# Uncertainty principle for position and hamiltonian

I found the uncertainty between delta x (position) and delta H (Hamiltonian) to be greater or equal to (h_bar*<p>)/ 2m.
Does this mean for stationary states, where <p>=0, the uncertainty can be zero? ie we can precisely measure the position and energy?

## Answers and Replies

olgranpappy
Homework Helper
I found the uncertainty between delta x (position) and delta H (Hamiltonian) to be greater or equal to (h_bar*<p>)/ 2m.
Does this mean for stationary states, where <p>=0, the uncertainty can be zero? ie we can precisely measure the position and energy?

So, in a stationary state <delta H>=0. You thus already know the energy exactly. But, unless the position eigenstate is also a stationary state then when you measure the position you collapse to a different state, an give up info about the momentum. Thus, because most hamiltonians include a kinetic term, it would seem like you can not measure both exactly. But, for certain cooked up systems maybe you can.

You should probably explain more about the particular problem/system you have in mind. Give us a few equations as well that further explain your idea. This will help us come to a reasonable answer.

Matterwave
As Olgran said, in a stationary state <H> is definite. So $$\Delta H\Delta x=0$$ already. Your uncertainty in x is not guaranteed to be zero. Remember that the uncertainty principle is a lower bound on your uncertainty, not an upper bound. It does, however, mean you have some "hope" of finding a definite H and X state, but it's not guaranteed that you can just from looking at the Uncertainty Principle.