Ch 4: S.E. Example1: Free Particle

As with the HJE, our first example is going to be a single “Free Particle”. The potential will be V = 0, and the Schrodinger Equation becomes

(1)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\frac{1}{2m}\nabla ^2\Psi =0 \end{equation*}

(2)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\frac{1}{2m}\nabla ^2\Psi =i\frac{\partial \Psi }{\partial t}+\frac{1}{2m}(\frac{\partial ^2\Psi }{\partial x^2}+\frac{\partial ^2\Psi }{\partial y^2}+\frac{\partial ^2\Psi }{\partial z^2})=0 \end{equation*}

For now, we’re going to simplify this even further by reducing the system down to 1 dimension:

(3)   \begin{equation*} i\frac{\partial \Psi }{\partial t}+\frac{1}{2m}\frac{\partial ^2\Psi }{\partial x^2}=0 \end{equation*}

As far as actually solving this thing, I’m once again going to refer you to an outside source (See Griffith’s Intro to QM book for lucid descriptions about how to solve all of the examples for the next couple of chapters). We’ll start by looking at separable (uniform energy/constant density) solutions. The solution follows the Separation of Variables procedure from the last section. The spatial part of the solution is

(4)   \begin{equation*} \psi (x)=A e^{i k x}+B e^{-i k x} \end{equation*}


(5)   \begin{equation*} k = \sqrt{2m E} \Longleftrightarrow  E=\frac{k^2}{2m} \end{equation*}

Apparently, k is the momentum of the particle. Or is it?! Let’s take a closer look. We’ll start by parameterizing the constants A and B with one variable, \alpha, to make \psi(x) look like this:

(6)   \begin{equation*} \psi (x)=Cos(\alpha *\pi /2)e^{i k x}+Sin(\alpha *\pi /2)  e^{-i k x} \end{equation*}

In the simplest case, \alpha = 0 or 1. Then,

(7)   \begin{equation*} \psi (x)=e^{\pm i k x}, \Longrightarrow  \nabla S=momentum=\pm k \end{equation*}

In this case, the momentum really is k, and all particles in the fluid move uniformly at the same speed. Also note that R, being the magnitude of \Psi, is just 1. It follows then that the Quantum Potential, being proportional to the 2nd derivative of R, is 0. In other words all of the system’s Energy is in the form of Kinetic Energy. Also note that the result is identical to the Classical HJE Free Particle from Chapter 2.1. This is as expected since the Quantum Potential vanishes.

Another interesting case is when \alpha = 1/2. In this case,

(8)   \begin{equation*} \psi (x)=\frac{e^{-i k x}}{\sqrt{2}}+\frac{e^{i k x}}{\sqrt{2}}=\sqrt{2} Cos(k x)=R(x) \end{equation*}

(9)   \begin{equation*} S(x)=0 \Longrightarrow  \nabla S=momentum=0 \end{equation*}

The Quantum Potential is

(10)   \begin{equation*} V_Q=\frac{1}{2m}\frac{\nabla ^2R}{R}=\frac{k^2}{2m}=E \end{equation*}

So the situation is reversed! Now we have no momentum and a uniform Quantum Potential equaling the total Energy, last time it was just the opposite. The system’s energy is “charged up” in the Quantum Potential, just like a compressed spring. Figure 3.1.1 shows an example of a density function that produces this constant Quantum Potential.

So what about a “mixed” case? Let’s try \alpha = 1/4. And let’s also set k = 1 for the sake of simplicity. Then,

(11)   \begin{equation*} \psi =e^{i x} Cos(\frac{\pi }{8})+e^{-i x} Sin(\frac{\pi }{8}) \end{equation*}

(12)   \begin{equation*} R^2=\psi (x)*\psi ^*(x)=1+\frac{Cos(2 x)}{\sqrt{2}} \end{equation*}

(13)   \begin{equation*} R=\sqrt{1+\frac{Cos(2 x)}{\sqrt{2}}} \end{equation*}

(14)   \begin{equation*} momentum=\nabla S=\frac{\sqrt{2}}{2+Sin(\frac{\pi }{4}-2 x)+Sin(\frac{\pi }{4}+2 x)} \end{equation*}

(15)   \begin{equation*} V_Q=\frac{1}{2m}\frac{\nabla ^2R}{R}=\frac{4 Cos(2 x)+\frac{3+Cos(4 x)}{\sqrt{2}}}{4 \sqrt{2} (1+\frac{Cos(2 x)}{\sqrt{2}})^2} \end{equation*}

Below are graphs showing some of the relevant features of this state.

Figure 4.1.1: Some quantities of interest for the "mixed" uniform energy state

Figure 4.1.1: Some quantities of interest for the “mixed” uniform energy state

And the animation below shows the Density, Momentum, Quantum Force (the derivative of the Quantum Potential), and a sample of actual particle trajectories.

Figure 4.1.2:  Particle motion for the 1-D "mixed state".

Figure 4.1.2: Particle motion for the 1-D “mixed state”.

Let’s bump back up to 3 dimensions and look at another important example: the Gaussian Distribution. This is another very standard problem, so I won’t dwell on it for long. I will say that it consists of an infinite sum of uniform energy (separable) solutions and has an initial density of

(16)   \begin{equation*} \Psi (t=0)=\sqrt{(\frac{2 \sqrt{2} a^{3/2}}{\pi ^{3/2}})}e^{-a\vec{Q}^2} \end{equation*}

Where ‘a’ is an arbitrary constant. Solving the Schrodinger Equation with this initial state gives us

(17)   \begin{equation*} \Psi (\vec{Q},t)=\frac{2^{3/4} a^{1/4} e^{-\frac{a m (\vec{Q}^2)}{m+2 i a t}} m}{\pi ^{3/4} (m+2 i a t) \sqrt{\frac{1}{a}+\frac{2 i t}{m}}} \end{equation*}

What is the velocity field for this system? Remember, the Gradient of the phase is the momentum. Is there a simple way to extract this quantity from \Psi without finding the phase first? It turns out there is. Note that

(18)   \begin{equation*} \frac{\nabla \Psi }{\Psi }=\frac{\nabla (R e^{i S})}{(R e^{i S})}=\frac{\nabla R e^{i S}+R(i\nabla S)e^{i S}}{(R e^{i S})}=\frac{\nabla R}{R} +i\nabla S \end{equation*}

Which implies

(19)   \begin{equation*} \vec{v}=\frac{\nabla S}{m}=\frac{1}{m}Im(\frac{\nabla \Psi }{\Psi }) \end{equation*}

(You can alternatively find \nabla S as \nabla (Im(Log(\Psi))), which is sometimes better computationally). Now we can compute the velocity field:

(20)   \begin{equation*} \vec{v}=\frac{4 a^2  t}{m^2+4 a^2 t^2}\vec{Q} \end{equation*}

To look at a particular example, let’s insert a = 1, m = 1837, and show the particle trajectories in an animation.

Chapter 5


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