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!