In this video, we will solve the potential step problem. In this problem, the potential profile is given as shown here. There is a boundary at z equals zero and to the left of it, for the negative z values, the potential is defined to be zero. To the right of this boundary for positive z values, the potential has a finite positive value, V naught. The particle with an energy E is incident from the left as shown here. Now, the energy E here is assumed to be greater than zero because if it is less than zero, then we won't have a propagating wave or an actual moving particle in that case, but the energy E can be either greater or smaller than the height of the potential barrier. Now, please note the difference between this problem and the harmonic oscillator or the potential well problem that we solved before. In those cases, the energy E is part of the solution that we had to find. We need to solve the eigenvalue equation, the Schrodinger's time-independent equation and had to find the energy E and the wave functions. In this case, energy E is given to us as a part of the initial condition and the difference here is because this here is a scattering problem. We're not talking about a bound or confined states. In the case of finite potential well or harmonic oscillator case, it's a bound system and energy is part of the solution that we have to find in this potential step or the potential barrier problem that we will discuss next it is an unbound system. It's a scattering problem. The energy of the particle is given as part of the initial condition. Now, in either case, we need to write down and solve the time-independent Schrodinger equation. We write down the equation for two separate regions; region 1 where z is negative and therefore the potential is zero, the Schrodinger's equation is like this. We can rewrite the same equation as shown here in a simpler form by defining a parameter k naught as shown here. For region 2, where z is positive and the potential is V naught, the Schrodinger's equation is shown here and we can again cast it into a simpler form by defining a parameter k, as shown here. You can notice that these two equations, the Schrodinger's equation for region 1 and region 2 are essentially identical. The only difference is that this number here, multiply to the wave function in the 2nd term of the equation is k naught square here and k squared in region 2. That's the only difference. Their solutions will have the same functional form. For region 1, the solution is given by sine and cosine or exponential plus or minus e^ik naught z, and we add constants A and B to be determined. For region 2, the solution has the same functional form, e^ikz or e^negative ikz, and once again, we multiply these unknown constants to it as a general solution and these constants are to be determined by the boundary conditions. Now, these wave functions represents propagating wave. Here, we're assuming that k_naught and k are both a real number. In that case, they represents a sine and cosine waves and assuming a time-dependence of exponential negative i Omega t. This exponential, ik_naught z and exponential ikz, both represents a wave propagating to the right, along the positive z direction, whereas exponential negative ik_naught z and exponential negative ikz represent the left-propagating wave, wave propagating towards the negative z direction. Now, the initial condition is given to us that the particle is coming from the left, and therefore, there shouldn't be any left-propagating wave in region 2. In region 2, only right-propagating wave is possible, which originates from the incident wave transmitting through the boundary. That means D equals 0, this term goes away. The wave function in region 2 has only one term, C constant to be determined times e^ikz. Now we apply the boundary condition at the interface, at the boundary to equals zero and we require, as usual, the continuity of the wave function Psi and the derivative. Those two conditions give you these two equations, six and seven. There are two equations and three unknowns, so we cannot determine all three constants unambiguously. Instead, we try to represent two of these constant in terms of the other. We choose constant A to be the free remaining parameter and express constant B and C in terms of A. In other words, determine the ratio B over A and C over A, and that gives you this equation eight here. Using that, we can write down the wave functions for region 1, negative z, and region 2, positive z, as shown here. We still have these constant A to be determined by the normalization condition. But as you will see, this is a scattering problem and we are interested in obtaining the reflection probability and transmission probabilities. To determine those quantities, we actually don't have to determine the constant A. Now we have found the general solution, we can consider two different cases. First, when the energy of the incident particle is greater than the potential step V_naught. In that case, the k_naught here, because our energy is always positive, k_naught is always real and positive. K, when energy is greater than V_naught, then k also is real and positive, and k_naught is always greater than k also. So in this case, the right-propagating wave in region 1, the first term in the solution that was given in the previous slide, represents the incident wave. The second term in the solution for region 1, which is the left-propagating wave, that represents the reflected wave. The right-propagating wave in region 2, and the solution for region 2 has only one term, if you remember, and that represents the transmitted wave. By comparing the amplitude of the reflected wave to the incident wave and also the amplitude of the transmitted wave to that of the incident wave, we can calculate the reflection and transmission coefficient, and from that, calculate the reflection and transmission probability. Define the reflection and transmission coefficient as the ratio of the amplitude of the reflected wave to the incident wave here for reflection coefficient, and the ratio of the amplitude of the transmitted wave to the incident wave for the transmission coefficient. We already found that ratio in the previous slide. The probability of reflection is simply given by the absolute value squared of the reflection coefficient, and that is given by this equation here. The transmission probability we calculate by 1 minus R squared. That is related to the absolute value square of the transmission coefficient, but has this additional term factor multiplied to it. This ratio, k over k naught, basically accounts for the fact that the wave travels at different speed in region 1 and region 2. If you plot the reflection probability and the transmission probability, the reflection probability simply decays monotonically as you increase your energy. The energy here is expressed as the ratio of E over V naught, the potential step barrier height. So the E over V naught being one, meaning that the energy is exactly at the barrier height. At that point, your reflection probability is one. As you increase the energy beyond that, it decreases and the transmission probability is just the opposite of it. Now, we can look at the other case where the energy of the incident particle is smaller than the height of the potential step here. In this case, k naught still remains real and positive, but k becomes purely imaginary. The quantity inside the square root becomes negative, and so k is imaginary. We rewrite k as this, i times Kappa. Kappa is a real positive quantity defined here, so V naught is this time larger than E. This quantity is positive, so Kappa here is a real positive quantity. That makes the transmitted wave an exponentially decaying wave as opposed to a sinusoidal propagating wave. The reflection coefficient can be written as this. If you take the absolute value square of this quantity, you will find that to be one. When the incident particle has energy is smaller than the barrier height, they always reflect if there is no transmission, 100 percent reflection probability, which is consistent with our physical intuition. What is not consistent with our classical physical intuition is the fact that there is a non-zero probability of finding the particle inside the barrier region. It does penetrate, it decays exponentially, and eventually reaches zero, but there is a non-zero probability of finding the particle inside the barrier region. In other words, the particle actually does penetrate into the barrier, and this is a non-classical behavior.
