Quantum Theory: Operator Exponentiation

[WisheDeom]

Homework Statement

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

The Attempt at a Solution

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

 

[gabbagabbahey]

Homework Statement

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

The Attempt at a Solution

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

Hint: Consider the action of the commutator [U^n, p] on the ket |x\rangle

 

[WisheDeom]

I think you meant [U^n,x], and if so, I got it! Thanks a lot.

 

[gabbagabbahey]

I think you meant [U^n,x], and if so, I got it! Thanks a lot.

I did, and you're welcome!