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Taylor Series Expansion About a Local Minimum

  1. Jun 25, 2014 #1
    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:

    [itex]V(x) = V(x_0) + V'(x_0) (x-x_0) + \frac{1}{2} V''(x_0)(x-x_0)^2[/itex]

    subtract [itex]V(x_0)[/itex] (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.
  2. jcsd
  3. Jun 25, 2014 #2


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    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.
  4. Jun 25, 2014 #3
    Which expression is informal, the one Griffiths used, or the one other authors use?
  5. Jun 25, 2014 #4


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    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 ...
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