you mean P_x to P_x + aQMrocks said: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] ??
Nope...........................seratend said:[itex] exp(- \frac{i}{\hbar}aX) [/itex]
: )
Seratend.
Thanks. Thats the answer i tempted to throw in too. But i have my doubts..seratend said:EDIT: sorry bad reading:
[itex] exp( \frac{i}{\hbar}aX^2/2) [/itex] is the answer to your question
Doubts, why? You just have to calculate.QMrocks said:Thanks. Thats the answer i tempted to throw in too. But i have my doubts..