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After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.

Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.

Part (c)

Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.

$\frac{k_2}{\omega}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore

$$E^2 = (k_3-eA)^2 - \omega + (2 n_1 + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.

Part (e)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$

This leads to

$$E^2 = (k_3-eA)^2 + \omega + (2 n_1' + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$

Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.

These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.

We have however missed a special case where $n_1=0$. In this case, $E = \pm(k_3 - eA)$, and (19.7a) and (19.7b) reduce to

$$ \begin{equation} \begin{split} \phi_1' &= (k_2 - \omega x^1) \phi_1 \\ \phi_2' &= i(E-k_3+eA)\phi_1- (k_2 - \omega x^1) \phi_2. \end{split} \end{equation} $$

Now $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes. We have

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$

So we reach the same conclusion, but through a very different argument.

After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.

Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.

Part (c)

Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.

$\frac{k_2}{\omega}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore

$$E^2 = (k_3-eA)^2 - \omega + (2 n_1 + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.

Part (e)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$

This leads to

$$E^2 = (k_3-eA)^2 + \omega + (2 n_1' + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$

Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.

These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.

We have however missed a special case where $n_1=0$. In this case, $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes.

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$

(Setting $\phi_1$ to zero won't give us any non-vanishing solution of $\phi_2$ that does not blow up as $x^1 \rightarrow \infty$.)

For any fixed $A$, only states with $n_3 > \frac{eAL}{2\pi}$ have positive energies. As $A$ increases by $\frac{2\pi}{eL}$, one such $n_3$ will now give us negative energy.

So we reach the same conclusion, but through very different reasoning.

After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.

Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.

Part (c)

Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.

$\frac{k_2}{\omega}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore

$$E^2 = (k_3-eA)^2 + 2 n_1 \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.

Part (e)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$

This leads to

$$E^2 = (k_3-eA)^2 + \omega + 2 n_1' \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$

Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.

These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.

We have however missed a special case where $n_1=0$. In this case, $E = \pm(k_3 - eA)$, and (19.7a) and (19.7b) reduce to

$$ \begin{equation} \begin{split} \phi_1' &= (k_2 - \omega x^1) \phi_1 \\ \phi_2' &= i(E-k_3+eA)\phi_1- (k_2 - \omega x^1) \phi_2. \end{split} \end{equation} $$

Now $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes. We have

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$

So we reach the same conclusion, but through a very different argument.

After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.

Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.

Part (c)

Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.

$\frac{k_2}{\omega}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore

$$E^2 = (k_3-eA)^2 - \omega + (2 n_1 + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.

Part (e)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$

This leads to

$$E^2 = (k_3-eA)^2 + \omega + (2 n_1' + 1) \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$

Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.

These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.

We have however missed a special case where $n_1=0$. In this case, $E = \pm(k_3 - eA)$, and (19.7a) and (19.7b) reduce to

$$ \begin{equation} \begin{split} \phi_1' &= (k_2 - \omega x^1) \phi_1 \\ \phi_2' &= i(E-k_3+eA)\phi_1- (k_2 - \omega x^1) \phi_2. \end{split} \end{equation} $$

Now $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes. We have

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$

So we reach the same conclusion, but through a very different argument.

After spending some more time, I have come up with my own explanation/solution which I share below.

Part (c)

Equation (19.8) is indeed the Schrödinger's equation for a standard harmonic oscillator. See this link.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{eB} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $\omega = e B$ and $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + eB\right)$.

$\frac{k_2}{eB}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$. Therefore

$$E^2 = (k_3-eA)^2 + 2 n_1 eB = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 eB.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution.

Part (d)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

Note also that the answer to (e) in Zhong-Zhi Xianyu's solution is actually wrong.

