Do you know what this particular unitary operator is?

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Discussion Overview

The discussion revolves around identifying a specific unitary operator that transforms the operator \( P_x + aX \) to \( P_x \). The context includes technical reasoning related to quantum mechanics and operator transformations.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that the unitary operator \( \exp\left(\frac{i}{\hbar}aX\right) \) transforms \( P_x + a \) to \( P_x \) and questions what operator achieves a similar transformation for \( P_x + aX \).
  • Another participant suggests that \( \exp\left(-\frac{i}{\hbar}aX\right) \) could be the answer, but later corrects themselves, proposing \( \exp\left(\frac{i}{\hbar}aX^2/2\right) \) instead.
  • There is an acknowledgment of potential confusion regarding the previous statements, with one participant expressing doubts about the proposed solution.
  • A hint is provided to explore the effect of the unitary operator \( \exp(i a(X)) \) on the momentum operator to clarify the situation.

Areas of Agreement / Disagreement

Participants express differing views on the correct unitary operator, with no consensus reached on the final answer. Doubts and corrections are present, indicating ongoing uncertainty.

Contextual Notes

Participants reference specific mathematical expressions and transformations without fully resolving the implications or calculations involved.

QMrocks
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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:
 
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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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