B.8 Chapter 17

B.8 Chapter 17

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra

Problem 1:

This is part of a proof of the Groenewold-van Hove theorem.

• Show that one can write $q^2{p}^2$ in two ways as a Poisson bracket

$$q^2{p}^2=\frac{1}{3}\{q^{2}p,p^{2}q\}=\frac{1}{9}\{q^3,p^3\}$$

• Assume that we can quantize any polynomial in $q,p$ by a Lie algebra homomorphism ${\pi }^{\prime }$ that takes polynomials in $q,p$ with the Poisson bracket to polynomials in $Q,P$ with the commutator, in a way that extends the standard Schr¨odinger representation

$${\pi }^{\prime }(q)=-iQ,{\pi }^{\prime }(p)=-iP,{\pi }^{\prime }(1)=-i1$$

Further assume that the following relations are satisfied for low degree polynomials (it actually is possible to prove that these are necessary):

$${\pi }^{\prime }(qp)=-i\frac{1}{2}(QP+PQ),{\pi }^{\prime }(q^2)=-i{Q}^2,{\pi }^{\prime }(p^2)=-i{P}^2$$ $${\pi }^{\prime }(q^3)=-i{Q}^3,{\pi }^{\prime }(p^3)=-i{P}^3$$

Then show that

$${\pi }^{\prime }(q^{2}p)=-i\frac{1}{2}(Q^{2}P+P{Q}^2)$$

(Hint: use $\{q^3,p^2\}=6q^{2}p)$

• Also show that

$${\pi }^{\prime }(q{p}^2)=-i\frac{1}{2}(Q{P}^2+P^{2}Q)$$

• Finally, show that

$${\pi }^{\prime }\left(\frac{1}{9}\{q^3,p^3\}\right)=-i\left(-\frac{2}{3}1-2iQP+Q^2{P}^2\right)$$

and

$${\pi }^{\prime }\left(\frac{1}{3}\{q^{2}p,p^{2}q\}\right)=-i\left(-\frac{1}{3}1-2iQP+Q^2{P}^2\right)$$

which demonstrates a contradiction.

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