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Quantum Theory: Operator Exponentiation

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Homework Statement



Let [itex]\left|x\right\rangle[/itex] and [itex]\left|p\right\rangle[/itex] denote position and momentum eigenstates, respectively. Show that [itex]U^n\left|x\right\rangle[/itex] is an eigenstate for [itex]x[/itex] and compute the eigenvalue, for [itex]U = e^{ip}[/itex]. Show that [itex]V^n\left|p\right\rangle[/itex] is an eigenstate for [itex]p[/itex] and compute the eigenvalue, for [itex]V = e^{ix}[/itex].

The Attempt at a Solution



I know that [itex](e^{ip})^{n} = e^{inp}[/itex], since p obviously commutes with itself; I also know that momentum is defined as the generator of translations, which leads to a translation operator [itex]T(x') = e^{(\frac{ipx'}{\hbar})}[/itex] with the property that [itex]T(x') \left| x \right\rangle = \left|x+x'\right\rangle[/itex], which is also an eigenstate of [itex]x[/itex]. Is the solution as simple as identifying [itex] U^n [/itex] with the translation operator, with [itex]n[/itex] in units of length/action?
 

Answers and Replies

  • #2
gabbagabbahey
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Homework Statement



Let [itex]\left|x\right\rangle[/itex] and [itex]\left|p\right\rangle[/itex] denote position and momentum eigenstates, respectively. Show that [itex]U^n\left|x\right\rangle[/itex] is an eigenstate for [itex]x[/itex] and compute the eigenvalue, for [itex]U = e^{ip}[/itex]. Show that [itex]V^n\left|p\right\rangle[/itex] is an eigenstate for [itex]p[/itex] and compute the eigenvalue, for [itex]V = e^{ix}[/itex].

The Attempt at a Solution



I know that [itex](e^{ip})^{n} = e^{inp}[/itex], since p obviously commutes with itself; I also know that momentum is defined as the generator of translations, which leads to a translation operator [itex]T(x') = e^{(\frac{ipx'}{\hbar})}[/itex] with the property that [itex]T(x') \left| x \right\rangle = \left|x+x'\right\rangle[/itex], which is also an eigenstate of [itex]x[/itex]. Is the solution as simple as identifying [itex] U^n [/itex] with the translation operator, with [itex]n[/itex] in units of length/action?
Hint: Consider the action of the commutator [itex][U^n, p][/itex] on the ket [itex]|x\rangle[/itex]
 
  • #3
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I think you meant [itex][U^n,x][/itex], and if so, I got it! Thanks a lot.
 
  • #4
gabbagabbahey
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I think you meant [itex][U^n,x][/itex], and if so, I got it! Thanks a lot.
I did, and you're welcome! :smile:
 

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