3

As far as I know, for a particle without spin the quantum state of the particle is described by a "point" in the Hilbert space of the equivalence classes of $L^2$ square-integrable functions $\lvert{\psi} \rangle$ defined on $\mathbb R^3$.

For complex-valued function, the square-integrable condition $f$ implies that its square $\lvert f \rvert ^2$ has finite Lebesgue integral on the measurable space $(\mathbb R^3, \mathcal A)$, where $\mathcal A$ is the Sigma-algebra of Lebesgue measurable sets $($or simply the Borel sigma algebra $\mathcal B(\mathbb{R}^3)$$)$.

Now, let us consider the eigenfunctions $\lvert \psi_j \rangle$ of momentum operator $\vec{P}$. The Lebesgue integral of each of their equivalence classes is not finite. So, momentum eigenfunctions are not square-integrable.

In this case, how do we express the quantum state of a particle in momentum eigenfunctions basis given that eigenfunctions of momentum are not square-integrable?

Qmechanic
  • 220,844
M. A.
  • 2,039
  • 5
  • 11
  • 28

2 Answers2

5

...the quantum state of the particle is described by a... square-integrable function[]...

Sounds reasonable to me.

... momentum eigenfunctions are not square-integrable.

So, a single momentum eigenfunction does not describe the quantum state of a particle.

In this case, how do we express the quantum state of a particle in momentum eigenfunctions basis...

We use an integral over many different momentum eigenfunctions, often referred to as a Fourier transform. Further, we set up the integral such that the result is square integrable. For example: $$ \psi(x) = \int_{-\infty}^{\infty} dp e^{ipx}e^{-\alpha p^2}\;, $$ where $\alpha>0$ is square integrable even though any individual $e^{ipx}$ is not.

hft
  • 27,235
5

A pure state of a spinless particle can be described by its position-space wave function $\psi(\mathbf{x})$, where $\psi \in L^2(\mathbb{R}^3)$, satisfying the normalization condition $\int_{\mathbb{R}^3} d^3x \, |\psi(\mathbf{x})|^2=1$.

Equivalently, this state can be described by its momentum-space wave function $\tilde{\psi}(\mathbf{p})$, where $\tilde{\psi} \in L^2(\mathbb{R}^3)$ with $\int_{\mathbb{R}^3} d^3p \,|\tilde{\psi}(\mathbf{p})|^2=1$.

$\psi$ and $\tilde{\psi}$ are related by the unitary transformation $\mathcal{F}: L^2(\mathbb{R}^3) \to L^2(\mathbb{R}^3)$ defined by ($\hbar=1$) $$ \tilde{\psi}(\mathbf{p})= (\mathcal{F} \psi)(\mathbf{p}):=\int\limits_{\mathbb{R}^3} \! \frac{d^3x}{(2\pi)^{3/2}} \, e^{-i \mathbf{p} \cdot \mathbf{x}} \, \psi(\mathbf{x}), \tag{1} \label{1}$$ where $\mathcal{F}$ is, of course, nothing else than the Fourier transformation. Given $\tilde{\psi}$, the position-space wave function $\psi$ can be recovered by the inverse mapping $$ \psi(\mathbf{x})=(\mathcal{F}^{-1}\tilde{\psi})(\mathbf{x})= \int\limits_{\mathbb{R}^3}\! \frac{d^3p}{(2\pi)^{3/2}} \, e^{i \mathbf{p} \cdot \mathbf{x}} \, \tilde{\psi}(\mathbf{p}). \tag{2} \label{2}$$ The momentum operator $\hat{\mathbf{p}}$ in the position-space representation, defined as the differential operator $(\hat{\mathbf{p}} \psi)(\mathbf{x})=-i \mathbf{\nabla} \psi(\mathbf{x})$ on a suitable dense domain $\mathcal{D}(\hat{\mathbf{p}})\subset L^2(\mathbb{R}^3)$ becomes a multiplication operator in the momentum-space representation, as can easily be seen by using eq. \eqref{2} for $\psi \in \mathcal{D}(\hat{\mathbf{p}})$: $$ (\hat{\mathbf{p}}\psi)(\mathbf{x})=-i \mathbf{\nabla} \psi(\mathbf{x})=-i \mathbf{\nabla} \int\limits_{\mathbb{R}^3} \! \frac{d^3p}{(2\pi)^{3/2}} \, e^{i \mathbf{p} \cdot \mathbf{x}} \, \tilde{\psi}(\mathbf{p})= \int\limits_{\mathbb{R}^3} \! \frac{d^3p}{(2\pi)^{3/2}} \, e^{i \mathbf{p} \cdot \mathbf{x}} \, \mathbf{p} \tilde{\psi}(\mathbf{p}). \tag{3} \label{3} $$ As the spectrum of the momentum operator $\hat{\mathbf{p}}$ is purely continuous ($\sigma_c(\hat{p}_k)=\mathbb{R}$ and $\sigma_p(\hat{p}_k)=\emptyset$ for $k=1,2,3$), the plane wave solutions $f_{\mathbf{p}}(\mathbf{x})=e^{i \mathbf{p} \cdot \mathbf{x}}$ of the eigenvalue equation $\hat{\mathbf{p}} f_{\mathbf{p}}(\mathbf{x})=\mathbf{p} f_{\mathbf{p}}(\mathbf{x})$ cannot be elements of the Hilbert space $L^2(\mathbb{R}^3)$. Note that normalizable eigenfunctions (i.e. proper eigenvectors) are only associated with the point spectrum of a selfadjoint operator. Nevertheless, as eq. \eqref{2} shows, the normalizable wave function $\psi(\mathbf{x})$ can be written as a linear combination (in fact an integral) of the plane wave solutions $e^{i \mathbf{p} \cdot \mathbf{x}}$ with the momentum-space wave function $\tilde{\psi}(\mathbf{p})$ as expansion coefficient.

Hyperon
  • 10,007