bhobba said:
I am saying exactly what I said. Look at the equation |u> = ∫|x><x|u> dx. All you are doing in integrating is multiplying something by a number and summing. Now if you obtain the same thing as what you multiply and sum ie a state what you multiply must be dimensionless.
But if you want to look at it as a probability density its the same thing. For example the 1/2 probability you assign to the heads and tails of a one dollar coin does not give it the dimensions of dollars. Probabilities are, by definition from the Kolmogorov axioms, dimensionless. They are a number assigned to an event.
Thanks
Bill
No! This is a very common mistake! The important point is that ##|x \rangle## is NOT a proper state but a distribution. By definition it's "normalized to a ##\delta## distribution, i.e.,
$$\langle x|x' \rangle=\delta(x-x').$$
This implies that the formal dimension of ##|x \rangle## is ##1/\text{length}^{1/2}##.
The proper state ##|\psi \rangle## is normalized to 1,
$$\langle \psi|\psi \rangle=1.$$
Thus the proper dimension is 1. Together, the wave function is
$$\psi(x)=\langle x|\psi \rangle,$$
which has dimension ##1/\text{length}^{1/2}##, as it must indeed be, because the normalization condition reads
$$1=\langle \psi|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle \psi|x \rangle \langle x|\psi=\int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \psi(x).$$
The ##\mathrm{d} x## has dimension ##\text{length}^1## and ##|\psi(x)|^2## the dimension ##\text{length}^{-1}##, which makes the dimensions correct.
Note that this is true for any wave function wrt. generalized eigenvectors for operators with the eigenvalues in the continuous part of their spectrum. The corresponding wave functions squared are probability DENSITIES not probabilities!