The last equation does not help, due to the two quartic terms on both sides. We will have to go back to the first-order differential equation.

$$\phi_1' = (k_2 - eBx^1)\phi_1 + i(E+k_3-eA)\phi_2 \tag{19.7a}$$

For simplicity, we take the case where $n_1=0$. The solution to $\phi_1$ is

$$\phi_1(x^1) = c e^{-\omega (x^1)^2 / 2},$$

where $c$ is a normalization factor.

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=-k_3+eA=-\frac{2\pi n_3}{L}+eA.$$

Now it is clear that $N_R$ actually increases by $\frac{eL^2B}{2\pi}$ as the background $A$ is adiabatically increased by $\frac{2\pi}{eL}$.

After spending some more time, I have come up with my own explanation/solution which I share below. For definitiveness, we take both $B$ and $A$ to be positive, and define $\omega \equiv eB$. Note that here $\omega$ and $B$ have the dimension of mass (or energy) squared.

Also note that $k_2$ and $k_3$ really should be $k^2$ and $k^3$. But I will not change the notation in the text below.

Part (c)

Equation (19.8) is formally the same as the Schrödinger's equation for a standard harmonic oscillator. See this link. We say "formally", because here $\omega$, as well as $E'$ have the dimension of energy-squared instead of energy.

It can be written as

$$-\frac{1}{2}\frac{\partial^2 \phi_1(x^1)}{(\partial x^1)^2} + \frac{\omega^2}{2} \left(x^1-\frac{k_2}{\omega} \right)^2 \phi_1(x^1) = E' \phi_1(x^1),$$

where $E' = \frac{1}{2}\left(E^2 - (k_3-eA)^2 + \omega\right)$.

$\frac{k_2}{\omega}$ simply shifts the center of oscillation.

For a harmonic oscillator, we have $E' = \left(\frac{1}{2} + n_1 \right)\omega$, where $n_1 = 0, 1, 2, \dots$. Therefore

$$E^2 = (k_3-eA)^2 + 2 n_1 \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 n_1 \omega \tag{A}.$$

This actually looks better than (19.9) in Zhong-Zhi Xianyu's solution. The RHS is always positive, and we can simply take its square root to get $E$.

Part (e)

For a better understanding of the problem, we actually also need an equation similar to (19.8) for $\phi_2$. Comparing (19.7a) and (19.7b), we see that we simply need to switch a bunch of signs.

$$\phi_2'' - \left[\omega^2 \left(x^1-\frac{k_2}{\omega} \right)^2 - E^2 + (k_3-eA)^2 + \omega \right]\phi_2 = 0 \tag{19.8'}.$$

This leads to

$$E^2 = (k_3-eA)^2 + \omega + 2 n_1' \omega = \left(\frac{2\pi n_3}{L}-eA\right)^2 + 2 (n_1'+1) \omega \tag{B}.$$

Comparing this to Equation (A), we see that $\phi_2$ is also a harmonic oscillator, but one energy level lower than $\phi_1$. So $n_1 = n_1' + 1$ where $n_1' = 0, 1, 2, \dots$.

These solutions will not lead to any non-conservation of fermions. As $A$ increases, eigenstates with $n_3 < \frac{eAL}{2\pi}$ will gain energy, while eigenstates with $n_3 > \frac{eAL}{2\pi}$ will lose energy. But even the latter will still have an energy level greater than $\sqrt{2(n_1'+1)\omega}$.

We have however missed a special case where $n_1=0$. In this case, $E = \pm(k_3 - eA)$, and (19.7a) and (19.7b) reduce to

$$ \begin{equation} \begin{split} \phi_1' &= (k_2 - \omega x^1) \phi_1 \\ \phi_2' &= i(E-k_3+eA)\phi_1- (k_2 - \omega x^1) \phi_2. \end{split} \end{equation} $$

Now $\phi_1$ is a still harmonic oscillator, while $\phi_2$ vanishes. We have

Substituting this into (19.7a), where we can set $k_2$ to zero, we have

$$E=k_3-eA=\frac{2\pi n_3}{L}-eA.$$

So we reach the same conclusion, but through a very different argument.