14. Example #14: Free Particle

We now turn to a case of free particle. In this case, the potential V (x) = 0 everywhere and the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                      (14.1)

Or, with $k\equiv {\sqrt{2mE}}/{\hbar }\;$ :

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $                                                      (14.2)

The general solution to this Schrödinger equation is:

$\psi (x)=A{{e}^{ikx}}+B{{e}^{-ikx}}$                                                (14.3)

It is the same as the case of the inside of the infinite square well. However, unlike the infinite square well, there are no boundary conditions to restrict the possible value of E, therefore, the free particle can carry any energy. The standard time-dependent solution is given by:

$\Psi (x,t)=A{{e}^{ik\left( x-\frac{\hbar k}{2m}t \right)}}+B{{e}^{-ik\left( x+\frac{\hbar k}{2m}t \right)}}$                                    (14.4)

This equation also represents wave of fixed profile, traveling with constant velocity in the positive or negative x-direction. Since every point on the waveform is moving along with constant velocity, its shape doesn’t change as it propagates. The first term in equation (14.4) represents a wave traveling to the right, and the second term represents a wave of the same energy traveling to the left. Let us say that k is negative for waves traveling to the left and k is positive for waves traveling to the right:

$k\equiv \pm \frac{\sqrt{2mE}}{\hbar }$                                                               (14.5)

And equation (14.4) can be rewritten:

${{\Psi }_{k}}(x,t)=A{{e}^{i\left( kx-\frac{\hbar {{k}^{2}}}{2m}t \right)}}$                                                     (14.6)

The velocity of the waves, i.e. the coefficient of t over the coefficient of x, is:

$v=\frac{\hbar |k|}{2m}=\sqrt{\frac{E}{2m}}$                                                       (14.7)

It is half of the velocity of a classical particle: ${{v}_{classical}}=\sqrt{\frac{2E}{m}}$ with energy $E=({1}/{2}\;)mv_{classical}^{2}$. This ‘classical velocity’ of the particle is in fact the group velocity of the particle, i.e. the velocity of the envelope, in contrast to the phase velocity of the individual ripples. Generally, the group velocity is given by: ${{v}_{group}}=\frac{d\omega }{dk}=\frac{\hbar k}{m}$ with $\omega ={{{\hbar }^{2}}k}/{2m}\;$ in our case. And the phase velocity is given by: ${{v}_{phase}}=\frac{\omega }{k}=\frac{\hbar k}{2m}$ which is half of group velocity, thus ${{v}_{classical}}={{v}_{group}}=2{{v}_{phase}}$ .

However, wave function (14.6) is non-normalizable:

$\int\limits_{-\infty }^{\infty }{\Psi _{k}^{*}{{\Psi }_{k}}dx=|A{{|}^{2}}\int\limits_{-\infty }^{\infty }{1}dx=|A{{|}^{2}}\infty }$                                  (14.8)

Physical interpretation of this characteristic is that: A free particle cannot exist in a stationary state, so there is no such thing as a free particle with a definite energy. Nevertheless, the separable solutions still play a mathematical role that is completely independent of their physical interpretation. The general solution to time-dependent Schrödinger equation is still a linear combination of separable solutions (in analogy to equation (10.3) in section 10):

$\Psi (x,t)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\phi (k){{e}^{i(kx-\frac{\hbar {{k}^{2}}}{2m}t)}}dk}$                                     (14.9)

The quantity ${1}/{\sqrt{2\pi }}\;$ usually appears in Fourier analysis and factored out for convenience. While the combination $\left( {1}/{\sqrt{2\pi }}\; \right)\phi (k)dk$ plays the ‘role’ of the coefficient c_n in equation (10.3). Now, the wave function (14.9) can be normalized for appropriate $\phi (k)$. The problem is now to determine $\phi (k)$ so as to fit the initial wave function which is given by:

$\Psi (x,0)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\phi (k){{e}^{ikx}}dk}$                                         (14.10)

The answer of this classical problem is provided by means of Fourier transform:

$f(x)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{F(k){{e}^{ikx}}dk}$ ↔ $F(k)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{f(x){{e}^{-ikx}}dx}$                  (14.11) 

Here, F (k) is the Fourier transform of f (x), and f (x) is the inverse Fourier transform of F (k). These integrals have to exist: the necessary and sufficient condition on f (x) is that $\int\limits_{-\infty }^{\infty }{|f(x){{|}^{2}}dx}$ be finite, in this case $\int\limits_{-\infty }^{\infty }{|F(k){{|}^{2}}dk}$ is also finite. For our purposes, the physical requirement that $\Psi (x,0)$ is normalizable, guarantees that the integrals exist. Therefore, the general solution of the Schrödinger equation for free particle is equation (14.9) with:

$\phi (k)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{\infty }{\Psi (x,0){{e}^{-ikx}}dx}$                                       (14.12)