91影视

If two (or more) distinct⁴⁴solutions to the (time-independent) Schr枚dinger equation have the same energy E . These states are said to be degenerate. For example, the free particle states are doubly degenerate-one solution representing motion to the right. And the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension⁴⁵ there are no degenerate bound states. Hint: Suppose there are two solutions, ${\mathit{蠄}}_{\mathbf{1}}$and ${\mathit{蠄}}_2$with the same energy E. Multiply the Schr枚dinger equation for ${\mathit{蠄}}_{\mathbf{1}}$by ${\mathit{蠄}}_2$and the Schr枚dinger equation for ${\mathit{蠄}}_{\mathbf{2}}$by ${\mathit{蠄}}_{\mathbf{1}}$and subtract, to show that $\left({\mathit{蠄}}_2\mathit{d}{\mathit{蠄}}_{\mathbf{1}}\mathbf{/}\mathit{d}\mathit{x}\mathbf{-}{\mathit{蠄}}_{\mathbf{2}}\mathit{d}{\mathit{蠄}}_{\mathbf{1}}\mathbf{/}\mathit{d}\mathit{x}\right)$is a constant. Use the fact that for normalizable solutions $\mathit{蠄}\mathbf{鈫�}\mathbf{0}\mathbf{}\mathbf{at}\mathbf{}\mathbf{卤}\mathbf{鈭�}$to demonstrate that this constant is in fact zero.Conclude that ${\mathit{蠄}}_2$s a multiple of _{${\mathit{蠄}}_{\mathbf{1}}$}and hence that the two solutions are not distinct.

Imagine a bead of mass m that slides frictionlessly around a circular wire ring of circumference L. (This is just like a free particle, except that $\mathbf{唯}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{L}\mathbf{)}\mathbf{=}\mathbf{唯}\left(x\right)$find the stationary states (with appropriate normalization) and the corresponding allowed energies. Note that there are two independent solutions for each energy En-corresponding to clockwise and counter-clockwise circulation; call them${\mathbf{唯}}_{\mathbf{n}}^{\mathbf{+}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and${\mathbf{唯}}_{\mathbf{n}}^{-}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ How do you account for this degeneracy, in view of the theorem in Problem 2.45 (why does the theorem fail, in this case)?

This is a strictly qualitative problem-no calculations allowed! Consider the "double square well" potential (Figure 2.21). Suppose the depth ${\mathit{V}}_{\mathbf{0}}$and the width a are fixed, and large enough so that several bound states occur.

(a) Sketch the ground state wave function ${\mathit{唯}}_{\mathbf{1}}$and the first excited state localid="1658211858701" ${\mathit{唯}}_{\mathbf{2}}$(i) for the case b = 0 (ii) for$\mathbf{b}\mathbf{鈮�}\mathbf{a}$and (iii) for $\mathbf{b}\mathbf{鈮�}\mathbf{a}$

(b) Qualitatively, how do the corresponding energies$\left(E_1\mathrm{and}{E}_2\right)$and vary, as b goes from 0 to ? Sketch ${\mathbf{E}}_{\mathbf{1}}\left(b\right)$and ${\mathbf{E}}_{\mathbf{2}}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$on the same graph.

(c) The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). If the nuclei are free to move, they will adopt the configuration of minimum energy. In view of your conclusions in (b), does the electron tend to draw the nuclei together, or push them apart? (Of course, there is also the internuclear repulsion to consider, but that's a separate problem.)

In Problem 2.7 (d), you got the expectation value of the energy by summing the series in Equation 2.39, but 1 warned you (in footnote 15 not to try it the "old fashioned way,"$<H>\mathbf{=}\mathbf{鈭�}\mathit{唯}{\left(x,0\right)}^{\mathbf{*}}\mathit{H}\mathit{唯}\left(x,0\right)\mathit{d}\mathit{x}$, because the discontinuous first derivative of$\mathit{唯}\mathbf{(}\mathbf{x}\mathbf{.}\mathbf{0}\mathbf{)}$renders the second derivative problematic. Actually, you could have done it using integration by parts, but the Dirac delta function affords a much cleaner way to handle such anomalies.

(a) Calculate the first derivative of $\mathit{唯}\mathbf{(}\mathbf{x}\mathbf{.}\mathbf{0}\mathbf{)}$(in Problem 2.7), and express the answer in terms of the step function, $\mathit{胃}\left(x-c1/2\right)$defined in Equation (Don't worry about the end points-just the interior region

(b) Exploit the result of Problem 2.24(b) to write the second derivative of $\mathit{唯}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}$in terms of the delta function.

(c) Evaluate the integral $\mathbf{鈭�}\mathit{唯}{\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}}^{\mathbf{*}}\mathit{H}\mathit{唯}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathit{d}\mathit{x}$and check that you get the same answer as before.

