# What is meant by taking the partial derivative of the Hamiltonian in this situation?

I'm doing a computation involving the quantum mechanical harmonic oscillator, and I have an expression of the form $\frac{\partial}{\partial \omega} \hat{H}$ where

$$\hat{H} = \frac{1}{2m} \left( - \hbar^2 \frac{\partial^2}{\partial x^2} + m^2 \omega^2 \hat{x}^2 \right) \ .$$

I've been told that $\frac{\partial}{\partial \omega} \frac{\partial^2}{\partial x^2}$ is "zero" and that the result is simply $m \omega \hat{x}^2$, but I can't see why this is true. The most obvious interpretation of $\frac{\partial}{\partial \omega} \frac{\partial^2}{\partial x^2}$ is one of composition of operators, but in this case $\frac{\partial}{\partial \omega} \frac{\partial^2}{\partial x^2} (\omega x^2 ) = 2 \neq 0$, for instance. So my questions are:

1. What does it mean to take the partial derivative of an operator?
2. Why is $\frac{\partial}{\partial \omega} \frac{\partial^2}{\partial x^2}$ zero?

Thanks.

When dealing with operators it is often easier to see what is going on by applying it to the appropriate object. In this case you have the Hamiltonian operator that acts on a wave function that I will denote by $\psi(x,y,z,t)$. Now if you apply $\hat{H}$ to $\psi(x,y,z,t)$ you see that at some point you will have to take the partial derivative of $\psi$ with respect to $\omega$ in which case you get zero because $\psi$ does not depend on $\omega$ explicitly. Therefore you can drop the term $\frac{\partial}{\partial \omega} \frac{\partial^2}{\partial x^2}$. This should answer the second part of your question and hopefully also shed some light on the first. To be more explicit, taking the derivative is an operator and if you have another operator you take the product, i.e. if you have an operator $\frac{\partial^2}{\partial x^2}$ then the partial derivative of this operator with respect to $\omega$ is $\frac{\partial }{\partial \omega} \frac{\partial^2}{\partial x^2}$. So you are composing/multiplying depending on how you like to call it.
• This would make sense except that the harmonic oscillator wave-functions (specifically the basis states) do depend on $\omega$ explicitly. If it helps I'm specifically looking at an application of the Hellmann–Feynman theorem. – FlagCapper Feb 3 '16 at 10:17