Quantum Mechanics 1
Position Operator and Momentum Operator.
1 Self-Adjoint Operators and Observables in Hilbert Space
In the mathematical formulation of quantum mechanics, a physical observable is represented by a self-adjoint operator \(A\) acting on a complex Hilbert space \(\mathbf{H}\). The duality between wave-like evolution and point-like measurement is captured by the spectral properties of these operators.
The distinction between a symmetric operator and a self-adjoint operator is of paramount importance in the spectral theorem.
In quantum mechanics, the requirement for an operator to be self-adjoint (rather than merely symmetric) ensures that its eigenvalues—which correspond to measurable physical values—are well-defined and real.
To prove the reality of the eigenvalue, consider a non-zero vector \(\psi \in \text{Dom}(A)\) satisfying \(A\psi = \lambda\psi\). By the definition of the inner product and the symmetry of \(A\), we have: \[\lambda \langle \psi, \psi \rangle = \langle \psi, \lambda\psi \rangle = \langle \psi, A\psi \rangle.\] Applying the symmetry condition \(\langle \phi, A\psi \rangle = \langle A\phi, \psi \rangle\) with \(\phi = \psi\), we obtain: \[\langle \psi, A\psi \rangle = \langle A\psi, \psi \rangle = \langle \lambda\psi, \psi \rangle = \bar{\lambda} \langle \psi, \psi \rangle.\] Equating the two expressions yields \(\lambda \langle \psi, \psi \rangle = \bar{\lambda} \langle \psi, \psi \rangle\). Since \(\psi \neq 0\), it follows that \(\|\psi\|^2 > 0\), thus \(\lambda = \bar{\lambda}\), which implies \(\lambda \in \mathbb{R}\). The case for \(A^m\) follows similarly: if \(A\) is symmetric and the higher-order domains are well-defined, \(\langle \psi, A^m\psi \rangle = \langle A\psi, A^{m-1}\psi \rangle = \dots = \langle A^{\lfloor m/2 \rfloor}\psi, A^{\lceil m/2 \rceil}\psi \rangle\), all of which must be real through recursive application of the symmetry property.
This theorem provides the rigorous mathematical justification for why observables like momentum \(\mathrm{P} = -i\hbar \mathrm{D}_x\) and position \(\mathrm{X} = x\), which can be shown to be essentially self-adjoint on appropriate domains, always yield real experimental data.
2 Position and the Position Operator
Now we consider a single particle moving along the real line \(\mathbb{R}\). The wave function of such a particle is a mapping \(\psi: \mathbb{R} \to \mathbb{C}\). Assuming time is fixed, the quantity \(|\psi(x)|^2\) is assumed to be the probability density of the particle’s location. This implies that the probability of the particle being in a measurable set \(E \subseteq \mathbb{R}\) is: \[\mathrm{I}(|\psi(x)|^2, E).\] For this interpretation to be consistent, we require \(\psi\) to be normalized such that: \[\mathrm{I}(|\psi(x)|^2, \mathbb{R}) = 1.\]
According to standard probability theory, the expectation value of the particle’s position is given by: 1 \[E(x) = \mathrm{I}(x |\psi(x)|^2, \mathbb{R}).\]
1 Naturally, we require this integral to be absolutely convergent.
2 Again, we assume absolute convergence.
The moments of the position are defined as the higher-order moments of the probability distribution: 2 \[E(x^m) = \mathrm{I}(x^m |\psi(x)|^2, \mathbb{R}).\]
Using this operator, the expectation value of position can be elegantly expressed in terms of the \(L^2\) inner product: \[\langle X \rangle_\psi = \langle \psi, X \psi \rangle = \mathrm{I}(\overline{\psi(x)} (x \psi(x)), \mathbb{R}) = \mathrm{I}(x |\psi(x)|^2, \mathbb{R}).\] Furthermore, the \(m\)-th moment corresponds to the expectation value of the \(m\)-th power of the operator: 3 \[\langle X^m \rangle_\psi = \langle \psi, X^m \psi \rangle = \mathrm{I}(x^m |\psi(x)|^2, \mathbb{R}).\]
3 It is worth noting that almost nothing here is guaranteed to be well-defined. Even if \(\psi\) is square-integrable, \(X\psi\) may not be; even if \(X\psi\) is square-integrable, \(X^m\psi\) may not be. Nevertheless, we believe this notation provides significant conceptual advantages.
