# Quantum mechanics stationary state

• I
• happyparticle
In summary: Physically, a superposition of states with different energies will result in a state that is not stationary, as it will evolve over time due to the phase differences between the different energy states.
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

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.

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.

vanhees71

## 1. What is a stationary state in quantum mechanics?

A stationary state in quantum mechanics refers to a state in which a system's properties, such as energy and momentum, do not change over time. This means that the system is in a stable and constant state.

## 2. How are stationary states related to energy levels?

Stationary states are directly related to energy levels in quantum mechanics. Each stationary state corresponds to a specific energy level, and the energy of a system can only change by transitioning between these states.

## 3. Can a system be in more than one stationary state at a time?

No, according to the principles of quantum mechanics, a system can only be in one stationary state at a time. This is known as the principle of superposition, which states that a system can exist in multiple states simultaneously, but when observed, it will collapse into a single state.

## 4. How do we determine the probability of finding a particle in a particular stationary state?

The probability of finding a particle in a particular stationary state is determined by the wave function of the system. The square of the wave function at a specific point represents the probability of finding the particle at that point. The higher the probability, the more likely the particle is to be found in that state.

## 5. Can a system's stationary states change over time?

No, stationary states are constant and do not change over time. However, the probabilities of finding a particle in different stationary states can change, which can lead to transitions between states and changes in the system's overall energy.

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