Texts on physics, maths and programming (Несколько текстов для зачёта), страница 9
Описание файла
Файл "Texts on physics, maths and programming" внутри архива находится в папке "3". Документ из архива "Несколько текстов для зачёта", который расположен в категории "". Всё это находится в предмете "английский язык" из 5 семестр, которые можно найти в файловом архиве МГТУ им. Н.Э.Баумана. Не смотря на прямую связь этого архива с МГТУ им. Н.Э.Баумана, его также можно найти и в других разделах. Архив можно найти в разделе "остальное", в предмете "английский язык" в общих файлах.
Онлайн просмотр документа "Texts on physics, maths and programming"
Текст 9 страницы из документа "Texts on physics, maths and programming"
Our second step is to regard χ (x) as a new trial function, which satisfies
(T+V(x)+w(x))χ(x)=E0χ(x) | (4.14) |
with w (x) being a step function,
| (4.15) |
and
| (4.16) |
We see that w (x) is now monotonic, with
w′(x) 0 | (4.17) |
for the entire range of x from −∞ to +∞. The hierarchy theorem can be applied again, and that will lead from χ (x) to ψ (x), as we shall see.
4.1. Construction of the first trial function
We consider first the positive x region. Following Section 2.1, we begin with the usual perturbative power series expansion for
ψ(x)=e-gS(x) | (4.18) |
with
gS(x)=gS0(+)+S1(+)+g-1S2(+)+ | (4.19) |
and
E=gE0(+)+E1(+)+g-1E2(+)+ | (4.20) |
in which Sn (+) and En (+) are g-independent. Substituting Figs. (4.18), (4.19) and (4.20) into the Schroedinger equation (4.2) and equating both sides, we find
| (4.21) |
| (4.22) |
etc. Thus, (4.21) leads to
| (4.23) |
Since the left side of (4.22) vanishes at x = 1, so is the right side; hence, we determine
E0(+)=1+λ, | (4.24) |
which leads to
S1(+)=(1+λ)ln(1+x). | (4.25) |
Of course, the power series expansion Figs. (4.19) and (4.20) are both divergent. However, if we retain the first two terms in (4.19), the function
| (4.26) |
serves as a reasonable approximation of ψ (x) for x > 0, except when x is near zero. By differentiating (+), we find (+) satisfies
(T+V(x)+u+(x)) (+)=gE0(+) (+) | (4.27) |
where
| (4.28) |
In order to construct the trial function (x) that satisfies the boundary condition (4.5), we introduce for x 0,
| (4.29) |
and
| (4.30) |
so that + (x) and its derivative +′ (x) are both continuous at x = 1, and in addition, at x = 0 we have +′ (0) = 0. For x 0, we observe that V (x) is invariant under
| (4.31) |
The same transformation converts + (x) for x positive to − (x) for x negative. Define
| (4.32) |
where
| (4.33) |
| (4.34) |
and
| (4.35) |
Both − (x) and its derivative −′ (x) are continuous at x = −1; furthermore, + (x) and − (x) also satisfy the continuity condition Figs. (4.4) and (4.5), as well as the Schroedinger equation Figs. (4.6a) and (4.6b), with the perturbative potentials v+ (x) and v- (x) given by
| (4.36a) |
and
| (4.36b) |
in which u+ (x) is given by (4.28),
| (4.37) |
| (4.38a) |
and
| (4.38b) |
In order that u+ (x), be positive for x > 0 and u- (x), positive for x < 0, we impose
| (4.39) |
in addition to the earlier condition λ > 0. From Figs. (4.28) and (4.37), we have
| (4.40a) |
and
| (4.40b) |
Likewise, from Figs. (4.38a) and (4.38b), we find
| (4.41a) |
and
| (4.41b) |
Furthermore, as x → ±1,
| (4.42a) |
and
| (4.42b) |
Thus, for x 0, we have
| (4.43) |
and, together with Figs. (4.36a) and (4.40a),
| (4.44a) |
for x positive. On the other hand for x 0, is not always positive; e.g., at x = 0,
|
which is positive for , but at x = −1+,
|
However, at x = −1, . It is easy to see that the sum can satisfy for x 0,
| (4.44b) |
To summarize: + (x) and − (x) satisfy the Schroedinger equation Figs. (4.6a) and (4.6b), with v± (x) given by Figs. (4.36a) and (4.36b),
| (4.45a) |
and
| (4.45b) |
and the boundary conditions Figs. (4.4) and (4.5). In addition, v± (x) satisfies
| (4.46) |
and the monotonicity conditions Figs. (4.7a) and (4.7b).
