# Lie group of Schrodinger Wave equation

In Ballentine's book on quantum mechanics (in 3rd chapter), he introduces the symmetry transformation of Galilean group associated with Schrodinger equation.

Now the Galilean group as such has 10 generators (3 rotations - $L_i$ , 3 translations - $P_i$, 3 boosts - $G_i$ and time translation - $H$). Apart from this the Schrodinger solution (the probability amplitude) is arbitrary upto to a phase factor ($e^{i\phi}$). Hence we include one more generator induced by the transformation of phase. With this the general Unitary transformation,

$$U = \sum_{i=1}^3\Big(\delta\theta_iL_i + \delta x_iP_i + \delta\lambda_iG_i +dtH\Big) + \delta\phi\mathbb{\hat1} = \sum_{i=1}^{10} \delta s_iK_i + \delta\phi\mathbb{\hat1}$$

The commutation relations of the group altogether can be given as,

$$[K_i,K_j] = i\sum_{n}C_{ij}^{\;\;n}K_n + ib_{ij}\mathbb{\hat1}$$.

Now, this commutation relation does not have the structure of Lie algebra. Since with Lie Algebra the elements are closed under commutation. This one has an extra element sitting with $\delta\phi\mathbb{\hat1}$.

What is really going on here ? Is this really a 11-parameter Lie group ? If so how do we convince about the algebra of generators ?

## 1 Answer

The multiplication by a phase of the wave function commutes with the action of the Galilean group.

It is always possible to add a generator, commuting with a Lie algebra generators, to form a Larger Lie algebra. In this case, the larger Lie algebra is called a central extension of the former. The origin of the name is that the added generator (or generators) commute with all the Lie algebra generators, thus they belong to the Lie algebra's center.

For example, The Heisenberg-Weyl algebra:

$$[x, p] = i \hbar \mathbb{1}$$

is a central extension of the two dimensional translation algebra $\mathbb{R}^2$:

$$[x, p] =0$$

If the central element does not appear in any commutator of the original Lie algebra generators, the central extension is trivial (all the $b_{ij}$s are zero) and the algebra is just a direct sum of the two algebras. This is the case in the specific example of the extension described in Ballentine's book. However, this is not the case in the Heisenberg-Weyl algebra, where the commutator of $x$ and $p$ produces the central element. In this case, the central extension is not trivial.

However, this is not the whole story yet. Ballentine is preparing the backgound for the description of a nontrivial central extension of the Galilean group:

It turns out that the Galilean algebra does not close in neither classical or quantum mechanics without a nontrivial central extension. The Poisson brackets in classical mechanics and the commutator in quantum mechanics of boosts and momenta is not trivial and has the form:

$$[G_i, P_j] = m \delta_{ij}$$

where: $m$ is the particle's mass. Note that the corresponding commutator in the Galilean algebra is vanishing. This result is due to V. Bargmann.

In the classical case, it can be seen quite easily. The Poisson brackets of the Noether charges computed from the free particle Lagrangian corresponding to the boosts and the momenta just satisfy the above relation and they do not Poisson-commute as in the unextended Galilean group algebra.

Finally, let me note that the central element is always represented by a unit matrix in an irreducible representation and a representation of the centrally extended algebra is called a ray representation of the original algebra.

• It would be of great help if you could also resolve why we need the idea of central extension in some cases, and not in some other cases. – user35952 Mar 20 '14 at 16:30
• @user35952 I'll try to answer your new question soon, I'll try to clarify the previous "Poincare group vs Galilean group" answer – David Bar Moshe Mar 20 '14 at 17:04
• @user35952: In some cases, every central extension is trivial. Thiis can be found out by calculating the corresponding cohomology group. – Arnold Neumaier Mar 21 '14 at 15:34