证明 . (a) In time-dependent perturbation theory, the probability of transition from an initial state $|\psi_i(t=-\infty)\rangle$ to a final state $|\psi_f(t=+\infty)\rangle$ is given by:

$P_{i \rightarrow f}=\frac{1}{\hbar^2}\left|\int_{-\infty}^{+\infty}\left\langle\psi_f(t)\left|H^{\prime}(t)\right| \psi_i(t)\right\rangle d t\right|^2$

where $H'(t)$ is the time-dependent perturbation, and $\langle \psi_f(t)|$ and $|\psi_i(t) \rangle$ are the time-dependent states of the system. In this case, the initial state is $|1\rangle$ and the final state is $|2\rangle$.

To lowest order in $v$, the perturbation to the Hamiltonian is given by $H'(t) = v(t) \left(\begin{smallmatrix} 0 & 1 \ 1 & 0 \end{smallmatrix}\right)$. Therefore, we have: \begin{align*} P_{1 \rightarrow 2} &= \frac{1}{\hbar^2} \left| \int_{-\infty}^{+\infty} \langle 2 | H'(t) | 1 \rangle e^{i(E_2-E_1)t/\hbar} , dt \right|^2 \ &= \frac{1}{\hbar^2} \left| \int_{-\infty}^{+\infty} v(t) e^{i2Et/\hbar} , dt \right|^2 \ &= \frac{4}{\hbar^2} \left| \int_{0}^{+\infty} v(t) \cos(2Et/\hbar) , dt \right|^2 \end{align*} where we used the fact that $E_2-E_1=2E$ and $\langle 2 | 1 \rangle = 0$.

Since $v \rightarrow 0$ for $t \rightarrow \pm \infty$, we can assume that $v(t)$ is non-zero only for a finite interval of time $T$ around $t=0$. Therefore, we have: \begin{align*} P_{1 \rightarrow 2} &= \frac{4}{\hbar^2} \left| \int_{-T/2}^{T/2} v(t) \cos(2Et/\hbar) , dt \right|^2 \ &\approx \frac{16}{\hbar^4} \left| \int_{0}^{T/2} v(t) \cos(2Et/\hbar) , dt \right|^2 \end{align*} where we used the fact that $v(t)$ is an even function and that the integral over $(-T/2,0)$ gives a contribution of order $v^2$, which we neglect to lowest order in $v$.

Therefore, the lowest-order probability of transition from $|1\rangle$ to $|2\rangle$ is proportional to the squared integral of $v(t)$ over a finite interval of time around $t=0$.

证明 .

When $E=0$, the Hamiltonian becomes

$H(t)=\left(\begin{array}{cc}0 & v(t) \ v(t) & 0\end{array}\right)$

The eigenstates of this Hamiltonian are given by

$| \pm\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1 \ \pm 1\end{array}\right)$,

with eigenvalues $\pm |v(t)|$. Note that these eigenstates do not depend on time.

Suppose the system starts out in the state $|1\rangle$. The probability of a transition from $|1\rangle$ to $|2\rangle$ is given by

$P_{1 \rightarrow 2}=\left|\int_{-\infty}^{\infty} d t\langle 2|v(t)| 1\rangle e^{i \omega_{21} t}\right|^2$

where $\omega_{21}$ is the energy difference between states $|2\rangle$ and $|1\rangle$. In this case, we have $\omega_{21}=2|v(t)|$.

Since $|1\rangle$ and $|2\rangle$ are not the eigenstates of the Hamiltonian, we need to use time-dependent perturbation theory to calculate the transition probability. The first-order perturbation theory gives

$P_{1 \rightarrow 2}^{(1)}=\frac{|\langle 2|v(t)| 1\rangle|^2}{\omega_{21}^2}$.

Substituting $|1\rangle$ and $|2\rangle$ and using the fact that they are orthonormal, we get

$P_{1 \rightarrow 2}^{(1)}=\frac{1}{2 \omega_{21}^2}\left|\left\langle 2\left|\sigma_x\right| 1\right\rangle\right|^2$,

where $\sigma_x$ is the Pauli matrix.

Using the fact that $|\pm\rangle$ are eigenstates of $\sigma_x$, we can write

$P_{1 \rightarrow 2}^{(1)}=\frac{1}{4 \omega_{21}^2}$

To compare the perturbative result with the exact result, we need to consider the condition for which the perturbation is small. The perturbation is small when $|v(t)|$ is small compared to $E$. This is because the unperturbed energy gap is $E$, and if the perturbation is much smaller than this, then the perturbative result will be a good approximation to the exact result. Therefore, the condition for the perturbative result to be a good approximation is

$|v(t)| \ll E$.

证明 .

In the classically allowed region, the semiclassical approximation to the radial wave function is given by $$u(r) \approx \frac{1}{\sqrt{p(r)}} \sin \left( \int_{0}^{r} p(r’) dr’ + \frac{\ell \pi}{2} \right)$$ where $p(r) = \sqrt{2m(E – V(r))}$ is the classical momentum.

In this case, we have $\ell = 0$, so the phase term reduces to $\frac{\ell \pi}{2} = 0$.

Since $V(r)$ vanishes at the origin, we have $p(0) = \sqrt{2mE}$. Near $r=0$, the potential $V(r)$ rises steadily, so we can approximate $V(r)$ by its first-order Taylor expansion around $r=0$: $$V(r) \approx V(0) + rV'(0) = 0$$ since $V(0) = 0$ and $V'(0)$ is finite. Therefore, we have $$p(r) \approx \sqrt{2mE} – \frac{1}{2}\sqrt{2m}V”(0)r + \mathcal{O}(r^2).$$

Integrating $p(r)$ from $0$ to $r$, we obtain $$\int_{0}^{r} p(r’) dr’ \approx \sqrt{2mE} r – \frac{1}{6}\sqrt{2m}V”(0)r^3.$$

Substituting this expression back into the semiclassical approximation for $u(r)$, we get $$u(r) \approx \frac{1}{\sqrt{\sqrt{2mE} – \frac{1}{2}\sqrt{2m}V”(0)r}} \sin \left( \sqrt{2mE} r – \frac{1}{6}\sqrt{2m}V”(0)r^3 \right)$$ in the classically allowed region $0 \leq r < a$.