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)

9. Quantum Mechanics: Further Interpretation of Wave Function

Born’s statistical interpretation of wave function says that: ${{\left| \Psi (x,t) \right|}^{2}}dx$ is the probability that the particle will be found at a coordinate between x and x+dx at time t. Consequently, the total probability of finding the particle somewhere should be equal to 1. Referring to the total probability for continuous distributions: $\int\limits_{-\infty }^{\infty }{\rho (x)dx}=1$ (please see Equation (8.10) in my previous post), the integral of ${{\left| \Psi (x,t) \right|}^{2}}dx$ must be equal to 1:

$\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx=1}$                                                    (9.1)

The wave function $\Psi (x,t)$ is a solution of Schrödinger equation:

$i\hbar \frac{\partial \Psi }{\partial t}=-\frac{\hbar }{2m}\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}+V\Psi$                                             (9.2)

And also, $A\Psi (x,t)$ is a solution of equation (9.2), where A is any (complex) constant. Since the condition in equation (9.1) has to be satisfied, we have to normalize the wave function to find this multiplicative constant. Please note that the wave function $\Psi (x,t)$  must go to zero as |x|→∞, this physically realizable states correspond to the ‘square-integrable’ solutions to Schrödinger’s equation.

Now, we are going to proof that a normalized wave function stays normalized for all future time, i.e. the normalization of the wave function is preserved.

Straightforwardly, by taking the first order derivative of the term on the left hand side of equation (9.1) with respect to time t, we obtain:

$\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx}=\int\limits_{-\infty }^{\infty }{\frac{\partial }{\partial t}{{\left| \Psi (x,t) \right|}^{2}}dx}$                                       (9.3)

The wave function $\Psi (x,t)$ is a function of place x as well as time t, therefore we take the partial derivative of the integrand on the right hand side of equation (9.3). Working out further the integrand on the right hand side of equation (9.3) with the help of product rule, we obtain:

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{\partial }{\partial t}$(Ψ^*Ψ) = Ψ^*$\frac{\partial \Psi }{\partial t}$ + $\frac{\partial {{\Psi }^{*}}}{\partial t}$Ψ                                         (9.4)

Taking the complex conjugate of Schrödinger equation (9.2):

$\frac{\partial {{\Psi }^{*}}}{\partial t}=-\frac{i\hbar }{2m}\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}+\frac{i}{\hbar }V{{\Psi }^{*}}$                                          (9.5)

Substituting the Schrödinger equation and its complex conjugate into equation (9.4):                                

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{i\hbar }{2m}$(Ψ^*$\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}$ - $\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}$Ψ)                                          (9.6)

Resolving this equation further, we get:

$\frac{\partial }{\partial t}$|Ψ|² = $\frac{\partial }{\partial x}$[$\frac{i\hbar }{2m}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ)]                                      (9.7)

                                    

By substituting equation (9.7) into (9.3),  the integral can then be analyzed explicitly: 

$\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{}}$|Ψ(x,t)|²dx = $\frac{i\hbar }{2m}$(Ψ^*$\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}$ - $\frac{{{\partial }^{2}}{{\Psi }^{*}}}{\partial {{x}^{2}}}$Ψ)|$_{-\infty }^{\infty }$                                      (9.8)

However, the wave function $\Psi (x,t)$ must go to zero as x goes to infinity, therefore we have: $\frac{d}{dt}\int\limits_{-\infty }^{\infty }{{{\left| \Psi (x,t) \right|}^{2}}dx=0}$  hence that the integral is constant, i.e. independence of time. Thus the wave function stays normalized for all future time. QED

Consider the expectation value of x for a particle in state Ψ given by:

$\left\langle x \right\rangle =\int\limits_{-\infty }^{\infty }{x}$|Ψ(x,t)|²dx                                                          (9.9)

Equation (9.9) gives the average of measurements performed on all particles in the state Ψ. Thus, the expectation value does not give the average of repeated measurements on one identical system. Rather it is the average of repeated measurements on a group of identical systems.

