Translation operator problem: Difference between revisions

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Source: "Modern Quantum Mechanics",Sakurai,problem1.30
'''(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>


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


a. Evaluate  <math>[x_{i},T(\mathbf{l}))]</math>
Let us now find the expectation value for the translated state <math>\hat{T}(\mathbf{l})|\alpha\rangle.</math>
b. Using (a) (or otherwise), demonstrate how the expectation value <math><\mathbf{x}></math>
changes under translation.


Solution:
<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>
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>


<math>\Rightarrow =[x_{i},T(\mathbf{l}))]=l_{i}T(\mathbf{l})</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>


b) <math><x_{i}>=<\alpha \mid x_{i}\mid \alpha ></math> ,<math>\mid \alpha ></math>  is a general ket
Back to [[Commutation Relations and Simultaneous Eigenvalues#Problems|Commutation Relations and Simultaneous Eigenvalues]]
 
 
<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>

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

Back to Commutation Relations and Simultaneous Eigenvalues