Modern Physics: Mastering Physics: The Finite Square-Well Potential: Bound States

Sunday, March 13, 2011

Mastering Physics: The Finite Square-Well Potential: Bound States

Learning Goal: To understand the qualities of the finite square-well potential and how to connect solutions to the Schrödinger equation from different regions.

The case of a particle in an infinite potential well, also known as the particle in a box, is one of the simplest in quantum mechanics. The closely related finite potential well is substantially more complicated to solve, but it also shows more of the qualities that are characteristic of quantum systems. The potential energy function for a finite square-well potential is

$U(x)= \Biggl \{ \matrix{ 0,\hfill & 0 \le x \le L, \hfill \cr U_0, & x<0 \hbox{ or } x>L, \cr}$

where

is a positive number that measures the depth of the potential well and

is the width of the well.

The figure is a graph of potential energy versus position, which shows why this is called the square-well potential. Inside the well (i.e., for $0 \le x \le L$ ) the solutions take the form $\psi(x) = A\cos kx + B\sin kx$ , where

and

are constants and $k=\sqrt{2mE}/\hbar$ . Outside the well, the solutions take the form $\psi(x)=Ce^{\kappa x}+De^{-\kappa x}$ , where

and

are constants and $\kappa=[2m(U_0-E)]^{1/2}/\hbar$ . In this problem, you will consider a particle in a state with energy E<U_0

. Such states are called bound states, because classically the particle would be trapped in the potential well.

Part A

For a one-dimensional wave function to be normalizable, it must go to zero as

goes to infinity or negative infinity. Consider the wave function $\psi(x)=C'e^{\kappa x}+D'e^{-\kappa x}$ in the region x>L

. As

goes to infinity, this must become zero. What does this imply about the constants

and

Notice that the wave function is nonzero in the entire domain x>L

, even though this region would be forbidden by classical mechanics since E < U_0

. This "tunneling" into the classically forbidden region is a key difference between classical and quantum mechanics. Also notice that, since $\psi(x)>0$ anywhere in the x>L

region, there is some small, but nonzero, probability of finding the particle hundreds of kilometers away from the potential well.

Part B

Now, consider the wave function $\psi(x)=Ce^{\kappa x}+De^{-\kappa x}$ in the region x<0

. As

goes to negative infinity, this must become zero. What does this imply about the constants

and

? (Be careful of signs.)

Again, you see that the wave function is nonzero throughout the entire classically forbidden region.

Part C

Recall that a physical solution to the Schrödinger equation must be continuous everywhere. Consider the two branches

$\psi(x)=\Biggl \{ \matrix{ Ce^{\kappa x},\hfill & x<0, \hfill \cr A\cos kx +B\sin kx, \hfill & 0 \le x \le L, \cr}$

that meet at the point x=0

. Use the continuity of the physical solution to find the value of

Express your answer in terms of

and

ANSWER:

Part D

The derivative must also be continuous everywhere to have a physical solution (unless there are points where the potential energy suddenly becomes infinite, which never happens for the finite square well). Take the derivative of the two branches that you looked at in Part C to find the value of

Express your answer in terms of

, and

ANSWER:

$\frac{C{\kappa}}{k}$

Part E

Since you found that C=A

in Part C, you can now write the equations for the wave functions as

$\psi(x)=\Biggl \{ \matrix{ A\cos kx + \frac{\kappa}{k} A \sin kx, \hfill & 0 \le x \le L, \hfill \cr D'e^{-\kappa x}, \hfill & x> L.\hfill \cr}$

These two branches meet at x=L

. Set them equal here to find an expression for $D'e^{-\kappa L}$ .

Express your answer in terms of

, and

ANSWER:

$D'e^{-\kappa L}$ =

$A{\cos}\left(kL\right)+\frac{{\kappa}}{k}A{\sin}\left(kL\right)$

Part F
The last boundary condition, continuity of the derivatives at , yields a similar equation: $-kA\sin kL + \kappa A \cos kL = -\kappa De^{-\kappa L}$ . Dividing this equation by your equation from Part E (and doing some algebra to simplify) gives $\tan(kL) = \frac{2\kappa k}{k^2 -\kappa^2}$ . This is a transcendental equation, which must be solved numerically or graphically. However, since and both depend on , the energy levels for the finite square-well bound states may be found from this equation. Instead of trying to do this, we will look at the behavior of this equation as $U_0 \rightarrow \infty$ . Solving for , you find $kL=\hbox{atan} [(2\kappa k)/(k^2-\kappa^2)]$ . In this limit, what value does the right-hand side of the equation approach? In other words, what is $\lim_{U_0 \to \infty} \hbox{atan}\left( \frac{2\kappa k}{k^2 -\kappa^2} \right)$ ?

Part F

The last boundary condition, continuity of the derivatives at x=L

, yields a similar equation:

$-kA\sin kL + \kappa A \cos kL = -\kappa De^{-\kappa L}$ .

Dividing this equation by your equation from Part E (and doing some algebra to simplify) gives

$\tan(kL) = \frac{2\kappa k}{k^2 -\kappa^2}$ .

This is a transcendental equation, which must be solved numerically or graphically. However, since

and

both depend on

, the energy levels for the finite square-well bound states may be found from this equation. Instead of trying to do this, we will look at the behavior of this equation as $U_0 \rightarrow \infty$ . Solving for

, you find $kL=\hbox{atan} [(2\kappa k)/(k^2-\kappa^2)]$ . In this limit, what value does the right-hand side of the equation approach? In other words, what is

$\lim_{U_0 \to \infty} \hbox{atan}\left( \frac{2\kappa k}{k^2 -\kappa^2} \right)$ ?

ANSWER:

Part G

This result tells you that, as U_0

goes to infinity, the equation reduces to $\tan(kL)=0$ . Substitute back in $k=\sqrt{2mE}/\hbar$ and solve for the energy levels E_n

in this limit. Use the fact that $\tan(n\pi)=0$ for any integer

Express your answer in terms of

, and

ANSWER:

$\frac{{\hbar}^{2}}{L^{2}}{\cdot}\frac{1}{2m}{\cdot}\left(n{\pi}\right)^{2}$

This result shows that the finite square-well solution becomes the particle-in-a-box solution as U_0

goes to infinity. Notice that n=0

seems to be allowed here, though it wasn't for the particle in a box. However, substituting n=0

gives a solution that cannot be normalized and thus is not a physical solution.

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