91影视

The adiabatic approximation can be regarded as the first term in an adiabatic series for the coefficients${\mathbf{C}}_{\mathbf{m}}\left(t\right)$in Equation. Suppose the system starts out in theth state; in the adiabatic approximation, it remains in theth state, picking up only a time-dependent geometric phase factor (Equation):

${\mathit{C}}_{\mathbf{m}}\left(t\right)\mathbf{=}{\mathit{未}}_{\mathbf{m}\mathbf{n}}{\mathit{e}}^{\mathbf{i}\mathbf{y}\mathbf{n}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}$

(a) Substitute this into the right side of Equationto obtain the "first correction" to adiabaticity:

${\mathit{C}}_{\mathbf{m}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{c}}_{\mathbf{m}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{-}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}<{唯}_m\left(t\text{'}\right)l\frac{鈭�}{鈭�t\text{'}}{唯}_n(t\text{'})>{\mathit{e}}^{\mathbf{i}\mathbf{y}\mathbf{n}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}}{\mathit{e}}^{\mathbf{i}\mathbf{\left(}{\mathbf{胃}}_{\mathbf{n}}\mathbf{\right(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}\mathbf{-}{\mathbf{胃}}_{\mathbf{t}\mathbf{!}}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{\left)}\mathbf{\right)}}\mathbf{dt}\mathbf{\text{'}}\mathbf{.}\mathbf{}\mathbf{}\left[10.95\right]$

This enables us to calculate transition probabilities in the nearly adiabatic regime. To develop the "second correction," we would insert Equationon the right side of Equation, and so on.

(b) As an example, apply Equationto the driven oscillator (Problem). Show that (in the near-adiabatic approximation) transitions are possible only to the two immediately adjacent levels, for which

$$\begin{gathered} {\mathbf{C}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\frac{\mathbf{尘蝇}}{\mathbf{2}\mathbf{h}}}\sqrt{\mathbf{n}\mathbf{+}\mathbf{1}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}\stackrel{\mathbf{藱}}{\mathbf{f}}\left(t\text{'}\right){\mathbf{e}}^{\mathbf{颈蝇迟}}\mathbf{dt}\mathbf{\text{'}} \\ {\mathbf{C}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\sqrt{\frac{\mathbf{尘蝇}}{\mathbf{2}\mathbf{h}}}\sqrt{\mathbf{n}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}\stackrel{\mathbf{藱}}{\mathbf{f}}\mathbf{(}\mathbf{t}\mathbf{\text{'}}\mathbf{)}{\mathbf{e}}^{\mathbf{颈蝇迟}}\mathbf{dt}\mathbf{\text{'}} \end{gathered}$$

The case of an infinite square well whose right wall expands at a constant velocity $\mathbf{\left(}\mathbf{v}\mathbf{\right)}$ can be solved exactly. A complete set of solutions is${\mathbf{桅}}_{\mathbf{n}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{鈮�}\sqrt{\frac{\mathbf{2}}{\mathbf{w}}}\mathbf{sin}\left(\frac{\mathrm{苍蟺}}{u}x\right){\mathbf{e}}^{\mathbf{i}\mathbf{(}{\mathbf{mvx}}^{\mathbf{2}}\mathbf{鈭�}\mathbf{2}{\mathbf{E}}_{\mathbf{n}}^{\mathbf{i}}\mathbf{nt}\mathbf{)}\mathbf{/}\mathbf{2}\mathbf{鈩�}\mathbf{w}}$

Where $\mathbf{w}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{鈮�}\mathbf{a}\mathbf{+}\mathbf{vt}$is the (instantaneous) width of the well and${\mathbf{E}}_{\mathbf{n}}^{\mathbf{i}}\mathbf{鈮�}{\mathbf{n}}^{\mathbf{2}}{\mathbf{蟺}}^{\mathbf{2}}{\mathbf{鈩�}}^{\mathbf{2}}\mathbf{/}\mathbf{2}{\mathbf{ma}}^{\mathbf{2}}$ is the $\mathbf{nth}$allowed energy of the original well (width). The general solution is a linear combination of the${\mathbf{桅}}^{\mathbf{\text{'}}}\mathbf{\text{s}}\mathbf{:}$

