In QFT we classify field operators according to how they transform under a given symmetry, i.e. in their being a basis for some representation of the symmetry group of the Hamiltonian/Lagrangian. This seems to hold for every degree of freedom (e.g. spin, angular momentum, 4-momentum).

In non-relativistic quantum mechanics we talk about "Schrödinger position representation" or "Schrödinger momentum representation", which tell us how the position and momentum operators act on the states. Are these representations in the group theoretical sense? Or does the word "representation" here mean something else?

  • $\begingroup$ It's a stupid use of the word representation, it only means that in the one, you pick momentum eigenstates as the basis of your Hilbert space, in the other, you pick position eigenstates. $\endgroup$ – ACuriousMind Nov 5 '14 at 14:00
  • $\begingroup$ @ACuriousMind Well, in some sense the "position" and "momentum" representations can be seen as unitarily equivalent (infinite dimensional) representations of the Heisenberg group...(I mean in the "position repr." the position is the multiplication operator and the momentum the derivation operator that realize the algebra of the Heisenberg group, in the "momentum repr." the opposite: the position is the derivation operator and the momentum the multiplication one). However I would not bet that the term "representation" in this context is used keeping this in mind ;-) $\endgroup$ – yuggib Nov 5 '14 at 14:25
  • $\begingroup$ @yuggib That actually seems like what I was asking for! Could you elaborate on that topic? $\endgroup$ – glS Nov 5 '14 at 19:16
  • $\begingroup$ I will write an answer elaborating it as soon as I can... $\endgroup$ – yuggib Nov 5 '14 at 20:16
  • $\begingroup$ related: physics.stackexchange.com/q/68981/58382 $\endgroup$ – glS Dec 31 '14 at 15:32

In quantum mechanics, the canonical commutation relations $[q,p]=i\mathbb{1}$, $[q,\mathbb{1}]=[p.\mathbb{1}]=0$ (I am taking $\hbar=1$) between position ( $q$ ), momentum ( $p$ ) and identity ($\mathbb{1}$) operators form an algebra, that is the Lie algebra of the Heisenberg group.

There are problems of domains because the operators are unbounded, so usually the commutation relations are rigorously expressed in the exponentiated form, the so-called Weyl relations (the blue boxed relation in the link).

Anyways, the Stone-von Neumann Theorem asserts, roughly speaking, that all the unitary irreducible representations of the Weyl relations (i.e. of the Heisenberg group) are all unitarily equivalent, and equivalent to the representation where one object of the algebra is the operator in $L^2$ that acts as multiplication by the variable, another is the derivation with respect to the same variable multiplied by $-i$ (i.e. $-i\partial_{(\cdot)}$) and the last is the identity operator.

The usual "Schrödinger position representation" is the identification of the Hilbert space of QM with $L^2$, where the position operator is the multiplication by the coordinate $x$. It follows that the momentum is the derivative $-i\partial_x$ (because they have to satisfy the canonical commutation relations), and that they form a representation of the algebra of the Heisenberg group.

The momentum representation instead is the representation of the Heisenberg group on $L^2$ where the momentum operator is the multiplication by the coordinate $k$ and the position operator is the derivation $-i\partial_k$.

As requested by the Stone-von Neumann theorem, the two representations are unitarily equivalent, and the unitary transformation that links the two is the Fourier transform (that is unitary on $L^2$).


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