91影视

Use appropriate connection formulas to analyze the problem of scattering from a barrier with sloping walls (Figurea).

Hint: Begin by writing the WKB wave function in the form

$\mathbf{蠄}\left(x\right)\mathbf{=}\{\begin{array}{l}\frac{1}{\sqrt{p\left(x\right)}}\left[{\mathrm{Ae}}^{\frac{i}{h}{鈭�}_x^{x^1}p\left(x^{\text{'}}\right){\mathrm{dx}}^{\text{'}}}+{\mathrm{Be}}^{\frac{-i}{h}{鈭�}_x^{r_1}p\left(x^{\text{'}}\right){\mathrm{dx}}^{\text{'}}}\right],\left(x<x_1\right)\\ \frac{1}{\sqrt{\left|p\left(x\right)\right|}}\left[{\mathrm{Ce}}^{\frac{i}{h}{鈭�}_{x_1}^{\text{'}}\left|p\left(x^{\text{'}}\right)\right|{\mathrm{dx}}^{\text{'}}}+{\mathrm{De}}^{-\frac{1}{h}{鈭�}_{x_1}^X\left|p\left(x^{\text{'}}\right)\right|{\mathrm{dx}}^{\text{'}}}\right],\left(X_1<X<X_2\right)\\ \frac{1}{\sqrt{p\left(x\right)}}\left[{\mathrm{Fe}}^{\frac{i}{h}{鈭�}_{x_2}^{x}p\left(x^{\text{'}}\right)\mathrm{dx}}\right].\left(x>x_2\right)\end{array}$

Do not assume C=0 . Calculate the tunneling probability, $\mathbf{T}\mathbf{=}{\left|F\right|}^{\mathbf{2}}\mathbf{/}{\left|A\right|}^{\mathbf{2}}$, and show that your result reduces to Equation 8.22 in the case of a broad, high barrier.

Use the WKB approximation to find the allowed energies of the general power-law potential:

$\mathit{V}\left(x\right)\mathbf{=}\mathit{伪}{\left|x\right|}^{\mathbf{v}}$where v is a positive number. Check your result for the case v=2 .

Use the WKB approximation to find the bound state energy for the potential in problem .

For spherically symmetrical potentials we can apply the WKB approximation to the radial part (Equation 4.37). In the case $\mathrm{I}=0$it is reasonable 15to use Equation 8.47in the form

${\mathbf{鈭�}}_{\mathbf{0}}^{{\mathbf{r}}_{\mathbf{0}}}\mathbf{p}\left(r\right)\mathbf{dr}\mathbf{=}\left(n-1/4\right)\mathbf{蟺丑}\mathbf{.}$

Where ${\mathbf{r}}_{\mathbf{0}}$is the turning point (in effect, we treat $\mathbf{r}\mathbf{=}\mathbf{0}$as an infinite wall). Exploit this formula to estimate the allowed energies of a particle in the logarithmic potential.

$\mathbf{V}\left(r\right)\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{In}\left(r/a\right)$

(for constant ${\mathbf{V}}_{\mathbf{0}}$and $\mathbf{a}$). Treat only the case $\mathbf{I}\mathbf{=}\mathbf{0}$. Show that the spacing between the levels is independent of mass

Use the WKB approximation in the form

${\mathbf{鈭�}}_{{\mathbf{r}}_{\mathbf{1}}}^{{\mathbf{r}}^{\mathbf{2}}}\mathbf{p}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{dr}\mathbf{=}\left(n-1/2\right)\mathbf{蟺丑}$

to estimate the bound state energies for hydrogen. Don't forget the centrifugal term in the effective potential (Equation ). The following integral may help:

${\mathbf{鈭�}}_{\mathbf{a}}^{\mathbf{b}}\frac{\mathbf{1}}{\mathbf{x}}\sqrt{\left(x-a\right)\left(b-x\right)}\mathbf{dx}\mathbf{=}\frac{\mathbf{蟺}}{\mathbf{2}}{\left(\sqrt{b}-\sqrt{a}\right)}^{\mathbf{2}}\mathbf{.}$

Note that you recover the Bohr levels when$\mathbf{n}\mathbf{鈮�}\mathbf{/}\mathbf{}\mathbf{and}\mathbf{}\mathbf{n}\mathbf{鈮�}\mathbf{1}\mathbf{/}\mathbf{2}$

