# Hamilton operator with moments of inertia : time - independence

1. Aug 16, 2011

### Juqon

1. The problem statement, all variables and given/known data
The Hamilton-operator is given as $$\hat{H}$$ and describes the movement of a free rigid object that has the moments of inertia $$I_{i}$$
Under what circumstances is
$$<\Psi|\hat{L_{1}}|\Psi>$$

time-independent?

2. Relevant equations
$$\hat{H}=\frac{\hat{L_{1}^{2}}}{2I_{1}}+\frac{\hat{L_{2}^{2}}}{2I_{2}}+\frac{\hat{L_{3}^{2}}}{2I_{3}}$$

$$[\hat{L_{j}},\hat{L_{k}}]=\iota\hbar\epsilon_{jkm}\hat{L_{m}}$$
$$<\Psi|\hat{L_{1}}|\Psi>$$

3. The attempt at a solution
If it wasn't in the brac-kets, I would just try $$\frac{dL_{1}}{dt}=0$$ Also, I thought maybe I could use another picture to have the time-indepence in it automatically, but I think SchrÃ¶dinger must be the right one as there the operators are constant.

2. Aug 16, 2011

### G01

HINT: Remember that the time dependence of the expectation value of an operator, O, is related to: $<[O,H]>$.

3. Aug 18, 2011

### Juqon

Thanks! I think a found a solution. What do you think about that?
I just need to know whether I can change the indices of the Levi-Civita tensor in a way so that I get a minus in front (other order of the indices) also with operators. If yes, this would not answer the question (see below).

If "otherwise": Is this the end result or can you transform that even more?
[PLAIN]http://img545.imageshack.us/img545/6031/timeindependencemomento.png [Broken]

Last edited by a moderator: May 5, 2017
4. Aug 18, 2011

### G01

No, you cannot switch the indices on the operators and just change the sign to compensate. That would mean the the angular momentum operators anti-commute, which they don't.

You end result is correct(when you don't switch operator indices), but you missed a sign, I think. Your result should be:

$$\frac{[L_2,L_3]_+}{I_2}=\frac{[L_2,L_3]_+}{I_3}$$

($[A,B]_+$ means anti-commutator) or when you simplify:

$$I_2=I_3$$

Physically, this means you have an "axially symmetric rotator."

You really went about this in the "brute-force" method! There's a much simpler way to get to the same result. You know you are looking for the condition when $L_1$ and $H$ commute. Now, you also know that you can choose any one angular momentum operator to commute with $L^2$. Thus, if you can write $H$ in terms of only $L^2$ and $L_1$.

You'll see that this is only possible in two cases: When $I_1=I_2=I_3$ and when $I_2=I_3$

5. Aug 19, 2011