Show that

$\mathit{唯}\left(x,t\right)\mathbf{=}{\left(\frac{m蝇}{蟺\sout{h}}\right)}^{\mathbf{1}\mathbf{/}\mathbf{4}}\mathbf{}\mathit{e}\mathit{x}\mathit{p}\left[-\frac{m蝇}{2\sout{h}}\left(x^2+\frac{a^2}{2}\left(1+e^{-2i蝇t}\right)+\frac{i\sout{h}t}{m}-2ax{e}^{{}^{-i蝇t}}\right)\right]$

satisfies the time-dependent Schr枚dinger equation for the harmonic oscillator potential (Equation 2.43). Here a is any real constant with the dimensions of length. 46

(b) Find$\mathbf{|}\mathbf{唯}\left(x,t\right){\mathbf{|}}^{\mathbf{2}}$ and describe the motion of the wave packet.

(c) Compute <x> and <p> and check that Ehrenfest's theorem (Equation 1.38) is satisfied.

Calculate $\left(x\right)\mathbf{,}\left(x^2\right)\mathbf{,}\left(p\right)\mathbf{,}\left(p^2\right)\mathbf{,}{\mathit{蟽}}_{\mathbf{x}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{蟽}}_{\mathbf{p}}$,for the nth stationary state of the infinite square well. Check that the uncertainty principle is satisfied. Which state comes closest to the uncertainty limit?

Consider the moving delta-function well: $\mathbf{V}\left(x,t\right)\mathbf{=}\mathbf{-}\mathbf{伪未}\left(x-\mathrm{vt}\right)$

where v is the (constant) velocity of the well. (a) Show that the time-dependent Schr枚dinger equation admits the exact solution $\mathbf{\text{蠄}}\left(x,t\right)\mathbf{=}\sqrt{\frac{\mathbf{尘伪}}{\sout{\mathbf{h}}}}{\mathbf{e}}^{\mathbf{-}\mathbf{尘伪}\left|x-\mathrm{vt}\right|}\mathbf{l}{\sout{\mathbf{h}}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{i}}\left[\left(E+\left(1/2\right){\mathrm{mv}}^2\right)t-\mathrm{mvx}\right]\mathbf{l}\sout{\mathbf{h}}$where $\mathbf{E}\mathbf{=}\mathbf{-}{\mathbf{尘伪}}^{\mathbf{2}}\mathbf{}\mathbf{l}\mathbf{}\mathbf{2}{\sout{\mathbf{h}}}^{\mathbf{2}}$ is the bound-state energy of the stationary delta function. Hint: Plug it in and check it! Use the result of Problem 2.24(b). (b) Find the expectation value of the Hamiltonian in this state, and comment on the result.

Consider the potential $\mathit{V}\left(x\right)\mathbf{=}\mathbf{-}\frac{\sout{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{m}}\mathit{s}\mathit{e}\mathit{c}{\mathit{h}}^{\mathbf{2}}\left(ax\right)$where a is a positive constant, and "sech" stands for the hyperbolic secant

(a) Graph this potential.

(b) Check that this potential has the ground state

${\mathbf{蠄}}_{\mathbf{0}}\left(x\right)$and find its energy. Normalize and sketch its graph.

(C)Show that the function ${\mathbf{蠄}}^{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{A}\left(\frac{\mathrm{ik}-a\mathrm{ta}\mathrm{nh}\left(\mathrm{ax}\right)}{\mathrm{ik}+a}\right){\mathbf{e}}^{\mathbf{kx}}$

(Where $\mathbf{k}\mathbf{=}\sqrt{\mathbf{2}\mathbf{mE}}\mathbf{}\mathbf{i}\sout{\mathbf{h}}$as usual) solves the Schr枚dinger equation for any (positive) energy E. Since$\mathbf{ta}\mathbf{}\mathbf{nh}\mathbf{}\mathbf{z}\mathbf{鈫�}\mathbf{-}\mathbf{1}\mathbf{as}$as This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e., no term . What is the asymptotic form${\mathbf{蠄}}_{\mathbf{k}}\left(x\right)$ of at large positive x? What are R and T, for this potential? Comment: This is a famous example of a reflectionless potential-every incident particle, regardless of its energy, passes right through.

The scattering matrix. The theory of scattering generalizes in a pretty obvious way to arbitrary localized potentials to the left (Region I),$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{0}$so$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Ae}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{ikx}}\mathbf{,}\mathbf{鈥�}\mathbf{where}\mathbf{}\mathbf{k}\mathbf{鈮�}\frac{\sqrt{\mathbf{2}\mathbf{mE}}}{\mathbf{鈩�}}$

To the right (Region III),$\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is again zero, so$\mathbf{蠄}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{Fe}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Ge}}^{\mathbf{-}\mathbf{ikx}}$In between (Region II), of course, I can't tell you what is until you specify the potential, but because the Schr枚dinger equation is a linear, second-order differential equation, the general solution has got to be of the form

