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)

Solved Example #1: The Infinite Square Well

Solved Example #1: The Infinite Square Well is attached.

11. Example #1 (Application of Schrödinger equation): The Infinite Square Well

Consider a particle with mass m, subjected in a potential:

V (x) = 0 for 0 ≤ x ≤ a, and V (x) → ∞ elsewhere.

The particle is completely free, except at the two ends x = 0 and x = a where an infinite force prevents the particle from escaping. In this case, we would like to find the general solution to the Schrödinger equation for this particular potential.

 Figure 11.1. Infinite square well potential.

The probability of finding the particle outside the well is zero, i.e. ψ (x) = 0. While inside the well, the time-independent Schrödinger equation is:

$-\frac{{{\hbar }^{2}}}{2m}\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=E\psi $                                                    (11.1)

Or:

$\frac{{{d}^{2}}\psi }{d{{x}^{2}}}=-{{k}^{2}}\psi $, with $k=\frac{\sqrt{2mE}}{\hbar }$                                        (11.2)

Equation (11.2) is the simple harmonic oscillator equation, and the general solution is given by:

$\psi (x)=A\sin kx+B\cos kx$                                           (11.3)

With A and B are arbitrary constants. These constants are fixed by boundary conditions of the problem. ψ is continuous, when the potential V (x) goes to infinity, i.e. ψ (a) =  ψ (0) = 0.

Substituting the boundary condition ψ (0) = 0 into equation (11.3), we obtain: B = 0, and hence:

$\psi (x)=A\sin kx$                                                (11.4)

The boundary condition ψ (a) = A sin ka = 0, this condition can be satisfied for A = 0, then ψ (x) = 0. However, if A = 0, we get a trivial, “non-normalizable” solution. Therefore, sin ka = 0 which means that:

ka = 0, ±π, ±2π, ±3π, …                                           (11.5)

The negative solution gives nothing new and the minus sign can be absorbed into A. So the distinct solution is given by:

${{k}_{n}}=\frac{n\pi }{a}$  with n = 1, 2, 3, …                                         (11.6)

With the help of boundary condition at x = a, the possible allowed values of energy E_n can be determined:

${{E}_{n}}=\frac{{{\hbar }^{2}}k_{n}^{2}}{2m}=\frac{{{n}^{2}}{{\pi }^{2}}{{\hbar }^{2}}}{2m{{a}^{2}}}$                                                   (11.7)

In order to fix the constant A, we have to normalize ψ:

$\int\limits_{0}^{a}{{}}$|A|² sin² (kx) dx = 1, so that: |A|² = ${}^{2}/{}_{a}$                                (11.8)

The simplest is to pick the positive real root for constant A: A = $\sqrt{{2}/{a}\;}$then, the solutions inside the potential well are:

${{\psi }_{n}}(x)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right)$                                              (11.9)

ψ₁ is the state with the lowest energy, the so-called ground state, and the other states with higher energies are the excited states. Notice that the energy increases in proportion to n² and each successive state has one more node as we go up in energy (see figure 11.2 below where first three initial states are depicted). And this is a general property, that applies for any shape of the potential.

 Figure 11.2. The first three states of infinite square well.

As can be seen in figure 11.2, the states are alternately even and odd, with respect to the center of the potential well. This property applies whenever the potential is an even function.

Another important property of the states ψ_n is that, they are mutually orthogonal. And this property can be shown:

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx}=0$                                        (11.10)

For m ≠ n. And for m = n, the integral in equation (11.10) is equal to 1. We can also expressed equation (11.10) in terms of Kronecker delta function which is defined by: δ_mn = 0 if m ≠ n and if m = n, δ_mn = 1. Therefore,

$\int{{{\psi }_{m}}{{(x)}^{*}}{{\psi }_{n}}(x)dx={{\delta }_{mn}}}$                                   (11.11)

We say that the ψ’s are orthonormal.

Please note that any other function f (x), can be expressed as linear combination of state in equation (11.9), this is called Dirichlet’s theorem:

$f(x)=\sqrt{\frac{2}{a}}\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sin \left( \frac{n\pi }{a}x \right)}$                                       (11.12)

This expression for f (x), is in fact the Fourier expansion. The expansion coefficients c_n can be evaluated for a given f (x) and by applying the orthonormality of ψ_n. Thus, the m-th coefficient in the expansion of f(x) is given by:

${{c}_{m}}=\int{{{\psi }_{m}}}{{(x)}^{*}}f(x)dx$                                      (11.13)

The stationary states for the infinite square well are:

${{\Psi }_{n}}(x,t)=\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}$                              (11.14)

And the most general solution to the time-dependent Schrödinger equation is a linear combination of stationary states:

$\Psi (x,t)=\sum\limits_{n=1}^{\infty }{{{c}_{n}}\sqrt{\frac{2}{a}}\sin \left( \frac{n\pi }{a}x \right){{e}^{-i({{{n}^{2}}{{\pi }^{2}}\hbar }/{2m{{a}^{2}}}\;)t}}}$                           (11.15)