Zwiebach Page 194 Q11.3: Understanding "Viewing" a Time-Dependent Heisenberg Op.

[ehrenfest]

Homework Statement

I am confused about Quick Calculation 11.3. What does he mean "by viewing x_0^I as the explicitly time-dependent Heisenberg operator defined by (11.38)"?

Specifically, what does "viewing" mean?

Homework Equations

The Attempt at a Solution

 

[Jimmy Snyder]

This one is not clear to me either. However, what I did was assume that x_0^I does have explicit time dependence by rewriting equation (11.38) on page 193 as

x_0^I(\tau) = x^I(\tau) - \frac{p^I}{m^2}\tau

and then showing that the derivative w.r.t. \tau is zero. It's not hard enough to justify a QC though.