$\mathit{唯}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\underset{n=1}{\overset{\mathrm{鈭�}}{鈭�}}{c}_n{\mathit{桅}}_{\mathbf{n}}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}$

the coefficients ${\mathit{c}}_{\mathbf{n}}$are independent of$\mathbf{t}$

(b) Suppose a particle starts outrole="math" localid="1659010978273" $\mathbf{(}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{)}$ in the ground state of the initial well:

role="math" localid="1659011031703" $\mathit{唯}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{蟺}}{\mathbf{a}}\mathbf{x}\mathbf{\right)}$

Show that the expansion coefficients can be written in the form

${\mathit{c}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{蟺}}{鈭�}_0^{蟺}{e}^{鈭�i{伪}_{鈭�}^2}\mathbf{sin}\mathbf{\left(}\mathit{n}\mathit{z}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathit{d}\mathit{z}$

Where $\mathit{伪}\mathbf{鈮�}\mathit{m}\mathit{v}\mathit{a}\mathbf{/}\mathbf{2}{\mathit{蟺}}^{\mathbf{2}}\mathit{鈩�}$is a dimensionless measure of the speed with which the well expands. (Unfortunately, this integral cannot be evaluated in terms of elementary functions.)

(c) Suppose we allow the well to expand to twice its original width, so the "external" time is given by$\mathit{w}\mathbf{\left(}{\mathbf{T}}_{\mathbf{e}}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathit{a}$ The "internal" time is the period of the time-dependent exponential factor in the (initial) ground state. Determine ${\mathbf{T}}_{\mathbf{e}}$and${\mathbf{T}}_{\mathbf{i}}$ show that the adiabatic regime corresponds to $\mathit{伪}\mathbf{鈮�}\mathbf{1}\mathbf{\text{sothat}}\mathbf{exp}\mathbf{(}\mathbf{鈭�}\mathbf{i}\mathbf{伪}{\mathbf{z}}^{\mathbf{2}}\mathbf{)}\mathbf{鈮�}\mathbf{1}$over the domain of integration. Use this to determine the expansion coefficients,${\mathit{C}}_{\mathbf{n}}$ Construct $\mathit{唯}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}$and confirm that it is consistent with the adiabatic theorem.

(d) Show that the phase factor inrole="math" localid="1659011579812" $\mathit{唯}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}$ can be written in the form

$\mathbf{胃}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{鈭�}\frac{\mathbf{1}}{\mathbf{鈩�}}{鈭�}_0^t{E}_1\mathbf{\left(}{\mathbf{t}}^{\mathbf{\text{'}}}\mathbf{\right)}{\mathbf{dt}}^{\mathbf{\text{'}}}\mathbf{.}$

Where${\mathit{E}}_{\mathbf{n}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{鈮�}{\mathit{n}}^{\mathbf{2}}{\mathit{蟺}}^{\mathbf{2}}{\mathit{鈩�}}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathit{m}{\mathit{w}}^{\mathbf{2}}$ is the instantaneous eigenvalue, at time$\mathit{t}$ Comment on this result.

Check the Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33, and show that the sum of the squares of the coefficients is 1, as required for normalization.

(a) Use Equation 10.42 to calculate the geometric phase change when the infinite square well expands adiabatically from width ${\mathit{w}}_{\mathbf{1}}$to width ${\mathit{w}}_{\mathbf{2}}$. Comment on this result.

(b) If the expansion occurs at a constant rate$\left(dw/dt=v\right)$, what is the dynamic phase change for this process?

(c) If the well now contracts back to its original size, what is Berry's phase for the cycle?

