I Should I always be careful about dimensional consistency?

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Will a dimensional inconsistency cause a problem?
I read an equation in a paper, $$\left | m \right >=\int G(\mathbf k) \left | \mathbf k \right > \frac {d^2 k}{4 \pi^2}$$ where ##G(\mathbf k)= \left < \mathbf k \right | \left . m \right >## is the momentum space wave function, ##k## is the two-dimensional frequency.

In this paper, ##\left | m \right >## is the transverse LG modes of a Gaussian beam, and it is dimensionless. Suppose ## \left | \mathbf k \right >## has a dimension of ##[m^l]##. Then from the definition of ##G(\mathbf k)##, it will have a dimension of ##[m^{-l}]##. But then the dimension of the equation will become ##[m^0]=[m^{-l+l-2}] ##, and that cause an inconsistency. Would it cause problems?
 
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If ##G(\mathbf{k}) = \langle \mathbf{k}|m\rangle## and ##|m\rangle## is dimensionless, then ##G(\mathbf{k})## has the same dimensions as ##|\mathbf{k}\rangle##
 
stevendaryl said:
If ##G(\mathbf{k}) = \langle \mathbf{k}|m\rangle## and ##|m\rangle## is dimensionless, then ##G(\mathbf{k})## has the same dimensions as ##|\mathbf{k}\rangle##
Thanks. I got confused because ##G(\mathbf{k})## given by the paper is clearly dimensionless, so I am trying to find a balance in that equation. However, I just find that ##G(\mathbf{k})## should have the dimension of ##[m^2]## and everything works out.

Thanks!
 
Hi, @stevendaryl , do ##\left | k \right > ## has the same dimension as ##\left < k \right |##?
 
It depends on how you normalize your states. Obviously in this case you have the HEP/QFT convention, i.e.,
$$\langle \vec{k}|\vec{k}' \rangle=(2 \pi)^2 \delta^{(2)}(\vec{k}-\vec{k}')$$
since you seem to work in 2D. That's because in this community you usually use natural units with ##\hbar=c=1## and in Fourier transforms you want for each energy or momentum integral a factor ##1/(2 \pi)##.

This implies that ##|\vec{k} \rangle## as well as ##\langle \vec{k}|## have dimension ##1/\text{momentum}##, and the completeness relation reads
$$\int_{\mathbb{R}^2} \frac{\mathrm{d}^2 k}{(2 \pi)^3} |\vec{k} \rangle \langle \vec{k}|=\hat{1}.$$
So you can expand all Hilbert space vectors in terms of these generalized momentum eigenvectors
$$|m \rangle=\int_{\mathbb{R}^2} \mathrm{d}^2 k \frac{1}{(2 \pi)^3} |\vec{k} \rangle \langle \vec{k}|m \rangle.$$
 
Haorong Wu said:
Hi, @stevendaryl , do ##\left | k \right > ## has the same dimension as ##\left < k \right |##?
It's usually the case that for any state ##|\psi\rangle##, the conjugate state ##\langle \psi |## has the same dimensions. You can think of ##|\psi\rangle## as a kind of column matrix (with maybe an infinite number of rows), and ##\langle \psi |## is the result of turning the column into a row (taking the transpose) and taking the complex-conjugate: ##\langle \psi| = (|\psi\rangle^T)^*##
 
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