Should I always be careful about dimensional consistency?

[Haorong Wu]

TL;DR Summary

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?

 

[stevendaryl]

If $G(\mathbf{k}) = \langle \mathbf{k}|m\rangle$ and $|m\rangle$ is dimensionless, then $G(\mathbf{k})$ has the same dimensions as $|\mathbf{k}\rangle$

 

[Haorong Wu]

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!

 

[Haorong Wu]

Hi, @stevendaryl , do $\left | k \right >$ has the same dimension as $\left < k \right |$?

 

[vanhees71]

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
$$|m \rangle=\int_{\mathbb{R}^2} \mathrm{d}^2 k \frac{1}{(2 \pi)^3} |\vec{k} \rangle \langle \vec{k}|m \rangle.$$

 

[stevendaryl]

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: $\langle \psi| = (|\psi\rangle^T)^*$