13.3 The Schr¨odinger representation

13.3 The Schr¨odinger representation

Concept links · terms present in this machine draft; source roles are unverified: Lie algebra · Lie algebra representation · group representation · irreducible representation · Hilbert space · Lie group

Since it can be defined in terms of 3 by 3 matrices, the Heisenberg group $H_3$ has an obvious representation on ${\mathbf{C}}^3$ , but this representation is not unitary and not of physical interest. What is of great interest is the infinite dimensional representation on functions of $q$ for which the Lie algebra version is given by the $Q,P,$ , and unit operators:

Definition (Schr¨odinger representation, Lie algebra version)

The Schr¨odinger representation of the Heisenberg Lie algebra ${\mathfrak{h}}_3$ is the representation $({\Gamma }_S^{\prime },L^2(\mathbf{R}))$ satisfying

$$\begin{array}{rl}{\Gamma }_S^{\prime }(X)\psi (q)& =-iQ\psi (q)=-iq\psi (q),{\Gamma }_S^{\prime }(Y)\psi (q)=-iP\psi (q)=-\frac{d}{dq}\psi (q)\\ {\Gamma }_S^{\prime }(Z)\psi (q)& =-i\psi (q)\end{array}$$

Factors of i have been chosen to make these operators skew-adjoint and the representation thus unitary. They can be exponentiated, giving in the exponential coordinates on $H_3$ of equation 13.2

$$\begin{array}{rl}{\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x}{0}\right),0\right)\right)\psi (q)& =e^{-xiQ}\psi (q)=e^{-ixq}\psi (q)\\ {\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{0}{y}\right),0\right)\right)\psi (q)& =e^{-yiP}\psi (q)=e^{-y\frac{d}{dq}}\psi (q)=\psi (q-y)\\ {\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{0}{0}\right),z\right)\right)\psi (q)& =e^{-iz}\psi (q)\end{array}$$

For general group elements of $H_3$ one has:

Definition (Schr¨odinger representation, Lie group version)

The Schr¨odinger representation of the Heisenberg Lie group $H_3$ is the representation $({\Gamma }_S,L^2(\mathbf{R}))$ 1 satisfying

$${\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\right)\psi (q)=e^{-iz}{e}^{i\frac{xy}{2}}{e}^{-ixq}\psi (q-y)\text{\hspace{1em}(13.3)}$$

To check that this defines a representation, one computes

$$\begin{array}{l}{\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\right){\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x^{\prime }}{y^{\prime }}\right),z^{\prime }\right)\right)\psi (q)\\ ={\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\right)e^{-i{z}^{\prime }}{e}^{i\frac{x^{\prime }{y}^{\prime }}{2}}{e}^{-i{x}^{\prime }q}\psi (q-y^{\prime })\\ =e^{-i(z+z^{\prime })}{e}^{i\frac{xy+x^{\prime }{y}^{\prime }}{2}}{e}^{-ixq}{e}^{-i{x}^{\prime }(q-y)}\psi (q-y-y^{\prime })\\ =e^{-i(z+z^{\prime }+\frac{1}{2}(x{y}^{\prime }-y{x}^{\prime }))}{e}^{i\frac{(x+x^{\prime })(y+y^{\prime })}{2}}{e}^{-i(x+x^{\prime })q}\psi (q-(y+y^{\prime }))\\ ={\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{x+x^{\prime }}{y+y^{\prime }}\right),z+z^{\prime }+\frac{1}{2}(x{y}^{\prime }-y{x}^{\prime })\right)\right)\psi (q)\end{array}$$

The group analog of the Heisenberg commutation relations (often called the “Weyl form” of the commutation relations) is the relation

$$e^{-ixQ}{e}^{-iyP}=e^{-ixy}{e}^{-iyP}{e}^{-ixQ}$$

This can be derived by using the explicit representation operators in equation 13.3 (or the Baker-Campbell-Hausdorf formula and the Heisenberg commutation relations) to compute

$$e^{-ixQ}{e}^{-iyP}=e^{-i(xQ+yP)+\frac{1}{2}[-ixQ,-iyP]}=e^{-i\frac{xy}{2}}{e}^{-i(xQ+yP)}$$

as well as the same product in the opposite order, and then comparing the results.

