Translation operator problem: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 1: Line 1:
a)  <math>[x_{i},T(\mathbf{l}))]=i\hbar\frac{\partial T(\mathbf{l})}{\partial p_{i}}=i\hbar(-i\frac{l_{i}}{\hbar})exp(-\frac{i\mathbf{p}.\mathbf{l}}{\hbar})</math>
a)  <math>[\hat{x}_{i},\hat{T}(\mathbf{l})]=i\hbar\frac{\partial T(\mathbf{l})}{\partial\hat{p}_{i}}=i\hbar\left (-i\frac{l_{i}}{\hbar}\right )\exp\left (-\frac{i\hat{\mathbf{p}}\cdot\mathbf{l}}{\hbar}\right )=l_{i}\hat{T}(\mathbf{l})</math>


<math>\Rightarrow =[x_{i},T(\mathbf{l}))]=l_{i}T(\mathbf{l})</math>
b) Given a general state <math>|\alpha\rangle,</math> the expectation value of <math>\hat{x}_{i}</math> is <math>\langle\hat{x}_{i}\rangle=\langle\alpha|\hat{x}_{i}|\alpha\rangle.</math>


b) <math><x_{i}>=<\alpha \mid x_{i}\mid \alpha ></math> ,<math>\mid \alpha ></math> is a general ket
Let us now find the expectation value for the translated state <math>\hat{T}(\mathbf{l})|\alpha\rangle.</math>


<math>\langle\alpha|\hat{T}^{\dagger}(\mathbf{l})\hat{x}_{i}\hat{T}(\mathbf{l})|\alpha\rangle=\langle\alpha|\hat{x}_{i}|\alpha\rangle+\langle\alpha|\hat{T}^{\dagger}(\mathbf{l})[\hat{x}_{i},\hat{T}(\mathbf{l})]|\alpha\rangle=\langle\hat{x}_{i}\rangle+\langle\alpha|\hat{T}^{\dagger}(\mathbf{l})l_{i}\hat{T}(\mathbf{l})|\alpha\rangle=\langle\hat{x}_{i}\rangle+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>
Therefore, the effect of the translation operator <math>\hat{T}(\mathbf{l})</math> is to shift the expectation value of the position operator <math>\hat{\mathbf{x}}</math> by the vector <math>\mathbf{l}.</math>


 
Back to [[Commutation Relations and Simultaneous Eigenstates]]
<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 14:33, 12 July 2013

a)

b) Given a general state the expectation value of is

Let us now find the expectation value for the translated state

Therefore, the effect of the translation operator is to shift the expectation value of the position operator by the vector

Back to Commutation Relations and Simultaneous Eigenstates