Just wanted to question how Hilbert space dimensions work. From what I've seen, they're used to generalize the mathematics of finite-dimensional Euclidean spaces into infinite-dimensional vectors. I've seen them used in quantum mechanics, and in some cases stated to "inherit finite dimensional space-time" despite the Hilbert space having infinite dimensions. So I'm basically asking how Hilbert spaces correlate to the physical spatial dimensions they represent. Are there Hilbert spaces that represent topological spaces with infinite spatial dimensions? same question for any function spaces with infinite dimensions.

  • 3
    $\begingroup$ stated to "inherit finite dimensional space-time" Source? $\endgroup$
    – Ghoster
    Commented May 28, 2023 at 19:01
  • $\begingroup$ sorry i wired it wrong. it’s from a paper that uses infinite dimensional function spaces, abss says space time “emerges” as finite ui.adsabs.harvard.edu/abs/2012AIPC.1424..149H/abstract $\endgroup$
    – Adithya
    Commented May 29, 2023 at 4:49

6 Answers 6


There’s no relation between the dimension of the Hilbert space and the actual physical space. If you work with a system of angular momentum $2$ - say hydrogen states with $n=3$ and $L=2$ - then your Hilbert space is of dimension $2L+1=5$ even if the physical space has dimension $3$.

The dimension of the Hilbert space is tied to the number of distinct outcomes of measurements of quantities.

  • $\begingroup$ Okay i think i understand. thanks. also whats L, and whats the formual 2L + 1 meant to represent? is it to calculate the hilberts space dimensionality? $\endgroup$
    – Adithya
    Commented May 29, 2023 at 4:45
  • $\begingroup$ also do you have any sources that i could read in to understand the difference between hilberts space dimensions and space time dimensions? $\endgroup$
    – Adithya
    Commented May 29, 2023 at 4:51
  • $\begingroup$ $2L+1$ is the number of states with angular momentum $L$ and energy $n=3$. There are $5$ such states because there are 5 possible outcomes of simultaneously measuring energy, total angular momentum, and the projection of angular momentum on the $\hat z$ axis. The corresponding quantum numbers are $(n,L,m)$, respectively. $\endgroup$ Commented May 29, 2023 at 12:08
  • $\begingroup$ for the same system can you consider higher or lower values for n and L or is this predetermined by the system? sorry if i’ve misunderstood the meaning of these parameters $\endgroup$
    – Adithya
    Commented May 30, 2023 at 12:54
  • $\begingroup$ You might be interested in states of a given energy (fixed but otherwise arbitrary $n$) in which case the dimension of the space is $n^2$ since it contains $L=n-1,n-2,\ldots, 0$ and for every $L$ with this $n$ you have $2L+1$ states. $\endgroup$ Commented May 30, 2023 at 13:26

I want to add the following to the good answer by ZeroTheHero.

In quantum mechanics Hilbert spaces emerge as state spaces. For example as the space the wave functions describing a particle live in. Those are infinite dimensional more often than not, because of the variety present in the wave functions. This is not unlike the phase space of a complicated mechanical system (e.g. $N$ pointlike bodies, each with three degrees of freedom for position, another three for motion, resulting in a $6N$-dimensional phase space).

The finite-dimensional subspaces ZeroTheHero mentioned can be thought of as finite dimensional subspaces of a suitable Hilbert space, stable under the action of the group of symmetries.


The link between the two spaces is a mathematical induction link (from simple to complex), the terms: vector, scalar product, orthogonality,..., have changed their application universe, we know for example that orthogonality of two vectors in the geometric sense is translated by $u.v=0$ , we can note the 'generalized' binary operation $u|v=0, u|u=1,v|v=1$, or in other domain (universe) like the Fourier series, we also have functions that verify $f|g=0, f|f=1,g|g=1$, the binary relation has changed the way of action: instead of 'a projection...., it becomes an integral..., we can imagine storing any kind of mathematical object defined in different 'universes' which have as results of a binary relation $x|y=0,x |x=1,...$, and we keep the terms, vector,...., instead of inventing a multitude of terms for each 'universe'.


This is the same as asking if the dimensions of a vector space are physical. A vector space is a mathematical construct. They are often used to model various physical or non-physical things. So it depends on the vector space.

Displacement vectors are physical where the relationship between our physical space and the vector space is very direct. You can easily say our physical space is a vector space. Velocity, forces, etc. are a bit different, but also physical in this sense.

By physical dimensions, I presume you mean x, y, and z axes of the space. You can say that the axes of displacement space are physical lines. The basis vectors are physical displacements in these directions. For velocity and force spaces, the axes represent velocities and forces parallel to the physical axes.

A Hilbert space is a vector space with an inner product that is complete with respect to the norm defined by the inner product.

A norm has been defined on the space of displacement vectors. The norm makes it possible to define convergence. The limit of every convergent sequence is a vector in the space, so the displacement vector space is complete. Thus it is a Hilbert space.

