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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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  • @jacob1729 Thanks. It is quite the same question but I fail to understand how members of the same vector space can have different physical dimensions as implied in the answer below. – Vinci DaLeo Nov 24 '21 at 18:21

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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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  • Ok, I have just one more question, if |r》and |Ψ》 have different dimensions, how can they be the members of the same vector space? – Vinci DaLeo Nov 24 '21 at 18:11
  • @Vinci if force and acceleration have different dimensions, how can they both be members of $\mathbb R^3$? – Emilio Pisanty Nov 24 '21 at 18:24
  • @Emilio Pisanty Fair point. But I was thinking along the following lines. Any vector in a vector space is a linear combination of the basis vectors. Assuming that the basis vectors all have same dimension(otherwise it doesn't make sense to add them), any linear combination of them MUST have that very same dimension. Right? – Vinci DaLeo Nov 24 '21 at 18:28
  • @Emilio Pisanty Also, a purist would argue that force and acceleration vectors are actually members of two different R^3 s, having different "units" – Vinci DaLeo Nov 24 '21 at 18:33
  • @VinciDaLeo Infinite-dimensional vector spaces simply don't behave like you think they should, and the notion of "basis" is ambiguous there. The continuous basis "vectors" like $\lvert x\rangle$ from which you assemble "actual" vectors $\lvert \psi\rangle$ don't actually live "in the same space" - their inner product is ill-defined/non-normalizable, after all, and so they are not members of a Hilbert space. See e.g. this answer of mine for the different notions of "basis" you might encounter. – ACuriousMind Nov 24 '21 at 18:49
  • @Vinci DaLeo You just saw that a consistent linear combination of vectors with dimensionfull coefficients has different dimension than the vectors: what more do you ask for? – Cosmas Zachos Nov 24 '21 at 18:50
  • @Vinci Indeed, you can argue that, and then you need to turn $\vec F=m\vec a$ into a relationship between two different vector spaces. (And similarly, basis vectors are typically unit-norm, so things like $\vec a = \sum_{j=1}^3 a_j \hat{\mathbf e}_j$ is also a relationship between different vector spaces.) There's nothing stopping you from going full-rigour on that formalization and it can indeed be done; it is just tedious. – Emilio Pisanty Nov 27 '21 at 14:20
  • (cont.) Regarding your question, there is nothing different in your OP w.r.t. what I just mentioned. Even infinite dimensionality does not bring anything novel there, beyond requiring basis functions with nontrivial dimensionality (as opposed to dimensionless $\hat{\mathbf e}_j$). – Emilio Pisanty Nov 27 '21 at 14:21