Free particle: Difference between revisions
imported>Michael Underwood No edit summary |
mNo edit summary |
||
(8 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
{{subpages}} | |||
{{TOC|right}} | |||
In [[physics]] a [[free particle]] is one whose motion is unaffected by any external factors. | In [[physics]] a [[free particle]] is one whose motion is unaffected by any external factors. | ||
That is, no net [[force]] is acting on the particle. | That is, no net [[force]] is acting on the particle. | ||
Line 4: | Line 6: | ||
==Classical mechanics== | ==Classical mechanics== | ||
In [[classical mechanics|classical]] or [[Newton]]ian mechanics the motion of a particle is governed by [[Newton's laws]] of motion. | In [[classical mechanics|classical]] or [[Newton]]ian mechanics the motion of a particle is governed by [[Newton's laws]] of motion. | ||
In particular, his second law states that the [[acceleration]] undergone by a particle is equal to the external forces acting on the particle, divided by the particle's mass. In other words, | In particular, his second law states that the [[acceleration]] undergone by a particle in an [[inertial frame of reference]] is equal to the external forces acting on the particle, divided by the particle's mass.<ref name=inertial> | ||
In an accelerating frame of reference the forces are augmented by the so-called [[inertial forces]] that are an artifact of the acceleration of the [[Frame of reference (physics)|frame of reference]] . | |||
</ref> In other words, | |||
:<math>a=\frac{F}{m}\ .</math> | :<math>a=\frac{F}{m}\ .</math> | ||
In the case of a free particle <math>F=0</math>, so the particle does not accelerate. The speed of the particle is therefore constant, and the position of the particle at any time <math>t</math> can be predicted with certainty if the the particle's velocity <math>v_0</math> is known along with its position <math>x_0</math> at a single instant <math>t_0</math>. In this case, | In the case of a free particle <math>F=0</math>, so the particle does not accelerate. The speed of the particle is therefore constant, and the position of the particle at any time <math>t</math> can be predicted with certainty if the the particle's velocity <math>v_0</math> is known along with its position <math>x_0</math> at a single instant <math>t_0</math>. In this case, | ||
Line 30: | Line 36: | ||
coincides with the level of the potential, i.e. we can always set | coincides with the level of the potential, i.e. we can always set | ||
<math>V(x)=0.</math> | <math>V(x)=0.</math> | ||
This means that in the | This means that in the case of one spatial dimension [[Schrödinger equation|Schrödinger's equation]] is given everywhere by | ||
:<math>i\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)\ .</math> | :<math>i\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)\ .</math> | ||
which admits [[plane wave]] solutions, | which admits [[plane wave]] solutions, | ||
:<math>\psi(x,t)=\exp\left(ikx-i\omega t\right)</math> | :<math>\psi(x,t)=A\exp\left(ikx-i\omega t\right) \ ,</math> | ||
for any frequency <math>\omega.</math> | for any frequency <math>\omega.</math> | ||
Here the [[wavenumber]] <math>k</math> is given by | Here ''A'' is a constant called the ''normalization factor'' used to adjust the probability of finding the particle,<ref name=normalization> | ||
For example, if the particle were not completely free, but confined to be within some finite region, the normalization constant would ensure that the probability of finding the particle inside the region was one. | |||
</ref> and the [[wavenumber]] <math>k</math> is given by | |||
:<math>k=\sqrt{\frac{2 m \omega}{\hbar}}\ .</math> | :<math>k=\sqrt{\frac{2 m \omega}{\hbar}}\ .</math> | ||
In the case of a particle in | In the case of a particle in three-dimensional space '''''r''''' = (''x, y, z'') the result is very similar. The equation to solve becomes | ||
:<math>i\hbar\frac{\partial}{\partial t}\psi(\ | :<math>i\hbar\frac{\partial}{\partial t}\psi(\mathbf{r},t) = -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r},t)</math> | ||
and the | and the solution is still a plane wave, now given by | ||
:<math>\psi(\ | :<math>\psi(\mathbf{r},t)=A\exp\left(i\mathbf{k\cdot r}-i\omega t\right)\ ,</math> | ||
where ''A'' is again a normalization constant used to adjust the probability of finding the particle, and '''''k''''' is the [[Wavevector| ''wavevector'']]. The magnitude of the wavevector '''''k''''' is now defined by the equation | |||
:<math>k=\sqrt{\ | :<math>k=\sqrt{\mathbf{k\cdot k}}=\sqrt{\frac{2 m \omega}{\hbar}}\ .</math> | ||
==Notes== | |||
<references/>[[Category:Suggestion Bot Tag]] |
Latest revision as of 17:01, 18 August 2024
![](http://s9.addthis.com/button1-share.gif)
In physics a free particle is one whose motion is unaffected by any external factors. That is, no net force is acting on the particle.
Classical mechanics
In classical or Newtonian mechanics the motion of a particle is governed by Newton's laws of motion. In particular, his second law states that the acceleration undergone by a particle in an inertial frame of reference is equal to the external forces acting on the particle, divided by the particle's mass.[1] In other words,
In the case of a free particle , so the particle does not accelerate. The speed of the particle is therefore constant, and the position of the particle at any time can be predicted with certainty if the the particle's velocity is known along with its position at a single instant . In this case,
It is quite common to take , simplifying this expression slightly to the more familiar
Quantum mechanics
There is a subtle difference between the classical and quantum interpretations of a free particle. Classically a particle is considered to be free whenever it is in a region of constant potential, This is because force is given by the gradient of the potential, so as required for a free particle. Quantum mechanically however, a particle does not possess a definite position and therefore cannot be said to be in a region of constant potential (unless that region is all space). This distinction is made clear by the particle in a box problem. Classically a particle in a box moves freely (i.e. as if the box wasn't there) until it hits a wall, at which point it around and then continues to move freely again. On the other hand a quantum particle in a box behaves differently than it would if the box did not exist.
The solution to Schrödinger's equation
As we have said, a free particle is one for which the potential term in the Hamiltonian is constant (but finite). When this is the case, the energy scale can always be shifted so that the zero of energy coincides with the level of the potential, i.e. we can always set This means that in the case of one spatial dimension Schrödinger's equation is given everywhere by
which admits plane wave solutions,
for any frequency Here A is a constant called the normalization factor used to adjust the probability of finding the particle,[2] and the wavenumber is given by
In the case of a particle in three-dimensional space r = (x, y, z) the result is very similar. The equation to solve becomes
and the solution is still a plane wave, now given by
where A is again a normalization constant used to adjust the probability of finding the particle, and k is the wavevector. The magnitude of the wavevector k is now defined by the equation
Notes
- ↑ In an accelerating frame of reference the forces are augmented by the so-called inertial forces that are an artifact of the acceleration of the frame of reference .
- ↑ For example, if the particle were not completely free, but confined to be within some finite region, the normalization constant would ensure that the probability of finding the particle inside the region was one.