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Do you know what this particular unitary operator is? 
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#1
Jun2105, 08:44 AM

P: 85

Recall that the unitary operator [itex] exp(\frac{i}{\hbar}aX) [/itex] transform the operator [itex] P_x +a [/itex] to [itex] P_x [/itex].
Now, what is the unitary operator that transform the operator [itex] P_x +aX [/itex] to [itex] P_x [/itex] ?? 


#2
Jun2105, 12:34 PM

P: 318

[itex] exp( \frac{i}{\hbar}aX) [/itex] : ) Seratend. EDIT: sorry bad reading: [itex] exp( \frac{i}{\hbar}aX^2/2) [/itex] is the answer to your question 


#3
Jun2105, 12:38 PM

P: 85




#4
Jun2105, 12:40 PM

P: 85

Do you know what this particular unitary operator is?
There is no typo.



#5
Jun2105, 12:40 PM

P: 318

Seratend. 


#6
Jun2105, 12:43 PM

P: 85




#7
Jun2105, 12:57 PM

P: 318

Hint: try to see what the unitary operator exp(i a(X)) does on the momentum operator. Seratend. 


#8
Jun2105, 12:57 PM

P: 85

My bad, Seratend. You are absolutely right! Case close :)



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