The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change whengradually increases from${\mathit{伪}}_{\mathbf{1}}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{伪}}_{\mathbf{2}}$. If the increase occurs at a constant rate, $\mathbf{(}\mathbf{诲伪}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{c}\mathbf{)}$what is the dynamic phase change for this process?

Show that if ${\mathbf{蠄}}_{\mathbf{n}}$ is real, the geometric phase vanishes. (Problems 10.3 and 10.4 are examples of this.) You might try to beat the rap by tacking an unnecessary (but perfectly legal) phase factor onto the eigenfunctions: ${\mathbf{蠄}}_{\mathbf{n}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{鈮�}{\mathbf{e}}^{\mathbf{颈蠁苍}}{\mathbf{蠄}}_{\mathbf{n}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$, where${\mathbf{蠒}}_{\mathbf{n}}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is an arbitrary (real) function. Try it. You'll get a nonzero geometric phase, all right, but note what happens when you put it back into Equation 10.23. And for a closed loop it gives zero. Moral: For nonzero Berry's phase, you need (i) more than one time-dependent parameter in the Hamiltonian, and (ii) a Hamiltonian that yields nontrivially complex eigenfunctions.

Work out to analog to Equation 10.62 for a particle of spin I.

(a) Derive the equation 10.67 from Equation 10.65.

(b) Derive Equation 10.79, starting with Equation 10.78.

A particle starts out in the ground state of the infinite square well (on the interval 0 鈮� x 鈮� a) .Now a wall is slowly erected, slightly off center:

$\mathit{V}\left(x\right)\mathbf{=}\mathit{f}\left(t\right)\mathit{未}\left(x-\frac{a}{2}-脪\right)$

where$\mathbf{f}\left(t\right)$rises gradually from $\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{鈭�}$According to the adiabatic theorem, the particle will remain in the ground state of the evolving Hamiltonian.

(a)Find (and sketch) the ground state at$\mathbf{t}\mathbf{}\mathbf{}\mathbf{鈫�}\mathbf{}\mathbf{鈭�}$ Hint: This should be the ground state of the infinite square well with an impenetrable barrier at$\mathbf{a}\mathbf{/}\mathbf{2}\mathbf{+}\mathbf{蔚}$ . Note that the particle is confined to the (slightly) larger left 鈥渉alf鈥� of the well.

(b) Find the (transcendental) equation for the ground state energy at time t.
Answer:

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{z}\mathbf{sin}\mathit{z}\mathbf{=}\mathit{T}\mathbf{[}\mathbf{cos}\mathbf{z}\mathbf{-}\mathbf{cos}\left(z未\right)\mathbf{]}\mathit{z}\mathbf{sin}\mathit{z}\mathbf{,} \\ \mathbf{where}\mathbf{}\mathit{z}\mathbf{鈮�}\mathit{k}\mathit{a}\mathbf{,}\mathbf{}\mathit{T}\mathbf{鈮�}\mathbf{maf}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{/}{\sout{\mathbf{h}}}^{\mathbf{2}}\mathbf{,}\mathit{未}\mathbf{鈮�}\mathbf{2}\mathbf{脪}\mathbf{/}\mathbf{a}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathit{k}\mathbf{鈮�}\sqrt{\mathbf{2}\mathbf{m}\mathbf{E}}\mathbf{/}\sout{\mathbf{h}}\mathbf{.} \end{gathered}$$ $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{z}\mathbf{sin}\mathit{z}\mathbf{=}\mathit{T}\mathbf{[}\mathbf{cos}\mathit{z}\mathbf{-}\mathbf{cos}\left(z未\right)\mathbf{]}\mathit{z}\mathbf{sin}\mathit{z}\mathbf{,} \\ \mathbf{where}\mathbf{}\mathit{z}\mathbf{鈮�}\mathit{k}\mathit{a}\mathbf{,}\mathbf{}\mathit{T}\mathbf{鈮�}\mathbf{maf}\left(t\right)\mathbf{/}{\sout{\mathbf{h}}}^{\mathbf{2}}\mathbf{,}\mathit{未}\mathbf{鈮�}\mathbf{2}\mathbf{脪}\mathbf{/}\mathbf{a}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathit{k}\mathbf{=}\sqrt{\mathbf{2}\mathbf{m}\mathbf{E}}\mathbf{/}\sout{\mathbf{h}} \end{gathered}$$