$\mathbf{蠄}$where $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}$and $\mathbf{}\mathit{g}\mathbf{\left(}\mathit{x}\mathbf{\right)}$are two linearly independent particular solutions. 48 There will be four boundary conditions (two joining Regions I and II, and two joining Regions II and III). Two of these can be used to eliminate C and D, and the other two can be "solved" for B and F in terms of \(A\) and G

$\mathbf{B}\mathbf{=}{\mathbf{S}}_{\mathbf{11}}\mathbf{A}\mathbf{+}{\mathbf{S}}_{\mathbf{12}}\mathbf{G}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}\mathbf{F}\mathbf{=}{\mathbf{S}}_{\mathbf{21}}\mathbf{A}\mathbf{+}{\mathbf{S}}_{\mathbf{22}}\mathbf{G}$

The four coefficients${\mathit{S}}_{\mathbf{i}\mathbf{j}}$which depend on k (and hence on E), constitute a$\mathbf{2}\mathbf{脳}\mathbf{2}\mathbf{}\mathbf{}$matrix s called the scattering matrix (or S-matrix, for short). The S-matrix tells you the outgoing amplitudes (B and F) in terms of the incoming amplitudes (A and G):

$\left(\begin{array}{c}B\\ F\end{array}\right)\mathbf{=}\left(\begin{array}{cc}{S}_{11}& S_{21}\\ S_{21}& S_{22}\end{array}\right)\left(\begin{array}{c}A\\ G\end{array}\right)$

In the typical case of scattering from the left, $\mathbf{G}\mathbf{=}\mathbf{0}$so the reflection and transmission coefficients are

${\mathbf{R}}_{\mathbf{l}}\mathbf{=}\frac{\mathbf{|}\mathbf{B}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{A}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}}_{\mathbf{G}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{11}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}{\mathbf{T}}_{\mathbf{I}}\mathbf{=}\frac{\mathbf{|}\mathbf{F}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{A}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}}_{\mathbf{G}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{2}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

For scattering from the right, and

${\mathbf{R}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{|}\mathbf{F}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{G}{\mathbf{|}}^{\mathbf{2}}}{\overline{)\mathbf{}}}_{\mathbf{A}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{22}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}\mathbf{鈥�}\mathbf{鈥�}\mathbf{鈥�}{\mathbf{T}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{|}\mathbf{B}{\mathbf{|}}^{\mathbf{2}}}{\mathbf{|}\mathbf{G}{\mathbf{|}}^{\mathbf{2}}}{\overline{)\mathbf{}}}_{\mathbf{A}\mathbf{=}\mathbf{0}}\mathbf{=}\mathbf{|}{\mathbf{S}}_{\mathbf{12}}{\mathbf{|}}^{\mathbf{2}}\mathbf{.}$

(a) Construct the S-matrix for scattering from a delta-function well (Equation 2.114). (b) Construct the S-matrix for the finite square well (Equation 2.145). Hint: This requires no new work, if you carefully exploit the symmetry of the problem.

The transfer matrix. The S- matrix (Problem 2.52) tells you the outgoing amplitudes (B and F)in terms of the incoming amplitudes (A and G) -Equation 2.175For some purposes it is more convenient to work with the transfer matrix, M, which gives you the amplitudes to the right of the potential (F and G)in terms of those to the left (A and b):

$\mathbf{\left(}\begin{array}{c}\mathbf{F}\\ \mathbf{G}\end{array}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{cc}{\mathbf{M}}_{\mathbf{11}}& {\mathbf{M}}_{\mathbf{12}}\\ {\mathbf{M}}_{\mathbf{21}}& {\mathbf{M}}_{\mathbf{22}}\end{array}\mathbf{\right)}\mathbf{\left(}\begin{array}{c}\mathbf{A}\\ \mathbf{B}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{2}\mathbf{.}\mathbf{178}\mathbf{]}$

(a) Find the four elements of the M-matrix, in terms of the elements of theS-matrix, and vice versa. Express$R_I,T_I,R_r\mathrm{and}{T}_r$(Equations 2.176and 2.177) in terms of elements of the M-matrix.,

(b) Suppose you have a potential consisting of two isolated pieces (Figure 2.23 ). Show that the M-matrix for the combination is the product of the twoM-matrices for each section separately: $\mathbf{M}\mathbf{=}{\mathbf{M}}_{\mathbf{2}}{\mathbf{M}}_{\mathbf{1}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{2}\mathbf{.}\mathbf{179}\mathbf{]}$

(This obviously generalizes to any number of pieces, and accounts for the usefulness of the M-matrix.)

FIGURE : A potential consisting of two isolated pieces (Problem 2.53 ).

(c) Construct the -matrix for scattering from a single delta-function potential at point $\mathit{V}\left(x\right)\mathbf{=}\mathbf{-}\mathit{伪}\mathit{未}\left(x-a\right)$ :

(d) By the method of part , find the M-matrix for scattering from the double delta function$\mathit{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathit{伪}\mathbf{[}\mathit{未}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{)}\mathbf{+}\mathbf{未}\left(X-a\right)\mathbf{]}$ .What is the transmission coefficient for this potential?