# I Particle in free space - what happens to the wave function after measurement?

#### saar321412

If I'm trying to solve the problem of a particle in free space (H = P2/2m ).
the eigenfunctions of the Hamiltonian cannot be normalized.
now assume I have a legitimate wave function expressed in terms of the eigenfunction of H and I want to measure its energy.
what will happen to the wavefunction after the measurement?

Related Quantum Physics News on Phys.org

#### vanhees71

Gold Member
You cannot say this without telling which specific measurement device you use. What happens depends on the specific interaction of the measured system with the measurement device.

In your case of the free particle, you should note that there are no energy eigenstates in the literal sense, because the energy eigenstates are uniquely given by the common eigenstates of thee three momentum components, and these are plane waves. They are not square-integrable and thus belong not to the Hilbert space (in wave mechanics the space of square-integrable functions) but to the dual of the nuclear space, where position and momentum operators are defined, i.e., they are distributions.

True states, coming close to energy eigenstates are square-integrable wave functions which peak quite sharply in momentum space. You can, e.g., choose a Gaussian $\tilde{\psi}_0(\vec{p}) \propto \exp[-(\vec{p}-\vec{p}_0)^2/(4 \sigma_p)]$ as an initial state. Then the solution of the Schrödinger equation in momentum space simply is
$$\tilde{\psi}(t,\vec{p})=\exp \left (-\frac{\mathrm{i} \vec{p}^2}{2m \hbar} t \right) \tilde{\psi}_0(\vec{p}).$$
The position representation then follows by Fourier transformation
$$\psi(t,\vec{x})= \frac{1}{\sqrt{2 \pi \hbar}} \int_{\mathbb{R}^3} \mathrm{d}^3 p \tilde{\psi}(t,\vec{x}),$$
which is again a Gaussian, which now however is rather broad, and the position uncertainty increases with time.

Always the Heisenberg uncertainty relation holds,
$$\Delta x_j \Delta p_j \geq \hbar/2$$
for each position-vector and momentum component, $j \in \{1,2,3\}$.

• Heikki Tuuri

#### Heikki Tuuri

To measure the energy of a free particle, you have to measure its momentum p. If you describe the particle with a wave packet, and want to improve your knowledge of the momentum p of the particle, then the uncertainty in the location of the particle has to grow. The wave packet becomes wider.

"Particle in free space - what happens to the wave function after measurement?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving