If a the Schrödinger equation for a free particle, give a matrix of probability higher in the center (like gravity), a particle like a photon must be moving not in a straight path if there is no forze that push it. So further than the Schrödinger probability there must be something that produces this propagation or movement in a stright path. Any ideas?

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    $\begingroup$ what do you men by "fire ball probability"? $\endgroup$
    – Chegon
    Commented Oct 31, 2019 at 21:33
  • $\begingroup$ You must be new to quantum physics . The photon follows a quantized Maxwell's equation for its wave function, electrons the Dirac equation, bosons the Klein Gordon. As the answer says a free particle is repersented by the plane wave solutions of the equations , i.e. no potential, so are spread all over the place. In order to model a real free particle one has to use the wave packet concept hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html in field theory $\endgroup$
    – anna v
    Commented Nov 1, 2019 at 5:41

1 Answer 1


Free particles are described by plane waves $$\psi(\vec{x},t)=e^{\vec{k} \cdot \vec{x} - \omega t}$$ where $\vec{p}=\hbar\vec{k}$.

Before I outline the explanation, recall the following equation about variance $\sigma$ $$\sigma(x)=\langle x^2 \rangle - (\langle x \rangle)^2.$$

The reason why free particles remain in a straight path can be elucidated by comparing them with bounded particles.

Bounded particles in potentials such as particle in the box and harmonic oscillators have wavefunctions which may or may not extend to infinity. Notice that $\langle \vec{p} \rangle =\frac{d}{dx} \langle \vec{x} \rangle =0$ for bounded particles since $\langle \vec{x} \rangle$ is constant. However this does not mean that every measurement will yield same constant position and zero momentum. It is the expectation that is constant, not each measurement if we take an ensemble of the system. Mathematically, this can be seen by non-zero variance citing the above equation, for particle in the box for example: $$\int_0^L dx \sin(kx) \hat{p} \sin(kx) =0 $$ $$\int_0^L dx \sin(kx) \hat{p^2} \sin(kx) \neq 0 $$ $$\implies \sigma(p) \neq 0.$$

Now let's consider the free particle with the plane wavefunction. It can be easily calculated that $\langle \vec{p} \rangle= \hbar \vec{k}$. Now, what is its variance when we measure infinitely many times with an ensemble? $\langle \vec{p}^2 \rangle= (\hbar \vec{k})^2$ and therefore it is zero. In other words, it is because the plane wave is an eigenstate of $\hat{p}$.

However, plane wave functions actually cannot exist in reality, hence there is no true free particle. Though theoretically, due to its usefulness up to approximations, free particles are described by wave functions i.e. eigenstates of momentum operator.

  • $\begingroup$ The wave packet formalism is used to describe free particles, i.e. a superposition of plane waves hyperphysics.phy-astr.gsu.edu/hbase/Waves/wpack.html $\endgroup$
    – anna v
    Commented Nov 1, 2019 at 5:38
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    $\begingroup$ @annav Indeed, that would be the case for non-monochromatic light. Actually, monochromatic light does not exist in reality due to the same reason why plane wave can't exist. $\endgroup$
    – Nugi
    Commented Nov 1, 2019 at 20:31

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