# What is the meaning of definite total energy of the wave function?

David J. Griffiths in Introduction to Quantum Mechanics asked:

What's so great about separable solutions of time independent Schroedinger equation?

They are states of definite total energy [...] the Hamiltonian $H(x,p) = \left(\frac{p^2}{2m} + V(x)\right)$

I don't get this.

You have undoubtably heard about Heisenberg's uncertainty principle:

$$\Delta x \, \Delta p \ge \frac{\hbar}{2}$$

There is a corresponding energy-time uncertainty principle:

$$\Delta E \, \Delta t \ge \frac{\hbar}{2}$$

The point about the separable wavefunctions is that they are time independant, which means that $\Delta t = \infty$ and therefore that $\Delta E = 0$. This is what Griffiths means by definite energy.

The above argument is rather handwaving, but if you read the part of Griffiths following the excerpt you quote he makes this rigorous. The energy uncertainty, $\Delta E$, is the standard deviation defined by:

$$(\Delta E)^2 = \langle H^2 \rangle - \langle H \rangle^2$$

Griffiths shows that this is equal to zero.

• The interpretation of the $\Delta t$ in the "energy-time uncertainty principle" is a bit difficult, see this question. Also, I don't how this answers OP's question why definite energy is a desirable property. Oct 21 '15 at 13:14
• @ACuriousMind: yes, I agree, and what Griffiths really means is $\langle H^2 \rangle - \langle H \rangle^2 = 0$. my aim is to give some intuitive understanding of why this is significant. Oct 21 '15 at 13:57