The position and momentum operators for a free particle in Heisenberg picture

[Haorong Wu]

Homework Statement

From Griffiths GM 3rd p.266

Consider a free particle of mass $m$. Show that the position and momentum operators in the Heisenberg picture are given by$$ {\hat x}_H \left( t \right) ={\hat x}_H \left( 0 \right) + \frac { {\hat p}_H \left( 0 \right) t} m $$ $$ {\hat p}_H \left( t \right) ={\hat p}_H \left( 0 \right) $$.
Hint: you will first need to evaluate the commutator $\left[ \hat x , {\hat H}^n \right]$; this will alow you to evaluate the commutator $\left[ \hat x , {\hat U} \right]$.

Homework Equations

$\hat U \left( t \right) =exp \left[ - \frac {it} \hbar \hat H \right]$
${\hat Q}_H \left( t \right) = {\hat U}^\dagger \left( t \right) \hat Q \hat U \left( t \right)$ is the Heisenberg-picture operator.

The Attempt at a Solution

Since $H=- \frac {\hbar^2} {2m } \frac {d^2} {dx^2}$ , then $\left[ \hat x , {\hat H}^n \right]=-2n { \left( - \frac {\hbar^2} {2m} \right) }^n \frac {d^{2n-1}} {dx^{2n-1}}$, and $\left[ \hat x , \hat U \right] = \left[ \hat x , exp \left(- \frac {it} \hbar \hat H \right) \right] = \left[ \hat x , \sum_{n=0}^\infty {\frac { { \left(\frac { -it} {\hbar} \hat H \right) }^{n} } {n!}} \right] = \sum_{n=0}^\infty \frac {\left( -2n \right) \left( \frac {it\hbar} {2m} \right)^n} {n!} \frac {d^{2n-1}} {dx^{2n-1}}$.

Then ${\hat x}_H \left( t \right) \psi_n \left( x \right) = {\hat U }^\dagger \hat x \hat U \psi_n \left( x \right) ={\left ( \hat x \hat U \right) } ^\dagger e^{\frac {-i \hat H t} {\hbar}} \psi_n \left( x \right) = {\left( \hat U \hat x + \left[ \hat x, \hat U \right] \right) } ^\dagger e^{\frac {-i E_n t} {\hbar}} \psi_n \left( x \right) = e^{\frac {-i E_n t} {\hbar}} \sum_{n=0}^\infty \left[ \frac {\left( -2n \right) \left( \frac {-it\hbar} {2m} \right)^n} {n!} \frac {d^{2n-1}} {dx^{2n-1}} + x e^{\frac {i E_n t} {\hbar}} \right] \psi_n \left( x \right)$

I don't know how to proceed with this monster. Perhaps, I'm heading a wrong direction?

 

[Orodruin]

You are off in the wrong direction. You should only need the commutation relation between x and p and the expression of the free Hamiltonian in terms of those. There is no need to involve any states.

You do not need to use the particular representation of the position and momentum operators. Knowing their commutator is sufficient.

 

[Haorong Wu]

You are off in the wrong direction. You should only need the commutation relation between x and p and the expression of the free Hamiltonian in terms of those. There is no need to involve any states.

You do not need to use the particular representation of the position and momentum operators. Knowing their commutator is sufficient.

Thanks! Perhaps the hint of the problem misguided me. I'll try the other way.