Do you know what this particular unitary operator is?

  • Thread starter QMrocks
  • Start date
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] ??

:confused:
 
QMrocks 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] ??

:confused:
you mean P_x to P_x + a

[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
 
Last edited:
seratend said:
[itex] exp(- \frac{i}{\hbar}aX) [/itex]

: )

Seratend.
Nope...........................
 
There is no typo.
 
QMrocks said:
There is no typo.
See corrected previous post.

Seratend.
 
seratend said:
EDIT: sorry bad reading:
[itex] exp( \frac{i}{\hbar}aX^2/2) [/itex] is the answer to your question
Thanks. Thats the answer i tempted to throw in too. But i have my doubts.. :confused:
 
QMrocks said:
Thanks. Thats the answer i tempted to throw in too. But i have my doubts.. :confused:
Doubts, why? You just have to calculate.
Hint: try to see what the unitary operator exp(i a(X)) does on the momentum operator.

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

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