Changing the representation of a ket from one basis to another
In \(S_{z}-\) basis, the state \(\mid \psi \rangle\) is expressed as:
\[\begin{equation} \mid \psi\rangle=\mid + \mathbf{z} \rangle\langle + \mathbf{z} \mid \psi \rangle + \mid- \mathbf{z} \rangle \langle - \mathbf{z} \mid \psi \rangle. \label{eq1} \end{equation}\]In \(S_{x}-\) basis, the same state is expressed as:
\[\begin{equation} \mid \psi\rangle=\mid + \mathbf{x} \rangle\langle + \mathbf{x} \mid \psi \rangle + \mid- \mathbf{x} \rangle \langle - \mathbf{x} \mid \psi \rangle. \label{eq2} \end{equation}\]In this post, we will learn how we can transform from a state like \(\mid \psi \rangle\) from \(S_{z}-\) to \(S_{x}-\) basis and vice versa. Specifically, we will learn to compute the amplites \(\langle \pm \mathbf{x} \mid \psi \rangle\) in \eqref{eq2} when the amplitudes \(\langle \pm \mathbf{z} \mid \psi \rangle\) in \eqref{eq1} are known.
Let’s first write eqns. \eqref{eq1} and \eqref{eq2} as column vectors:
\[\begin{equation} \left|\psi\right\rangle \underset{S_{z} \text { basis }}{\longrightarrow}\left(\begin{array}{c} \left\langle+\mathbf{z} \mid \psi\right\rangle \\ \left\langle-\mathbf{z} \mid \psi\right\rangle \end{array}\right) \label{eq3} \end{equation}\] \[\begin{equation} \left|\psi\right\rangle \underset{S_{x} \text { basis }}{\longrightarrow}\left(\begin{array}{c} \left\langle+\mathbf{x} \mid \psi\right\rangle \\ \left\langle-\mathbf{x} \mid \psi\right\rangle \end{array}\right). \label{eq4} \end{equation}\]Consider the first element in the column vector in eqn. \eqref{eq4}. Since the identity operator leaves a state unchanged, we can write:
\[\begin{equation} \langle + \mathbf{x} \mid \psi \rangle = \langle + \mathbf{x} \mid \hat{I} \mid \psi \rangle \\ \label{eq5} \end{equation}\]In \(S_{z}-\) basis, the identity operator can be written as:
\[\begin{equation} \hat{I} = \mid + \mathbf{z} \rangle \langle + \mathbf{z} \mid + \mid -\mathbf{z} \rangle \langle - \mathbf{z} \mid. \label{eq6} \end{equation}\]By combining equations \eqref{eq5} and \eqref{eq6}, we can write,
\[\begin{equation} \begin{aligned} \langle + \mathbf{x} \mid \psi \rangle &= \langle + \mathbf{x}\mid\left(\mid +\mathbf{z}\rangle\langle+\mathbf{z}|+|-\mathbf{z}\rangle\langle-\mathbf{z}| \right) \mid \psi \rangle \\ &= \langle + \mathbf{x} \mid + \mathbf{z} \rangle \langle + \mathbf{z} \mid \psi \rangle + \langle + \mathbf{x} \mid - \mathbf{z} \rangle \langle - \mathbf{z} \mid \psi \rangle. \end{aligned} \label{eq7} \end{equation}\]Similarly,
\[\begin{equation} \langle - \mathbf{x} \mid \psi \rangle = \langle - \mathbf{x} \mid + \mathbf{z} \rangle \langle + \mathbf{z} \mid \psi \rangle + \langle - \mathbf{x} \mid - \mathbf{z} \rangle \langle - \mathbf{z} \mid \psi \rangle \label{eq8} \end{equation}\]Equations \eqref{eq7} and \eqref{eq8} can be nicely written in the form of matrices:
\[\begin{equation} \left(\begin{array}{l} \langle + \mathbf{x} \mid \psi\rangle \\ \langle - \mathbf{x} \mid \psi\rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{x} \mid +\mathbf{z} \rangle & \langle + \mathbf{x} \mid -\mathbf{z} \rangle \\ \langle - \mathbf{x} \mid +\mathbf{z} \rangle & \langle - \mathbf{x} \mid -\mathbf{z} \rangle \end{array}\right) % \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right) \label{eq9} \end{equation}\]We know that,
\[\begin{equation} \langle \pm \mathbf{x} \mid=\langle \pm \mathbf{z} \mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \label{eq10} \end{equation}\]Substituting eqn. \eqref{eq10} in eqn. \eqref{eq9}, we get:
\[\begin{equation} \left(\begin{array}{l} \langle + \mathbf{x} \mid \psi\rangle \\ \langle - \mathbf{x} \mid \psi\rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{z} \mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \mid +\mathbf{z} \rangle & \langle + \mathbf{z}\mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \mid -\mathbf{z} \rangle \\ \langle - \mathbf{z} \mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \mid +\mathbf{z} \rangle & \langle - \mathbf{z} \mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \mid -\mathbf{z} \rangle \end{array}\right) % \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right) \label{eq11} \end{equation}\]The \(2\times2\) matrix in eqn. \eqref{eq11} is the matrix representatation of the rotation operator \(\hat{R}^{\dagger}\), or more precisely, of \(\hat{R}^{\dagger}\left(\frac{\pi}{2}\mathbf{j}\right)\). Let’s denote it by \(\mathbb{\hat{S}}^{\dagger}\).
