Translation operator problem: Difference between revisions
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a) <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>\ | 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> | ||
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> | 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]] | |||
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