(c) Setting 未 = 0 , solve graphically for z, and show that the smallest z goes from 蟺 to 2蟺 as T goes from 0 to 鈭�. Explain this result.

(d) Now set 未 = 0.01 and solve numerically for z, using

$\mathbf{T}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{}\mathbf{,}\mathbf{}\mathbf{5}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{20}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{100}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{1000}$

(e) Find the probability ${\mathit{P}}_{\mathbf{r}}$that the particle is in the right 鈥渉alf鈥� of the well, as a function of z and 未. Answer:

${\mathit{P}}_{\mathbf{r}}\mathbf{=}\mathbf{1}\mathbf{/}\left[1+\left(I_{+}I{I}_{-}\right)\right]\mathit{P}\mathit{r}\mathbf{=}\mathbf{1}\mathbf{/}\left[1+\left(I_{+}II?\right)\right]\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}{\mathit{I}}_{\mathbf{+}}\mathbf{鈮�}\mathbf{[}\mathbf{1}\mathbf{卤}\mathit{未}\mathbf{-}\left(z\left(1卤未\right)\right)\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left[z\left(1鈭�未\right)/2\right]$

. Evaluate this expression numerically for the T鈥檚 and 未 in part (d). Comment on your results.

(f) Plot the ground state wave function for those same values of T and 未.
Note how it gets squeezed into the left half of the well, as the barrier grows.

The driven harmonic oscillator. Suppose the one-dimensional harmonic oscillator (mass m, frequency 蝇) is subjected to a driving force of the form F(t) = m 蝇虏 f(t) , where f(t) is some specified function. (I have factored out m 蝇虏 for notational convenience; f(t) has the dimensions of length.) The Hamiltonian is

$\mathit{H}\left(t\right)\mathbf{=}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{鈭�}}^{\mathbf{2}}}{\mathbf{鈭�}{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{蝇}}^{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{m}{\mathit{蝇}}^{\mathbf{2}}\mathit{x}\mathit{f}\left(t\right)$ (10.90).

Assume that the force was first turned on at time $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{:}\mathbf{}\mathit{f}\left(t\right)\mathbf{=}\mathbf{0}\mathbf{}\mathbf{for}\mathbf{}\mathit{t}\mathbf{鈮�}\mathbf{0}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}$This system can be solved exactly, both in classical mechanics and in quantum mechanics.

(a)Determine the classical position of the oscillator, assuming it started from rest at the origin $\left(x_c\left(0\right)=\stackrel{藱}{x_c}\left(0\right)=0\right)$. Answer:

${\mathit{x}}_{\mathbf{c}}\left(t\right)\mathbf{=}\mathit{蝇}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}\mathit{f}\left(t\text{'}\right)\mathit{s}\mathit{i}\mathit{n}\left[\left(t-t\text{'}\right)\right]\mathit{d}\mathit{t}\mathbf{\text{'}}\mathbf{.}$(10.91).

