the set of infinitesimal rotations.

Consider instead an operator such as

$1 + \frac{i\alpha}{\hbar} \phi L_z$, $\alpha$ infinitesimal.

If this operator acts on a wave function it does not perform a simple rotation. Instead it does some sort of weird smearing, taking $\psi(\theta, \phi) \to \psi(\theta, \phi + \alpha \phi)$. Locally, at any given point $(\theta, \phi)$, this looks like a rotation by an angle $\alpha \phi$ around the z axis. But since the angle $\alpha \phi$ varies throughout space the overall effect of the operator is not a rotation but something more complicated. So multiplying angular momentum operators by things like $\phi$ instead of real numbers results in operators like $\phi L_z$ which do not generate rotations.

This expresses the fact that if you only look at one tiny region of space, there are only two independent things rotations can do to that tiny region. For example, if you look at a tiny patch of a sphere, all small-angle rotations of the sphere look like translations of that patch, and there are only two independent directions in which you can translate something along a 2D surface. Nevertheless there are three independent rotation directions for the sphere as a whole; it's just that some rotations which really are different look the same to someone who is confining their attention to the behavior of a particular tiny patch of the sphere.

9. Mar 31, 2014

### Matterwave

Ok, that makes sense to me. Thanks!