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I was reading a booklet on formalism in quantum mechanics. Therein, I learnt that the wavefunction $\psi(\mathbf{r})$ is but just one way of representing the state of system which is a vector in Hilbert space in general, $|\psi \rangle$. This was called the position space representation of the state.

Then, it further stated that $\psi(\mathbf{r})$ is like coefficient in the expansion of $|\psi \rangle$ in the orthonormal basis $| \mathbf{r} \rangle$ (which were basically eigenfunctions of position operator):

$$|\psi \rangle = \int \textrm{d}^3r \ \psi(\mathbf{r}) | \mathbf{r} \rangle.$$

Now, since $|\psi \rangle$ is expanded in terms of $|\mathbf{r}\rangle$, both should be on the same footing, i.e, both should have the same units if any at all. But the formula suggests otherwise:

On the RHS, inside the integral, $\textrm{d}^3 \mathbf{r}$ has unit metre$^3$, $\psi(\mathbf{r})$ has unit metre$^{-3/2}$ as can be obtained from normalization condition. That leaves $|\mathbf{r}\rangle$. Now, no matter what unit you assign to $|\mathbf{r}\rangle$, you cannot at the same time, assign the same unit to $|\psi\rangle$ on the LHS. But the two seem to need to have same units.

Can someone please explain what's going on here?

Thomas Fritsch
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Vinci DaLeo
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1 Answers1

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The state $|\psi\rangle$ is dimension-less. This can be seen from its normalization condition $$\langle\psi|\psi\rangle=1.$$

On the other hand, the basis states such as $|\mathbf{r}\rangle$ are not dimension-less. This can be seen from their normalization condition $$\langle\mathbf{r}|\mathbf{r}'\rangle=\delta(\mathbf{r}-\mathbf{r}').$$

Because the $\delta$-function has dimension $\frac{1}{\text{volume}}$, a state $|\mathbf{r}\rangle$ has dimension $\frac{1}{\sqrt{\text{volume}}}$.

Furthermore, the wave function $\psi(\mathbf{r})$ has dimension $\frac{1}{\sqrt{\text{volume}}}$ because of its normalization condition $$\int d^3r\ |\psi(\mathbf{r})|^2=1.$$

So putting all the above together, the equation $$|\psi\rangle=\int d^3r\ \psi(\mathbf{r})|\mathbf{r}\rangle$$ becomes dimensional consistent. Left and right side are both dimension-less.

Thomas Fritsch
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