# Change of Basis

In the article on representations we found that the probability amplitude in the momentum basis is given by

\begin{align} \tilde{\psi}(p,t)= \langle p | \psi(t) \rangle \end{align} We can insert the unit operator $\mathbb{I}=\int dx \, |x \rangle \langle x|$ \begin{align} \tilde{\psi}(p,t)&= \int dx \, \langle p |x \rangle \langle x| \psi(t) \rangle\\ \end{align} but $\langle x| \psi(t) \rangle$ is the position space wave function, so we find \begin{align} \tilde{\psi}(p,t)&= \int dx \, \langle p |x \rangle \, \psi(x,t) \\ \end{align} where $\langle p |x \rangle$ is the probability amplitude, that when the position of the particle is $x$ the momentum is $p$. In another article we show that the $\langle p |x \rangle$ is given by $e^{i \frac{p \cdot x}{\hbar}}$. As the notation suggests, the change of basis from $\psi(x,t)$ to $\tilde{\psi}(p,t)$ is given by the fouier transform.

\begin{align} \tilde{\psi}(p,t)&= \int dx \, e^{i \frac{p \cdot x}{\hbar}} \, \psi(x,t) \\ \end{align}

Similarly we find $\psi(x,t)$ in the position basis from $\tilde{\psi}(p,t)$ by \begin{align} \psi(x,t)&= \int dp \, \langle x |p \rangle \, \langle p | \psi(t) \rangle \\ &= \int dp \, \langle x |p \rangle \, \tilde{\psi}(p,t)\\ \end{align} $\langle x |p \rangle$ is the complex conjugate of $\langle p |x \rangle$. Thus the change of basis is given by the inverse fourier transform.

\begin{align} \psi(x,t)&= \int dp \, e^{-i \frac{p \cdot x}{\hbar}} \, \tilde{\psi}(p,t) \\ \end{align}

For a discrete basis the change of basis can be performed in the same manner by inserting the unit operator and represent it in the desired basis. A state vector represented in the basis $\{ u_i \}$ is \begin{align} \langle u_i | \psi(t) \rangle \end{align} if we want to represent $| \psi(t) \rangle$ in the another basis $\{ |v_k \rangle \}$ we insert the unit operator $\sum_k |v_k \rangle \langle v_k |$ \begin{align} \langle u_i | \psi(t) \rangle &=\langle u_i |\sum_k |v_k \rangle \langle v_k | \psi(t) \rangle \\ &=\sum_k \langle u_i |v_k \rangle \langle v_k | \psi(t) \rangle \\ \end{align} where \begin{align} S_{ik}=\langle u_i |v_k \rangle \end{align} is the transformation matrix. The inverse transformation is associated with the hermitian conjugate \begin{align} S^{\dagger}_{ki}=\langle v_k | u_i \rangle \end{align} and \begin{align} \langle v_k | \psi(t) \rangle &=\langle v_k |\sum_i |u_i \rangle \langle u_i | \psi(t) \rangle \\ &=\sum_i \langle v_k |u_i \rangle \langle u_i | \psi(t) \rangle \\ \end{align} \begin{align} \langle v_k | \psi(t) \rangle &=\sum_i S^{\dagger}_{ki} \langle u_i | \psi(t) \rangle \\ \end{align}

Video Lectures:

• V. Balakrishnan - Quantum Physics Lec 7 [1]