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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'''(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>T(\mathbf{l})=exp(-\frac{i\mathbf{p} | |||
a | '''(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>\ | <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> | ||
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 Eigenvalues#Problems|Commutation Relations and Simultaneous Eigenvalues]] | |||
Latest revision as of 13:25, 18 January 2014
(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