Responsible party: Esquire, David Hilbert's Hat

To go back to the lecture note list, click lec_notes
previous lecture note: lec_notes_1028
next lecture note: lec_notes_1102

Main class wiki page: home

<math>\sigma_{A}^2\sigma_B^2\geq|\frac{[\hat{A},\hat{B}]}{2i}|^2</math>

1-Dimensional Example: Let <math>\hat{A}=\hat{x}=x</math> and <math>\hat{B}=\hat{p}=-i\hbar(\frac{\partial}{\partial x})</math>

<math>[\hat{x},\hat{p}]=i\hbar </math> ⇒ <math>\sigma_{x}^2\sigma_p^2\geq|\frac{i\hbar}{2i}|^2=(\frac{\hbar}{2})^2</math> ⇒ <math>\sigma_{x}\sigma_p\geq\frac{\hbar}{2}</math>

<math>\sigma_A^2=<(\hat{A}-<A>)^2>_{\psi} </math>

<math><\psi|(\hat{A}-<A>)^2\psi>=<\psi|(\hat{A}-<A>)(\hat{A}-<A>)|\psi> </math>

Assuming <math>\hat{A}</math> is Hermitian,

<math>(\hat{A}-<A>)(\hat{A}-<A>)|\psi>=(|(\hat{A}-<A>)\psi>)^2 </math>

Now Let f and g be functions of A and B and psi respectively,

<math>\sigma_A^2=<f|f> </math> where <math>|f>=(\hat{A}-<A>)\psi </math>

<math>\sigma_B^2=<g|g> </math> where <math>|g>=(\hat{B}-<B>)\psi </math>

<math>\sigma_A^2\sigma_B^2=<f|f><g|g>\geq|<f|g>|^2 </math>

This works if you think of |f> and |g> as vectors with an angle (<math>\theta </math>) separating them, in this case

<math>|f|^2|g|^2\geq|f|^2|g|^2cos^2\theta </math>

Keep in mind, f and g can be (and often are) complex. Any complex number can be expressed as follows:

<math> Z=x+iy </math> so <math>|z|^2 \geq x^2+y^2 ⇒ y=\frac{z-z^*}{2i}</math>

This is analogous to <math>\frac{<f|g>-<g|f>}{2i}\geq\frac{[\hat{A},\hat{B}]}{2i} </math>

To go back to the lecture note list, click lec_notes
previous lecture note: lec_notes_1028
next lecture note: lec_notes_1102