What Are the Dimensions of Hilbert Space Elements in Quantum Mechanics?

[Ravi Mohan]

TL;DR

Quick basics revision

Hi Fellas! My first post after a long hiatus from forums. Feeling nostalgia (this is the place where it all began, my fuel for quantum fascination so to speak).

I am revisiting the mathematical formulation of quantum mechanics with the dimensional (MLT) perspective. I want to understand what are the dimensions of Hilbert space elements, within the context of quantum mechanics. I was thinking of the elements as dimensionless but am stuck in the logical reasoning.

Consider an element of (rigged?) Hilbert space $\mathcal{H}$ with operators having continuous spectral decomposition, in the sense the insertion or completeness is represented by
$$\hat{\mathbb{1}} = \int dx|x\rangle\langle x|,$$
were $x$ is continuous and all those usual gory specifications.
I am interested in knowing the role of high school dimensional analysis played here. For that I consider a unit norm state vector $|\Psi\rangle$. According to the premise I started with, $|\Psi\rangle$ should have the dimensions of $[M^0L^0T^0I^0\Theta^0J^0]$ (symbols have usual meaning).

Now I start the quantum gymnastics
\begin{align}
|\Psi\rangle &= \hat{\mathbb{1}}|\Psi\rangle \\
&= \int dx|x\rangle\langle x| \cdot |\Psi\rangle\\
&= \int dx \psi(x)|x\rangle

\end{align}
I see no issue in first equation because $|\Psi\rangle$ and $\hat{\mathbb{1}}$ are dimensionless.
In second equation, I don't know how to distribute dimensions. My common sense senses that since $\psi(x)$ is to be interpreted as "wavefunction" its units need be $L^{-0.5}$. But then the inner product looses its value, in the sense that $\psi(x) = \langle x|\Psi\rangle$ where $|\Psi\rangle$ and the element of dual vector space $\langle x|$ are supposed to be dimensionless. How do I reconcile?

Maybe it is nothing, and maybe I need jolting back to undergrad, a little lesson in dimensional analysis should be helpful!
Thanks!

H

 

[anuttarasammyak]

Your argument suggests that $|\Psi>$ has no physical dimension.
<\Psi|\Psi>=1 when normalized. Do you have an uneasiness with it ?

 

[Ravi Mohan]

Are you suggesting dual vector space has inverse dimensions?

 

[anuttarasammyak]

No, both $<\Psi|$ and $|\Psi>$ have no dimension.

 

[Ravi Mohan]

And what about $\psi(x)$?

 

[anuttarasammyak]

$L^{-1/2}$ as you said because <x| has dimension $L^{-1/2}$.
\int |x&gt;dx&lt;x|=1 corresponding
L^{-1/2} L^1 L^{-1/2} =L^0=1

[EDIT]
Above is 1 D case. $|\psi|^2$ is probability density per unit length which has dimension $L^{-1}$.
For 3D case
\int\int\int |\psi|^2 dxdydz=1 and $|\psi|^2$ is probability density per unit volume which has dimension $L^{-3}$.

 

[Ravi Mohan]

So you are suggesting $\langle x|$, and element of the dual of same hilbert space $\mathcal{H}$ is now equipped with some dimension?

For clarity $|\Psi\rangle, |x\rangle \in\mathcal{H}$.

 

[Ravi Mohan]

Or maybe, something like
$$\int_{x}^{x+dx} dx|x\rangle\in \mathcal{H}.$$

But yeah, no.

 

[anuttarasammyak]

<x| ,|x> have dimension $L^{-1/2}$
<p|, |p> have dimension $[ML/T] ^ {-1/2}$ or so.
They do not belong to ordinally Hilbert Space as you said in OP.

For confirmation
&lt;x|x&#039;&gt;=\delta(x-x&#039;)
has dimension $L^{-1}$ because
\int \delta(x-x&#039;)dx&#039;=1
corresponding
L^{-1}L^1=L^0

 

[Ravi Mohan]

Yeah so what is it?
About the rigged Hilbert space that I need more understanding on?
I don't see any reconciliation, yet!

 

[anuttarasammyak]

I have read R. de la Madrid, "Quantum Mechanics in Rigged Hilbert Space Language," PhD Thesis (2001)
referred in Wikipedia https://en.wikipedia.org/wiki/Rigged_Hilbert_space
and got more insight on Rigged Hilbert Space. It may be helpful also to you.

 

[Ravi Mohan]

Well there is a particular reason for me to post here. Reading vague literature is not my cup of tea, anymore ...

 

[anuttarasammyak]

Both $<\Psi|$ and $|\Psi>$ have no dimension.
Inner product $<\Psi|\Psi>$ has no dimension. Value 1 when normalized.

Both $<x|$ and $|x>$ have dimension $L^{-1/2}$.
Inner product $<x|x'>=\delta(x-x')$ has dimension $L^{-1}$.

$\psi(x)=<x|\Psi>$
$\psi^*(x)=<\Psi|x>$ so from the above
Both $\psi(x)$ and $\psi^*(x)$ have dimension $L^{-1/2}$.
Component product $\psi(x)\psi^*(x)$ has dimension $L^{-1}$.
Inner product $\int \psi(x)\psi^*(x) dx$ has no dimension, Value 1 when normalized.

Do you find any problems on the above ?

 

[PeterDonis]

Moderator's note: Thread level changed to "I". This is not "B" level material.

 

[Ravi Mohan]

Do you find any problems on the above ?

NO