I (think) I've found the Heisenberg Lie algebra representation through quantization. Where we have $q \mapsto q$ and $p \mapsto -i \hbar \frac{\partial}{\partial q}$.

So this is only a Lie algebra representation. $\rho_*: \mathfrak{h} \to \text{End}(V) $.

Where $V$ is the hilbert space (representation space). And $\mathfrak{h}$ is the Heisenberg Lie algebra.

How do I lift this to a lie group representation $\rho: H \to \text{Gl}(V)$ . Where $H$ is the Heisenberg group? And how do I show it's unitary?


You are representing the three algebra elements $q, p, 1\!\!1$ by hermitian operators acting on functions of q. For simplicity, non-dimensionalize $\hbar=1$--only the clueless keep it unscaled.

Multiplying linear combinations of such operators by i and exponentiating yields the generic unitary group element representation for you , $$ e^{ic + ibq + a\partial_q} ~ f(q) = e^{ic + iab/2} e^{ibq} e^{a\partial_q} ~f(q)= e^{ic + iab/2 +ibq}~ f(q+a), $$ for real coefficients a, b, c, the standard form used routinely in applications.

The first equality is a bland application of the Zassenhaus formula ("inverse Campbell-Baker-Haussdorf"), and the second equality is using up the Lagrange shift operator involved.

  • $\begingroup$ is f(q) the element of the hilbert space? Where the action of exponential is simply multiplying? $\endgroup$ – AkatsukiMaliki Jun 11 at 1:58
  • $\begingroup$ And what is the degree of the representation? $\endgroup$ – AkatsukiMaliki Jun 11 at 2:00
  • $\begingroup$ Yes your algebra rep acts on functions of q. Yes, surviving exponential is mere multiplication. Infinite degree. $\endgroup$ – Cosmas Zachos Jun 11 at 2:48
  • $\begingroup$ thank you! What about the irreducibility. How would you prove it? $\endgroup$ – AkatsukiMaliki Jun 11 at 3:59
  • $\begingroup$ Oohhhh.. it is a long story... real long... another question... $\endgroup$ – Cosmas Zachos Jun 11 at 14:15

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