10  Dirac vs Von Neumann

Separable Hilbert spaces have countable dimensions. But the positions of particles are quantified with real numbers, whose cardinality is \(\mathfrak{c}\). This is not in itself a problem. To see why, consider the specific case of the infinite potential well:

\[V\left(x\right)=\begin{cases} \begin{array}{c} 0,\\ \infty, \end{array} & \begin{array}{c} 0\le x\le L\\ \text{otherwise} \end{array}.\end{cases}\]

Another way to view this is as a free particle with fixed boundary conditions. Since the potential is zero within the well we have

\[\hat{H}=\frac{\hat{p}^{2}}{2m}\]

and hence the energy eigenstates should also be eigenstates of squared-momentum. Working in the position basis, the time independent Schroedinger equation reads

\[\hat{H}\phi_{n}\left(x\right)=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}\phi_{n}\left(x\right)=E_{n}\phi_{n}\left(x\right)\]

and we find

\[\begin{aligned} E_{n} & =\frac{\pi^{2}\hbar^{2}n^{2}}{2mL^{2}}\\ \phi_{n}\left(x\right) & =\frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right). \end{aligned}\]

Since the eigenenergies are non-degenerate, \(\phi_{n}\left(x\right)\) form a complete orthonormal basis for the Hilbert space. Its dimension is \(\aleph_{0}\) since the eigenstates are labelled by integers. Vitally, we can note that any (sensibly behaved) function of position \(f\left(x\right)\), \(x\in\left[0,L\right]\) can be decomposed into this basis:

\[\begin{aligned} f\left(x\right) & =\sum^{\aleph_{0}}_{n=1}f_{n}\phi_{n}\left(x\right) \end{aligned} \tag{10.1}\]

where

\[f_{n}\triangleq\int^{L}_{0}\phi^{*}_{n}\left(x\right)f\left(x\right)\text{d}x. \tag{10.2}\]

So we can use a countable basis to describe functions of position, even though positions themselves have the cardinality of the continuum. This is not a problem or paradox. Dirac and Von Neumann agree on it, and indeed Von Neumann uses these results as the motivation for his own formalism.

The point of disagreement comes when Dirac introduces his famous notation. He says we can work in a basis-independent way by saying

\[\hat{H}|\phi_{n}\rangle=E_{n}|\phi_{n}\rangle\]

and defining

\[\phi_{n}\left(x\right)\triangleq\langle x|\phi_{n}\rangle.\]

This is appealing, as we can rewrite Equation 10.1 in a basis-independent way (i.e. not in the position basis, or momentum basis, or any other basis) by similarly defining

\[\langle x|f\rangle=f\left(x\right)\]

to give

\[\begin{aligned} |f\rangle & =\sum^{\aleph_{0}}_{n=1}f_{n}|\phi_{n}\rangle \end{aligned}\]

where Equation 10.2 becomes

\[f_{n}=\langle\phi_{n}|f\rangle. \tag{10.3}\]

Combining these equations we find

\[|f\rangle=\sum^{\aleph_{0}}_{n=1}f_{n}|\phi_{n}\rangle=\sum^{\aleph_{0}}_{n=1}\langle\phi_{n}|f\rangle|\phi_{n}\rangle=\sum^{\aleph_{0}}_{n=1}|\phi_{n}\rangle\langle\phi_{n}|f\rangle\]

and since this holds for all \(|f\rangle\) it follows that

\[\mathbb{I}=\sum^{\aleph_{0}}_{n=1}|\phi_{n}\rangle\langle\phi_{n}|. \tag{10.4}\]

The issue is that the notation suggests position eigenstates \(|x\rangle\) are themselves elements of a Hilbert space. For example, comparing Equation 10.2 and Equation 10.3:

\[\begin{aligned} \langle\phi_{n}|f\rangle & =\int^{L}_{0}\phi^{*}_{n}\left(x\right)f\left(x\right)\text{d}x\\ & =\int^{L}_{0}\langle\phi_{n}|x\rangle\langle x|f\rangle\text{d}x\\ & =\langle\phi_{n}|\left(\int^{L}_{0}|x\rangle\langle x|\text{d}x\right)|f\rangle \end{aligned}\]

and since this holds for all \(\langle\phi_{n}|\) and all \(|f\rangle\), it follows that

\[\mathbb{I}=\int^{L}_{0}|x\rangle\langle x|\text{d}x. \tag{10.5}\]

By construction this resembles Equation 10.4. But Equation 10.5 assigns a basis element \(|x\rangle\) to each point on the real interval \(x\in\left[0,L\right]\), and hence the basis has dimension \(\mathfrak{c}>\aleph_{0}\)! With this, Von Neumann had a big problem, calling it a ‘mathematical fiction’.

