11.2 The group R and the Fourier transform

11.2 The group R and the Fourier transform

Concept links · terms present in this machine draft; source roles are unverified: adjoint operator · unitary representation · irreducible representation · U(1)

In the previous section, we imposed periodic boundary conditions, replacing the group $R$ of translations by the circle group $S^1$ , and then used the fact that unitary representations of this group are labeled by integers. This made the analysis relatively easy, with $\mathcal{H}=L^2(S^1)$ and the self-adjoint operator $\begin{array}{r}P=-i\hslash \frac{\partial }{\partial \varphi }\end{array}$ behaving much the same as in the finite dimensional case: the eigenvectors of $P$ give a countable orthonormal basis of H and $P$ can be thought of as an infinite dimensional matrix.

Unfortunately, in order to understand many aspects of quantum mechanics, one can’t get away with this trick, but needs to work with $R$ itself. One reason for this is that the unitary representations of R are labeled by the same group, $R$, and it will turn out (see the discussion of the Heisenberg group in chapter 13) to be important to be able to exploit this and treat positions and momenta on the same footing. What plays the role then of $|n⟩=e^{in\varphi },n\in \mathbf{Z}$ will be the $|k⟩=e^{ikq},k\in \mathbf{R}$ These are functions on $R$ that are one dimensional irreducible representations under the translation action on functions (as usual using equation 1.3)

$$\pi (a)e^{ikq}=e^{ik(q-a)}=e^{-ika}{e}^{ikq}$$

One can try and mimic the Fourier series decomposition, with the coeficients $c_n$ that depend on the labels of the irreducibles replaced by a function $\tilde{f}(\mathbf{k})$ depending on the label k of the irreducible representation of $\mathbf{R}:$

Definition (Fourier transform)

The Fourier transform of a function $\psi$ is given by a function denoted $\mathcal{F}\psi$ or ${\psi }_{.}$ , where

$$(\mathcal{F}\psi )(k)=\tilde{\psi }(k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{e}^{-ikq}\psi (q)dq\text{\hspace{1em}(11.3)}$$

This integral is not well-defined for all elements of $L^2(\mathbf{R})$ , so one needs to specify a subspace of $L^2(\mathbf{R})$ to study for which it is well-defined, and then extend the definition to $L^2(\mathbf{R})$ by considering limits of sequences. In our case a good choice of such a subspace is the Schwartz space $S(\mathbf{R})$ of functions $\psi$ such that the function and its derivatives fall of faster than any power at infinity. We will not try and give a more precise definition of $S(\mathbf{R})$ here, but a good class of examples of elements of $S(\mathbf{R})$ to keep in mind are products of polynomials and a Gaussian function. The Schwartz space has the useful property that we can apply the momentum operator $P$ an indefinite number of times without leaving the space.

Just as a function on $S^1$ can be recovered from its Fourier series coeficients $c_n$ by taking a sum, given the Fourier transform $\tilde{\psi }(k)$ of $\psi ,\psi$ itself can be recovered by an integral, with the following theorem

Theorem (Fourier Inversion)

For $\psi \in S(\mathbf{R})$ one has $\tilde{\psi }\in S(\mathbf{R})$ and

$$\psi (q)=\tilde{\mathcal{F}}\tilde{\psi }=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{ikq}\tilde{\psi }(k)dk\text{\hspace{1em}(11.4)}$$

Note that $\tilde{\mathcal{F}}$ is the same linear operator as ${\mathcal{F}}_{:}$ with a change in sign of the argument of the function it is applied to. Note also that we are choosing one of various popular ways of normalizing the definition of the Fourier transform.

In others, the factor of $2\pi$ may appear instead in the exponent of the complex exponential, or just in one of $\mathcal{F}$ or $\tilde{\mathcal{F}}$ and not the other.

The operators $\mathcal{F}$ and $\tilde{\mathcal{F}}$ are thus inverses of each other on $S(\mathbf{R})$ . One has

Theorem (Plancherel)

$\mathcal{F}$ and $\tilde{\mathcal{F}}$ extend to unitary isomorphisms of $L^2(\mathbf{R})$ with itself. In particular

$${\int }_{-\infty }^{\infty }|\psi (q){|}^{2}dq={\int }_{-\infty }^{\infty }|\tilde{\psi }(k){|}^{2}dk\text{\hspace{1em}(11.5)}$$

Note that we will be using the same inner product on functions on $\mathbf{R}$

$$⟨{\psi }_1,{\psi }_2⟩={\int }_{-\infty }^{\infty }\overline{{\psi }_1(q)}{\psi }_2(q)dq$$

both for functions of $q$ and their Fourier transforms, functions of $k,$ with our normalizations chosen so that the Fourier transform is a unitary transformation.

An important example is the case of Gaussian functions where

$$\begin{array}{rl}\mathcal{F}{e}^{-\alpha \frac{q^2}{2}}& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}{e}^{-\alpha \frac{q^2}{2}}dq\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{\alpha }{2}((q+i\frac{k}{\alpha }{)}^2-(\frac{ik}{\alpha }{)}^2)}dq\\ & =\frac{1}{\sqrt{2\pi }}{e}^{-\frac{k^2}{2\alpha }}{\int }_{-\infty }^{+\infty }{e}^{-\frac{\alpha }{2}{q}^{\prime 2}}d{q}^{\prime }\\ & =\frac{1}{\sqrt{\alpha }}{e}^{-\frac{k^2}{2\alpha }}\end{array}\text{\hspace{1em}(11.6)}$$

A crucial property of the unitary operator $\mathcal{F}$ on $\mathcal{H}$ is that it diagonalizes the diferentiation operator and thus the momentum operator $P.$ . Under the Fourier transform, constant coeficient diferential operators become just multiplication by a polynomial, giving a powerful technique for solving diferential equations. Computing the Fourier transform of the diferentiation operator using integration by parts, we find

$$\begin{array}{rl}\widetilde{\frac{d\psi }{dq}}& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}\frac{d\psi }{dq}dq\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }\left(\frac{d}{dq}(e^{-ikq}\psi )-\left(\frac{d}{dq}{e}^{-ikq}\right)\psi \right)dq\\ & =ik\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}\psi dq\\ & =ik\tilde{\psi }(k)\end{array}\text{\hspace{1em}(11.7)}$$

$\mathrm{So}.$ under Fourier transform, diferentiation by $q$ becomes multiplication by ik. This is the infinitesimal version of the fact that translation becomes multiplication by a phase under the Fourier transform, which can be seen as follows. If

${\psi }_a(q)=\psi (q+a)$ then

$$\begin{array}{rl}\widetilde{{\psi }_a}(k)& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ikq}\psi (q+a)dq\\ & =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }{e}^{-ik(q^{\prime }-a)}\psi (q^{\prime })d{q}^{\prime }\\ & =e^{ika}\tilde{\psi }(k)\end{array}$$

Since $p=\hslash k$ , one can easily change variables and work with $p$ instead of $k.$ As with the factors of 2$\pi$, there’s a choice of where to put the factors of ℏ in the normalization of the Fourier transform. A common choice preserving symmetry between the formulas for Fourier transform and inverse Fourier transform is

$$\tilde{\psi }(p)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{+\infty }{e}^{-i\frac{pq}{\hslash }}\psi (q)dq$$ $$\psi (q)=\frac{1}{\sqrt{2\pi \hslash }}{\int }_{-\infty }^{+\infty }{e}^{i\frac{pq}{\hslash }}\tilde{\psi }(p)dp$$

We will however mostly continue to set $\hslash =1$ , in which case the distinction between k and $p$ vanishes.

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