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!

 

[paranormal]

I appreciate your attempt at a solution and your understanding of the properties of the operators involved. However, your solution is not quite complete. To fully show that U^n\left|x\right\rangle is an eigenstate of x, we need to show that it satisfies the eigenvalue equation xU^n\left|x\right\rangle = \lambda U^n\left|x\right\rangle. This can be done by using the commutation relation [x,p]=i\hbar and the fact that U=e^{ip} to show that U^n\left|x\right\rangle is indeed an eigenstate of x with eigenvalue \lambda = x+nh. Similarly, for V^n\left|p\right\rangle to be an eigenstate of p, we need to show that it satisfies the eigenvalue equation pV^n\left|p\right\rangle = \mu V^n\left|p\right\rangle, which can be done using the commutation relation [x,p]=i\hbar and the fact that V=e^{ix}. The eigenvalue in this case is \mu = p+nh, where n is again in units of length/action. Therefore, U^n\left|x\right\rangle and V^n\left|p\right\rangle are eigenstates of x and p, respectively, with eigenvalues that depend on the initial state and the number of times the operator is applied. This shows the power of operator exponentiation in quantum theory.