In quantum mechanics the eigenfunctions of the position are dirac delta functions, $A\delta(x-x_0)$, where $A$ is some constant. Eigenfunctions of the position are usually normalized with a "Delta function normalization". This means that $\langle x | x' \rangle = \delta(x-x')$, which can be understood in a formal way if one applies the shifting property of one of the delta functions to the other in the integral $$ \langle x | x' \rangle = \int_\mathbb{R}\mid A \mid^2\delta(y-x)\delta(y-x')~\mathrm{d}y.$$

This gives as a result $\mid A \mid^2 \delta(x-x')$, so the coefficient $A$ should be set to $1$. Up until today I thought I understood this kind of fine, but I realized now that there is a problem with dimensions somewhere here. In order for an expression like $ \int_\mathbb{R} \psi^*(x) \psi(x)~\mathrm{d}x$ to be dimensionless, position space wavefunctions should have dimensions of $\frac{1}{\sqrt{L}}$. This means that an expansion in terms of position eigenfunctions, like $$ \mid \psi \rangle = \int_\mathbb{R} \psi(x) \mid x \rangle~\mathrm{d} x$$ would only make sense if position eigenfunctions themselves had units of $\frac{1}{\sqrt{L}}$.

This is a bit disturbing to me, the inner product of two position eigenfunctions should, if one looks at the integral, $$\langle x | x' \rangle = \int_\mathbb{R}\delta(y-x)\delta(y-x')~\mathrm{d}y$$ be dimensionless (two things with dimensions $\frac{1}{\sqrt{L}}$ inside and de $\mathrm{d}x$ factor, with dimensions $\frac{1}{L}$) but delta function normalization implies it is a new delta function... thus it has dimensions of $\frac{1}{\sqrt{L}}$! How can this be made right?

It is also strange to me that in other areas of physics (electromagnetism for instance) delta functions are usually given dimensions of $\frac{1}{D}$, where $D$ is the dimension of the argument. This seems not to be the case here, and I'm a bit weirded out by it. When changing from cartesian to polar coordinates, for instance, the natural thing in electromagnetism is to consider $$\delta(x-x_0) \delta(y-y_0) = \frac{\delta(r-r_0)\delta(\phi-\phi_0)}{r},$$ but this messes up units in quantum mechanics, because in quantum mechanics the left hand side would have different dimensions than the right hand side... One could perhaps consider $$\frac{\delta(r-r_0)\delta(\phi-\phi_0)}{\sqrt{r}}$$ as the right thing to write down in QM, but there should be only one way to write delta functions in different variables... shouldn't there? the Jacobian determinant should go downstairs. How does this make sense in QM?

  • 1
    $\begingroup$ Related, possible duplicate : physics.stackexchange.com/q/33760 $\endgroup$
    – user167453
    Aug 27, 2017 at 19:51
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Aug 27, 2017 at 20:00

1 Answer 1


No, the inner product of two position eigenfunctions shouldn't be dimensionless. You chose to normalize them such that $\langle x | x' \rangle = \delta(x-x')$; therefore, the inner product has the dimensions of $\delta$, i.e., $1/L$. Don't confuse the state with the wavefunction: the wavefunction corresponding to $|a\rangle$, $\delta(x-a)$ is not $|a\rangle$ but $\langle x | a \rangle$, so it has a different dimension.


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