Quantum mechanics stationary state

[happyparticle]

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

 

[vanhees71]

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.

 

[PeroK]

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.