I couldn't really find a fitting title for this question. I'm still relatively new to QM and am trying to get the basics down. I understand that a physical system is associated with a Hilbert Space, where the vectors of $H$ represent the possible "states" of the physical system. I've often seen in texts that this function contains all there is to know about a system.

So for a particle moving through space $\mathbb{R}^3$, as far as I can tell, the Hilbert space is $L_2(\mathbb{R}^3)$, the space of square-integrable complex valued functions (called wave functions).

Operators represent observables, these operators act on wavefunctions, so if the wave function contains all information about the particle, there should be, say, a spin operator acting on this function. But spin seems to warrant its own Hilbert space, (it is a discrete variable) and so is not represented in $L_2(\mathbb{R}^3)$.

So in what way does the wave function contain all possible information about a particle? I have to analyse the same particle or system in at least two different Hilbert spaces depending on which observable I want to consider.

  • $\begingroup$ physics.stackexchange.com/q/105799 $\endgroup$
    – lemon
    Feb 4 '15 at 7:59
  • $\begingroup$ Sorry but that's a completely different question. I'm not asking whether two different physical systems can share the same Hilbert Space, I'm asking why we say the wave function fully represents a particle when we use different Hilbert Spaces in analysing different properties $\endgroup$
    – Lammey
    Feb 4 '15 at 8:03
  • $\begingroup$ given two hilbert spaces, you can construct another one either by direct sum or direct product. In this case you should use the second, so the total hilbert space of a spinning particle would be the tensor product of the $L^2$ and $\mathbb{C}$ (spin degrees of freedom). $\endgroup$
    – yuggib
    Feb 4 '15 at 8:27
  • $\begingroup$ spin space is $\mathbb{C}^2$ sorry (for 1/2 spin particles) $\endgroup$
    – yuggib
    Feb 4 '15 at 8:59
  • $\begingroup$ Oh ok, thanks. Just to make sure, say I have a spin-1/2 particle, and I construct the Hilbert space $L^2 \otimes C^2$. Does that exhaust all the properties of the particle? Is there anything other than spin I have to account for? Like the final Hilbert Space might be $L^2 \otimes C^2 \otimes H \otimes H' \otimes \dots$ $\endgroup$
    – Lammey
    Feb 4 '15 at 10:22

The wave function only contains all the information about the system im so far as you consider it. Meaning each qualitatively different physical system needs its a modified Hilbert space to fit what can happen with the system.

In case you have something like spin on its own in $ H_{Spin}$ and you want to look at a freely moving particle in $H_{free}$ that has spin aswell. You can combine the Hilbert spaces with the tensor product to look at the combined properties . The new space $ H_{free} \otimes H_{Spin}$ then contains all the possible states for a moving spin carrying particle.

Each possible (measurable) state coresponds to an eigenvector of an operator which are used as base vectors. So the pure non moving spin particle Hilbert space (for spin 1/2) has two dimensions. While the freely moving particle has infinite possible states, that is why people say it needs an infinite dimensional vector space.

You asked in a comment if there could be more to the space, anything more you have to build into the space. And of course you could. But it depends what you want to look at.

Compare it to this: A particle of Newtonian physics will never show signs of the curvature of space-time unless you build the possibility for that into the theory. It is the same here, just that we have to taylor the theory a bit closer to each case because absolutely generall considerations would be too difficult.

I recommend Sharkar: "Principles of Quantum Mechanics" on Spin for the Spin particle example and especially "1.10. Generalization to Infinite Dimensions" on the topic on infinite dimensional function space, the beste explanation on an intuitive level i have yet seen.

And "Consistent Quantum Theory" Chapter 6 by Giffith for a qualitative understanding of tensor produts: http://quantum.phys.cmu.edu/CQT/


The general question is quite hard to tackle I think, because a rigorous motivation of Hilbert space would end up in the theory of operator algebras (see e.g. this answer) and the OP is probably not interested in these aspects at the moment.

As for the example of spin, the Hilbert space in this case is still an $L^2$ space, but the functions are no longer taking values in $\mathbb C$, as for the case of scalar particles, but in the representation space $\mathbb C^2$, so that the Hilbert space is $L^2(\mathbb R^3,\mathbb C^2)$. If $V$ is any other representation space for an irreducible representation of $SU(2)$, then the Hilbert space that describes particle transforming under this irreducible representation of the group would be elements in $L^2(\mathbb R^3, V)$ (e.g. $V=\mathbb R^3$ for the matrices $SO(3)$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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