12.2 Momentum space representation

12.2 Momentum space representation

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We began our discussion of the state space $H$ of a free particle by taking states to be wavefunctions $\psi (q)$ defined on position space, thought $\mathrm{of}$ variously as being in $\mathcal{S}(\mathbf{R}),{\mathcal{L}}^2(\mathbf{R})$ or $S^{\prime }(\mathbf{R})$ . Using the Fourier transform, which takes such functions to their Fourier transforms

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

in the same sort of function space, we saw in section 11.5 that the state space $H$ can instead be taken to be a space of functions $\tilde{\psi }(k)$ on momentum space. We will call such a choice of $\mathcal{H}.$ , with the operator $P$ now acting as

$$P\tilde{\psi }(k)=k\tilde{\psi }(k)$$

the momentum space representation, as opposed to the previous position space representation. By the Plancherel theorem (11.2) these are unitarily equivalent representations of the group $R$, which acts in the position space case by translation by a in the position variable, in the momentum space case by multiplication by a phase factor $e^{ika}$

In the momentum space representation, the eigenfunctions of $P$ are the distributions $\delta (k-k^{\prime })$ , with eigenvalue $k^{\prime }.$ , and the expansion of a state in terms of eigenvectors is

$$\tilde{\psi }(k)={\int }_{-\infty }^{+\infty }\delta (k-k^{\prime })\tilde{\psi }(k^{\prime })d{k}^{\prime }\text{\hspace{1em}(12.2)}$$

The position operator is

$$Q=i\frac{d}{dk}$$

which has eigenfunctions

$$\frac{1}{\sqrt{2\pi }}{e}^{-ik{q}^{\prime }}$$

and the expansion of a state in terms of eigenvectors of $Q$ is just the Fourier transform formula 11.3.

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