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Do you know what this particular unitary operator is?

by QMrocks
Tags: operator, unitary
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QMrocks
#1
Jun21-05, 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] ??

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seratend
#2
Jun21-05, 12:34 PM
P: 318
Quote Quote by QMrocks
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] ??

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
QMrocks
#3
Jun21-05, 12:38 PM
P: 85
Quote Quote by seratend
[itex] exp(- \frac{i}{\hbar}aX) [/itex]

: )

Seratend.
Nope...........................

QMrocks
#4
Jun21-05, 12:40 PM
P: 85
Do you know what this particular unitary operator is?

There is no typo.
seratend
#5
Jun21-05, 12:40 PM
P: 318
Quote Quote by QMrocks
There is no typo.
See corrected previous post.

Seratend.
QMrocks
#6
Jun21-05, 12:43 PM
P: 85
Quote Quote by seratend
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..
seratend
#7
Jun21-05, 12:57 PM
P: 318
Quote Quote by QMrocks
Thanks. Thats the answer i tempted to throw in too. But i have my doubts..
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.
QMrocks
#8
Jun21-05, 12:57 PM
P: 85
My bad, Seratend. You are absolutely right! Case close :)


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