(b) Show that the solution to the (time-dependent) Schr枚dinger equation for this oscillator, assuming it started out in the nth state of the undriven oscillator $\left(蠄\left(x,0\right)={蠄}_n\left(x\right)\right)\mathbf{}\mathbf{where}\mathbf{}{\mathbf{蠄}}_{\mathbf{n}}\left(x\right)$is given by Equation 2.61), can be written as

${\mathbf{蠄}}_{\mathbf{n}}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{A}}_{\mathbf{n}}\left({\hat{a}}_{+}\right)}^{\mathbf{n}}\mathbf{}{\mathbf{蠄}}_{\mathbf{0}}\left(x\right)\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{with}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{E}}_{\mathbf{n}}\mathbf{=}\left(n+\frac{1}{2}\right)\sout{\mathbf{h}}\mathit{蝇}$ (2.61).

localid="1656143246748" ${\mathbf{蠄}\mathbf{(}\mathbf{x}\mathbf{,1)}\mathbf{=}{\mathbf{蠄}}_{\mathbf{n}}\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{c}}\mathbf{)}\mathbf{e}}^{\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathbf{[}\mathbf{-}\left(n+\frac{1}{2}\right)\sout{\mathbf{h}}\mathbf{蝇}\mathbf{t}\mathbf{+}\mathbf{m}{\stackrel{\mathbf{藱}}{\mathbf{x}}}_{\mathbf{c}}\left(x-\frac{x_c}{2}\right)\mathbf{+}\frac{\mathbf{m}{\mathbf{蝇}}^{\mathbf{2}}}{\mathbf{2}}{\mathbf{鈭�}}_{\mathbf{0}}^{\mathbf{t}}\mathbf{f}\left(t\text{'}\right){\mathbf{x}}_{\mathbf{x}}\left(t\text{'}\right)\mathbf{d}\mathbf{t}\mathbf{\text{'}}\mathbf{]}}$(10.92).

(c) Show that the Eigen functions and Eigenvalues of H(t) are

${\mathit{蠄}}_{\mathbf{n}}\left(x,t\right)\mathbf{=}{\mathit{蠄}}_{\mathbf{n}}\left(x-f\right)\mathbf{;}\mathbf{}\mathbf{}{\mathit{E}}_{\mathbf{n}}\left(t\right)\mathbf{=}\left(n+\frac{1}{2}\right)\sout{\mathbf{h}}\mathit{蝇}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{蝇}}^{\mathbf{2}}{\mathit{f}}^{\mathbf{2}}$ (10.93).

(d) Show that in the adiabatic approximation the classical position (Equation 10.91) reduces to ${\mathit{x}}_{\mathbf{c}}\left(t\right)\mathbf{鈮�}\mathit{f}\left(t\right)$State the precise criterion for adiabaticity, in this context, as a constraint on the time derivative of f. Hint: Write $\mathit{s}\mathit{i}\mathit{n}\left[蝇\left(t-t\text{'}\right)\right]\mathbf{}\mathit{a}\mathit{s}\mathbf{}\left(1/dt\text{'}\right)\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left[蝇\left(t-t\text{'}\right)\right]$and use integration by parts.

(e) Confirm the adiabatic theorem for this example, by using the results in (c) and (d) to show that

$\mathbf{蠄}\left(x,t\right)\mathbf{鈮�}{\mathbf{蠄}}_{\mathbf{n}}\left(x,t\right){\mathit{e}}^{\mathbf{i}{\mathbf{胃}}_{\mathbf{n}}\left(t\right)}{\mathit{e}}^{\mathbf{i}{\mathbf{纬}}_{\mathbf{n}}\left(t\right)}$(10.94).

Check that the dynamic phase has the correct form (Equation 9.92). Is the geometric phase what you would expect?

${\mathit{e}}^{\mathbf{i}{\mathbf{胃}}_{\mathbf{n}}\left(t\right)}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{where}\mathbf{}{\mathit{胃}}_{\mathbf{n}}\left(t\right)\mathbf{鈮�}\frac{\mathbf{1}}{\sout{\mathbf{h}}}{\mathbf{鈭�}}_{\sout{\mathbf{h}}}^{\mathbf{t}}\mathbf{}{\mathit{E}}_{\mathbf{n}}\left(t\text{'}\right)\mathit{d}\mathit{t}\mathbf{\text{'}}$(9.92).