Time-Independant Schrodinger Equation: Free Particle and Particle in One-Dimensional Box

Free Particle

The eigenvalues \(E_i\) of the energy operator are the possible measurable values of the total energy of a quantum system. For a nonrelativistic (moving at speeds much less than the speed of light), massive particle that is an isolated system the total energy of the particle is just its kinetic energy:

$$E=\frac{1}{2}mv^2=\frac{p^2}{2m}.\tag{1}$$

The energy operator \(\hat{E}\) associated with this particle is the one with eigenvalues \(E_i=\frac{p^2}{2m}\). This operator is given by

$$\hat{E}=\frac{\hat{p}^2}{2m}.\tag{2}$$

In this example we’ll use Schrödinger’s time-dependent equation to solve for the wavefunction \(\psi(x,t)\) which will allow us to calculate the probability (as a function of position and time) of measuring any physical quantity:

$$iℏ\frac{∂}{∂t}\psi(x,t)=\hat{P}^2/2m\psi(x,t).\tag{3}$$

What exactly is \(\hat{P}^2\)? We can find out the answer to this question by letting \(-iℏ\frac{∂}{∂x}\) act on \(-iℏ\frac{∂}{∂x}\) to get \(\hat{P}^2=-ℏ^2\frac{∂^2}{∂x^2}\). Thus the energy operator is

$$\hat{E}=\frac{-ℏ^2}{2m}\frac{∂^2}{∂x^2}.\tag{4}$$ 

Schrödinger’s time-dependent equation for a nonrelativistic, massive, free particle is given by

$$iℏ\frac{∂}{∂t}\psi(x,t)=\frac{-ℏ^2}{2m}\frac{∂^2}{∂x^2}\psi(x,t).\tag{5}$$

 "A  wave function  that satisfies the nonrelativistic Schrödinger equation with  V  = 0. In other words, this corresponds to a particle traveling freely through empty space. The  real part  of the  wave function  is plotted here."\(^{[1]}\)

"A wave function that satisfies the nonrelativistic Schrödinger equation with V = 0. In other words, this corresponds to a particle traveling freely through empty space. The real part of the wave function is plotted here."\(^{[1]}\)

 

We can solve this differential equation to obtain the wavefunction \(\psi(x,t)\) shown in the animation on the right. The wavefunction moves to the right and becomes more dispersed as time goes on. Thus, initially we have a decent idea of where the particle is; if we let the particle move away, without disturbing it by not introducing any potential \(V(x)\), as time goes on the measurement of its \(x\) position will become more and more uncertain.


Particle in One-Dimensional Box

This video was produced by DrPhysicsA\(^{[2]}\).

 

Figure 1 (click to expand): Illustration of a free particle moving in a "one-dimensional box" which is trapped inside of an infinite potential well.

The time-independent Schrodinger equation is \(\frac{-ℏ^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi\) where \(E\) is the total energy of the system and \(V\) is the potential. We’ll start by considering a “free particle.” This is just a single particle as an isolated system. For simplicity we’ll imagine that the particle is confined to only moving along the x-axis and we’ll just think about the probability amplitudes \(\psi_x\) at each value of \(x\). Since the particle is an isolated system, \(V=0\) at each value of \(x\). (Potentials only exist when there is an external field present produced by some particles in the outside environment.)

Schrodinger’s equation simplifies to

$$\frac{-ℏ^2}{2m}\frac{d^2{\psi_x}}{dx^2 }=E\psi_x.$$

Let’s solve for the solution \(\psi_x\) of this differential equation by doing some algebra and making some substitutions; then once we solve for the probability amplitude \(\psi_x\) it’ll be straightforward to determine the probability distribution \(P(x)\) of the particle. Let’s multiply both sides of this equation by \(\frac{-2m}{ℏ^2}\) to get

$$\frac{d^2{\psi_x}}{dx^2}=\frac{-2m}{ℏ^2}=E\psi_x.$$

Since the free particle is an isolated system it does not have any potential energy. However it can have kinetic energy. Thus the total energy \(E\) of the system (which in this case is just the particle) is given by \(E=KE=\frac{1}{2}mv^2=\frac{P^2}{2m}\). The equation \(p=ℏk\) is the momentum of a quantum particle; substituting this into \(E\) we get \(E=ℏ^2\frac{k^2}{2m}\). Let’s substitute this into Schrodinger’s equation to get

$$\frac{d^2\psi_x}{dx^2}=\frac{-2m}{ℏ^2}\biggl(\frac{ℏ^2k^2}{2m}\biggl)\psi_x=-k^2\psi_x.$$ 

