Particles incident from the left in Figure P 28.76 are confronted with a step in potential energ

Particles incident from the left in Figure P 28.76 are...

Key Concepts

Step Potential

The step potential is a fundamental model in quantum mechanics where the potential energy of a particle changes abruptly at a certain point in space. This model helps illustrate how particles behave when encountering sudden changes in potential energy, providing insights into phenomena that have no classical analogue, such as partial reflection when the particle energy exceeds the potential step.

Quantum Scattering

Quantum scattering involves the study of particles interacting with potential barriers or steps, leading to phenomena like reflection and transmission that are described probabilistically. It is a key concept for understanding how a quantum particle’s wave nature influences its behavior, contrasting with classical expectations by allowing non-zero reflection probability even in cases where the particle has sufficient energy to overcome the potential change.

Wave Function Continuity and Boundary Conditions

In quantum mechanics, the solutions to the Schrödinger equation must satisfy continuity conditions at boundaries where the potential changes. Both the wave function and its first derivative are required to be continuous, ensuring the physical acceptability of the solutions. These matching conditions are used to determine the amplitudes of the reflected and transmitted waves, and are essential for deriving expressions like the reflection coefficient.

Reflection Coefficient

The reflection coefficient is defined as the ratio of the probability density flux reflected back from the potential step to the incident probability density flux. It quantifies the proportion of particles that do not transmit across the step and instead are reflected, encapsulating the quantum mechanical deviation from classical predictions. The derivation of its closed form involves solving for the wave amplitudes on either side of the potential discontinuity using the boundary conditions.

Transcript

00:01 In this question, we want to find what is the reflection coefficient, basically described as r goes to b square over a square.

00:15 In the case when we have an incoming particle with some wave function, and it meets a barrier, over here, this is our potential barrier, u where u is actually less than e the total energy of our particle right and somebody gets reflected someone it gets transmitted as side 2 and we're going to find how we're going to get this reflection coefficient which is described as b square way square why b square because b over here this part of the wave function describes a wave a plain wave that is traveling in the opposite direction right to the left whereas a on the other hand this is describing a plane wave that is traveling to the right this is our initial incoming wave so by taking b square over a square we're actually taking the absolute square of the incoming, sorry, this is the reflected, reflected size square of the reflected wave function, divided by the wave function of the incoming square.

01:53 Right, and this gives us the amount of wave function, it's a bit hard to describe, but amount of probability that would be reflected when we have a certain amount of incoming wave.

02:17 So it is the reflection coefficient.

02:22 Now we're going to start off by first checking that this wave functions actually satisfy the schrodinger's equation.

02:33 And using that schrodinger's equation, we can actually tell what it is our k1 and what is our k2.

02:41 So first off, the schrodinger's equation, let me just write that.

02:45 Down.

02:59 So for the first part, the wave function, we will have to differentiate this twice with respect to x and we should get negative k1 square.

03:20 All right, and this is our second derivative putting it into the expression.

03:50 It is actually just negative square times si get negative hbar square over 2m times negative k1 square plus u when in this case u is 0 right when it is still in this left region right so u is 0 is equate this to e si therefore from here we can see that it actually sorry i'm missing a psi over here, there should be k1 square times si.

04:37 So we see over here that it actually satisfies our schrodinger's equation because it is just a constant multiplied by the wave function, the original wave function, right? well, constant is over here, which implies that the energy is just k1 square, hbar square, over 2m.

04:59 So k1 square, hb a square over 2m e right so we square root both sides so be our expression for k1 now we can do the sorry this is slightly wrong i should this needs to be flipped right this is actually 2m e over h bar square square root so k1 you do the same process for k2.

05:45 So now we are looking at the second wave function, which is equals to c, e .3 power ik, k2x, differentiating it twice, respect to x, get negative k2 square times the original wave function.

06:18 Substituting this into the shrewangler sequence equation, this is what we get.

06:32 But in this case, our u is no longer zero, right? you have to account for this u.

06:37 We can actually bring you to the right -hand side.

06:52 It's equals to e -minus u times side.

06:57 So they satisfy the schrodinger's equation if our e minus u is equal to hbars square k2 square over 2m.

07:07 And from this we can actually derive for this k2 square or k2 just bring 2m e minus u tibre by hbar square take the square root of the entire expression now what you want to do in order to actually infer what is our c what is our a and what's our b by matching the boundary conditions so what you mean by matching the boundary conditions is that because we expect the wave function to be a continuous function throughout all space, it means that at this boundary between here, the wave functions must agree with each other at this particular point.

08:07 The same idea also goes for the gradient, right? so it must be a smooth, differentiable equation for the wave function and therefore at this particular point the gradient because it is smooth they must have the same gradient at this boundary again so these two different conditions that we all have to satisfy and we are given that this point over here is when x equals to zero right this x equals to zero point for more convenience actually you choose x equals to 0 to be the boundary point so at that point we just have to equate side 1 when x equals to 0 must be goes to side 2 when x equals to 0 from this we can actually get a plus b must be equals to c very simple for the second condition for it to be smooth implies that the differentiation, or the gradient of the wave function, x equals to 0, 3 equals to the gradient from side 2, and x equals to 0.

09:51 So if you differentiate the first function like this, you will get ik1.

10:45 Now substituting in x equals to 0 for all these terms.

10:50 It means that the exponential terms will all go to just 1, because when e to the power of 0 is 1.

11:00 So this would be, and we can divide by the common factor i.

11:05 You see that i appears in order of the terms.

11:08 So this would be a minus b times k1, it's equals to c times k2.

11:20 So this is our second equation.

11:29 Now in order to find our ratio, remember that we want to find the ratio of b square over a square, it means that we want to get rid of the c factor...

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