3 Momentum and the Momentum Operator
While the wave function \(\psi(x)\) is expressed as a function of position, the information regarding the particle’s momentum is encoded in the oscillatory behavior of the wave.
4 The proof of this proposition is physically intuitive but mathematically subtle, as it involves the transition from plane waves to square-integrable wave packets.
More precisely, this implies that a wave function of the form \(\psi(x) = e^{ikx}\) represents a particle with a definite momentum \(p = \hbar k\). Since \(e^{ikx} \notin L^2(\mathbb{R})\), we first consider a particle on a circle to avoid technicalities. We treat \(\psi\) as a \(2\pi\)-periodic function on \(\mathbb{R}\) satisfying: \[\mathrm{I}(|\psi(x)|^2, 0, 2\pi) = 1.\]
For any \(k \in \mathbb{Z}\), the normalized wave function \(\psi(x) = \frac{e^{ikx}}{\sqrt{2\pi}}\) represents a state with definite momentum \(p = \hbar k\). In this case, a measurement of momentum yields \(\hbar k\) with probability 1. Since \(\left\{ \frac{e^{ikx}}{\sqrt{2\pi}} \right\}_{k \in \mathbb{Z}}\) forms an orthonormal basis for \(L^2([0, 2\pi])\), any typical wave function can be expressed as: \[\psi(x) = \sum_{k=-\infty}^{\infty} a_k \frac{e^{ikx}}{\sqrt{2\pi}}, \quad \sum_{k=-\infty}^{\infty} |a_k|^2 = 1.\]
In such a superposition, the momentum is no longer deterministic. A measurement yields \(\hbar k\) with probability \(|a_k|^2\). Thus, the expectation value and higher moments of momentum are: \[E(p) = \sum_{k=-\infty}^{\infty} \hbar k |a_k|^2, \quad E(p^m) = \sum_{k=-\infty}^{\infty} (\hbar k)^m |a_k|^2. \]5
5 We assume these series converge absolutely.
We seek an operator \(P\) such that \(E(p^m) = \langle \psi, P^m \psi \rangle\). This requires \(P e^{ikx} = \hbar k e^{ikx}\). The unique linear operator satisfying this is: \[P = -i\hbar \mathrm{D}_x.\]
Returning to the real line \(\mathbb{R}\), we maintain the definition \(P = -i\hbar \mathrm{D}_x\). Although \(e^{ikx}\) is not square-integrable, the Fourier transform allows us to represent any \(\psi \in L^2(\mathbb{R})\) as a linear combination of these “momentum eigenfunctions”: \[\psi(x) = \frac{1}{\sqrt{2\pi}} \mathrm{I}(e^{ikx} \hat{\psi}(k), \mathbb{R})\] where the Fourier transform \(\hat{\psi}(k)\) is defined as: \[\hat{\psi}(k) = \frac{1}{\sqrt{2\pi}} \mathrm{I}(e^{-ikx} \psi(x), \mathbb{R}).\]
Since the Fourier transform is a unitary operator on \(L^2(\mathbb{R})\), we have \(\|\psi\|^2 = \|\hat{\psi}\|^2 = 1\). We can now state the properties of \(P\) strictly within the Hilbert space context.
6 For example, the Schwartz space \(\mathcal{S}(\mathbb{R})\) of rapidly decreasing functions.
To prove this for \(m=1\), we use the definition of the inner product and the inverse Fourier transform of \(\psi\): \[\langle \psi, P \psi \rangle = \mathrm{I}(\overline{\psi(x)} (-i\hbar \mathrm{D}_x \psi(x)), \mathbb{R}).\] Differentiating the integral representation \(\psi(x) = \frac{1}{\sqrt{2\pi}} \mathrm{I}(\hat{\psi}(k) e^{ikx}, \mathbb{R}_k)\) with respect to \(x\) under the integral sign gives \(\mathrm{D}_x \psi(x) = \frac{1}{\sqrt{2\pi}} \mathrm{I}(ik \hat{\psi}(k) e^{ikx}, \mathbb{R}_k)\). Thus: \[P \psi(x) = \frac{1}{\sqrt{2\pi}} \mathrm{I}(\hbar k \hat{\psi}(k) e^{ikx}, \mathbb{R}_k).\] This shows that \(P\psi\) is the inverse Fourier transform of \(\hbar k \hat{\psi}(k)\). By Plancherel’s Theorem (unitarity of the Fourier transform), the inner product in \(x\)-space equals the inner product in \(k\)-space: \[\langle \psi, P \psi \rangle = \langle \hat{\psi}(k), \hbar k \hat{\psi}(k) \rangle = \mathrm{I}(\hbar k |\hat{\psi}(k)|^2, \mathbb{R}).\] The general case for \(P^m\) follows by recursive application, noting that \(P^m \psi\) corresponds to \((\hbar k)^m \hat{\psi}(k)\) in the Fourier domain.
