Translation operator problem: Difference between revisions

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(New page: Source: Problem: The translation operator for a finite (spatial) displacement is given by , where '''p''' is the momentum operator. <math>T(\mathbf{l})=exp(-\frac{i\mathbf{p}.\mathbf{l}...)
 
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<math>\Rightarrow =[x_{i},T(\mathbf{l}))]=l_{i}T(\mathbf{l})</math>
<math>\Rightarrow =[x_{i},T(\mathbf{l}))]=l_{i}T(\mathbf{l})</math>


b) <math><x_{i}>=<\alpha \mid x_{i}\mid \alpha ></math>
b) <math><x_{i}>=<\alpha \mid x_{i}\mid \alpha ></math> ,<math>\mid \alpha ></math>  is a general ket




<math><\alpha \mid \ T^{+}(\mathbf{l})[x_{i},T(\mathbf{l}))]\mid \alpha >=<\alpha \mid T^{+}(\mathbf{l})l_{i}T(\mathbf{l})\mid \alpha >=l_{i}</math>
<math><\alpha \mid \ T^{+}(\mathbf{l})[x_{i},T(\mathbf{l}))]\mid \alpha >=<\alpha \mid T^{+}(\mathbf{l})l_{i}T(\mathbf{l})\mid \alpha >=l_{i}</math>
<math><\alpha \mid \ T^{+}(\mathbf{l})[x_{i},T(\mathbf{l}))]\mid \alpha >=<\alpha \mid T^{+}x_{i}T\mid \alpha\ >-<\alpha \mid T^{+}Tx_{i}\mid \alpha ></math>
<math>\Rightarrow <x_{i}>_{translated}=<x_{i}>+l_{i}\Rightarrow <\mathbf{x}>_{translated}=<\mathbf{x}>+\mathbf{l}</math>

Revision as of 00:22, 11 December 2009

Source: Problem: The translation operator for a finite (spatial) displacement is given by , where p is the momentum operator.

a. Evaluate b. Using (a) (or otherwise), demonstrate how the expectation value changes under translation.

Solution: a)

b) , is a general ket