Consider the case of a symmetrical double well, such as the one pictured in Figure. We are interested in Figure 8.13bound states with E<V(0) .
(a) Write down the WKB wave functions in regions (i) $\mathit{x}\mathbf{>}{\mathit{x}}_{\mathbf{2}}$ , (ii) ${\mathit{x}}_{\mathbf{1}}\mathbf{<}\mathit{x}\mathbf{<}{\mathit{x}}_{\mathbf{2}}$ , and (iii) $\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}{\mathit{x}}_{\mathbf{1}}$. Impose the appropriate connection formulas at and (this has already been done, in Equation $\mathbf{8}\mathbf{.}\mathbf{46}$, for ${\mathit{x}}_{\mathbf{2}}$ ; you will have to work out ${\mathit{x}}_{\mathbf{1}}$ for yourself), to show that

$$\begin{gathered} \mathit{蠄}\left(x\right)\mathbf{=}\{\begin{array}{l}\frac{D}{\sqrt{P\overline{)\left(x\right)}}}exp\left[-\frac{1}{h}\overline{){鈭�}_{x_2}^x}p\left(x\text{'}\right)dx\text{'}\right] \\ \frac{2D}{\sqrt{P\overline{)\left(x\right)}}}sin\left[\frac{1}{h}\overline{){鈭�}_{x_2}^x}p\left(x\text{'}\right)dx\text{'}+\frac{蟺}{4}\right] \\ \frac{D}{\sqrt{P\overline{)\left(x\right)}}}\left[2cos胃\overline{){}^{\frac{1}{h}{鈭�}_v^{x_1}}}p\overline{)\left(x\right)}dx+sin胃e\frac{1}{h}{鈭�}_v^{x_1}\overline{)p}\overline{)\left(x\right)}dx\right]\\ \end{array} \end{gathered}$$

FIGURE 8.13: Symmetrical double well; Problem 8.15.
Where

$\mathit{胃}\mathbf{鈮�}\frac{\mathbf{1}}{\mathbf{h}}{\mathbf{鈭�}}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathit{p}\left(x\right)\mathit{d}\mathit{x}$ .

(b) Because v(x) is symmetric, we need only consider even $\mathbf{(}\mathbf{+}\mathbf{)}$and odd (-)wave functions. In the former case $\mathit{蠄}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ , and in the latter case $\mathit{蠄}\mathbf{=}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Show that this leads to the following quantization condition:

$\mathit{t}\mathit{a}\mathit{n}\mathit{胃}\mathbf{=}\mathbf{卤}\mathbf{2}{\mathit{e}}^{\mathbf{f}}$.

Where

$\mathit{f}\mathbf{鈮�}\frac{\mathbf{1}}{\mathbf{h}}{\mathbf{鈭�}}_{\mathbf{-}{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}^{\mathbf{1}}}\overline{)\mathbf{p}\overline{)\left(x\right)}\mathbf{d}\mathbf{x}}$

Equation 8.59determines the (approximate) allowed energies (note that Ecomes into 1 and ${\mathit{x}}_{\mathbf{2}}$, so $\mathit{胃}$ and $\mathit{蠒}$ are both functions of E ).

(c) We are particularly interested in a high and/or broad central barrier, in which case $\mathit{蠒}$ is large, and ${\mathit{e}}^{\mathbf{蠒}}$is huge. Equation 8.59 then tells us thatA $\mathit{胃}$must be very close to a half-integer multiple of $蟺$ . With this in mind, write localid="1658823154085" $\left(胃=n+1/2\right)\mathit{蟺}\mathbf{+}\mathit{o}$ . where localid="1658823172105" $|鈭�|\mathbf{<}\mathbf{<}\mathbf{1}$, and show that the quantization condition becomes

localid="1658823190448" $\mathit{胃}\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{蟺}\mathit{m}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{f}}$.

(d) Suppose each well is a parabola: 16

localid="1658823244259" $$\begin{gathered} \mathit{v}\left(x\right)\mathbf{=}\{\begin{array}{l}\frac{1}{2}m{蝇}^2{\left(x+a\right)}^2,ifx<0 \\ \frac{1}{2}m{蝇}^2{\left(x-a\right)}^2,ifx<0\\ \end{array} \end{gathered}$$ .

