I am trying to understand a relation between the two theorems stated in the title. What I observed so far is that since $H^{2}(\mathbb{R},U(1))=\{e\}$, using Bargmann's theorem, we have that projective unitary representations of the additive group $(\mathbb{R},+)$ are in one-to-one correspondence with unitary representations of the additive group $(\mathbb{R},+)$, which will lead to the Schrödinger equation via Stone's theorem.

On the other hand, we have that $H^2(\mathbb{R}^{2n},U(1))=\mathbb{R}$, so the unitary projective representations of $\mathbb{R}^{2n}$ are not in one-to-one correspondence with its unitary representations, rather in one-to-one correspondence with the unitary representations of the Heisenberg group, a central extension of $\mathbb{R}^{2n}$. This is what we get from Bargmann's theorem.

Finally, the Stone-Von-Neumann theorem says that all the irreducible unitary representations of the Heisenberg group are unitarily equivalent to the Schrödinger representation. I stated these theorems without all their technical details, but my question is:

Can the two theorems be composed to obtain the following theorem, which, let's say, we call (Stone-Von-Neumann-Bargmann):

There exists a unique (up to unitary equivalence) strongly continuous projective representation of $\mathbb{R}^{2n}$ for all $n \in \mathbb{N}$ (including zero) on a separable Hilbert space.


1 Answer 1


Consider the strongly-continuous projective unitary representations of $\mathbb{R}^2$ on $L^2(\mathbb{R}, dx)$ given by $$\mathbb{R}^2\ni (a,b)\mapsto U(a,b) := e^{i(\overline{aX - i b \frac{d}{dx}})}$$ and $$\mathbb{R}^2\ni (a,b)\mapsto V(a,b) := e^{i(\overline{aX - i 2b \frac{d}{dx}})}$$ where the bar denotes the closure of the (essentially selfadjoint) operators initially defined on the Schwartz' space. It holds (please check my computations, I am not sure on coefficients of the phases) $$U(a,b)U(a',b')= e^{-i(ab'-ba')} U(a+a',b+b')\:,$$ whereas $$V(a,b)V(a',b')= e^{-i2(ab'-ba')} V(a+a',b+b')\:.\tag{1}$$ En passant, we observe that the projective nature of the representations is evident above in view of the cocycles (the exponential functions) as factors in the right-hand sides.

If a unitary operator existed $K: L^2(\mathbb{R}, dx)\to L^2(\mathbb{R}, dx)$ such that $$KU(a,b)K^{-1}= V(a,b)\tag{2}$$ (if I correctly interpret your notion of unitary equivalence) we would obtain a contradiction. In fact, we have $$KU(a,b)U(a',b')K^{-1}= e^{-i(ab'-ba')} KU(a+a',b+b')K^{-1}\:,$$ that is $$KU(a,b)K^{-1}KU(a',b')K^{-1}= e^{-i(ab'-ba')} KU(a+a',b+b')K^{-1}\:.$$ From (2), $$V(a,b)V(a',b') = e^{-2i(ab'-ba')} V(a+a',b+b')\:.$$ Comparing with (1) and using the fact that all operators are bijective as they are unitary, we would obtain $$e^{-i(ab'-ba')} =e^{-2i(ab'-ba')} \quad \forall a,a',b,b' \in \mathbb{R}$$ that is evidently false.

Another possibility, in the spirit of projective representations, is to make weaker condition (2) into $$KU(a,b)K^{-1}= \chi(a,b) V(a,b)\tag{3}$$ where $\chi(a,b)$ is a unit complex number. That is equivalent to saying that the two representations are in the same class of projective representations. With some computation the condition above leads to $$e^{-i(ab'-ba')} =e^{-2i(ab'-ba')} \chi(a,b)^{-1} \chi(a',b')^{-1} \chi(a+a',b+b')\:.$$ That is $$e^{i(ab'-ba')} = \chi(a,b)^{-1} \chi(a',b')^{-1} \chi(a+a',b+b')\quad \forall a,a',b,b' \in \mathbb{R}$$ The right-hand side is invariant under swapping $(a,b) \leftrightarrow (a',b')$, whereas the left-hand side changes the sign of the exponent under that transformation. As a consequence $$e^{i2(ab'-ba')} = 1 \quad \forall a,a',b,b' \in \mathbb{R}$$ that is impossible as well.

In summary, the answer is NO: there are infinitely many inequivalent (according to (2) or (3)) projective unitary representations of $\mathbb{R}^2$.

The result easily extends to $\mathbb{R}^{2n}$.

Physically speaking, the root of the problem is here the value of the Planck constant, the central charge which fixes the coefficient in front of the exponent $i(ab'-a'b)$. It is an unfixed parameter when dealing with strongly-continuous projective unitary representations of $\mathbb{R}^{2n}$. However it must be fixed as soon as one sees these representations as properly unitary representations of the Heisenberg group.

Actually $\hbar$ also takes place in the Heisenberg group, but usually is taken to be $1$.

As a final comment, I stress that separability is not necessary in the Stone-von Neumann theorem as it is a consequence of a general result: an irreducible strongly continous unitary representation of a (finite-dimensional) Lie group is ncessarily defined in a separable Hilbert space. (see e.g. Proposition 7.36 of this book of mine.)

  • $\begingroup$ Thank You Valter, you answered my question. As an aside: is there something we can say about strongly continuous unitary projective representations of $R^{2n}$? Using Bargmann's theorem, they are in one-to-one correspondence to strongly continuous unitary representations of which group? $\endgroup$
    – ProphetX
    Sep 18, 2023 at 21:41
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    $\begingroup$ I do not know, or better, I am not sure. I have to think about it. I am not completely sure about my previous version of this comment. $\endgroup$ Sep 18, 2023 at 21:47
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    $\begingroup$ Yes, I think that my previous comment, now cancelled, was correct. The group you are looking for is the trivial generalisation of the Heisenberg group in 2n+1 dimensions. $\endgroup$ Sep 18, 2023 at 21:56
  • $\begingroup$ Yes, I think that is it. Thanks! $\endgroup$
    – ProphetX
    Sep 18, 2023 at 22:01

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