I Quantum mechanics stationary state

happyparticle
Messages
490
Reaction score
24
Hi,
I have hard time to really understand what's a stationary state for a wave function.
I know in a stationary state all observables are independent of time, but is the energy fix?
Is the particle has some momentum?
If a wave function oscillates between multiple energies does it means that the wave function is a sum of stationary states and then it is not a stationary state?

Thanks
 
Physics news on Phys.org
For a Hamiltonian that does not explicitly depend on time, the stationary states are the energy eigenstates. Let ##u_E(\vec{x})## be a solution of the time-independent Schrödinger equation (i.e., the eigenvalue equation for the Hamiltonian),
$$\hat{H}u_E=E u_E,$$
and take this as initial state for your wave function, you get
$$\psi(t,\vec{x})=\exp(-\mathrm{i} \hat{H} t/\hbar) u_E(\vec{x})=u_E(\vec{x}) \exp(-\mathrm{i} E t/\hbar),$$
which means that ##\psi## is always ##u_E## multiplied by a phase factor, i.e., the state always stays the one described by ##u_E(\vec{x})## and thus is stationary. Indeed the probability distribution for finding the particle at position ##\vec{x}## is
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2=|u_E(\vec{x})|^2,$$
which is time-independent.

In general the particle in an energy eigenstate does not have a determined momentum. That's only possible, if the momentum operators ##\hat{\vec{p}}=-\mathrm{i} \hbar \vec{\nabla}## commute with the Hamiltonian, i.e., if the Hamiltonian doesn't depend on the position operators ##\hat{\vec{x}}##. That's fulfilled for the free-particle Hamiltonian,
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2.$$
In this case a complete set of compatible observables are the three momentum components, and the corresponding eigenfunctions,
$$u_{\vec{p}}(\vec{x})=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar),$$
form a complete set of energy eigenfunctions. The energy eigenvalues are given by
$$\hat{H} u_{\vec{p}}(\vec{x})=-\frac{\hbar^2}{2m} \Delta u_{\vec{p}}(\vec{x})=\frac{1}{2m} \vec{p}^2 u_{\vec{p}}(\vec{x}), \quad E=\frac{1}{2m} \vec{p}^2.$$
A wave function doesn't oscillate between energy values. The value of the wave function is a probability amplitude, i.e., ##|\psi(t,\vec{x})|^2## is the probability distribution for finding the particle at position ##\vec{x}## when measured at time ##t##.

A general wave function is a superposition of energy eigenfunctions.
$$\psi(t,\vec{x})=\sum_E \exp(-\mathrm{i} E t/\hbar) c_E u_E(t,\vec{x}),$$
where the ##c_E## are coefficients determined by the initial state, ##\psi(t=0,\vec{x})=\psi_0(\vec{x})##,
$$c_E =\langle u_E|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x u_E^*(\vec{x}) \psi_0(\vec{x}),$$
where for simplicity I assumed that the energy eigenvalues are non-degenerate.
 
  • Like
Likes happyparticle, BvU, topsquark and 1 other person
happyparticle said:
If a wave function oscillates between multiple energies does it means that the wave function is a sum of stationary states and then it is not a stationary state?
A linear combination (superposition) of stationary states is not a stationary state. This is fundamental to QM.

Mathematically, a linear combination of eigenvectors of an operator (with different eigenvalues) is not an eigenvector of that operator.
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Back
Top