Sketch this potential, find (Equation 8.58 ), and show that

localid="1658823297094" ${\mathit{E}}_{\mathbf{n}}^{\mathbf{卤}}\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{h}\mathit{蝇}\mathit{m}\frac{\mathbf{h}\mathbf{蝇}}{\mathbf{2}\mathbf{蟺}}{\mathit{e}}^{\mathbf{-}\mathbf{f}}$

Comment: If the central barrier were impenetrable localid="1658823318311" $\left(f鈫�鈭�\right)$ , we would simply have two detached harmonic oscillators, and the energies, localid="1658823339386" ${\mathit{E}}_{\mathbf{n}}\mathbf{=}\left(n+1/2\right)\mathit{h}\mathit{蝇}$ , would be doubly degenerate, since the particle could be in the left well or in the right one. When the barrier becomes finite (putting the two wells into "communication"), the degeneracy is lifted. The even (localid="1658823360600" ${\mathit{蠄}}_{\mathbf{n}}^{\mathbf{+}}$) states have slightly lower energy, and the odd ones ${蠄}_n^{.}$ have slightly higher energy.
(e) Suppose the particle starts out in the right well-or, more precisely, in a state of the form

localid="1658823391675" $\mathit{蠄}\left(x,0\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left({蠄}_n^{+}+{蠄}_n\right)$

which, assuming the phases are picked in the "natural" way, will be concentrated in the right well. Show that it oscillates back and forth between the wells, with a period

localid="1658823416179" $\mathit{蟿}\mathbf{=}\frac{\mathbf{2}{\mathbf{蟺}}^{\mathbf{2}}}{\mathbf{蝇}}{\mathit{e}}^{\mathbf{.}}$

(f) Calculate , for the specific potential in part (d), and show that for , localid="1658823448873" $\mathit{V}\left(0\right)\mathbf{.}\mathbf{>}\mathbf{>}\mathit{E}\mathbf{,}\mathit{蠒}\mathbf{~}\mathit{m}\mathit{蝇}{\mathit{a}}^{\mathbf{2}}\mathbf{/}\mathit{h}$.

As an explicit example of the method developed inProblem 7.15, consider an electron at rest in a uniform magnetic field$\mathbf{B}\mathbf{=}{\mathbf{B}}_{\mathbf{2}}\stackrel{\mathbf{铃�}}{\mathbf{K}}$for which the Hamiltonian is (Equation 4.158):

$\mathit{H}\mathbf{=}\mathbf{-}\mathit{纬}\mathit{B}$ (4.158).

${\mathit{H}}^{\mathbf{0}}\mathbf{}\mathbf{=}\frac{\mathbf{e}{\mathbf{B}}_{\mathbf{z}}}{\mathbf{m}}{\mathit{S}}_{\mathbf{z}}$ (7.57).

The eigenspinors localid="1656062306189" ${\mathbf{x}}_{\mathbf{a}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{b}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{the}\mathbf{}\mathbf{corresponding}\mathbf{}\mathbf{energies}\mathbf{,}{\mathbf{E}}_{\mathbf{a}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{E}}_{\mathbf{b}}$, are given in Equation 4.161. Now we turn on a perturbation, in the form of a uniform field in the x direction:

$\{\begin{array}{l}x+,\mathrm{with}\mathrm{energy}{E}_{+}=-\left({\mathrm{纬叠}}_0魔\right)/2\\ x-,\mathrm{with}\mathrm{energy}{E}_{-}=-\left({\mathrm{纬叠}}_0魔\right)/2\end{array}$ (4.161).

${\mathit{H}}^{\mathbf{\text{'}}}\mathbf{=}\frac{\mathbf{e}{\mathbf{b}}_{\mathbf{x}}}{\mathbf{m}}{\mathit{S}}_{\mathbf{x}}$ (7.58).

(a) Find the matrix elements of H鈥�, and confirm that they have the structure of Equation 7.55. What is h?

(b) Using your result inProblem 7.15(b), find the new ground state energy, in second-order perturbation theory.

(c) Using your result inProblem 7.15(c), find the variation principle bound on the ground state energy.

About how long would it take for a (full) can of beer at room temperature to topple over spontaneously, as a result of quantum tunneling? Hint: Treat it as a uniform cylinder of mass m, radius R, and height h. As the can tips, let x be the height of the center above its equilibrium position (h/2) .The potential energy is mgx, and it topples when x reaches the critical value ${\mathbf{X}}_{\mathbf{0}}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathbf{h}\mathbf{/}\mathbf{2}\mathbf{)}}^{\mathbf{2}}}\mathbf{-}\mathit{h}\mathbf{/}\mathbf{2}$. Calculate the tunneling probability (Equation
8.22), for E = 0. Use Equation 8.28, with the thermal energy $\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{)}{\mathbf{mv}}^{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathbf{\left)}{\mathbf{k}}_{\mathbf{B}}\mathbf{T}\mathbf{\right)}$to estimate the velocity. Put in reasonable numbers, and give your final answer in years.