If we use the “guess-and-check” method to solve for a solution \(\psi_x\) to this differential equation, we’ll see that there are many different solutions. For example \(Acoskx\) and \(Bsinkx\) are different solutions. The general solution is given by \(\psi_x=Ae^{ikx}+Be^{-ikx}\) which, using trigonometry, can also be written as \(\psi_x=Acoskx+Bsinkx\) (\(A\) and \(B\) are different in this equation). If we differentiate this solution with respect to x twice and get \(-k^2\psi_x\), then we know that we guessed the right solution. Doing this we get

$$\frac{d\psi_x}{dx}=A(ik)e^{ikx}+B(-ik)e^{-ikx}⇒\frac{d^2ψ_x}{dx^2}=A(ik)^2e^{ikx}+B(ik)^2e^{-ikx}=-Ak^2e^{ikx}-Bk^2e^{-ikx}=-k^2(Ae^{ikx}+Be^{-ikx})=-k^2\psi_x.$$

If \(\psi_x=Ae^{ikx}+Be^{-ikx}\) then when we plug this into \(\frac{d^2\psi_x}{dx^2}=-k^2\psi_x\) both sides of the equation are indeed equal. Therefore this must be the solution. Using trigonometry we can rewrite the solution as \(\psi_x=A(coskx+isinkx)+B(cos⁡(-kx)+isin(-kx))=(A-B)coskx+(A-B)isinkx=Ccoskx+Dsinkx\) where \(C\) and \(D\) are just numbers. To save letters we’ll just rename \(C\) and \(D\) to \(A\) and \(B\).
Let’s now consider a different situation. Suppose that along the x-axis there is a constant potential \(V_0\). At every \(x\) the particle will have a constant potential energy given by \(PE=\frac{V_0}{m}\). Let’s use Schrodinger’s equation to solve for the probability amplitudes \(\psi_x\) for a particle in the presence of a constant potential \(V_0\). First let’s subtract \(V_0\psi_x\) from both sides to get

$$\frac{-ℏ^2}{2m}\frac{d^2\psi}{dx^2}=(E-V_0)\psi_x.$$ 

Next multiply both sides by \(\frac{-2m}{ℏ^2}\) to get

$$\frac{d^2\psi}{dx^2}=\frac{2m}{ℏ^2}(V_0-E)\psi_x.$$

Let \(k^2=\frac{2m}{ℏ^2}(V_0-E)\); then we have

$$\frac{d^2\psi}{dx^2}=k^2\psi_x.$$

Let’s use the “guess-and-check” method to solve for \(\psi_x\). Let’s guess \(\psi_x=Ae^{-kx}\) and plug it into the equation to see if both sides are equal:

$$\frac{d\psi_x}{dx}=A(-k)e^{-kx}⇒\frac{d^2\psi_x}{dx^2}=Ak^2e^{-kx}=k^2\psi_x.$$

Figure 2 (click to expand): The probability amplitude of measuring the particle at a position \(x\) in the presence of a potential \(V_0\) decreases exponentially with increasing \(V_0\). 

We see that this is indeed a solution. From this solution \(\psi_x=Ae^{-kx}\), we see that as the x distance increases \(\psi_x\) decreases exponentially. Since \(P(x)=\psi_x\psi_x^*\) the probability of finding the particle decreases exponentially with increasing \(x\). The bigger the value of \(k\) is (where \(k=\sqrt{2m\frac{(V_0-E)}{ℏ^2}}\)), the more rapidly \(\psi_x\) decreases with increasing \(x\). By increasing the constant value of \(V_0\), we see that k gets bigger; therefore the bigger \(V_0\) is the more rapidly \(\psi_x\) decreases with increasing \(x\). As \(V_0\) approaches \(∞\), \(\psi_x\) decreases from \(\psi_x=A\) (at \(x=0\)) to \(0\) immediately with increasing \(x\).