4 The Position and Momentum Operators
In the transition from classical to quantum mechanics, the fundamental observables of position and momentum are elevated to the status of linear operators acting on the Hilbert space \(\mathbf{H} = L^2(\mathbb{R})\).
Neither the position nor the momentum operator is defined on the entire Hilbert space \(L^2(\mathbb{R})\), as the resulting functions \(x\psi(x)\) or \(\mathrm{D}_x \psi(x)\) may fail to be square-integrable. Thus, \(X\) and \(P\) are unbounded operators defined on suitable dense subspaces \(\text{Dom}(X)\) and \(\text{Dom}(P)\).
One of the most profound departures from classical physics is the fact that these operators do not commute.
To prove this, we apply the commutator to a test function \(\psi(x)\) in a sufficiently smooth domain (such as the Schwartz space \(\mathcal{S}(\mathbb{R})\)): \[(XP - PX)\psi = X(P\psi) - P(X\psi) = x(-i\hbar \mathrm{D}_x \psi) - (-i\hbar \mathrm{D}_x (x\psi)).\] Applying the Leibniz rule (product rule) to the second term: \[\mathrm{D}_x (x\psi) = \psi + x \mathrm{D}_x \psi.\] Substituting this back into the commutator expression: \[(XP - PX)\psi = -i\hbar x \mathrm{D}_x \psi + i\hbar (\psi + x \mathrm{D}_x \psi) = -i\hbar x \mathrm{D}_x \psi + i\hbar \psi + i\hbar x \mathrm{D}_x \psi = i\hbar \psi.\] Since this holds for any \(\psi\) in the dense domain, we conclude \(XP - PX = i\hbar \mathrm{I}\).
We observe a parallel between this result and the classical Poisson bracket \(\{x, p\} = 1\). This suggests a deep analogy where the commutator of two quantum operators \([A, B]\) corresponds to \(i\hbar\) times the Poisson bracket \(\{f, g\}\) of their classical counterparts.
The physical validity of measurement requires these operators to be symmetric (and ultimately self-adjoint).
7 In formal language, this proposition states that \(X\) and \(P\) are symmetric operators on their dense subspaces. In fact, \(X\) and \(P\) are essentially self-adjoint on these domains, though the proof of essential self-adjointness is deferred.
For the position operator \(X\): \[\langle \phi, X\psi \rangle = \mathrm{I}(\overline{\phi(x)} x \psi(x), \mathbb{R}) = \mathrm{I}(\overline{x \phi(x)} \psi(x), \mathbb{R}) = \langle X\phi, \psi \rangle.\] For the momentum operator \(P\), we use integration by parts: \[\langle \phi, P\psi \rangle = \mathrm{I}(\overline{\phi(x)} (-i\hbar \mathrm{D}_x \psi(x)), \mathbb{R}) = -i\hbar \left( [\overline{\phi(x)}\psi(x)]_{-\infty}^{\infty} - \mathrm{I}(\overline{\mathrm{D}_x \phi(x)} \psi(x), \mathbb{R}) \right).\] For functions in \(L^2(\mathbb{R})\) (or the Schwartz space), the boundary term vanishes at infinity. Thus: \[\langle \phi, P\psi \rangle = i\hbar \mathrm{I}(\overline{\mathrm{D}_x \phi(x)} \psi(x), \mathbb{R}) = \mathrm{I}(\overline{-i\hbar \mathrm{D}_x \phi(x)} \psi(x), \mathbb{R}) = \langle P\phi, \psi \rangle.\] Note that the complex conjugate \(\overline{-i} = i\) is crucial for the symmetry of \(P\).
This symmetry ensures that the expectation values of position and momentum are always real-valued, consistent with experimental observation.