Taylor Series Expansion About a Local Minimum

[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.

Could someone please help me understand this excerpt. Thank you.

 

[UltrafastPED]

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.

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.

 

[Bashyboy]

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

 

[UltrafastPED]

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

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 ...