# 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."$$^{}$$

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$$^{}$$.

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.

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.

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.