Representing dimensions in Dirac delta function results

Question

The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

How do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$ or $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$

Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

P.S. The problem comes when trying to normalize the momentum operator.

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-\frac{1}{2}}$.

Normalizing it, $$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$ $$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$ $$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$ Therefore, ignoring consistency of units, and assuming $A$ is positive, $$A=\frac{1}{\sqrt{2\pi\hbar}}$$ Notice that the units do not match

Answer 1

The dimension of the Dirac delta function is the inverse of the dimension of its argument. So if $x$ is a length then $\delta(x)$ has the dimension of inverse length.

In your example, the momentum eigenstate in position representation has the wavefunction $$ \psi_p(x) = A \mathrm{e}^{i p x / \hbar} $$ The interpretation of the wavefunction is that $|\psi|^2$ is a probability density, which is dimension of 1/Length. Therefore, $A$ has dimension of $1/\sqrt{\mathrm{Length}}$.

As you say, calculating the scalar product of two momentum eigenfunction with eigenvalues $p_1$ and $p_2$ gives $$ \int_{-\infty}^\infty \mathrm{d}x \, \psi_{p_1}^*(x) \psi_{p_2}(x) = |A|^2 \int_{-\infty}^\infty \mathrm{d}x \mathrm{e}^{i(p_2-p_1)x/\hbar} = |A|^2 2\pi \hbar \delta(p_2 - p_1) $$ The left-hand side is dimensionless, and therefore so is the right-hand side. $\hbar \delta(p_2-p_1)$ has dimension of Length, so again we get that the dimension of $A$ is $1/\sqrt{\mathrm{Length}}$.

Answer 2

The units do add up in the example you provide:

You've correctly established that the delta distribution has the dimension of the inverse dimension of its argument. This can be seen in a variety of ways, e.g. by looking at $\int f(x)\delta(x-x_0) dx = f(x_0)$, where the LHS has to have whatever dimensions $f$ has.

With this, we look at the dimensions of $\lvert A \rvert^2 2\pi\hbar \;\delta(p_1-p_2)$: $$ \left[\lvert A \rvert^2 2\pi\hbar\; \delta(p_1 - p_2) \right] = \textsf{L}^{-1} \left[ \hbar \delta(p_1-p_2) \right] = \textsf{L}^{-1} \frac{\left[ \hbar \right]}{[(p_1-p_2)]} = \textsf{L}^{-1} [x] = \textsf{L}^{-1} \textsf{L}^{1} = 1, $$where I have used that $\frac{p}{\hbar}$ has to have inverse dimensions of $x$ to make the argument of the exponential dimensionless.

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_{NB: For $\psi_p(x)=A\exp(\frac{ipx}{\hbar})$, we want $\int \lvert \psi \rvert^2 dx$ to represent a probability, which is dimensionless. From this it follows that $A$ has to have dimensions $\textsf{L}^{-\frac12}$.}

Answer 3

I am not entirely sure what you mean with the notation e.g. $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}.$$

The momentum eigenstates are $\psi_p(x) = \frac{1}{2\pi\hbar} e^{ipx/\hbar}$, this is true in any unit system. You don't need to multiply with any units. In the SI system, this has units of $$ \frac{1}{\sqrt{kg\, m^2\, s^{-1}}} . \tag{*}$$

Define $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$ Over here, $A$ has units of $m^{-1}$.

Note that $\int |\psi_p(x)|^2\, dx \neq 1$ (the state is not normalizable), therefore there's no reason why the dimension of $A$ should be inverse length! What we use instead of normalization is the following: $$ \langle p' \mid p \rangle = \int \psi_p(x) \psi_{p'}(x)^\ast dx = \delta(p'-p) , $$ here the $\delta$-function has units of $kg^{-1}\, m^{-1}\, s$ which is fully compatible with (*).

Or maybe it helps you to consider $$ \delta(x') = \langle x' | x=0 \rangle = \int \psi_p(0)^\ast \psi_p(x')\, dp = \frac{1}{2\pi\hbar} \int e^{ipx'/\hbar}\, dp , $$ where the units also match (of course).

Answer 4

tl;dr- Some of the common "definitions" of the Dirac delta function aren't mathematically rigorous, but rather conceptual. The confusion about units seems to come from these functions being taken more literally than intended. In reality, it's a unitless factor.

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The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$ From this we see that the dirac function has units of $x^{-1}$.

This doesn't follow. The Dirac delta function produces a unitless value, typically used as a factor for some other term to zero out that term's value for most values of $x$.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.

They're not assigning units because the Dirac delta function itself lacks them.

On the misconception about units cancelling

Below in the comments, @BySymmetry explained the confusion as resulting from the observation that $$ \int \text{d}x{\delta}\left(x\right)f\left(x\right)=f\left(0\right), $$ where if we want the units to cancel, then we need ${\delta}\left(x\right)$ to have units of $x^{-1}$.

As Wikipedia notes:

Consequently, the delta measure has no Radon–Nikodym derivative — no true function for which the property $$ \int _{-\infty }^{\infty }f(x)\delta (x)\,dx=f\left(0\right) $$ holds.^[21] As a result, the latter notation is a convenient abuse of notation, and not a standard (Riemann or Lebesgue) integral.

-Dirac delta function, Wikipedia

In short, this equation is an "abuse of notation", not the actual definition of a Dirac delta function. So, the idea that it should have units of $x^{-1}$ based on this equation's just a misunderstanding.

Conceptually, the Dirac delta's just a device to make a function zero everywhere but at one point. It's fundamentally a $0$-or-$1$ multiplication factor, so it just lacks units. You can write up that definition however you like to make it fit into conventional mathematical notation, but at the end of the day, that's it.

Example of the fallacy

Having units implies that, if you change the units - say from meters to lightyears - then you need to insert a multiplication factor. But, if you calculate some value with a Dirac delta, then switch your unit of length, do you actually multiply by such a conversion factor? Doing so would be a mathematical mistake.

The issue's that a Dirac delta's meant to be over an infinitely small space, so assigning it proportionality through units doesn't make sense.