However, Hilbert spaces are most useful for other types of space, such as state spaces in quantum mechanics. Here the vectors do not represent things that relate simply to physical lines. A vector may represent the state of a photon.

Suppose a photon comes from a distant monochromatic star. Light from the star is an expanding spherical wave. But the sphere is so big we cannot distinguish it from an infinite plane wave. This infinite space-filling function is the state of the photon: $\psi = Ae^{\vec k \cdot \vec x - \omega t}$. It is a vector in the state space. (I am glossing over the fact that if you normalize a truly infinite plane wave, the amplitude comes out zero. There are some approximations here. You have to use some mathematical care to use a plane wave as a state. See Normalizing the solution to free particle Schrödinger equation)

It is not like a displacement vector. If you had a flood of photons in this state, $\psi$ would represent the electric field. $\psi * \overline \psi d \vec x$ represents the number of photons per sec in a volume $d \vec x$. If you have just one photon, it represents the probability of detection.

A photon from a different star would have a $\vec k$ in a different direction, and would be in a different state. Likewise a photon with a different wavelength would be a different state.

The fact that quantum mechanical states are vectors is a bit counterintuitive. If you add two vectors, you get a vector. A photon can have a state that is a sum of different plane waves.

In the starlight example, we have a well defined $\vec k$ with no uncertainty of direction. The transverse component of $\vec k$ is $\vec 0$. This is only possible if we have an infinite uncertainty of transverse position. And we do have an infinitely broad plane wave.

If the plane wave passes though a pinhole, the transverse uncertainty of position is reduced. The transverse uncertainty of momentum grows. There is now a range of possible $\vec k$ values. The photon state now is a linear combination of states with different $\vec k$ values. If the photon hits a screen, you cannot predict which $\vec k$ state will be measured.

The dot product for the state vector space is defined by

$$<\psi | \phi> = \int_{All\,space} \psi \overline \phi d \vec x$$

I will gloss over normalization again. Plane waves oscillate. Two different plane waves with different $\vec k$'s oscillate at different rates or in different directions. The product over all space averages to $0$. If the integral is $0$, two different plane waves are orthogonal as vectors. Each value of $\vec k$ generates a different dimension. This Hilbert space is infinite dimensional.

I started this question when there were no answers. Now there are several, and this looks like overkill in the light of @ZeroTheHero. Hope it helps never the less.

  • $\begingroup$ thanks this was very helpful $\endgroup$
    – Adithya
    Commented May 29, 2023 at 17:47

"Space-time" is the 4-dimensional space of the Lie Group parameters $(\vec{x},t)$ that specify the space and time transformations done to kets (which stand for particles) in a (not space-time) large dimensional Hilbert Space. The space-time parameters $(\vec{x},t)$ along with rotation and boost parameters $(\vec{\theta},\vec{\lambda})$ label the transformations of the Poincare Group, which are the transformations we can do to objects in the Hilbert Space. Please see this Physics Stack answer.

**Added Edit: We are very accustomed to saying (and seeing) "I moved the chair from the origin $(\vec{x}=0,t=0)$ in space-time to $(\vec{x},t)$ in space-time", and classical mechanics and relativity use objects being at different places in space-time. Historically, space-time is a very real place to our senses. We walk around in it, and we coordinatize locations in it with $(\vec{x},t)$.

However, in Quantum Mechanics the statement becomes "I transformed the ket (chair) to the ket' (translated_chair) in the Hilbert space of kets." The parameters $(\vec{x},t,\vec{\theta},\vec{\lambda})$ just specify the transformation on Hilbert space which turns ket(chair) into ket'(translated_ chair). If you translate/rotate/boost ket enough, the amplitude <ket|ket'> to still detect ket may be very small.

The Lie Group paramters $(\vec{x},t,\vec{\theta},\vec{\lambda})$ do form a topological space. The Poincare group will almost certainly be updated to a better one in the future (eg: non-abelian translations and gauge transformations.


Hilbert spaces are used in quantum mechanics as spaces of possibilities. They need to be large enough to contain all possibilities for what something might be.

For example if you have a closed box with some particles inside then the inside of the box might have 1 particle, 2 particles, and so on. Each number of particles will correspond to a dimension in the Hilbert space. So it is infinite dimensional.

I have no clue what "inherit finite dimensional space-time" means. Perhaps it means a Hilbert space is similar to normal 3-dimensional space in some ways. Where did you read it?

  • 2
    $\begingroup$ Sorry but this is incorrect. The Hilbert space for a single particle in a box, or for a single particle in a harmonic well, is infinite-dimensional in both cases. $\endgroup$ Commented May 29, 2023 at 12:09
  • $\begingroup$ @ZeroTheHero In my example it certainly is. $\endgroup$
    – Daron
    Commented May 30, 2023 at 8:28
  • $\begingroup$ so its taken as infinite to account for all possibilities because we cant be sure of how many there are? or are there actually infiinite possibilities $\endgroup$
    – Adithya
    Commented Jun 1, 2023 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.