Ok so I will provide the following example, which I am choosing at random from Sabio et al(2010):

$$\psi(r,\phi)~=~\left[ \begin{array}{c} A_1r\sin(\theta-\phi)\\ A_2\frac{K}{2}r^3\sin^3(\theta-\phi)\\ A_2r^2\sin^2(\theta-\phi)\\ -A_1r\sin(\theta-\phi)\\ \end{array} \right].$$

Obviously, this is a 4-component vector, but the author calls it a wavefunction. In fact, calling such things wavefunctions is fairly common in my little corner of condensed matter theory, and I'm still unsure whether this is sloppy language, or I am missing something. Perhaps the thing on the LHS should in fact be the ket $|\psi\rangle$ because my understanding of a wavefunction is that it can not be a vector. I.e., it is defined:

$$\psi(x)~=~\langle x|\psi\rangle.$$

This should just be a number correct? I.e. at any point (let's say $x=1$ for a 1-dimensional system) that wavefunction should just be a number, not a vector?

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    $\begingroup$ $| \psi \rangle$ contains all necessary info. If you really need a configuration-space representation, you indeed project onto $x$, as $\psi(x) = \langle x | \psi \rangle.$ There are no interesting examples, everything is trivial. You can take a particle in a box with wavenumber $k$, so you have a state $| k \rangle$ (which is a momentum/energy eigenstate), and then $\psi(x) = \langle x | k \rangle \propto e^{ikx}.$ I don't understand your example with the commutator. The LHS is a state in Hilbert space, the RHS is a commutator of operators. $\endgroup$ – Vibert Mar 3 '13 at 0:58
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    $\begingroup$ Ask your bra if he knows anything about quantum mechanics. If he does, buy him a beer, and ask him what the wavefunction is. $\endgroup$ – KDN Mar 3 '13 at 1:10
  • $\begingroup$ Thanks for the answer. Firstly I should explain: that isn't a commutator, simply a vector. It should be a column vector, but I don't know the syntax to construct such an object on the forum, hence the reason it is transposed. Anyway you say that $|\psi>$ contains all necessary information, which I understand, however say I want to find $|\psi(x)|^2$ then I need to know the wavefunction, as this (to the best of my knowledge) cannot be calculated directly from $|\psi>$ (even though the integral of $|\psi(x)|^2$ can be). Thanks again. $\endgroup$ – Lachy Mar 3 '13 at 1:14
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    $\begingroup$ Eh, doesn't $|\psi \rangle = \int dx |x\rangle \langle x|\psi \rangle=\int \psi(x) |x\rangle$ make sense? $\endgroup$ – alexarvanitakis Mar 3 '13 at 2:01
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    $\begingroup$ Ah, I think I'm starting to see what you're asking about. If I may make a suggestion: it might help to go back and edit your question to center around this specific example, and to bring up the issues you mentioned in this last comment. Don't add on an "Edit:" block at the end, just rewrite the question entirely. What you're really asking, it seems, is something like "what does it mean for something to be a ket?" Meanwhile I will see if I can develop an answer. $\endgroup$ – David Z Mar 3 '13 at 2:52

An inner product doesn't necessarily have to produce a scalar. For example, consider matrix multiplication. When you take the inner product of a $M\times N$ matrix and an $N\times P$ matrix, you get another matrix which is $M\times P$. The inner product only "collapses" one dimension of the first thing with one dimension of the second thing.

You can even consider "generalized matrices" which can have numbers of dimensions other than 2. Say you have a matrix of dimensions $A\times B \times C$, and another of dimensions $B\times C\times D$. There are actually two different ways you can take the inner product of these two matrices; you can combine the second dimension of the first one with the first dimension of the second one, and get a result with dimensions $A\times C\times C\times D$. Or you can combine the third dimension of the first one with the second dimension of the second one, and get a result that is $A\times B\times B\times D$. Or you could do both (that would be two inner products), and get an $A\times D$ result.

In the case of wavefunctions, you know that a wavefunction is defined from the inner product $\langle \vec{x}\vert\psi\rangle$, where $\lvert\psi\rangle$ and $\lvert \vec{x}\rangle$ are quantum states. These quantum states are abstract objects, and in particular, they're not constrained to having only one dimension. In your example, the state $\lvert\psi\rangle$ is some abstract object that probably has four dimensions: three of the dimensions correspond to spatial position, and are labeled by the index $\vec{x}$ (or $r$, $\theta$, $\phi$, given how it's written), but there is another dimension has four components. You can think of $\lvert\psi\rangle$ as a matrix which has dimensions $A\times B\times C\times D$, where $A$, $B$, and $C$ just happen to be infinity (and $D = 4$). When you take the three inner products of a matrix of dimension $A\times B\times C$ with a matrix of dimension $A\times B\times C\times D$, you get a vector of dimension $D$, and that's exactly what's happening here.

If you want to get a single component out of this state $\lvert\psi\rangle$, you will need to take four inner products: the three that are involved in $\lvert \vec{x}\rangle$, and one with a basis vector along the other dimension, let's call it $e_i$. These basis vectors would be of the form

$$e_0 = \begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}$$

or similar. A single component of the state might be written $\psi_i(x)$, and it would be defined as the quadruple inner product

$$\psi_i(\vec{x}) = e_i\cdot \psi(\vec{x}) = e_i\cdot\langle\vec{x}\vert\psi\rangle$$

If you already have a representation of $\psi$ in terms of its components, you could also make use of this identity:

$$\psi_i(\vec{x}) = \sum_j \iiint \mathrm{d}^3\vec{x}' \underbrace{(e_i)_j}_{\text{component of basis vector}}\overbrace{\delta(\vec{x}' - \vec{x})}^{\text{component of basis function}}\psi_j(\vec{x}')$$

which really just tells you how to express the same object in a different basis.

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  • $\begingroup$ Thanks so much mate, that's super helpful! I guess that in the end my problem was primarily one of misunderstanding notation. ie I thought the symbol $|\rangle$ qualified an object as a column vector, so knowing they can have high dimensions pretty much tells me the rest of the story. Thanks again! $\endgroup$ – Lachy Mar 3 '13 at 9:03
  • $\begingroup$ Yeah, the use of ket notation has nothing to do with whether an object is a vector or not. $\endgroup$ – David Z Mar 3 '13 at 11:09

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