Rotation operators are unitary

11 Sep 2022

Suppose we have a spin-\(\frac{1}{2}\) particle with its intrinsic spin angular momentun vector in the \(+z\) direction. This state is denoted by the ket \(|+z \rangle\). If we rotate the vector by an angle of \(90^{\circ}\) about \(+y\) axis, then it will point in the \(+x\) direction. This new state is represented by the ket \(|+x \rangle.\) Mathematically, this can be written as,

\[\begin{equation} |+x\rangle=\hat{R}\left(\frac{\pi}{2} j\right)|+z\rangle \label{eq: one} \end{equation}\]

where, \(\hat{R}\) is called the rotation operator. The term \(\frac{\pi}{2}\) indicates the angle of rotation, and \(j\) indicates the axis about which the ket is rotated.

What could be the bra equivalent of equation \eqref{eq: one}? Our first guess could be:

\[\begin{equation} \langle+x|= \langle +z| \hat{R}\left(\frac{\pi}{2} j\right) \label{eq: two} \end{equation}\]

Let’s do a simple calculation to check if equation \eqref{eq: two} makes sense or not. We have,

\[\begin{equation} \begin{aligned} \langle+x \mid+x\rangle &=\left\langle+z\left|\hat{R}\left(\frac{\pi}{2} j\right) \hat{R}\left(\frac{\pi}{2} j\right)\right|+z\right\rangle \\ &=\left\langle+z\left|\hat{R}\left(\frac{\pi}{2} j\right)\right|+x\right\rangle \\ &=\langle+z \mid-z\rangle \\ \end{aligned} \end{equation}\]

This is an absurd result because, \(\langle+x \mid+x\rangle =1\) and \(\langle+z \mid-z\rangle =0\). Therefore, the bra equivalent of equation \eqref{eq: one} can not be of the form given in equation \eqref{eq: two}.

Let’s try a different bra equation ¹:

\[\begin{equation} \langle+x| =\langle+z| \hat{R}^{\dagger}\left(\frac{\pi}{2} j\right) \end{equation}\]

where, we have introduced a new operator \(\hat{R}^{\dagger}\), called the adjoint operator of the operator \(\hat{R}\). Now, the normalisation condition can be expanded as,

\[\begin{equation} \langle+x \mid+x\rangle =\left\langle+z \left| \hat{R}^{\dagger}\left(\frac{\pi}{2} j\right) \hat{R}\left(\frac{\pi}{2} j\right) \right| +z\right\rangle = 1. \end{equation}\]

This can be true only if the adjoint operator \(\hat{R}^{\dagger}\) is the inverse of the operator \(\hat{R}\), such that

\[\begin{equation} \hat{R}^{\dagger}\hat{R} = \hat{I}, \label{eq: unitary} \end{equation}\]

where \(\hat{I}\) is the identity operator. The action of an identity operator is to leave the ket unchanged.

Specifically, if \(\hat{R}\left(\frac{\pi}{2} j\right)\) rotates a ket by \(90^{\circ}\) in counterclockwise direction about the \(y-\) axis, then \(\hat{R}^{\dagger}\left(\frac{\pi}{2} j\right)\) rotates the ket by \(90^{\circ}\) in clockwise direction about the \(y-\) axis, so that the net rotation is zero. In other words, the successive action of \(\hat{R}^{\dagger}\left(\frac{\pi}{2} j\right)\) and \(\hat{R}\left(\frac{\pi}{2} j\right)\) is equivalent to the action of an identity operator on the ket, as the identity operator leaves the ket unchanged.

In general, an operator \(U\) satisfying \(\hat{U}^{\dagger}\hat{U} = \hat{I}\) is called unitary. Clearly, rotation operators are unitary operators.

1. How can we visualise this rotation, say by using a Cartesian coordinate system? ↩