# Taylor Series Expansion About a Local Minimum

1. Jun 25, 2014

### Bashyboy

Hello everyone,

I am currently reading chapter two, section 3 of Griffiths Quantum Mechanics textbook. Here is an excerpt that is giving me some difficulty:

"Formally, if we expand V(x) in a Taylor series about the minimum:

$V(x) = V(x_0) + V'(x_0) (x-x_0) + \frac{1}{2} V''(x_0)(x-x_0)^2$

subtract $V(x_0)$ (you can add a constant to V(x) with impunity, since that doesn't change the force),..."

Okay, I understand that is does not change the force field for which V(x) is a potential energy, but doesn't it change the potential energy function itself? If I recall correctly, I have seen other authors simply define the zero point of the potential energy function to be at the minimum, which seems to be a better argument to me, as the potential energy is a relative quantity, depending upon the reference frame, unlike, say, distance.

2. Jun 25, 2014

### UltrafastPED

That's exactly what Griffiths is doing - only the expression is informal. In this case he is focused on the force, so the choice of constant is irrelevant - it will vanish when the gradient is taken.

3. Jun 25, 2014

### Bashyboy

Which expression is informal, the one Griffiths used, or the one other authors use?

4. Jun 25, 2014

### UltrafastPED

Griffith's statement ... he refers to the physics, and leaves out the mathematical reasoning. "Informal" in the mathematical sense that the details are not presented ...