# What should we use to express free particle in Quantum Mechanics?

1. Oct 21, 2011

### Karmerlo

What should we use to express free particle in Quantum Mechanics? Wave packet or plane wave? Can free particle be localized in Quantum Mechanics?

2. Oct 22, 2011

### vanhees71

Plane waves do not represent states. They are generalized eigenfunctions of the momentum operator,

$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{x} \cdot \vec{p}).$$

Here and in the following, I use natural units with $\hbar=1$. The plane waves do not represent (pure) states since they are not square integrable, but belong to the dual space of the domain, where momentum and position operators and arbitrary powers of these operators are defined, i.e., they are distributions over this space. They are "normalized to a $\delta$ distribution",

$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}').$$

True states must be represented by square integrable functions normalized to 1. Then their modulus squared gives the probability distribution for the particle's position.

You can localize a particle as precise as you wish, but then momentum will be very undertermined and vice versa. That's the content of Heisenberg's Uncertainty Relation,

$$\Delta x_j \Delta p_j \geq \frac{1}{2}, \quad j \in \{1,2,3\}.$$

3. Oct 22, 2011

### tom.stoer

It's true that plane waves are generalized eigenfunctions of the momentum operator, nevertheless e.g. in scattering theory (in QM or QFT) you use plane waves a free, incoming particles. So strictly speaking they cannot not realized in nature, but as mathematical idealizations they are useful b/c you get physically correct results.

4. Oct 23, 2011

### juanrga

Plane waves describe quantum states for free particles. A plane wave is delocalized over a volume of size L^3 with L --> infinite.

5. Oct 23, 2011

### nonequilibrium

I think no one has specifically answered the question, which was if a free particle can be localized:

Yes it can. The Schrödinger equation for a free particle has wave packets as a solution (and plane waves, which can be made into 'exact'/mathematically sound solutions by using heavier artillery than regular hilbert spaces).

I hope this helps.

6. Oct 24, 2011

### dextercioby

Good question in post #1. I would say that there's no such thing as a free particle, from the practical point of view. We cannot separate a system from its surroundings, i.e. we cannot neglect its interactions. However, the free particle approximation (because we can take this as an idealization for a particle very far from its 'neighbors' with which it could interact) works with the other assumptions of scattering theory.

However, trying to put a free particle in a mathematical formalism based only on separable inf-dim. Hilbert spaces fails, for all its observables such as energy, position, momentum have a purely continuous spectrum, thus an energy/momentum/position eigenstate would 'live' outside a separable inf-dim. Hilbert space.

Wave packets are means to bring such states back within the Hilbert space, but their theory is based on distribution theory as well, because the function e^(ipx) which accounts for the nucleus of the Fourier transformation is a tempered distribution, eigenvector of the momentum operator in the position representation, or eigenvector of the position operator in the momentum representation.

7. Oct 25, 2011

### vanhees71

Yes, it cannot be stressed enough: Plane waves do not represent states of particles in $\mathbb{R}^3$.

In scattering theory you don't use plane waves but asymptotically free wave packets to describe the in and out states, supposed these provide the correct asymptotics, i.e., if the interaction potential between the scatterers falls faster than $1/r$. E.g., in Coulomb scattering the free-particle states are not the apprriate asymptotic states due to the long-range nature of the Coulomb force. There you use "distorted waves" as a generalized basis for the asymptotically free scattering states, i.e., the unbound solutions of the Coulomb-Hamiltonian eigenvalue problem.

8. Oct 25, 2011

### tom.stoer

Sorry, but I disagree.

In QM one uses plane waves |p> for incoming particles in the ansatz (Lippmann-Schwinger equation)

$$|\psi\rangle = |p\rangle + G^+_p V\,|\psi\rangle$$

$$G^+_p = \frac{1}{E_p-H_0+i\epsilon} = \int \frac{d^3k}{(2\pi)^3} \frac{|k\rangle\langle k|}{E_p - E_k + i\epsilon}$$

and the iteration

$$|\psi\rangle = \left[1 + (G^+_p V) + (G^+_p V)^2 + \ldots\right]|p\rangle$$

...

Going to perturbative QFT one uses the S-matrix which is expressed in terms of plane wave states |in> and |out>.

$$S_{\text{out},\text{in}}=\langle\text{out}|S|\text{in}\rangle$$

I know that it's mathematically save to formulate scattering theory in terms of wave packets; I know that distorted waves are appropriate for long-range potentials.

But there are numerous examples where these 'mathematically ill-defined' and 'physically unreasonalbe' plane waves are used to derive correct results.

9. Oct 25, 2011

### vanhees71

That's of course true, but to derive the formulae you quote properly, you should use wave packets or some other regulator, e.g., introducing a finite volume ("quantization volume") with periodic boundary conditions. Otherwise, you have the difficulty to define the squares of S-matrix elements moduli, which define the physically relevant quantities like decay widths for particle decays or cross sections for scattering processes. Afterwards you let the volume to infinity.

For the wave-packet treatment see Weinberg, The quantum theory of fields, vol. 1 or Peskin/Schroeder Introduction to quantum field theory.

For the finite-volume regularization see, e.g., my qft manuscript on my homepage,

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

10. Oct 25, 2011

### tom.stoer

Beeindruckend :-)

As I said, 'mathematically ill-defined' and 'physically unreasonable' plane waves are often useful to derive physically correct results; but of course there are situations where these 'physically reasonable' plane waves lead to mathematical nonsense ;-)

11. Oct 26, 2011

### vanhees71

Well, but you avoid the "physical nonsense" to a certain extent already in the $\mathrm{i} \epsilon$'s in defining your Green's function (which is retarded in this non-relativistic context).

12. Oct 26, 2011

### tom.stoer

Of course :-)

What I want to say is that strictly speaking there are no plane wave states in nature, nevertheless we can calculate rather accurately how nature behaves using plane wave states.

Strange but true ...

13. Nov 12, 2011

### Karmerlo

14. Nov 12, 2011

### nonequilibrium

Thank you for sharing that article :)