- #1
Yoran91
- 37
- 0
Hi everyone,
I'm stuck on the concept of the rotation operator in QM.
From what I understand, one constructs a representation of SO(3) on a Hilbert space by mapping a rotation matrix [itex]R\in SO(3)[/itex] specified by an angle [itex]\phi[/itex] and a unit vector [itex]\vec{n}[/itex] to
[itex]D(R) = \exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}] [/itex].
However, I know that
[itex] R = exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}] [/itex],
which is just the exponential map from [itex]\mathfrak{so}(3)[/itex] to [itex]SO(3)[/itex].
This would amount to saying
[itex]D(R)=R[/itex],
which confuses me. What is going on here? Are we viewing [itex]R[/itex] as an operator on the Hilbert space?
I'm stuck on the concept of the rotation operator in QM.
From what I understand, one constructs a representation of SO(3) on a Hilbert space by mapping a rotation matrix [itex]R\in SO(3)[/itex] specified by an angle [itex]\phi[/itex] and a unit vector [itex]\vec{n}[/itex] to
[itex]D(R) = \exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}] [/itex].
However, I know that
[itex] R = exp[-\frac{i \phi}{\hbar}\vec{J}\cdot \vec{n}] [/itex],
which is just the exponential map from [itex]\mathfrak{so}(3)[/itex] to [itex]SO(3)[/itex].
This would amount to saying
[itex]D(R)=R[/itex],
which confuses me. What is going on here? Are we viewing [itex]R[/itex] as an operator on the Hilbert space?