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Example 1: quantum particle in an infinitely narrow/deep well

The 1D non-dimensional Schrödinger equation is

,)( Exu

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

).()( xxu Let

Thus, Eq. (1) becomes

(3)

.as0 x

We’ll be interested in bound states, for which

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(Solutions that don’t decay as x → 0 describe free particles.)

.)( Ex

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...

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

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,dd)(d

xExxx

To figure out how the delta function affects the solution, integrate Eq. (3) from –ε to +ε, which yields

hence,

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,d)0()()(

xE

.0)0()0()0(

hence, taking the limit ε → 0,

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.dd)(d

xxExxxxx

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

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.d0)]()([)()()(

xxE

.0)0()0(

Finally, taking the limit ε → 0, we obtain

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Eqs. (4)-(5) are often called the matching conditions, as they ‘match’ (connect) the regions x < 0 and x > 0. These conditions describe how the delta function affects the solution.

Now we can solve Eq. (3) for x < 0 and x > 0 separately and then match the two solutions across the point x = 0 (without actually dealing with the delta function).

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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

,0if

xeAeA xExE

0if0)( xE

where A– and A+ are undetermined constants.

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Imposing the boundary condition (2), we obtain

Imposing the matching conditions (4)-(5), we obtain

,0)( AEAEA

.0 AA

Solving for A± , one can see that a solution exists only if

,4

1E

.0if

,0if)(

xeA

xeAx

xE

xE

which is the energy of the (only existing) bound state.

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where S(x) is the ‘source density’, L is the half-size of the resonator, and c is the speed of sound.

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Example 2: the Green’s function of a boundary-value problem

,sin)(2

22

2

2

txSx

pc

t

p

,at0 Lxx

p

Waves generated in a resonator by a distributed monochromatic source of frequency ω are described by

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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

,sin)(),( txPtxp

and Eqs. (6)-(7) yield

),(22 xSPcP

.at0 LxP

(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...

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,d),()()( 000

L

L

xxxGxSxP

where G(x, x0) satisfies

(10)

the Green’s function

),( 02

222 xx

x

GcG

.at0 Lxx

G

(11)

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Physically, (11)-(12) describe the wave field from a point source of unit amplitude, located at x = x0. 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.

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L

L

L

L

xx

xxGxScxxxGxS 02

02

02

0002 d

),()(d),()(h.s.-l.

As follows from (11), the [...] in the above equality equals δ(x – x0) – hence,

It is now evident that the l.-h.s. of (8) does equal its r.-h.s.

.d)()(h.s.-l. 000

L

L

xxxxS

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.

.d),(

),()(

d),(

)(d),()(h.s.-l.

020

22

02

0

020

2

02

0002

L

L

L

L

L

L

xx

xxGcxxGxS

xx

xxGxScxxxGxS

P P

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.1d00

0

0

22

xxxx

x

x x

G

x

GcxG

As in Example 1, we need to understand how the delta function affects G. To do so, integrate (11) from x0 – ε to x0 + ε,

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Next, multiply (11) by (x – x0) and repeat the same

procedure again. Integrating the term involving ∂2G/∂x2 by parts, we obtain

.0])()[(

)()(

d)(

00

00

0

0

2

002

02

xxxx

xxxx

x

x

GGc

x

Gxx

x

Gxxc

xGxx

(14)

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As before, assume that G is continuous, but ∂G/∂x involves a jump at x = x0. Thus, taking in (13)-(14) the limit ε → 0, we obtain

,1000

2

xxxx x

G

x

Gc

,0])()[(0000

2 xxxx GGc

hence,

(16) and (15) confirm our assumption that G is indeed continuous, whereas its derivative ‘jumps’ (both, at x = x0).

,1

200 00

cx

G

x

G

xxxx

.)()( 00 00 xxxx GG

(15)

(16)

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Conditions (15)-(16) ‘match’ the solutions for x < x0 and for

x > x0. The delta function doesn’t contribute to either of

these regions (as it’s ‘localised’ at x = x0), so we can forget about it.For x < x0, we have

,02

222

x

GcG

,at0 LxG

hence,

Similarly,

where k = ω/c, and A– is an arbitrary constant.

,for)(cos 0xxLxkAG

.for)(cos 0xxLxkAG

(17)

(18)

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Use (15)-(16) to match (17) and (18) at x = x0:

,1

)(sin)(sin 200 cLxkkALxkkA

).(cos)(cos 00 LxkALxkA

Solve these equations for A±:

where we have used the formula

,2sin

)(cos,

2sin

)(cos2

02

0

kLkc

LxkA

kLkc

LxkA

)sin(sincoscossin bababa

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with a = x0 – L and b = x0 + L.

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.if)(cos)(cos

,if)(cos)(cos

2sin

1),(

00

00

20

xxLxkLxk

xxLxkLxk

kLkcxxG

Summarising Eqs. (17)-(19), we obtain

Observe that G(x, x0) is symmetric with respect to

interchanging x and x0,

which is typically the case with Green’s functions.

),,(),( 00 xxGxxG

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.

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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.

.d)()](,[(b)

],function givena is )([d)()()](,[(a)

2

xxfxfG

xgxxfxgxfG

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,orif0)( bxaxxf

۞ 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

,if0

),,(if)(

1

)(

1exp

,if0

)( 22

bx

baxbxax

ax

x

where a < b.

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۞ 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):

).()](),([ 00 xfxfxx

(b) A ‘regular’ function g(x) can be treated as a GF:

.d)()()](,[

xxfxgxfG

the ‘regular’ function associated

with G

the functional the test function

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۞ 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,

.d)()()](,[

xxfxgxfG

)].(,[)](,[ xfGxfG

Integrating the r.-h.s. by parts and recalling that f(x) is a function with compact support, we obtain...

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,d)()(0)](,[

xxfxgxfG

as required. █

Example 6: a GF unrelated to the delta function

Consider the GF corresponding to the following functional:

.d)(

d)(

lim)(,1

0

a

a

xax

xfx

ax

xfxf

axP

For example,

.1

1,

1

12

12

xxP

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Example 7: the Sokhotsky formula

),(1

0

1axi

axiax

P

where the GF on the l.-h.s. is defined by

.d)(

lim)(,0

10

x

iax

xfxf

iax