0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $1/\sqrt{\mathrm{Length}}$.

$\endgroup$
  • $\begingroup$ The question is how to we represent A. A has a value of $\frac{1}{2\pi\hbar}$ along with units that comes from the delta function $\endgroup$ – Ariana Jun 15 '17 at 16:42
0
$\begingroup$

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.


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 $\textsf{L}^{-\frac12}$.

$\endgroup$
  • $\begingroup$ Sorry, I've edited the error in the question, I meant for it to be $m^{-\frac{1}{2}}$ $\endgroup$ – Ariana Jun 14 '17 at 18:05
  • $\begingroup$ @ArianaGrande Nice, we are on the same page, then :) $\endgroup$ – Wojciech Morawiec Jun 14 '17 at 18:06
  • $\begingroup$ However, in the example, $A^2$ is shown to be $\frac{1}{2\pi\hbar}$ for normalization purposes, so how we represent the units of the delta function in the original wave function $Ae^{\frac{\iota px}{\hbar}}$ $\endgroup$ – Ariana Jun 14 '17 at 18:12
  • $\begingroup$ @ArianaGrande Ok, I'm confused: In the edit to your question you say something along the lines of "ignoring unit consistency, the units do not match". What exactly is it you are asking? $A$ doesn't have dimensions of $(2\pi\hbar)^{-1/2}$, it has dimensions of $(2\pi\hbar)^{-1/2} (\mathrm{kg}\mathrm{\frac{m}{s}})^{1/2}$ because of the units of the delta distribution, which in this case has the dimension of inverse momentum. $\endgroup$ – Wojciech Morawiec Jun 14 '17 at 18:17
  • $\begingroup$ Yes, there is a problem there, the question is how should I express the extra units, oh and the $kg\frac{m}{s}$ should go inside the inverse root.(oh and the recent edits are just formatting changes, no content added) $\endgroup$ – Ariana Jun 14 '17 at 18:20
0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ $\psi(x)$ is the probablity density of position, which should have units of $m^{-\frac{1}{2}}$, however, A have units of $kg^{-\frac{1}{2}}m^{-1}s^{\frac{1}{2}}$ $\endgroup$ – Ariana Jun 14 '17 at 18:17
  • $\begingroup$ @ArianaGrande Now I see where you're coming from. I edited my reply: $\int |\psi_p(x)|^2\, dx$ is not one! $\endgroup$ – Noiralef Jun 14 '17 at 18:25
  • $\begingroup$ If that isn't one there are massive implications on quantum mechanics! $\endgroup$ – Ariana Jun 14 '17 at 18:30
  • $\begingroup$ @ArianaGrande Momentum and position eigenstates are not part of the Hilbert space. The topic is mathematically a bit tricky. Have a look for example here (which is about position eigenstates, but it's the same problem): physics.stackexchange.com/questions/339177/… $\endgroup$ – Noiralef Jun 14 '17 at 18:32
  • $\begingroup$ Good point, in the short section I've presented, many many justification steps are ignored. However, the wave function still should have units of $m^{-\frac{1}{2}}$ and total probablity of measuring a particle at any position is 1. P.S. really the momentum eigenvectors are just the position being completely undefined $\endgroup$ – Ariana Jun 14 '17 at 18:39
-6
$\begingroup$

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.


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.

$\endgroup$
  • $\begingroup$ Could downvoters explain? $\endgroup$ – Nat Jun 14 '17 at 16:45
  • 1
    $\begingroup$ The Dirac delta does indeed have units of $x^{-1}$. This can clearly be seen from the defining relation of the delta function $\int dx\; \delta(x) f(x) = f(0)$. Since $dx$ has units of $x$, $\delta(x)$ must have units of $x^{-1}$ $\endgroup$ – By Symmetry Jun 14 '17 at 16:46
  • $\begingroup$ @BySymmetry Ah, I get the confusion - I'll update the answer. $\endgroup$ – Nat Jun 14 '17 at 16:54
  • 2
    $\begingroup$ @Nat Saying the dirac delta function does not have units fail to the identity $\delta(ax)=\frac{1}{|a|}\delta(x)$ for $a\neq 0$ Furthermore, another defination of the the dirac delta function is the fourier transform of 1(units). Which also imples the delta function has the inverse of the units of its input $\endgroup$ – Ariana Jun 14 '17 at 17:23
  • 1
    $\begingroup$ @Nat What you are basically saying is that $\delta(x)$ does not have a meaning on its own, only $\int \bullet \delta(x) dx$ which is unitless. But it is often helpful to treat $\delta(x)$ as if it were a function, and then it has units of $x^{-1}$. Also, there's no reason why the identity Ariana mentioned should only hold for scalar $\alpha$. $\endgroup$ – Noiralef Jun 14 '17 at 17:28

protected by Qmechanic Jun 14 '17 at 18:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.