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.


2 Answers 2


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.

  • $\begingroup$ 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). $\endgroup$
    – The Vee
    Mar 4, 2014 at 0:33
  • $\begingroup$ 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). $\endgroup$ Mar 5, 2014 at 4:03
  • $\begingroup$ Thanks Rod, I was a bit sloppy. Indeed the C* norm is unique as you wrote. Bye, Valter $\endgroup$ Mar 5, 2014 at 6:22

We can realize the displacement operator as


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.


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