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Is it the same as the space of all possible descriptions of a single electron? If not, how do they differ?

Please give the mathematical name or specification of this space or these spaces.

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Yes it is indeed. The Hilbert space of a single electron describes all possible states one electron can be in. It is (if we neglect spin) $L^2(R^3)$. (If we include spin, it becomes spinor valued wave functions instead of complex valued wave functions.)

There is a technicality: the zero vector of the Hilbert space does not describe a possible state, and, two vectors which differ by a scalar, describe the exact same physical state.

Amusingly, this Hilbert space is abstractly isomorphic to the Hilbert space of a deuteron, or a proton, or a meson, or.... actually, almost all Hilbert spaces of a finite system of distinguishable particles are abstractly isomorphic, but this has no practical importance, nor is it a theoretical nuisance, nor does it have much to do with your question.

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Electrons have spin. You want $L^2(\mathbb{R}^3) \otimes \mathbb{C}^2$. –  user1504 Feb 4 '12 at 23:33
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I evasively avoided saying whether they were complex valued functions or spinor valued functions to avoid this complication....but you are quite right, of course. And, either space is abstractly isomorphic to the other, as I mentioned. –  joseph f. johnson Feb 4 '12 at 23:38
    
Well, I think you guys have basically answered my question, but I am confused about this abstract isomorphism thing. Are you saying that L^2(R^3) is abstractly isomorphic to $L^2(\mathbb{R}^3) \otimes \mathbb{C}^2$ ? –  Jim Graber Feb 5 '12 at 4:16
    
Also I am surprised that you say all particles have the same Hilbert space. I was expectting to ask a follow on question about the difference between the Hilbert space of a photon and the Hillbert space of an electron or a massive spin one particle or a graviton or a spin zero particle. –  Jim Graber Feb 5 '12 at 4:21
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The photon is an exception: now I admit I formulated my answer primarily with non-Relativistic Quantum Mechanics in mind. But even so, yes it is an amusing surprise: all separable Hilbert spaces which are infinite dimensional are abstractly isomorphic and I way wanted to forestall your follow-up question by getting that in first. But the abstract isomorphism is unintuitive and has no physical significance. Even worse, $L^2(R)$ is abstractly isomorphic to $L^2(R^3)$ which is, in turn, isomorphic to $L^2(R^6)$ so the space for one particle is isomorphic to the space for two particles... –  joseph f. johnson Feb 5 '12 at 4:46
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I think Joseph f. johnson mixed up something. Not every two Hilbert spaces are isomorphic! For example, take $C^2$ and $C^3$, which are finite dimensional Hilbert spaces but not isomorphic.

What Joseph f. johnson might had in mind, was the following theorem: Let $H$ be a infinite dimensional Hilbert space (with some "nice" properties, eg separable), than one can always find a set $M$ and a measure $d \mu$ such that $H$ is isomorphic to $L^2(M, d \mu)$. So to say $L^2(M, d \mu)$ is the prototype for all Hilbert spaces. Examples: $L^2(R)$ is trivially isomorphic to $L^2(R)$, $L^2(R^3) \otimes L^2(R^3)$ is isomorphic to $L^2(R^6)$. The last one is the Hilbert space for 2 particles.

For the relativistic case one can create one-particle Hilbert spaces for particles with mass $m>0$. The base is than the space-like hyperboloid $k_\mu k^\mu = m, p_0 >0$ and the measure is $d^3 k / (2 k^0)$ in momentum representation.

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Is the two particle Hilbert space the Hilbert space for a hydrogen atom or a single instance of positronium? Or must it be something like two electrons or two protons? –  Jim Graber Feb 6 '12 at 0:26
    
I was expecting $L^2(R^3) \otimes L^2(R^3)$ isomorphic to $L^2(R^9)$, not $L^2(R^6)$ ?? –  Jim Graber Feb 6 '12 at 0:27
    
Some mathematical, technical, (mostly) useless details: Even nonseparable Hilbert spaces are isomorphic to $L^2(M, d\mu)$ for some measure space $M$. In this case, $d\mu$ won't be $\sigma-$finite. Anyway you don't need this to prove that any two infinite dimensional, separable Hilbert spaces are isomorphic: this is simply because they both admit a countable orthonormal basis. Since most infinite-dimensional Hilbert spaces that occur in pratice are separable, they are all isomorphic. It may be seen as another version of Hilbert's paradox. –  Giuseppe Negro Feb 6 '12 at 12:08
    
I know about Hilberts Hotel. So $L^2 (R^m)$ is isomorphic to $L^2(R^n)$ for any n and m? Is there some more distinguishing categorization that captures something more like what we usually think of as dimension? –  Jim Graber Feb 7 '12 at 1:13
    
As to your first query: The two particle Hilbert space we have been writing down here has to be for two distinguishable particles. So, the hydrogen atom, or positronium, but not two electrons. Because the two electrons are not distinguishable, the system can only have states which lie inside a certain subspace of $L^2(R^3) \otimes L^2(R^3)$ instead of the whole space. As to your fourth query. There is no simple categorisation that captures our physical intuition about the dimension of the underlying variable space. –  joseph f. johnson Feb 8 '12 at 15:55
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