While it is tempting to suggest that these states simply exist in a non-separable Hilbert space, in fact they do not. Dirac recognised that they cannot live in a Hilbert space at all, and so rejected the idea that a Hilbert space was what we really use in quantum mechanics. Instead he formalised his orthonormal basis using

\[\langle x|y\rangle=\delta\left(x-y\right)\]

defining The Dirac delta function:

\[\delta\left(x-y\right)=\begin{cases} \begin{array}{c} \infty,\\ 0, \end{array} & \begin{array}{c} x=y\\ x\ne y \end{array}\end{cases}\]

normalised such that

\[\forall\epsilon>0\qquad\int^{y+\epsilon}_{y-\epsilon}\delta\left(x-y\right)\text{d}x=1.\]

This is to be thought of as a formal extension of the Kronecker delta, Equation 9.2. If we accept Dirac’s notation, we can do neat tricks such as instantly rewrite into the momentum basis:

\[\widetilde{f}\left(p\right)=\langle p|f\rangle\]

which naturally agrees with the usual Fourier transform:

\[\begin{aligned} \widetilde{f}\left(p\right) & =\langle p|f\rangle\\ & =\langle p|\mathbb{I}|f\rangle\\ & =\langle p|\int^{L}_{0}|x\rangle\langle x|\text{d}x|f\rangle\\ & =\int^{L}_{0}\langle p|x\rangle\langle x|f\rangle\text{d}x\\ & =\int^{L}_{0}\exp\left(-ipx/\hbar\right)f\left(x\right)\text{d}x \end{aligned}\]

where the momentum eigenstate is a plane wave:

\[\langle x|p\rangle=\exp\left(ipx/\hbar\right).\]

10.1 Square-summable functions

A function \(f:\left[a,b\right]\in\mathbb{R}\rightarrow\mathbb{C}\) is square summable (aka square integrable) iff

\[\int^{b}_{a}\left|f\left(x\right)\right|^{2}\text{d}x<\infty. \tag{10.6}\]

Here, \(a<b\), but neither need be finite. Square summable functions form an inner product space, conventionally denoted \(\mathcal{L}^{2}=\left(L^{2},\langle\cdot|\cdot\rangle_{2}\right)\), where \(L^{2}\) (pronounced L-two) is the set of square summable functions under the norm induced by the inner product \(\langle\cdot|\cdot\rangle_{2}\). It can be shown that any such space is complete. Hence, \(\mathcal{L}^{2}\) is a Hilbert space¹.

Following Dirac, we can rewrite Equation 10.6 as

\[\begin{aligned} \int^{b}_{a}\langle f|x\rangle\langle x|f\rangle\text{d}x & <\infty\\ & \downarrow\nonumber \\ \langle f|\left(\int^{b}_{a}|x\rangle\langle x|\text{d}x\right)|f\rangle & <\infty. \end{aligned}\]

Therefore square-summability in Dirac notation reads

\[\langle f|f\rangle<\infty.\]

But in quantum mechanics states are normalised:

\[\langle f|f\rangle=1\]

so this is automatically true. The point is that you would be assuming square-summability naturally as a physicist.

We can similarly define \(\mathcal{L}^{p}\) spaces, \(p>2\), as sets of \(p\)-summable functions with a suitably defined metric:

\[\left\Vert f\right\Vert _{p}\triangleq\int^{b}_{a}\left|f\left(x\right)\right|^{p}\text{d}x<\infty.\label{eq:L2-1}\]

However, \(p=2\) is the only case in which the metric \(\left\Vert \cdot\right\Vert _{p}\) is compatible with an inner product. Hence, \(\mathcal{L}^{2}\) is the only \(\mathcal{L}^{p}\) space which is also a Hilbert space.

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1. Prof. Robbins comments: “provided integration and square-integrability are understood in the sense of Lebesgue. If integration is in the sense of Riemann, then the space of square-integrable functions is not complete.”