# I Kinetic energy operator

1. Dec 13, 2016

### Konte

Hello everybody,

A special problem constrain me to make a variable change in my Hamiltonian operator, so with the kinetic energy operator, I have a doubt.
The variable change is: $\theta \longrightarrow (\theta + k)$ (where $k$ is a constant).
And the kinetic energy operator change as :
$$\hat{T}_{old}=\frac{-\hbar^2}{2m} \frac{\partial}{\partial \theta^2} \,\, \longrightarrow \,\, \hat{T}_{new}=\frac{-\hbar^2}{2m} \frac{\partial}{\partial (\theta+k)^2}$$

When $\hat{T}_{old}$ and $\hat{T}_{new}$ operate respectively onto a function, say $\psi(\theta)$, will I have two different results?

Thank you very much.

2. Dec 13, 2016

### hilbert2

If $\theta$ is an angular variable, such as the $\theta$ in cylindrical coordinates, then that is not the correct expression of kinetic energy. If $\theta$ is a normal cartesian position coordinate, then those operators produce the same result when operating on a function, as you can deduce from the equation $\frac{d}{d\theta} = \frac{d(\theta+k)}{d\theta}\frac{d}{d(\theta + k)}=\frac{d}{d(\theta + k)}$

3. Dec 13, 2016

### Konte

This hamiltonian is for a plane rotator (polar coordinate whith fixed-$r$). I should have mentioned inertia $I$ instead of mass $m$.
So it is about an angular variable $\theta$. Is something change with respect to your answer in the Cartesian coordinate case?

Thanks.

4. Dec 13, 2016

### hilbert2

Yes, if you use $I$ instead of $m$, the expression is correct. The derivative $\frac{d}{d\theta}$ is equal to $\frac{d}{d(\theta + k)}$ for any variable $\theta$ and constant $k$.