4.2. Construction of the second trial function
To construct the second trial function χ (x) introduced in (4.9), we define f± (x) by
| (4.47a) |
and
| (4.47b) |
To retain flexibility it is convenient to impose only the boundary condition (4.11) first, but not (4.10); i.e., at x = 0
χ+′(0)=χ-′(0)=0, | (4.48) |
but leaving the choice of the overall normalization of χ+ (0) and χ− (0) to be decided later. We rewrite the Schroedinger equations Figs. (4.12) and (4.13) in their equivalent forms
| (4.49a) |
and
| (4.49b) |
where
| (4.50a) |
and
| (4.50b) |
Because at x = 0, +′(0) = −′(0) = 0, in accordance with (4.5), we have, on account of (4.48),
f+′(0)=f-′(0)=0. | (4.51) |
So far, the overall normalization of f+ (x) and f− (x) are still free. We may choose
| (4.52) |
From Figs. (4.6a), (4.47a), (4.49a) and (4.50a), we see that f+ (x) satisfies the integral equation (for x 0)
| (4.53a) |
Furthermore, from Figs. (4.6a) and (4.49a), we also have
| (4.54a) |
Likewise, f− (x) satisfies (for x 0)
| (4.53b) |
and
| (4.54b) |
The function f+ (x) and f− (x) will be solved through the iterative process described in Section 1. We introduce the sequences and for n = 1, 2, 3, … , with
| (4.55a) |
for x 0, and
| (4.55b) |
for x 0, where satisfies
| (4.56a) |
and
| (4.56b) |
Thus, Figs. (4.55a) and (4.55b) can also be written in their equivalent expressions
| (4.57a) |
for x 0, and
| (4.57b) |
for x 0. For n = 0, we set
| (4.58) |
through induction and by using Figs. (4.55a), (4.55b), (4.56a) and (4.56b), we derive all subsequent and . Because v± (x) satisfies Figs. (4.44a), (4.44b) and (4.46), the Hierarchy theorem proved in Section 3 applies. The boundary conditions f+ (∞) = f− (−∞) = 1, given by (4.52), lead to , in agreement with Figs. (4.55a) and (4.55b). According to Figs. (3.24), (3.25), (3.26) and (3.27) of Case (A) of the theorem, we have
| (4.59a) |
| (4.59b) |
| (4.60a) |
at all finite and positive x, and
| (4.60b) |
at all finite and negative x. Since
| (4.61a) |
and
| (4.61b) |
with both v± (0) finite,
| (4.62) |
both exist. Furthermore, by using the integral equations Figs. (4.55a) and (4.55b) for and by following the arguments similar to those given in Section 5 of [3], we can show that
| (4.63) |
also exist. This leads us from the first trial function (x) given by (4.3) to f+ (x) and f− (x) which are solutions of
| (4.64a) |
and
| (4.64b) |
with
| (4.65) |
and the boundary conditions at x = 0,
′(0)=f+′(0)=f-′(0)=0. | (4.66) |
An additional normalization factor multiplying, say, f− (x) would enable us to construct the second trial function χ (x) that satisfies Figs. (4.9), (4.10), (4.11), (4.12) and (4.13).
4.3. Symmetric vs asymmetric potential
As we shall discuss, the general description leading from the trial function χ (x) to the final wave function ψ (x) that satisfies the Schroedinger equation (4.2) may be set in a more general framework. Decompose any potential V (x) into two parts
| (4.67) |
Next, extend the functions Va (x) and Vb (x) by defining
| (4.68) |
Thus, both Va (x) and Vb (x) are symmetric potential covering the entire x-axis. Let χa (x) and χb (x) be the ground state wave functions of the Hamiltonians T + Va and T + Vb:
(T+Va(x))χa(x)=Eaχa(x) | (4.69a) |
and
(T+Vb(x))χb(x)=Ebχb(x). | (4.69b) |
The symmetry (4.68) implies that
| (4.70) |
and at x = 0
| (4.71) |
Choose the relative normalization factors of χa and χb, so that at x = 0
χa(0)=χb(0). | (4.72) |
The same trial function (4.9) for the specific quartic potential (4.1) is a special example of
| (4.73) |
with
| (4.74) |
In general, from Figs. (4.69a) and (4.69b) we see that χ (x) satisfies
| (4.75) |
Depending on the relative magnitude of Ea and Eb, we define, in the case of Ea > Eb
| (4.76a) |
and
| (4.77a) |
otherwise, if Eb > Ea, we set
| (4.76b) |
and
| (4.77b) |
Thus, we have either
| (4.78a) |
at all finite x, or
| (4.78b) |
at all finite x. A comparison between Figs. (4.9), (4.10), (4.11), (4.12), (4.13), (4.14), (4.15), (4.16) and (4.17) and (4.73)–(4.77a) shows that w (x) of (4.14) and the above differs only by a constant.