# Angular momentum operator

1. Jan 30, 2014

### Hazzattack

Hi guys,

this might be a stupid question but if I wanted a general expression for the time evolution of the angular momentum operator is it just the same as Hamiltonian?

i.e ih ∂/∂t ψ = L2 ψ

Solving this partial differential gives the time evolution of the angular momentum operator?

2. Jan 30, 2014

### vanhees71

No! You should very (!) carefully read your quantum-mechanics textbook with regard to time evolution in quantum mechanics. It's not an easy subject. There is some arbitrariness in the choice of how you make the operators, representing observables ("observable operators") and the Statistical Operator representing the state of you system ("state operator") time-dependent. All these choices are physically equivalent and thus lead to the same conclusions about observable quantitities (expectation values of operators, cross sections, transition probabilities,...). A particular choice of the distribution of the time dependence on observable and state operators is called the choice of a picture of the time evolution, and there is always a time-dependent unitary transformation to switch from one picture to another one. In your book you might find this under the keywords "Heisenberg picture", "Schrödinger picture", "Dirac/interaction picture".

3. Jan 30, 2014

### pellman

No. You are apparently comparing this to the Schrodinger eq ih∂/∂t ψ = H ψ . However, the schrodinger equation does not give the time evolution of the Hamiltonian operator. This equation gives the time evolution of the state ψ. The operator on the RHS is necessarily the Hamiltonian because the Hamiltonian, by definition, is the operator which gives the time-evolution of the state. To write ih ∂/∂t ψ = L2 ψ is equivalent to setting the Hamiltonian equal to L2.

There is no time-evolution of the Hamiltonian. The Hamiltonian is constant. Unless one is working with a system which does not conserve energy.

4. Jan 30, 2014

### Hazzattack

thanks for the response guys. In summary, it was a stupid question ha. So how would one go about working out the time evolution of the angular momentum operator for a 2 level system for example?

5. Jan 30, 2014

### WannabeNewton

$H$ is the generator of time translations whereas $J$ is the generator of spatial rotations. In what you've written you're considering the Schrodinger picture in which the time dependence is carried by the state vectors and not the operators themselves. The picture in which the operators themselves carry time dependence is also important (and called the Heisenberg picture) but your post reflects the former not the latter. In such a case...

Say we have a state $|\psi(t) \rangle$. If we translate in time then $|\psi(t) \rangle$ transforms under a unitary representation of the Galilei group according to $|\psi (t)\rangle \rightarrow e^{i \varepsilon H} |\psi(t) \rangle = |\psi(t - \varepsilon)\rangle$ where $\varepsilon$ is a time translation parameter. From this it is easy to see that $H |\psi(t) \rangle = i\frac{d}{dt}|\psi(t) \rangle$. This gives us the equation of motion for $|\psi(t)\rangle$.

On the other hand, as already stated, $J$ is the generator of spatial rotations. In this case $|\psi \rangle$ transforms again under a unitary representation of the Galilei group but according to $|\psi \rangle \rightarrow e^{-i \varepsilon \cdot J} |\psi \rangle$ where now $\varepsilon$ is a rotation angle parameter combined with a unit vector for the rotation axis. In order to make intuitive sense of such a state transformation we would like to make use of how the spatial coordinates are transforming under rotations so we go to the coordinate representation. Then $\psi(x) = \langle x|\psi \rangle \rightarrow \psi(R^{-1}x)$ where $R$ is a 3x3 rotation matrix. If we go through similar motions as above then $\langle x|J |\psi\rangle = -i (\mathbf{x} \times \nabla) \psi(x)$ for a single-component wave function. But this is not an equation of motion for the state vector, it just tells us how (in the coordinate representation which is most intuitive for the purpose of rotations in the single-component case) state functions transform under an infinitesimal spatial rotation. Don't confuse the two.

In the Heisenberg picture, wherein operators carry time dependence, there is an equation of motion for each operator (i.e. for each observable). To have an operator inherit the time dependence we simply make use of the unitary representation of time translations. Then our angular momentum operator $J$ would inherit time dependence as per $J(t) = e^{iHt}J e^{-iHt}$ and we can write down the equation of motion for $J(t)$ as $\frac{d}{dt}J(t) = i [H,J(t)]$. This is called the Heisenberg equation of motion (to contrast with the Schrodinger equation of motion, or just Schrodinger equation, which is $H |\psi(t) \rangle = i\frac{d}{dt}|\psi(t) \rangle$ as mentioned above and tells us the evolution of a state vector as opposed to that of an operator/observable).

Last edited: Jan 30, 2014
6. Jan 30, 2014

### Jilang

7. Jan 30, 2014

### WannabeNewton

Here's a standard textbook exercise which might be instructive to work through: http://www.physics.uc.edu/~belhormp/academics/presentations/Sakurai P2.1.pdf

It's for the spin operator as opposed to the orbital angular momentum operator but both belong to the angular momentum operator so it should still do the trick as far as getting down the mechanics. You can then straightforwardly apply the result to spin precession in an external homogenous magnetic field by writing down the appropriate Hamiltonian and choosing a coordinate system in which the magnetic field is aligned with the $z$ axis.

8. Jan 30, 2014

### strangerep

No. "Stupid" would have been not asking the question at all (fearing embarrassment, say), thus retaining the misconception. (Actually, this is probably a good practical definition of "stupid" in the real world.)

First, write down the Hamiltonian of the system...

9. Jan 31, 2014

### Hazzattack

Thanks for all the help guys, I'm going to read through some of the suggested texts. I'll get back to you with any problems. But as always, thanks for the great advice.