The wave function in the position representation is $\langle\ x\rvert\psi\rangle$ = $ \psi (x) $ , where $ \psi (x) $ are the continuous coefficients that multiply the orthonormal basis vectors, i.e, $\rvert\psi\rangle$ = $ \psi (x) $$\rvert\ x \rangle$, and $\rvert\psi\rangle$ is the generic state vector . However in some cases I see the $\rvert\psi\rangle$ represented as a column vector as \begin{pmatrix} \psi_1(x,t)\ \\ \psi_2(x,t)\ \\ \psi_3(x,t) \end{pmatrix} $\psi_1 $, $\psi_2 $, $\psi_3 $ may themselves be continuous, but the column vector seems to indicate that the state vector is a discrete sum of all individual $\psi $ s times the basis vectors. Is this to indicate that a stationary state is not possible, only a linear combination of a bunch of stationary states is physically acceptable? In that case isn't it an integral rather than a sum?

The wave function in the position representation is $\langle\ x\rvert\psi\rangle$ = $ \psi (x) $ , where $ \psi (x) $ are the continuous coefficients that multiply the orthonormal basis vectors, i.e, $\rvert\psi\rangle$ = $ \psi (x) $$\rvert\ x \rangle$, and $\rvert\psi\rangle$ is the generic state vector . However in some cases I see the $\rvert\psi\rangle$ represented as a column vector as \begin{pmatrix} \psi_1(x,t)\ \\ \psi_2(x,t)\ \\ \psi_3(x,t) \end{pmatrix} $\psi_1 $, $\psi_2 $, $\psi_3 $ may themselves be continuous, but the column vector seems to indicate that the state vector is a discrete sum of all individual $\psi $ s times the basis vectors. First of all I find it difficult to grasp that the state vector is a discrete sum of continuous wave functions. Is this done to indicate that a stationary state is not possible, only a linear combination of a bunch of stationary states is physically acceptable? In that case isn't it an integral rather than a sum?