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?

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    $\begingroup$ I've never seen a column vector like that unless the author was talking about a spinor. $\endgroup$
    – DanielSank
    Commented Jan 20, 2016 at 5:58
  • $\begingroup$ I came across something like this in an elementary QM book - $\rvert\psi\rangle$ = \begin{pmatrix} 1/6. e^{ikx-i\omega.t}\\ \sqrt2/6. e^{2ikx-2i\omega.t} \\ \sqrt3/6.e^{3ikx-3i\omega.t} \\ . \\ . \\ .\\ \end{pmatrix} $\endgroup$ Commented Jan 20, 2016 at 7:16
  • $\begingroup$ The author writes "...more commonly, however, each component of the state vector represents a function of position and time, something like this - " and then writes the equation I wrote in the comment above. My question is - If the author is wrong then that's the end of the story. If he is correct then is he trying to represent a linear combination of a number of stationary states? $\endgroup$ Commented Jan 20, 2016 at 7:24
  • $\begingroup$ Can you specify where you read this? It looks like the writer is trying to make a point... $\endgroup$
    – Adi Ro
    Commented Jan 20, 2016 at 7:57
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    $\begingroup$ note that $|\psi\rangle =\psi(x)|x\rangle$ is wrong. $\endgroup$ Commented Jan 20, 2016 at 21:49

1 Answer 1


Such a vector describes the quantum state of a spin-1 particle on a line, or any other particle with a position degree of freedom and 3 internal states.

To start with, you can expand wavefunctions in a basis. E.g., if you have a wavefunction $\vert\psi\rangle$ which depends on position, you can expand it in the position basis $\vert x \rangle_p$, i.e., write $$ \vert\psi\rangle = \int \psi(x)\vert x\rangle_\mathrm{p} \mathrm{d}x\ , $$ where $\psi(x)=\langle x\vert\psi\rangle$. One also often just writes $\psi(x)$ in that case.

Now if you have a quantum state which does not depend on position, but e.g. on some internal degree of freedom $s=0,...,S-1$ (e.g., a spin being $+1$, $0$, or $-1$), then you might write $$ \vert\psi\rangle = \sum_{\sigma=0}^{S-1} \psi_\sigma \vert\sigma\rangle_\mathrm{s}\ , $$ where $\psi_\sigma = \langle \sigma \vert\psi \rangle$. (Clearly, the spin basis $\vert \sigma \rangle_s$ has nothing to do with the position basis $\vert x \rangle_\mathrm{p}$ -- one has to be careful here, and notation is often sloppy.) Given that $\sigma$ has a discrete number of settings, this is often written as a vector with $S$ components, e.g. for $S=3$ $$ \vert\psi\rangle = \left(\begin{matrix} \psi_0\\\psi_1\\\psi_2 \end{matrix}\right)\ . $$ Now imagine you have an object which has both position and spin (e.g., a spin-1 particle, which has 3 spin states). Then, it can be written as $$ \vert\psi\rangle = \int\mathrm{d}x \sum_{\sigma=0}^{S-1} \psi_\sigma(x) (\vert\sigma\rangle_\mathrm{s}\otimes \vert x\rangle_\mathrm{p})\ , $$ or as $$ \vert\psi\rangle = \left(\begin{matrix} \psi_0(x)\\\psi_1(x)\\\psi_2(x) \end{matrix}\right)\ . $$ Thus, your vector describes the quantum state of a particle with some position and 3 internal states (e.g., a spin-1 particle on a line).

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    $\begingroup$ Do you know where I can read about the mathematically rigorous definition of the very first integral expansion in your answer (if $|x\rangle$ had some finite representation as a column vector I would interpret this as an ordinary componentwise integral, however, since this is not the case there is probably some other definition to this)? $\endgroup$
    – Richard
    Commented Sep 28, 2022 at 21:23

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