量子物理学基础代写|FOUNDATIONS OF QUANTUM PHYSICS PHYS104 University of Liverpool Assignment

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The experiments in question, such as the double-slit experiment and the photoelectric effect, challenged classical physics and led to the development of quantum mechanics, which has since become a fundamental theory in physics. It explains the behavior of matter and energy at the atomic and subatomic level, where classical mechanics fails. Quantum mechanics has led to the development of many technologies, including lasers, transistors, and quantum computing.

Consider a two-state system with Hamiltonian $$H(t)=\left(\begin{array}{cc} +E & v(t) \\ v(t) & -E \end{array}\right)$$ where $v(t)$ is real and where $v \rightarrow 0$ for $t \rightarrow \pm \infty$. (a) Suppose that at $t=-\infty$ the system is in the state $|1\rangle$. Use time dependent perturbation theory to determine the probability that at $t=+\infty$ the system is in the state $|2\rangle$, to lowest order in $v$.

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

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 $t=0$.

(b) If $E=0$, the eigenstates of $H(t)$ do not depend on $t$. Use this fact to calculate the probability of a transition from $|1\rangle$ to $|2\rangle$ exactly, in this case. What is the result obtained from time-dependent perturbation theory in this case? What is the condition that the perturbative result is a good approximation to the exact result?

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

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

Consider $s$-wave scattering for a particle of mass $m$ off a potential $V(r)$ which vanishes at the origin, rises steadily as $r$ increases from zero, reaches a maximum at $r=c$, and then goes quickly to zero as $r$ increases further.

For $\ell=0$, the radial wave function $u(r)$ satisfies the same Schrödinger equation as that for a particle in one dimension with potential $V$, subject to the boundary condition $u(0)=0$.

Consider scattering with energy $E$ where $0 \ll E \ll V(c)$. The classical turning points are at $r=a$ and $r=b$ with $a<c<b$.
(a) What is the semiclassical approximation to the wave function in the classically allowed region, $0 \leq r<a$ ?

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 $0 \leq r < a$.