Due to the time dependence of Ψ, the expectation value of x will change and it is interesting in knowing how fast $\left\langle x \right\rangle $ will change. By taking the derivative of equation (9.9) with respect to time, we get:

$\frac{d\left\langle x \right\rangle }{dt}=\int{x\frac{\partial }{\partial t}}$|Ψ|² dx                                                       (9.10)

By replacing the term $\frac{\partial }{\partial t}$|Ψ|² for $\frac{\partial }{\partial x}$[$\frac{i\hbar }{2m}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ)] (referring to equation (9.7)), equation (9.10) becomes:

$\frac{d\left\langle x \right\rangle }{dt}=\frac{i\hbar }{2m}$∫ x$\frac{\partial }{\partial x}$(Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ) dx                                   (9.11)

Using integration by parts: $\int\limits_{a}^{b}{f\frac{dg}{dx}dx}=fg$|$_{a}^{b}-\int\limits_{a}^{b}{\frac{df}{dx}gdx}$; and under the condition that the ∫ |Ψ|² dx  must be finite: Ψ and its derivative ${}^{\partial \Psi }/{}_{\partial x}$ must go to zero at ± ∞, equation (9.11) can be simplified:

$\frac{d\left\langle x \right\rangle }{dt}=-\frac{i\hbar }{2m}$ ∫ (Ψ^*$\frac{\partial \Psi }{\partial x}$ - $\frac{\partial {{\Psi }^{*}}}{\partial x}$Ψ) dx                                   (9.12)

 

Carrying out another integration by parts, we conclude that:

$\frac{d\left\langle x \right\rangle }{dt}=-\frac{i\hbar }{m}\int{{{\Psi }^{*}}\frac{\partial \Psi }{\partial x}dx}$                                                 (9.13)

Equation (9.13) is the velocity of the expectation value of expectation value of x which is not the same thing as the velocity of the particle. Since the particle doesn’t even have a determinate position prior to the measurement, it is not even clear what velocity means in quantum mechanics. It is just the probability of getting a particular value of velocity. However, equation (9.13) tells us how to calculate the expectation value of velocity straightforwardly from Ψ and for our present purposes, we adequately could postulate that the expectation value of the velocity <v> is equal to the time derivative of the expectation value of position <x>:

$\left\langle v \right\rangle =\frac{d\left\langle x \right\rangle }{dt}$                                                           (9.14)

 Figure 9.1. Working out of expectation value of velocity v

In quantum mechanics it is more convenient to work with momentum (p=mv) rather that velocity:

$\left\langle p \right\rangle =m\frac{d\left\langle x \right\rangle }{dt}=-i\hbar \int{\left( {{\Psi }^{*}}\frac{\partial \Psi }{\partial x} \right)dx}$                                    (9.15)

To summarize, we rewrite the expressions for <x> and <p>:

$\left\langle x \right\rangle =\int{{{\Psi }^{*}}(x)\Psi dx}$                                                    (9.16)

$\left\langle p \right\rangle =\int{{{\Psi }^{*}}(\frac{h}{i}\frac{\partial }{\partial x})\Psi dx}$                                                 (9.17)

The term x that represents position and the term $\left( \frac{\hbar }{i}\frac{\partial }{\partial x} \right)$ that represents momentum respectively in equation (9.16) and (9.17) are the operators in quantum mechanics.

All dynamical variables can be expressed in terms of position and momentum, for instance, kinetic energy: $T={}^{1}/{}_{2}m{{v}^{2}}={}^{{{p}^{2}}}/{}_{2m}$ . Generally, to calculate the expectation value of a quantity, the following equation can be applied:

$\left\langle Q(x,p) \right\rangle =\int{{{\Psi }^{*}}Q(x,\frac{\hbar }{i}\frac{\partial }{\partial x})\Psi dx}$                                   (9.18)

And for kinetic energy:

$\left\langle T \right\rangle =\frac{-{{\hbar }^{2}}}{2m}\int{{{\Psi }^{*}}\frac{{{\partial }^{2}}\Psi }{\partial {{x}^{2}}}dx}$                                               (9.19)