Note that, for the Schr¨odinger representation, we have

$${\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{0}{0}\right),z+2\pi \right)\right)={\Gamma }_S\left(\left(\left(\genfrac{}{}{0ex}{}{0}{0}\right),z\right)\right)$$

so the representation operators are periodic with period 2$\pi$ in the z-coordinate. Some authors choose to define the Heisenberg group $H_3$ as not ${\mathbf{R}}^2\oplus \mathbf{R}$ , but ${\mathbf{R}}^2\times S^1$ , building this periodicity automatically into the definition of the group, rather than the representation.

We have seen that the Fourier transform $\mathcal{F}$ takes the Schr¨odinger representation to a unitarily equivalent representation of $H_3$ , in terms of functions of $p$ (the momentum space representation). The equivalence is given by a change

$${\Gamma }_S(g)\to {\tilde{\Gamma }}_S(g)=\mathcal{F}{\Gamma }_S(g)\tilde{\mathcal{F}}$$

in the representation operators, with the Plancherel theorem (equation 11.5 ensuring that $\mathcal{F}$ and $\tilde{\mathcal{F}}={\mathcal{F}}^{-1}$ are unitary operators.

In typical physics quantum mechanics textbooks, one often sees calculations made just using the Heisenberg commutation relations, without picking a specific representation of the operators that satisfy these relations. This turns out to be justified by the remarkable fact that, for the Heisenberg group, once one picks the constant with which $Z$ acts, all irreducible representations are unitarily equivalent. By unitarity this constant is $-ic,c\in \mathbf{R}$ . We have chosen $c=1$ 2 but other values of c would correspond to diferent choices of units.

In a sense, the representation theory of the Heisenberg group is very simple: there’s only one irreducible representation. This is very diferent from the theory for even the simplest compact Lie groups $(U(1)$ and $SU(2))$ ) which have an infinity of inequivalent irreducibles labeled by weight or by spin. Representations of a Heisenberg group will appear in diferent guises (we’ve seen two, will see another in the discussion of the harmonic oscillator, and there are yet others that appear in the theory of theta-functions), but they are all unitarily equivalent, a statement known as the Stone-von Neumann theorem. Some good references for this material are [91], and [41]. In depth discussions devoted to the mathematics of the Heisenberg group and its representations can be found in [51], [26] and [95].

In these references can be found a proof of the (not dificult)

Theorem

The Schr¨odinger representation ${\Gamma }_S$ described above is irreducible.

and the much more dificult

Theorem (Stone-von Neumann)

Any irreducible representation $\pi$ of the group $H_3$ on a Hilbert space, satisfying

$${\pi }^{\prime }(Z)=-i1$$

is unitarily equivalent to the Schr¨odinger representation $({\Gamma }_S,L^2(\mathbf{R}))$ .

Note that all of this can easily be generalized to the case of d spatial dimensions, for d finite, with the Heisenberg group now $H_{2d+1}$ and the Stone-von Neumann theorem still true. In the case of an infinite number of degrees of freedom, which is the case of interest in quantum field theory, the Stone-von Neumann theorem no longer holds and one has an infinity of inequivalent irreducible representations, leading to quite diferent phenomena. For more on this topic see chapter 39.

It is also important to note that the Stone-von Neumann theorem is formulated for Heisenberg group representations, not for Heisenberg Lie algebra representations. For infinite dimensional representations in cases like this, there are representations of the Lie algebra that are “non-integrable”: they aren’t the derivatives of Lie group representations. For such non-integrable representations of the Heisenberg Lie algebra (i.e., operators satisfying the Heisenberg commutation relations) there are counter-examples to the analog of the Stone von-Neumann theorem. It is only for integrable representations that the theorem holds and one has a unique sort of irreducible representation.

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