$\mathbf{T}\mathbf{鈮�}{\mathbf{e}}^{\mathbf{-}{\mathbf{2}}_{\mathbf{T}}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{with}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{纬}\mathbf{鈮�}\frac{\mathbf{1}}{\sout{\mathbf{h}}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{a}}\left|p\left(x\right)\right|\mathbf{dx}\mathbf{.}\mathbf{.}\mathbf{.}$ (8.22).

tau=$\frac{\mathbf{2}{\mathbf{r}}_{\mathbf{1}}}{\mathbf{V}}{{\mathbf{e}}^{\mathbf{2}}}_{\mathbf{T}}$ (8.28).

Use the WKB approximation to find the allowed energies $\left(E_n\right)$of an infinite square well with a 鈥渟helf,鈥� of height${\mathit{V}}_{\mathbf{0}}$, extending half-way across

role="math" localid="1658403794484" $\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}{\mathbf{v}}_{\mathbf{0}\mathbf{,}}& \left(0<x<a/2\right)\\ \mathbf{0}\mathbf{,}& \left(a/2<x<a\right)\\ \mathbf{鈭�}\mathbf{,}& \left(\mathrm{otherwise}\right)\end{array}$

Express your answer in terms ofrole="math" localid="1658403507865" ${\mathbf{V}}_{\mathbf{0}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{E}}_{\mathbf{n}}^{\mathbf{0}}\mathbf{鈮�}{\left(\mathrm{苍蟺魔}\right)}^{\mathbf{2}}\mathbf{/}\mathbf{2}{\mathbf{ma}}^{\mathbf{2}}$(the nth allowed energy for the infinite square well with no shelf). Assume that, but do not assume that ${\mathit{E}}_{\mathbf{1}}^{\mathbf{0}}\mathbf{>}{\mathit{V}}_{\mathbf{0}}$. Compare your result with what we got in Section 7.1.2, using first-order perturbation theory. Note that they are in agreement if either${\mathit{V}}_{\mathbf{0}}$is very small (the perturbation theory regime) or n is very large (the WKB鈥攕emi-classical鈥攔egime).

Question:

An illuminating alternative derivation of the WKB formula (Equation) is based on an expansion in powers of$\sout{\mathbf{h}}$. Motivated by the free particle wave function $\mathbf{蠄}\mathbf{=}\mathbf{Aexp}\mathbf{}\mathbf{(}\mathbf{卤}\mathbf{ipx}\mathbf{/}\sout{\mathbf{h}}\mathbf{)}$,, we write

$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{if}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{/}\mathbf{h}}$

Wheref(x)is some complex function. (Note that there is no loss of generality here-any nonzero function can be written in this way.)

(a) Put this into Schr枚dinger's equation (in the form of Equation8.1), and show that

. ${\mathbf{ihf}}^{\cancel{\mathbf{c}}\cancel{\mathbf{c}}}\mathbf{-}{\left(f^{\cancel{c}}\right)}^{\mathbf{2}}\mathbf{+}{\mathbf{p}}^{\mathbf{2}}\mathbf{+}\mathbf{0}\mathbf{}\mathbf{.}$

(b) Write f(x)as a power series in$\sout{\mathbf{h}}$:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{f}}_{\mathbf{0}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\sout{\mathbf{h}}{\mathit{f}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathit{f}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$And, collecting like powers of$\sout{\mathbf{h}}$, show that

${\left({\stackrel{藱}{o}}_0\right)}^{\mathbf{2}}\mathbf{=}{\mathbf{p}}^{\mathbf{2}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{i}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{0}}\mathbf{=}\mathbf{2}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{0}}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{i}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{1}}\mathbf{=}\mathbf{2}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{0}}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{2}}\mathbf{+}{\mathbf{\left(}{\stackrel{\mathbf{藱}}{\mathbf{o}}}_{\mathbf{1}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

(c) Solve for${\mathbf{f}}_{\mathbf{0}}\left(x\right)$and${\mathbf{f}}_{\mathbf{1}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, and show that-to first order inyou recover Equation8.10.