In the region 0 ≤x ≤a, a particle is described by the wave function ψ1(x)=-b(x^2-a^2) . In the r

In the region 0 ≤x ≤a, a particle is described by the wave...

Key Concepts

Continuity of the Wave Function

In quantum mechanics, the wave function must be continuous across the boundaries between regions, ensuring that the probability density does not exhibit any unphysical jumps. This condition is applied at the interfaces of piecewise-defined potentials to connect the different solutions smoothly.

Continuity of the Derivative

If the potential is well-behaved (not singular), the first derivative of the wave function is also required to be continuous at the boundaries. This ensures that the kinetic energy operator, which involves second derivatives of the wave function, does not introduce inconsistencies, and it leads to additional equations that help determine the unknown parameters in the solution.

Boundary Conditions in Piecewise Problems

Many quantum systems have potentials defined in segments, leading to solutions that are valid only in specified regions. The application of boundary conditions—both for the function and its derivative—is essential in connecting these segmental solutions and in determining relationships between constants in the overall solution.

Transcript

00:01 Hello, and in this question here, we're going to be looking at matching boundary conditions.

00:06 So we have a particle, and for x greater than zero and x less than a, the particle is described by the wave function, psi 1, a function of x, which is equal to minus b times x squared minus a squared, where b and a are unknown constants.

00:26 For the region of space, where x is greater than a, but less than w, the wave function is given by si 2, and this is equal to x minus d, all to be squared minus c, and for x greater than w, we have the wave function is equal to psi 3, which is equal to 0.

00:49 So by matching boundary conditions at x equals a and x equals w, we'd want to find out these unknown constants, and c and w in terms of a and b.

01:05 So let us first look at the boundary condition x equals a.

01:11 And what are our two conditions that we want to have? well first of all we want our wave function, si, a function of x, to be continuous at x equals a.

01:28 And similarly we also want the derivative of si, with respect to x to be continuous continuous at x equals a okay so if we implement the if we sub in si into if we look at our wave function of these two conditions we'll hope to find c and d in terms of a and b so what does this continuous mean well let's say we have some x -axis here and this is the point x equals a here to be continuous we'd hope that as we approach x equals a from left to right we will get the same value for si so we'd have if we're to evaluate si at point a okay so if you evaluate si at point a and we evaluate it from going from left to right we'd hope we will get the same value if we evaluated si of a approaching from right to left.

02:53 So what does this mean? well, in this region here, approaching from left to right, the wave function is given by si one of x.

03:03 And in this region here, from going from right to left, the wave function is si 2 of x.

03:12 So this means we want si i.

03:16 Evaluated at x equal to a to be equal to si 2 evaluated at x equals a.

03:24 Now we have defined what si 1 and side 2 are up at the beginning of the question.

03:30 So filling in for that, filling in x equals a there, we get minus b times a squared minus a squared is equal to a minus d to be squared minus c well the left -hand side is equal to zero so this means we get a minus d to be squared minus c is equal to zero or that c is equal to a minus d to be squared so this is our first equation and our second equation we'll get from the derivatives so so that the fact that the derivatives be equal at x equals a.

04:13 So the derivative of d -sy -1, d -x, is equal to, just by calculating the derivative, we get minus 2bx, and d -sy -2 -d -x is equal to 2 times x minus d.

04:33 So for the first derivative to be continuous, we want d -sy -1, evaluated at x -equals a to be equal to the derivative of si -2 with respect to x, evaluated at x equals a...

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