- #1

EE18

- 28

- 5

In Problem 3.7, Ballentine says:

The unitary operator ##U(v) = exp(iv·G)## describes the instantaneous ##(t = 0)## effect of a transformation to a frame of reference moving at the velocity ##v## with respect to the original reference frame. Its effects on the velocity and position operators are:

##UVU^{−1} =v−vI, UQU^{−1}=Q##. Find an operator ##G_t## such that the unitary operator ##U(v, t) = exp(iv·G_t)## will yield the full Galilei transformation: ##UVU^{−1} =V−vI, UQU^{−1} =Q−vtI##. Verify that ##G_t## satisfies the same commutation relation with P, J, and H as does G.

This seems to imply that the development in the text itself (Chap 3.4) was fine. But why should this be? Surely we've just shown that G in general actually takes a different form and G=MQ is wrong?

Edit: I am having some trouble using MathJax -- isn't ## the delimiter here?

The unitary operator ##U(v) = exp(iv·G)## describes the instantaneous ##(t = 0)## effect of a transformation to a frame of reference moving at the velocity ##v## with respect to the original reference frame. Its effects on the velocity and position operators are:

##UVU^{−1} =v−vI, UQU^{−1}=Q##. Find an operator ##G_t## such that the unitary operator ##U(v, t) = exp(iv·G_t)## will yield the full Galilei transformation: ##UVU^{−1} =V−vI, UQU^{−1} =Q−vtI##. Verify that ##G_t## satisfies the same commutation relation with P, J, and H as does G.

This seems to imply that the development in the text itself (Chap 3.4) was fine. But why should this be? Surely we've just shown that G in general actually takes a different form and G=MQ is wrong?

Edit: I am having some trouble using MathJax -- isn't ## the delimiter here?

Last edited: