# # Application of the Schrödinger Theory

## # Particle Inside an Infinite Potential Well

A 1D well of length of $a$ with infinitely high walls,

$E_{p}(x)=\left\{\begin{array}{l}\infty \; \text { for } \; 0>x>a \\ 0 \; \text { for } \; 0 \leq x \leq a\end{array}\right.$

Since $E_p$ outside ($E_{po}$) is $\infty$, one must provide an infinite amount of energy to pull the particle out of the well. This means that the particle cannot get out.

To find all the possible wavefunctions that can be associated with the particle inside such a well. Find the position and momentum expectations.

$\psi(x, t) = \chi(x) \Gamma(t) = \chi(x) e^{\frac{-iG}{\hbar}t}$

#### # Step 1 Outside of the well

Before finding all the possible wavefunctions that can be associated with the particle inside such a well, let us answer the question: What is the eighenfunction $\chi$ outside? The answer is $\chi = 0$. For a particle to get out, and infinite energy must be provided. This is physically impossible, and therefore $|\chi|^2 = 0$ outside the well,

$\chi = 0 \; \text{for} \; 0 \geq x \geq a$

#### # Step 2 General solutions

To evaluate $\chi$ inside, we must solve $-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi=E \chi$ for the case $E_p = 0$,

$-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}=E \chi$

Divide both sides by $-\frac{\hbar}{2m}$ and put both terms on the left side ot get

$\frac{d^2\chi}{dx^2} + \frac{2mE}{\hbar^2} \chi = \frac{d^2\chi}{dx^2} + k^2 \chi = 0$

where $k^2 = \frac{2mE}{\hbar^2}$. The equation is a second-order differential equation with constant coefficients. Try a solution of the form,

$\chi = e^{\alpha x}$

where $\alpha$ is a constant as yet undetermined. We obtain

$\frac{d\chi}{dx} = \alpha e^{\alpha x}$

$\frac{d^2\chi}{dx^2} = \alpha^2 e^{\alpha x}$

$\alpha^2 e^{\alpha x} + k^2 e^{\alpha x} = 0$

$\alpha ^2 + k^2 = 0 \Rightarrow \alpha = \pm ik$

The two solutions are therefore $e^{ikx}$ and $e^{-ikx}$. The general solution will be the sum of these two

$\chi(x) = a e^{ikx} + b e^{-ikx} = (a + b) \cos(kx) + i(a-b) \sin(kx)$

where $a$ and $b$ are arbitrary constants. Because $a$ and $b$ are arbitrary constantd, we simplify the notation by intruducing new constatns $A$ and $B$,

$\chi(x) = A \cos(kx) + B\sin(kx)$

$A$ and $B$ are now the arbitrary constants to be determined from physical considerations.

#### # Step3 Well-behaved

Thus far, no restrictions have been found as to the values that $k$ and therefore $E$ can take. However, as soon as we require that $\chi$ be well-behaved, the restrictions on $k$ and $E$ will appear. For a well-behaved $\chi$, three requirments must be satisfied for $\chi$ and $d\chi/dx$: 1. Finite 2. Single value 3. Continous.

The conditions of 1 and 2 are satisfied by $\chi$. The condition that $\chi$ be continuous requires that $\chi$ be 0, at $x=0$ and at $x=a$, because $\chi$ is zero outside the well.

At the left side of the well,

$\chi(0) = A\times 1 + B \times 0 = A = 0$

Therefore

$\chi = B\sin(kx)$

At the right side of the well,

$\chi(a) = B\sin(ka) = 0$

There are two ways to satisfy. Either $B=0$ or $\sin(kx) = 0$. If $B=0$, the $\chi =0$ everywhere. This means that the paricle is not in the well. The only meaningful way to satisfy the condition of continuity is when $\sin ka = 0$, that $ka=0, \pi, 2\pi, \cdots$ or

$k = n\frac{\pi}{a}, \; n = 1, 2, 3, \cdots$

Notice that $n=0$ is not an acceptable choice. If $n=0$, then $k=0$ and $\chi$ would be zero everywhere; that is the particle could not be in the well. The set of eigenfunctions are

$\chi_n = B \sin(n \frac{\pi}{a} x)$

By substituding $k$ to $k^{2}=\frac{2 m E}{\hbar^{2}}$, we get a set of eigenvalues

$k^2 = \frac{2mE}{\hbar^2} \Rightarrow \frac{n^2 \pi^2}{a^2} = \frac{2mE}{\hbar^2}$

$E_n = n^2 \frac{\pi^2 \hbar^2}{2ma^2} = n^2 E_0, \; \text{where} \; E_0 = \frac{\pi^2 \hbar^2}{2ma^2}$

Notice

We should not that the first derivative of the eigenfunctions in this case is not continuous at $x=0$ and at $x=a$.

