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>[\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>
'''(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>


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)''' 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>
Let us now find the expectation value for the translated state <math>\hat{T}(\mathbf{l})|\alpha\rangle.</math>

Revision as of 13:39, 8 August 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 Eigenvalues