Equations \eqref{eq9} and \eqref{eq11} provide a means of transforming a let \(\mid \psi \rangle\) from \(S_{z}-\) to \(S_{x}-\) basis.
To transform from \(S_x-\) to \(S_z-\) basis, we start from eqn. \eqref{eq3} and follow the same steps, but express the identity operator in the \(S_x-\) basis. The corresponding transformation relation is given by:
\[\begin{equation} \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{z} \mid +\mathbf{x} \rangle & \langle + \mathbf{z} \mid -\mathbf{x} \rangle \\ \langle - \mathbf{z} \mid +\mathbf{x} \rangle & \langle - \mathbf{z} \mid -\mathbf{x} \rangle \end{array}\right) % \left(\begin{array}{l} \langle + \mathbf{x} \mid \psi\rangle \\ \langle - \mathbf{x} \mid \psi\rangle \end{array}\right) \label{eq12} \end{equation}\]Since,
\[\mid \pm \mathbf{x}\rangle=\hat{R}\left(\frac{\pi}{2} \mathbf{j}\right)\mid \pm \mathbf{z}\rangle,\]Equation \eqref{eq12} can be written as,
\[\begin{equation} \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{z} \mid \hat{R}\left(\frac{\pi}{2} \mathbf{j}\right) \mid +\mathbf{z} \rangle & \langle + \mathbf{z}\mid \hat{R}\left(\frac{\pi}{2} \mathbf{j}\right) \mid -\mathbf{z} \rangle \\ \langle - \mathbf{z} \mid \hat{R}\left(\frac{\pi}{2} \mathbf{j}\right) \mid +\mathbf{z} \rangle & \langle - \mathbf{z} \mid \hat{R}\left(\frac{\pi}{2} \mathbf{j}\right) \mid -\mathbf{z} \rangle \end{array}\right) % \left(\begin{array}{l} \langle + \mathbf{x} \mid \psi\rangle \\ \langle - \mathbf{x} \mid \psi\rangle \end{array}\right) \label{eq13} \end{equation}\]The \(2 \times 2\) matrix in eqn.\eqref{eq13} is the matrix representation of the operator \(\hat{R}\), or more precisely, of \(\hat{R}\left(\frac{\pi}{2}\mathbf{j}\right)\). Since we have denoted the matrix representation of \(\hat{R}^{\dagger}\) by \(\mathbb{\hat{S}}^{\dagger}\), the \(2 \times 2\) matrix in eqn. \eqref{eq13} should be denoted by \(\mathbb{\hat{S}}\).
Equations \eqref{eq12} and \eqref{eq13} provide a means of transforming a ket \(\mid \psi \rangle\) from \(S_x-\) to \(S_z-\) basis.
Since \(\langle a \mid b \rangle = \langle b \mid a \rangle^{*}\), the \(2 \times 2\) matrix in eqn. \eqref{eq12} can be written as:
\[\begin{equation} \left(\begin{array}{ll} \langle + \mathbf{z} \mid +\mathbf{x} \rangle & \langle + \mathbf{z} \mid -\mathbf{x} \rangle \\ \langle - \mathbf{z} \mid +\mathbf{x} \rangle & \langle - \mathbf{z} \mid -\mathbf{x} \rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{x} \mid +\mathbf{z} \rangle^{*} & \langle + \mathbf{x} \mid -\mathbf{z} \rangle^{*} \\ \langle - \mathbf{x} \mid +\mathbf{z} \rangle^{*} & \langle - \mathbf{x} \mid -\mathbf{z} \rangle^{*} \end{array}\right). \label{eq14} \end{equation}\]Clearly, the square matrices in equations \eqn{eq9} and \eqn{eq12} are adjoints of each other. This is in tune with fact that these two matrices also represent the rotation operator \(\hat{R}\left(\frac{\pi}{2}\mathbf{j}\right)\) and its adjoint \(\hat{R}^{\dagger}\left(\frac{\pi}{2}\mathbf{j}\right)\).
By plugging in eqn. \eqref{eq9} into eqn. \eqref{eq12}, we can write:
\[\begin{equation} \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right) = % \left(\begin{array}{ll} \langle + \mathbf{z} \mid +\mathbf{x} \rangle & \langle + \mathbf{z} \mid -\mathbf{x} \rangle \\ \langle - \mathbf{z} \mid +\mathbf{x} \rangle & \langle - \mathbf{z} \mid -\mathbf{x} \rangle \end{array}\right) % \left(\begin{array}{ll} \langle + \mathbf{x} \mid +\mathbf{z} \rangle & \langle + \mathbf{x} \mid -\mathbf{z} \rangle \\ \langle - \mathbf{x} \mid +\mathbf{z} \rangle & \langle - \mathbf{x} \mid -\mathbf{z} \rangle \end{array}\right) % \left(\begin{array}{l} \langle + \mathbf{z} \mid \psi\rangle \\ \langle - \mathbf{z} \mid \psi\rangle \end{array}\right). \end{equation}\]In the above equation, the product of the two square matrices forms an operator. Since the product is leaving the ket unchanged, it must be an identity operator. By replacing the matrices by their respective symbols, we can write:
\[\begin{equation*} \mathbb{\hat{S}}\,\mathbb{\hat{S}}^{\dagger} = \mathbb{\hat{I}} \end{equation*}\]meaning that the \(\mathbb{S}-\) matrices are unitary. This result should not be surprising because the \(\mathbb{S}-\) matrices are matrix representations of rotation operators, and the rotation operators are unitary.