Left side of the well:

$0 \neq \frac{n\pi B}{a}\cos (n\frac{\pi}{a} x) \lvert_{x=0} = \frac{n\pi B}{a}$

Right side of the well:

$0 \neq \frac{n\pi B}{a}\cos (n\frac{\pi}{a} x) \lvert_{x=a} = \frac{n\pi B}{a} \cos (n\pi)$

As a result, then $d^2\chi/dx^2$ would be infinite at $0$ and $a$. From the time-independent Schrödinger Equation, this would imply that either $E$ or $E_p$ is infinite. In this idealized example $E_p = \infty$.

The solution of the Schrödinger Equation has given us a set of eigenfunctions $\chi$ that can be used to describe a particle in the potential well. It does not tell us what particular $\chi$ is associated withe the particle. Which particular $\chi$ one assigns to it depends on how the particle was palced in the well and what is done to the particle afterward. If we leave the particle alone, it will, following the tendency of all physical systems, tend to go to the lowest energy state available, which is called the ground state or $E = E_0$.

In the present case of lowest energy, the eigenfunction representing the x-position of the particle, will be,

$\chi(x) = B \sin\left(\frac{\pi}{a} x\right)$

When the $\chi(x)$ is multiplied by the time part of the wavefunction,

$\Gamma(t)=e^{\frac{-i E_0}{\hbar} t}$

the resulting wavefunction

$\psi_1(x,t) = B\sin\left(\frac{\pi}{a} x\right) e^{\frac{-i E_0}{\hbar} t}$

TIP

$A = a + bi \Rightarrow |A| = \sqrt {a^2 + b^2} \Rightarrow |A|^2 = a^2 + b^2$ $A^* = a - bi \Rightarrow A^*A = (a-bi)(a+bi) = a^2 + b^2$. We obtain

$|A|^2 = A^* A$

Then

$|\psi_1|^2 = \psi_1^* \psi_1 = B^2 \sin^2(\frac{\pi}{a}x) \left(e^{\frac{i E_0}{\hbar} t}\right) \left(e^{\frac{-i E_0}{\hbar} t}\right) = B^2 \sin^2(\frac{\pi}{a}x)$

#### # Step 4 Normalization

The probability of finding the particle somewhere in space must be 1. In mathematical terms this fact is stated as follows:

$\int_{-\infty}^{\infty} |\psi|^2 dx= \int_{0}^{a} |\psi|^2 dx = \int_{0}^{a} B^2 \sin^2(\frac{\pi}{a}x) dx = 1$

From the equation 64 in the integration tabel (opens new window).

$B^2\left[\frac{x}{2} - \frac {\sin2\frac{\pi}{a}x}{4\frac{\pi}{a}} \right]_0^a = 1$

$\Rightarrow B^2 \left((\frac{a}{2}) - (0)\right) =1$

$B = \sqrt{\frac{2}{a}}$

Therefore, when the particle goes to the ground state, the eigenvalue is

$E = E_{0}=\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}$

the eigenfunction is

$\chi_1(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi}{a}x\right)$

the wavefunction is

$\psi_1(x,t) = \chi_1(x) \Gamma_1(t) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi}{a}x\right) e^{\frac{-i E_0}{\hbar} t}$

and

$|\psi_1(x,t)|= |\psi_1| = \frac{2}{a} \sin^2\left(\frac{\pi}{a}x\right)$

#### # Step 5 Expectations

Consider a particle in the ground state. Average position

$\bar{x} = \int_{-\infty}^{\infty} \psi_1^{*}x\psi_1 dx = \frac{2}{a} \int_0^a \sin^2 \left(\frac{\pi}{a}x\right) x dx = \frac{a}{2}$

$\bar{p} = \int_{-\infty}^{\infty} \psi^* ( -i \hbar \frac{\partial}{\partial x}) \psi dx = 0$

$\bar{E} = \int_{-\infty}^{\infty} \psi^{*} E \psi dx = \int_{-\infty}^{\infty} \psi^{*} \left( i \hbar \frac{\partial \psi}{\partial t} \right) dx = E_0$

## # Harmonic Oscillator

### # Classical

Suppose a body os mass $m$ is connected to a massless spring, with a spring constant $k$, and the body is free to oscillate on the frictionless surface. At its rest, or equilibrium position, the position coordinate is $x=0$. If the body is pushed to compress the distance $x_0$, or pulled to stretch it a distance $x_0$, and then released, the body will then begin to oscillate. We may calculate its subsequent motion from Newton's second law. If you pull on a spring with force $F$ it pulls in the opposite direction with force $-F$. Thys, the force that the spring exerts on the body is $-kx$. The acceleration is not constant, we use the fundamental definition $a_x=d^2x/dt^2$. Applying Newton's second law to the body, we obtain

$-kx = m \frac{d^2x}{dt^2} \;\;\; \text{(10.10)}$

We may guess a solution.

