Quantum Theory: Operator Exponentiation

[WisheDeom]

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?

 

[gabbagabbahey]

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]

 

[WisheDeom]

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

 

[gabbagabbahey]

I think you meant [itex][U^n,x][/itex], 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.