- #1
phreak
- 134
- 1
I saw a rather easy proof of the Heisenberg Uncertainty Principle in a PDE textbook the other day, but I'm not sure if it's correct. The proof goes as following:
Note that [tex] \left| \int xf(x) f'(x) \right| \le \left[ \int |xf(x)|^2 dx \right]^{1/2} \left[ \int |f'(x)|^2 dx \right]^{1/2} [/tex], by the Cauchy-Schwartz inequality. Now, the Fourier transform of df/dx is ipF(p), so along with Parseval's Equality, the right side of the above equation equals:
[tex]\overline{x} \cdot \left[ \int |ipF(p)|^2 \frac{dp}{2\pi} \right]^{1/2} = \overline{x} \cdot \overline{p}. [/tex]
Integrating the left side by parts, we then get that it is 1/2, so that [tex]\overline{x} \cdot \overline{p} \ge 1/2[/tex].
Now, here is my problem with this proof. I don't understand why [tex]\int |ip F(p)|^2 \frac{dp}{2\pi} = \overline{p}^2[/tex]. Using Parseval's equality, this would be fine if the Fourier transform of xf(x) is kF(k), but as far as I know this isn't true. Can anyone point me in the right direction?
Note that [tex] \left| \int xf(x) f'(x) \right| \le \left[ \int |xf(x)|^2 dx \right]^{1/2} \left[ \int |f'(x)|^2 dx \right]^{1/2} [/tex], by the Cauchy-Schwartz inequality. Now, the Fourier transform of df/dx is ipF(p), so along with Parseval's Equality, the right side of the above equation equals:
[tex]\overline{x} \cdot \left[ \int |ipF(p)|^2 \frac{dp}{2\pi} \right]^{1/2} = \overline{x} \cdot \overline{p}. [/tex]
Integrating the left side by parts, we then get that it is 1/2, so that [tex]\overline{x} \cdot \overline{p} \ge 1/2[/tex].
Now, here is my problem with this proof. I don't understand why [tex]\int |ip F(p)|^2 \frac{dp}{2\pi} = \overline{p}^2[/tex]. Using Parseval's equality, this would be fine if the Fourier transform of xf(x) is kF(k), but as far as I know this isn't true. Can anyone point me in the right direction?