On the units of the wavefunction

Question

I have an embarrassigly basic question I cannot answer myself. The wavefunction in QM is defined as $\Psi(x) = \langle x \vert \Psi \rangle$, where we chose a 1-dimensional physical space for simplicity. We know that a quantum state is represented by a normalized vector. $$ 1 = \langle \Psi \vert \Psi \rangle = \int \text d x\; \langle \Psi \vert x \rangle \langle x \vert \Psi \rangle = \int \text{d}x\; |\Psi(x)|^2 \tag 1 $$ From this requirement we deduce that the units of $\Psi(x)$ are $[L]^{-1/2}$. Now, we consider a state of definite position $$\vert \Psi \rangle = \vert x_0 \rangle.$$ It follows from the definition that $$\Psi(x) = \langle x \vert x_0 \rangle = \delta(x - x_0).$$ We apply the known property of the Dirac delta distribution. $$ 1 = \int \text d x\; \delta(x - x_0) \tag 2 = \int \text d x\; \Psi(x) $$ Equations (1) and (2) seem to be in contraddiction. I get that the square of a delta distribution is not defined but the two equations do not even agree on the units of $\Psi(x)$, infact equation (2) implies that $\Psi(x)$ has units of $[L]^{-1}$. I'm pretty sure the correct units are $[L]^{-1/2}$ but

how can this mismatch in units be explained?

Related posts

• Units of a dirac delta function in quantum mechanics: the question is verbose but very similar to mine, that being said I didn't really understand the answer. I get that different vectors can have different units (for example $$\vert p \rangle = \int \frac{\text dp}{\sqrt{2\pi\hbar}} \exp\{ipx/\hbar\} \vert x \rangle\;$$ has different units from $\vert x \rangle$) and that different scalar products can have different units but in terms of dimensions equation (1) is supposed to hold for any choise of $\vert \Psi \rangle$, even for non normalizable vectors such as $\vert x \rangle$ and $\vert p \rangle$.

Answer 1

The correct answer leaves the units in the $|\vec{r}\rangle$ basis vectors, not in $|\psi\rangle$. I'll work in 3 dimensions.

$$\int |\langle \vec{r}|\psi\rangle|^2 dV=1$$

$[dV]=[L]^3$, so $$\left[ \langle \vec{r}|\psi\rangle \right]=\left[\psi(\vec{r})\right]=\left[ \left(\frac{1}{L}\right)^{3/2}\right]$$

But if we repeat the same thing with momentum for example, we get that $\psi(\vec{p})$ has different units. Same with energy or any other variable.

We must conclude that the units are not in $|\psi\rangle$, but in $\langle \vec{r}|, \langle\vec{p}|$, or otherwise.

So while it's true that $$[\psi(\vec{r})]=[\left(\frac{1}{L}\right)^{3/2}]$$ it's only because $[ \langle \vec{r}|]=[\left(\frac{1}{L}\right)^{3/2}]$. The vector $|\psi\rangle$ is unitless.

Now I'll address the apparent contradiction with dirac delta functions. You have said that $\psi(x)=\delta(x)$ in an position eigenstate. But that means $|\psi(x)|^2$ should be something else, like a delta squared. That's true, but people essentially ignore that and just replace it with a delta function. The framework for perfect eigenstates is idealized and not suitable for the real world. It's impossible in practice to actually create a perfect position or momentum eigenstate, so it will never actually get integrated over like in your example. It's just a mathematical tool for breaking down other states into components.

Answer 2

Your wave function $\Psi_{X}(x) = \delta(x-X)$ is not normalizable. Therefore, you can't really use your equation to find its dimensions. The canonical way to normalize this is $$ \big( \Psi_{X} , \Psi_{Y} \big) \equiv \int dx \Psi^*_{X}(x) \Psi_{Y}(x)= \delta(X-Y) . $$ However, non-normalizable wavefunctions can be given any units you want since they are non-normalizable, so any rescaling is allowed.