                                                                                   Figure 3 (click to expand): Illustration of a free particle moving in a one-dimensional box which is "pinned down" by a finite well. The probability amplitude of finding the particle outside of the box decreases exponentially as a function of distance.

                                                                                  Figure 3 (click to expand): Illustration of a free particle moving in a one-dimensional box which is "pinned down" by a finite well. The probability amplitude of finding the particle outside of the box decreases exponentially as a function of distance.

Suppose that a particle is moving along the x-axis in the presence of the potential \(V(x)\) as shown in figure 3. In the region where \(x≤0\) and \(x≥L\) there is a constant potential \(V_0\). We showed in the previous example that \(\psi_x\) for values of \(x\) where there is a constant potential \(V_0\) is given by \(\psi_x=Ae^{-kx}\). Suppose that \(V_0→∞\) in the regions where \(x≤0\) and \(x≥L\). If this happens then \(\lim_{V_0\to∞} \psi_x=0\) and we can say that \(\psi_x=0\) at all those values of \(x\). When the particle is in the region \(0<x<L\) where \(V=0\) it is an isolated system. Thus, in this region, the solution \(ψ_x=Acoskx+Bsinkx\) describes the probability amplitude associated with every \(x\) position. Let’s solve for the constants \(A\) and \(B\). To do this we’ll use the two boundary conditions \(\psi_x(0)=0\) and \(\psi_x(L)=0\). Using the first boundary condition \(\psi_x\) becomes

$$\psi_x(0)=0=Acos(k·0)+Bsin(k·0)=A.$$

Since \(A=0\) our solution \(\psi_x\) simplifies to \(\psi_x=Bsinkx\). Applying the second boundary condition, we get

$$\psi_x(L)=0=Bsin(kL).$$ 

Notice that although \(B=0\) would satisfy this boundary condition it would correspond to a solution \(\psi_x=0\) for all values of \(x\) within the range \(0<x<L\). Another way to satisfy this boundary condition (which as we shall see will lead to much less trivial solutions \(ψ_x\) is if \(sin(kL)=0\). \(sin(kL)=0\) when \(kL=nπ\)where \(n=0,1,2,…\). In other words when \(k=nπ/L\) the solution \(\psi_x\) satisfies \(\psi_x(L)=0\). If we substitute this value of \(k\) into the solution we get

$$\psi_x(x)=Bsin\biggl(\frac{nπ}{L x}\biggl)\text{                  (0<x<L).}$$

We see that inside the potential well the function \(\psi_x(x)\) is a standing wave. If we substitute \(k=nπ/L\) into \(E=\frac{ℏ^2k^2}{2m}\) and express \(ℏ\) in terms of Plank’s constant; we get

$$E_n=\frac{h^2}{8mL^2}n^2.$$ 

Schrodinger’s equation predicts that the total energy of a particle trapped in a potential well is quantized and comes in discrete values; in other words the energy distribution is not continuous as shown in Figure 4 below.

 Figure 4: Illustration shows the allowed energy levels of a particle trapped inside of a one-dimensional box.

Figure 4: Illustration shows the allowed energy levels of a particle trapped inside of a one-dimensional box.

The lowest possible energy level for a particle in a box is \(E_1=\frac{h^2}{8mL^2}\). Thus Schrodinger's equations predict that the energy \(E_n\) of this particle can never be zero; therefore the particle can never be at rest. The lowest possible energy level, given by \(E_1=\frac{h^2}{8mL^2}\), is called the ground-state energy. The allowed energy levels of the particle in the box are shown in Figure 4.


This article is licensed under a CC BY-NC-SA 4.0 license.

References

1. Wikipedia contributors. "Schrödinger equation." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 20 May. 2017. Web. 22 May. 2017.

2. DrPhysicsA. "Schrodinger Equation for Free Particle and Particle in a Box Part 2". Online video clip. YouTube. YouTube, 12 November 2010. Web. 05 May 2017.

3. Friedman, Art; Susskind, Leonard. Quantum Mechanics: The Theoretical Minimum. Basic Books, 2015. Print.