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# 1 Example 1: quantum particle in an infinitely narrow/deep well The 1D non-dimensional Schrödinger equation is where ψ is the “wave function” (such that.

## Presentation on theme: "1 Example 1: quantum particle in an infinitely narrow/deep well The 1D non-dimensional Schrödinger equation is where ψ is the “wave function” (such that."— Presentation transcript:

1 Example 1: quantum particle in an infinitely narrow/deep well The 1D non-dimensional Schrödinger equation is where ψ is the “wave function” (such that | ψ(x) | 2 is the probability of finding the particle at a point x ), E is the particle’s energy, and u(x) is the potential of an external force. If, for example, the particle is an electron in an atom, u(x) would be the potential of the force of attraction exerted by the nucleus. (1) Week 12 3. Differential equations involving generalised functions

2 Let Thus, Eq. (1) becomes (3) We’ll be interested in bound states, for which (2) (Solutions that don’t decay as x → 0 describe free particles.) We assume – and will verify once we’ve found the solution – that ψ(x) is continuous, but ψ'(x) has a discontinuity at x = 0 [caused by the presence of δ(x) in Eq. (3)]. Thus, ψ(x) should look like...

3...and the derivative, ψ'(x), should look like...

4 To figure out how the delta function affects the solution, integrate Eq. (3) from –ε to +ε, which yields hence, (4) hence, taking the limit ε → 0,

5 Next, multiply Eq. (3) by x and, again, integrate from –ε to +ε : Integrating the first term by parts and evaluating the second one, we obtain (5) Finally, taking the limit ε → 0, we obtain

6 Eqs. (4)-(5) are often called the matching conditions, as they ‘match’ (connect) the regions x 0. These conditions describe how the delta function affects the solution. Now we can solve Eq. (3) for x 0 separately and then match the two solutions across the point x = 0 (without actually dealing with the delta function).

7 Re-write the Eq. (3) for x ≠ 0 as follows: Note that we expect E to be negative (as it should be for bound states) – hence, it’s convenient to use – E instead of E. The general solution of the above equation is where A – and A + are undetermined constants.

8 Imposing the boundary condition (2), we obtain Imposing the matching conditions (4)-(5), we obtain Solving for A ±, one can see that a solution exists only if which is the energy of the (only existing) bound state.

9 where S(x) is the ‘source density’, L is the half-size of the resonator, and c is the speed of sound. (6) Example 2: the Green’s function of a boundary-value problem Waves generated in a resonator by a distributed monochromatic source of frequency ω are described by (7)

10 Physically, one should expect that, regardless of the initial condition, all of the wave field will sooner or later have the frequency of the source. Thus, let’s seek a solution of the form and Eqs. (6)-(7) yield (8) (9) Eqs. (8)-(9) are what we are going to work with (the preceding equations were included just to explain the problem’s physical background). It can be demonstrated that the solution of (8)-(9) is...

11 where G(x, x 0 ) satisfies (10) the Green’s function (11) (12) Physically, (11)-(12) describe the wave field from a point source of unit amplitude, located at x = x 0. Then, solution (10) implies that we treat the original (distributed) source as a continuous distribution of point sources and just sum up (integrate) their contributions.

12 As follows from (11), the [...] in the above equality equals δ(x – x 0 ) – hence, It is now evident that the l.-h.s. of (8) does equal its r.-h.s. To prove that (10) is indeed a solution, substitute (10) into Eq. (8) and show that its l.-h.s. equals its r.-h.s.

13 As in Example 1, we need to understand how the delta function affects G. To do so, integrate (11) from x 0 – ε to x 0 + ε, (13) Next, multiply (11) by (x – x 0 ) and repeat the same procedure again. Integrating the term involving ∂ 2 G/∂x 2 by parts, we obtain (14)

14 As before, assume that G is continuous, but ∂G/∂x involves a jump at x = x 0. Thus, taking in (13)-(14) the limit ε → 0, we obtain hence, (16) and (15) confirm our assumption that G is indeed continuous, whereas its derivative ‘jumps’ (both, at x = x 0 ). (15) (16)

15 Conditions (15)-(16) ‘match’ the solutions for x x 0. The delta function doesn’t contribute to either of these regions (as it’s ‘localised’ at x = x 0 ), so we can forget about it. For x < x 0, we have hence, Similarly, where k = ω/c, and A – is an arbitrary constant. (17) (18)

16 Use (15)-(16) to match (17) and (18) at x = x 0 : Solve these equations for A ± : where we have used the formula (19) with a = x 0 – L and b = x 0 + L.

17 Summarising Eqs. (17)-(19), we obtain Observe that G(x, x 0 ) is symmetric with respect to interchanging x and x 0, which is typically the case with Green’s functions. This property is usually called the “mutuality principle”. Physically, it means that, if the wave source and the ‘measurer’ swap positions, the measurement doesn’t change.

18 4. The formal definition of generalised functions We’ll formalise GFs by putting them in correspondence to linear functionals defined for some unproblematic (well-behaved) test functions. ۞ An functional G assigns to a function f(x) a number (which we shall denote [G, f(x)] ). Example 3: (a) is a linear functional, (b) is a nonlinear functional.

19 ۞ A function f(x) such that ۞ The space T of all analytic functions with compact support is called “the space of test functions”. where a < b, is said to have compact support. Example 4: an analytic function with compact support where a < b.

20 ۞ A generalised function is a linear functional defined on T. Example 5: To avoid confusion, distinguish in the above the functional G and the corresponding function g(x) [and the test function f(x) ]. (a) The delta function is a linear functional (and, thus, a GF): (b) A ‘regular’ function g(x) can be treated as a GF: the ‘regular’ function associated with G the functional the test function

21 ۞ The derivative of a GF G is the GF G' such that Theorem 1: If a ‘regular’ function g(x) corresponds to a generalised function G, then g'(x) corresponds to G'. Proof: By definition, Integrating the r.-h.s. by parts and recalling that f(x) is a function with compact support, we obtain...

22 as required. █ Example 6: a GF unrelated to the delta function Consider the GF corresponding to the following functional: For example,

23 Example 7: the Sokhotsky formula where the GF on the l.-h.s. is defined by

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