-1
$\begingroup$

I am a little bit confused about which is really the energy operator. During the lectures, the professor told us that the energy operator is simply the Hamiltonian $\hat{H}$ and that the eigenvalues represent the energy of the system. On Wikipedia [1], though, there is written that the energy operator is

$$ i \hbar \frac{\partial}{\partial t}. $$

On this website I rode that $\hat{H}$ and $\hat{E}$ are the same thing so now I do not understand which one is the right energy operator.

[1] https://en.wikipedia.org/wiki/Energy_operator

$\endgroup$

marked as duplicate by AccidentalFourierTransform, stafusa, Phonon, Chris, sammy gerbil Feb 1 '18 at 2:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

To answer your question I will draw an analogy with a simple question about classical mechanics. Hopefully this will clarify the confusion.

Imagine someone coming to you asking what is the correct expression for acceleration. Somewhere she read the correct expression was:

$$a=\frac{\partial^2 x}{\partial t^2}$$

Somewhere else she read

$$a=\frac{F}{m}$$

She is wondering, which of these two is the correct expression. My answer would be that the first expression is the definition of the acceleration, while the second expression is the value acceleration might take in a given experiment. The equation of motion is thus:

$$\frac{\partial^2 x}{\partial t^2}=\frac{F}{m}$$

The question you are asking is similar.

$$\hat E =i\hbar \frac{\partial}{\partial t}$$

is the operator that is defined as the energy operator, while

$$\hat H = \frac{-\hbar^2}{2m}\nabla^2 +V$$

is the energy operator for a given experiment. The equation of motion is thus:

$$i\hbar \frac{\partial}{\partial t}=\frac{-\hbar^2}{2m}\nabla^2 +V$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.