# How is the set of displacement operators best called?

Displacement operators $\hat D(x,p), \ \ x,p\in\mathbb{R},$ follow a composition rule $$D(x,p) D(x',p') = \exp\frac{i(px'-xp')}2 D(x+x',p+p').$$ Because of the extraneous phase factor, the set of all displacements is not a group. I could take a set of all displacements multiplied by all possible phase factors and I believe that would be isomorphic to $U(1) \rtimes \mathbb{R}^2$, but if there was an algebraic name for the kind of object that the set of displacements alone forms, that would suit my purpose better.

The set of all possible elements of the form $e^{i\alpha}D(x,p)$ with $\alpha, x,p \in \mathbb R$ verifying the commutation relations you wrote in addition to: $$D(x,p)^* = D(-x,-p)\:,\quad D(0,0)=I$$ is a group and it is called Heisenberg group, it is homeomorphic (diffeomorphic) to $U(1) \times \mathbb R^2$ but not isomorphic as a Lie group. It is a real three-dimensional simply-connected Lie group with Lie algebra generated by three elements, one commuting with the others which, in turn, verify the standard position momentum relation commutations.

The *-algebra of all possible complex linear combinations of elements $D(x,y)$ with $x,p \in \mathbb R$ verifying the commutation relations you wrote in addition to: $$D(x,p)^* = D(-x,-p)$$ is called Weyl *-algebra (actually there is only one way to equip it with a norm making it a $C^*$algebra.)

The objects $D(x,p)$ are called Weyl generators.

• Oh yes, of course :-) I knew Heisenberg Lie algebra and did not realize that this was exactly the group generated thereby (and thus would obviously be called Heisenberg group). Mar 4, 2014 at 0:33
• I edited your second last sentence so that it sounds more like a statement of the uniquenss theorem you're alluding to. Please check sense carefully (i.e. I think you're saying that the norm is unique). Mar 5, 2014 at 4:03
• Thanks Rod, I was a bit sloppy. Indeed the C* norm is unique as you wrote. Bye, Valter Mar 5, 2014 at 6:22

We can realize the displacement operator as

$$\tag{1}\hat{D}(x,p)~=~e^{x\hat{P}+p\hat{X}},$$

where the elements $\hat{X}$, $\hat{P}$ and ${\bf 1}$ generates the Heisenberg algebra

$$\tag{2} [\hat{X},\hat{P}]=i{\bf 1}.$$

These elements can be realizes as differential operators in the Schrödinger representation. (See also the Stone-von Neumann theorem.) Under operator composition $\circ$, we get a 2-cocycle

$$\tag{3} \hat{D}(x,p) \circ \hat{D}(x^{\prime},p^{\prime})~=~ e^{\frac{i}{2}(x^{\prime}p-p^{\prime}x)}\hat{D}(x+x^{\prime},p+p^{\prime}),$$

due to a simple special case of the BCH formula. The operation composition (3) is associative. The formula for three operators reads

$$\tag{4} \hat{D}(x,p) \circ \hat{D}(x^{\prime},p^{\prime})\circ \hat{D}(x^{\prime\prime},p^{\prime\prime})$$ $$~=~ e^{\frac{i}{2}(x^{\prime}p+x^{\prime\prime}p+x^{\prime\prime}p^{\prime}-p^{\prime}x-p^{\prime\prime}x-p^{\prime\prime}x^{\prime})} \hat{D}(x+x^{\prime}+x^{\prime\prime},p+p^{\prime}+p^{\prime\prime}).$$

The displacement operators are closely related to the Heisenberg group, cf. this paragraph.