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| classes:2009:fall:phys4101.001:lec_notes_1030 [2009/11/02 22:21] – myers | classes:2009:fall:phys4101.001:lec_notes_1030 [2009/11/04 01:56] (current) – Filling in a few steps with explination, adding a few more equations to make the derivation more explicit. mbryan |
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| ===Generalized Uncertainty Principle:=== | ===Generalized Uncertainty Principle:=== |
| | The general formula is given by Griffiths as: |
| | <math>\sigma_{A}^2\sigma_B^2\geq ( \frac{<[\hat{A},\hat{B}]>}{2i} )^2</math> |
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| <math>\sigma_{A}^2\sigma_B^2\geq|\frac{[\hat{A},\hat{B}]}{2i}|^2</math> | For example, use the canonical commutation relation [x, p] = i<math>\hbar</math>. Then, |
| | <math>\sigma_{x}^2\sigma_{p}^2\geq (\frac{i\hbar}{2i})^2 </math> => <math>\sigma_{x}\sigma_{p}\geq\hbar/2</math>. In general, the commutation relation between any non-commuting operators has an imaginary term, which will cancel the imaginary part of the 2i term in the generalized uncertainty principle formula. Recall that this only works with x and <math> p_{x} </math>; for example, <math> [x, p_{y}] = 0 </math> so the uncertainty in x and <math> p_{y} </math> is zero. That is, you can measure x and <math> p_{y} </math> simultaneously. |
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| 1-Dimensional Example: Let <math>\hat{A}=\hat{x}=x</math> and <math>\hat{B}=\hat{p}=-i\hbar(\frac{\partial}{\partial x})</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> |
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| ===Derivation:=== | ===Derivation:=== |
| | Start with the general form for any observable A with operator <math> \hat{A} </math>. |
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| <math>\sigma_A^2=<(\hat{A}-<A>)^2>_{\psi} </math> | <math>\sigma_A^2=<(\hat{A}-<A>)^2>_{\psi} </math> |
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| <math><\psi|(\hat{A}-<A>)^2\psi>=<\psi|(\hat{A}-<A>)(\hat{A}-<A>)|\psi> </math> | <math><\psi|(\hat{A}-<A>)^2\psi>=<(\hat{A}-<A>)\psi|(\hat{A}-<A>)\psi> </math> |
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| | Which will be true for any observable A since all observables are Hermitian. |
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| Assuming <math>\hat{A}</math> is Hermitian, | Assuming <math>\hat{A}</math> is Hermitian, |
| <math>(\hat{A}-<A>)(\hat{A}-<A>)|\psi>=(|(\hat{A}-<A>)\psi>)^2 </math> | <math>(\hat{A}-<A>)(\hat{A}-<A>)|\psi>=(|(\hat{A}-<A>)\psi>)^2 </math> |
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| Now Let f and g be functions of A and B and psi respectively, | Now Let f and g be functions of A, <math> \psi </math> and B, <math> \psi </math> respectively, |
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| <math>\sigma_A^2=<f|f> </math> where <math>|f>=(\hat{A}-<A>)\psi </math> | <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_B^2=<g|g> </math> where <math>|g>=(\hat{B}-<B>)\psi </math> |
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| <math>\sigma_A^2\sigma_B^2=<f|f><g|g>\geq|<f|g>|^2 </math> | Then invoking the Schwartz Inequality, which is true for any vectors in an inner product space such as Hilbert space: |
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| This works if you think of |f> and |g> as vectors with an angle (<math>\theta </math>) separating them, in this case | <math>\sigma_A^2\sigma_B^2=<f|f><g|g> </math> <math>\geq</math> <math> |<f|g>|^2 </math> |
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| | Geometrically this is similar to a simple case of |f> and |g> as vectors in 3D space with an angle (<math>\theta </math>) separating them, such as |
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| <math>|f|^2|g|^2\geq|f|^2|g|^2cos^2\theta </math> | <math>|f|^2|g|^2\geq|f|^2|g|^2cos^2\theta </math> |
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| Keep in mind, f and g can be (and often are) complex. Any complex number can be expressed as follows: | Keep in mind, f and g can be (and often are) complex when you have <f|g>. Any complex number can be expressed as follows: |
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| | <math> z=x+iy </math> |
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| | As having a real part and an imaginary part, where x and y are both real. |
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| | <math>|z|^2 = x^2+y^2 \geq y^2 </math> |
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| | Now we know that the eigenvalues of any observable <math> \hat{A} </math> and <math> \hat{B} </math> are real, and that the commutation relation between A and B is complex. So we can solve for the imaginary part of <math> |z|^2 </math> to obtain y: |
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| | <math> x + iy - (x - iy) = 2iy = z - z* </math> and solving for y, <math> y = \frac{z - z*}{2i} </math>. |
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| | Now we can let z be the inner product of our two vectors, and y will be the commutation relation between them; |
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| | <math> z = <f|g> </math>, <math> z* = <g|f> </math>, and <math> \frac{z - z*}{2i} = \frac{<f|g> - <g|f>}{2i} </math> |
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| | and this numerator is nothing but <math> [\hat{A}, \hat{B}] </math>, |
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| | and |
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| | <math>\sigma_A^2\sigma_B^2 </math> <math>\geq</math> <math> |<f|g>|^2 </math> <math> \geq </math> <math> y^2 = (\frac{z - z*}{2i})^2 </math>, so |
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| | <math>\sigma_{A}^2\sigma_B^2\geq ( \frac{[\hat{A},\hat{B}]}{2i} )^2</math> |
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| | This relation will hold for any general pair of observables A and B which do not commute - these are the so called "incompatible observables." A rather famous example outside of x and p is the pair E and t. Incompatible observables do not share a complete set of eigenfunctions, meaning that any two of these observables cannot be expressed with one complete set of eigenfunctions, so you can't measure the two observables at once. |
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| <math> Z=x+iy </math> so <math>|z|^2 \geq x^2+y^2 => y=\frac{z-z^*}{2i}</math> | |
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| This is analogous to <math>\frac{<f|g>-<g|f>}{2i}\geq\frac{[\hat{A},\hat{B}]}{2i} </math> | |
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