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I have seen the following two descriptions of the position basis:

$$\tag{1}| x\rangle=\delta(x-x_0)$$

and also

$$\tag{2}\langle x_0| x\rangle=\delta(x-x_0),$$

which (if either) of these is correct? Perhaps they are equivalent under change of notation? The first seems right to me, as it would solve the operator equation $$\tag{3}x| x\rangle=x_0| x\rangle,$$ but I would like to be sure. If we are working in the momentum representation, is $$\tag{4}|p\rangle=\delta(p-p_0)$$ also valid?

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The first is wrong. – David H Mar 2 '14 at 17:08
Comment to the question (v2): The third and fourth eqs. are also wrong. – Qmechanic Mar 2 '14 at 17:17
up vote 0 down vote accepted

If you put $\chi_{r_0}(r)= \delta (r-r_0 )$ then $[ \chi_{r_0}(r)]$ forms a basis. In Dirac's notation: $\chi_{r_0}(r) \rightarrow |r_0 \rangle$ and you can verify that this set is a basis because it satisfy:

Orthonormality: $\langle r_0 | r'_{0} \rangle = \delta (r_0 - r'_0 )$

Closure relation: $\int d^3 r_0 \ \ |r_0 \rangle \langle r_0|=$ 1

where 1 is the identity operator.

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So, if I understand correctly, equations 1 and 2 in my original question are both correct (but the notation is messed up making them inconsistent)? – Lachy Mar 2 '14 at 17:32
Yes! there are a lot of different notations: in general r represent the point in the space, r_0 the eigenvalue of the operator R and |r_0 > the eigenket relative to r_0 – LC7 Mar 2 '14 at 17:52
Another thing: for the momentum my comment is still valid (you just have to change r in p and so on) – LC7 Mar 2 '14 at 17:54

The latter description is correct (as is described in Sakurai, Gasiorowicz, Griffiths, and probably some other books that I don't own). What it is saying is that the inner product between $|x\rangle$ and $|x_0\rangle$ is either 0 if $x\neq x_0$ or 1 if $x=x_0$. That is, the states are orthogonal.

The momentum space description $$ \langle p|p'\rangle=\delta\left(p-p'\right) $$ is also valid.

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Thanks! Is it actually possible to calculate $|x\rangle$ or $|p\rangle$? That is, given $\psi(p)$, is it actually possible to calculate $|\psi\rangle=\int\psi(p)|p\rangle dp$ ? – Lachy Mar 2 '14 at 17:20
That integral doesn't make a lot of sense to me. Think of Fourier transforms, if you integrate over $p$, then you have $\psi(x)$ and not $\psi(p)$. To get $\psi(p)$, then you need to compute $\int dx'\langle p|x'\rangle\langle x'|\alpha\rangle$ for some arbitrary state $|\alpha\rangle$. – Kyle Kanos Mar 2 '14 at 17:35
The equation you just gave would (correctly) give $\psi(p)$ if I already had $|\psi \rangle$ (or $|\alpha\rangle$, as you used). On the contrary I already have the wavefunction $\psi(p)$ and want to calculate the representation free ket $|\psi\rangle$ – Lachy Mar 2 '14 at 17:41
Oh, now I see what you're trying to do. Normally you'd write it as $|\psi\rangle=\int |p\rangle\psi(p)dp=\int |p\rangle\langle p|\psi\rangle dp$ as $\psi(p)=\langle p|\psi\rangle$ and $\int|p\rangle\langle p|dp=1$. So I suppose as long as you know what $|p\rangle$ is, you should be able to do that. – Kyle Kanos Mar 2 '14 at 17:52

$ | x \rangle $ is a position eigenstate, the state for a particle with definite location $x$. This is an abstract vector.

$\delta (x - x_0)$ is a wavefunction (or distribution) for a particle with definite location $x_0$. It is the state $| x_0 \rangle$ on the position basis

$$\langle x | x_0 \rangle = \delta (x - x_0)$$

If $\hat x$ is the position operator then the third equation should be

$$\hat x | x_0 \rangle = x_0 | x_0 \rangle $$

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