rafitiki said:
Hello everyone, I'm having moonlight questions.
I used to think that supposing that I had a finite dimensional representation of the canonical commutation relations and then taking the trace was enough to find a contradiction but actually, if we try to represent a Lie algebra with three elements that satisfy [P,X]=ih1, why would 1 have to be represented by the identity matrix ? It seems to me that we can only assume that it is a central element of the image of the representation. In this case, if the image of 1 by the representation is traceless, then there is no contradiction coming from taking the trace in this equation. Nevertheless, in a finite dimensional setup, its representation has to be a subalgebra of gl_r for a certain r and there it looks like asking [P,X]=ih1 AND [1,X]=[1,P]=0 is impossible. I can't find a contradiction though. Any help is welcome.
Goodnight
This is a mess, my friend. The canonical commutation relation of QM has
no finite-dimensional representation. I think you are mixing between the finite-dimensional (
non-hermitian) representations of the Heisenberg Lie algebra, which are
useless in physics, with the irreducible
infinite-dimensional hermitian representation of the algebra on the space L^{2}(\mathbb{R}) of square-integrable functions (the so-called Schrödinger representation).
The (3-dimensional) Heisenberg Lie algebra \mathfrak{h}_{3} is the vector space \mathbb{R} \oplus \mathbb{R} \oplus \mathbb{R} with the Lie bracket
[ X , Y ] = Z , \ \ \ \ [ X , Z ] = [ Y , Z ] = 0 .
Note that all Lie brackets of Z with anything else are zero. All Lie brackets of Lie brackets are also zero. So, the Heisenberg algebra is
minimally non-trivial and
nilpotent Lie algebra.
In a local (real) coordinates (x,y,z), a general element A of \mathfrak{h}_{3} can be expanded in the basis (X,Y,Z) as
A = xX + yY + zZ and the Lie bracket becomes
\left[ (x,y,z) , (\bar{x}, \bar{y},\bar{z}) \right] = ( 0 , 0 , x\bar{y} - y\bar{x}) .
Like any other Lie algebra, \mathfrak{h}_{3} has faithful matrix representation. In fact it is
isomorphic to the Lie algebra of 3 \times 3 traceless upper triangular real matrices with Lie bracket realized by matrix commutator:
<br />
(x,y,z) \leftrightarrow \begin{pmatrix} 0 & x & z \\<br />
0 & 0 & y \\<br />
0 & 0 & 0<br />
\end{pmatrix}<br />
Under mild assumptions, \mathfrak{h}_{3} can be exponentiated (using the BCH formula) to the Heisenberg Lie group H_{3}, the space \mathbb{R}^{2} \oplus \mathbb{R} with multiplication
<br />
\left( \begin{pmatrix}x \\ y \end{pmatrix}, z \right) \left( \begin{pmatrix}\bar{x} \\ \bar{y} \end{pmatrix} , \bar{z} \right) = \left( \begin{pmatrix} x + \bar{x} \\ y + \bar{y} \end{pmatrix}, z + \bar{z} + \frac{1}{2} ( x \bar{y} - y \bar{x}) \right) .<br />
Note that H_{3} has one-dimensional centre, spanned by (0,0,1): Each one of the Lie algebra basis (X,Y,Z) generates a subgroup of H_{3} isomorphic to \mathbb{R}; the elements of the first two subgroups generate the full group, while the elements of the third one
commute with all group elements, i.e.
central.
Again, H_{3} is isomorphic to the Lie group of upper triangular 3 by 3 real matrices with 1 on the diagonal:
<br />
\left( \begin{pmatrix}x \\ y \end{pmatrix}, z \right) \leftrightarrow \begin{pmatrix} 1 & x & z + \frac{1}{2}xy \\<br />
0 & 1 & y \\<br />
0 & 0 & 1<br />
\end{pmatrix}<br />
This isomorphism means that H_{3} has natural
finite-dimensional representation on \mathbb{C}^{3}. However, this representation is of
no use in physics, because it is
not unitary. Indeed, almost all of the structure of QM is determined by the infinite-dimensional
unitary representation of the Heisenberg group H_{3}, the Schrödinger representation \Phi_{S} on the space of square-integrable functions L^{2}(\mathbb{R}). The Schrödinger representation (\Phi_{S}, L^{2}(\mathbb{R})) of H_{3} is obtained by exponentiating the following hermitian representation of \mathfrak{h}_{3} on L^{2}(\mathbb{R})
d_{e}\Phi_{S}(X) = - i Q = -i q , \ \ d_{e}\Phi_{S}(Y) = - i P = - \frac{d}{dq}, d_{e}\Phi_{S}(Z) = - i \ \mbox{I}_{L^{2}(\mathbb{R})} = - i . Namely
d_{e}\Phi_{S}\left( (x,y,z) \right) = x Q + y P - i z \ \mbox{I}_{L^{2}(\mathbb{R})}.
For general group elements (x,y,z) \in H_{3} and all \Psi(q) \in L^{2}(\mathbb{R}), the Schrödinger representation (\Phi_{S}, L^{2}(\mathbb{R})) of H_{3} is given by
\Phi_{S}\left( (x,y,z) \right) ( \Psi ) (q) = e^{- i (xq + z - \frac{1}{2}xy)} \Psi (q - y) .
One can easily show that
1) \Phi_{S} is a group homomorphism, i.e. representation.
2) the Schrödinger representation (\Phi_{S}, L^{2}(\mathbb{R})) of H_{3} is irreducible.
And not so easy is the proof of the following famous
Stone-von Neumann theorem:
3) Every unitary representation \pi of H_{3} on a Hilbert space \mathcal{H} on which the central element acts as e^{-i}\ \mbox{I}_{\mathcal{H}}, is isomorphic to a direct sum of copies of the Schrödinger representation \Phi_{S} on L^{2}(\mathbb{R})
( \pi , \mathcal{H}) = \bigoplus ( \Phi_{S}, L^{2}(\mathbb{R})) .
Namely,
any irreducible representation \pi of the Heisenberg group H on a Hilbert space \mathcal{H}, satisfying d_{e}\pi (Z) = - i \ \mbox{I}_{\mathcal{H}} is
unitarily equivalent to the infinite-dimensional representation \Phi_{S} on L^{2}(\mathbb{R}) of Schrödinger.
The so-called Weyl quantization procedure is obtained by extending \Phi_{S} to the
associative group algebra of H_{3}. Precisely speaking, for a phase-space function f(q,p), Weyl takes the associated operator to be
\Phi_{S}(f) = \int d\alpha \ d\beta \ \hat{f}(\alpha , \beta) \ \Phi_{S}\left( e^{i (\alpha Q + \beta P)} \right) ,
where
\hat{f}(\alpha , \beta) = (\frac{1}{2 \pi})^{2} \int dq \ dp \ f(q,p) \ e^{- i (\alpha q + \beta p)} , is the Fourier transform of f. The operator \Phi_{S}(f) is, at least formally, self-adjoint if f is real-valued, and is well-defined and of trace class if f \in \mathcal{S}(\mathbb{R}^{2}), i.e., if f is C^{\infty} and decreases, together with all its derivatives, faster than the reciprocal of any polynomial at infinity. If f is C^{\infty} and grows at most polynomially at infinity, then \Phi_{S}(f) will in general be a densely defined unbounded operator. And I should just stop here.
