2
$\begingroup$

This question comes from physics but I think it fits better here. On page 488 of Sakurai's Modern Quantum Mechanics he says that in order to find an infinite series of higher order derivatives of a function, i.e. $$ i\frac{\partial}{\partial t}\psi(x,t) = \left[m - \frac{1}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{8m^3}\frac{\partial^4}{\partial x^4}-\cdots\right]\psi(x,t), $$ you would have to specify the function at "farther and farther away from the origin" (or whatever point you are evaluating at).

The problem (in the physics context) is that this implies nonlocality. Ideally, you should be able to specify the time derivative and spacial derivatives at a point, solving the equation. However, I think Sakurai means to imply that knowing all derivatives at a point is equivalent to knowing the value of the function everywhere.

My question is is it possible to take all derivatives $\frac{\partial^n}{\partial x^n}\psi$ for $n \in \mathbb{N}$ only knowing the value of the function on a finite interval? (thanks to Roland for the wording). If so, can't we do the same thing we do with the first derivative (making the interval smaller and smaller) in order to make the interval vanishingly small? Could we then recover locality?

$\endgroup$
  • $\begingroup$ I don't see a connection between taking a derivative at a point $x_0$ (which is calculating a limit with $x \rightarrow x_0)$ and the size of the interval $x_0$ is in Can you clarify this? I also don't get the 'farther way from the origin' part. $\endgroup$ – Roland Jan 27 '17 at 21:32
  • 1
    $\begingroup$ I think as long as the function is defined on a dense set around a point, then you can take the infinity-th derivative, if it exists. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 21:32
  • $\begingroup$ @Roland I think he's thinking that if you have a finite amount of points, the derivative is estimated by two points near a point (since this comes from a physics context), so with each derivative, you have less points to work with. $\endgroup$ – Simply Beautiful Art Jan 27 '17 at 21:33
  • $\begingroup$ Another noteworthy point: I think it's more important to take all derivatives $\frac{\partial^n}{\partial x^n}\psi$ for $n\in \mathbb{N}$, because the infiniy-th derivative is rather boring $\endgroup$ – Roland Jan 27 '17 at 21:37
  • $\begingroup$ Perhaps you can try to understand the series in terms of the Fourier transform of the wave function. The series of derivatives is then transformed into a series in the wave number which in turn is just a function of the wave number. $\endgroup$ – Dr. Wolfgang Hintze Jan 27 '17 at 21:37
3
$\begingroup$

Think of the limit definition of the derivative in the form ...

$$ f(x)=\lim_{h \to 0}\frac1h\left[ f(x+h)-f(x) \right] $$

this can be generalized to ... $$ f^{(n)}(x)=\lim_{h \to 0}\frac1h \displaystyle \sum_{k=0}^{n}(-1)^{n-k}\binom nk f(x+kh) $$

The crucial point being that $f$ needs to be defined on the interval $[x,x+nh]$ , this is not an issue for finite $n$ because the interval has zero length when you take $h\to 0$ but if you take the limit as $n \to \infty$ you get an interval of undefined length.

$\endgroup$
  • $\begingroup$ There is a factor $\binom{n}{k}$ missing in the summation. $\endgroup$ – Count Iblis Jan 28 '17 at 0:57
  • $\begingroup$ Interesting argument. Is it necessary to use this definition of the $n^{th}$ derivative instead of $f^{(n)}(x)=\lim_{h \to 0}[f^{(n-1)}(x+h)-f^{(n-1)}(x)]$ $\endgroup$ – Hugh Jan 28 '17 at 1:18
  • 1
    $\begingroup$ I think it comes down to the same thing once you make substitutions like $$f^{(n-1)}(x+h)=\lim_{h \to 0}\frac 1h[f^{(n-2)}(x+2h)-f^{(n-2)}(x+h)] $$ $\endgroup$ – WW1 Jan 28 '17 at 1:31
  • $\begingroup$ It makes me think that perhaps my formula for $f^{(n)}(x)$ should have $(\frac 1h)^n$ instead of $\frac 1h$ $\endgroup$ – WW1 Jan 28 '17 at 1:34
  • 1
    $\begingroup$ Where is it written that you must use the same values of $h$ for every value of $n$? If your interval has width $d,$ take $h<d/(2n).$ (Factor of two thrown in for good measure because we can.) The difficulty isn't mathematical; it has to do with physics. $\endgroup$ – David K Jan 28 '17 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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