$x = A \sin(\omega t + \phi) \;\;\; \text{(10.9)}$

Then, we obtain

$\frac{dx}{dt} = A\omega\cos(\omega t + \phi)$

$\frac{d^2x}{dt^2} = -A\omega^2\sin(\omega t + \phi)$

Substitute these into the main equantion

$- k A \sin (\omega t + \phi) = - m A \omega^2 \sin(\omega t + \phi)$

$\Rightarrow k = m \omega^2$

$\Rightarrow \omega = \sqrt{\frac{k}{m}} \;\;\; \text{(10.12)}$

Therefore, Eq. 10.9 is a solution when the constants have the relation of Eq. 10.12. Using $\omega = 2\pi\nu$, we obtain the frequency of osillation

$\nu = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

and the period

$T = \frac{1}{\nu} = 2\pi \sqrt{\frac{m}{k}}$

...

### # Quantum world

$E_p = \frac{1}{2} kx^2$

Eigenvalues,

$E_n = \left( n + \frac{1}{2} \right) \hbar \sqrt{\frac{k}{m}}$

We have $\omega = \sqrt{\frac{k}{m}}$, we obtain

$E_n = \left( n + \frac{1}{2} \right) \hbar \omega = \left( n + \frac{1}{2} \right) h \nu$

1. $\omega = \sqrt{\frac{k}{m}}$ 是通過牛頓得來的，為什麼要代在這裡來使用?
2. 量子簡協振子又不會像我們認為的樣子去振動，這裡的$\omega, \nu$ 這些究竟又代表什麼意思？

#### # Some important aspects

1. The difference between adjacent energy levels is a constant $h\nu$. The is consistent with Planck's blackbody theory.
2. $E_0 = \frac{1}{2} h \nu$, which is not equal to 0. This is different from Plank's blackbody theory, i.e,, $E = nh\nu, E_0 = 0)$
3. The eigenfunction on the ground state is

$\chi_0 = C e^{-\frac{\sqrt{mk}x^2}{2\hbar}}$

WARNING

This equation seems incorrect.

where $C$ can be decided through

$\int_{-\infty}^\infty |\chi|^2 dx = 1 \Rightarrow C =$

1. In classical mechanics, if $E = \frac{1}{2} h\nu = \frac{1}{2}kx_\text{max}^2$

$x_\text{max} = \sqrt{\frac{2E}{k}} = \sqrt{\frac{h\nu}{k}}$

In classical mechanics, the particle can not exceed $x_\text{max}$. But the quantum mechanics, the particle may exceed $x_\text{max}$ with low probabilities.

Todo

Draw a image using Python

Quantum Tunneling

Quantum tunneling has been used in Scanning Tunneling Microscope (STM). You should try to understand how does it work.

Eigenfunctions,

$\chi_n$

Wavefunctions,

$\psi_n(x) = \frac{1}{\sqrt{2^nn!}\pi^{1/4}} H_n(x) e^{-x^2/2}$

## # Schrödinger equation for H atom

Process: 1. Change equation from xyz to spherical coordinates. 2. Separation of variables. 3. Solve three differential equations with three requirments.

In this process, we obtain three numbers(we call them quantum numbers) $n$, $l$, $m_l$.

Because the energy

$E_n = - \frac{Z^2e^4m}{8\epsilon_0^2h^2} \frac{1}{n^2}$

depends only on the quantum number $n$, it is called the principal quantum number.

TIP

The restrictions on the these quantum numbes are

$n = 1, 2, 3, ..$

$l = 0, 1, 2, ..., n-1$

$m_l = 0, \pm 1, \pm 2, ..., \pm l$

where $n > l \geq |m_l|$, and

• $l$ is called the orbital quantum number, because $l$ determines the magnitude of the angular momentum $\bold{L}$ of the atom.

$L = \sqrt{l(l+1)} \hbar$

• $m_l$ is called the magnetic quantum number, because $m_l$ determines the orientation of the angular momentum $\bold{L}$ in a magnetic field. If an atom is placed in a magnetic field directed along the $z$ direction, the $z$ conponent of the angular momentum $\bold{L}$ of the atom is given by

$L_z = m_l \hbar \;\;\; (21.10)$

The vector $\bold{L}$ is perpendicular to the plane of rotation. The result of Eq. 21.10 tells us that in an atom, $\bold{L}$ cannot have any arbitrary orientation with respect to the $z$ axis, but rather it can have only certain discrete orientations. This is known as space quantization.

Suppose $l=2$, then $m_l$ can be 2, 1, 0, -1, -2. Thus $L_z = 2\hbar, \hbar, 0, -\hbar, -2\hbar$. By the way the magnitude to $\bold{L}$ can be calculated as $L = \sqrt{2(2+1)} \hbar = \sqrt{6} \hbar$. The spacial variation of $\chi$ depends, on the three quntum numbers, and the wavefunciton is written with the quantum numbers sunscripts $\chi_{nlm_l}$. Because for a given $n$ the other two numbers can take several values, this means that it is possible for the electron to have quite different characteristics while maintaining the same energy. States $\chi$ habing the same energy but different values for the quantum numbers $l$ and $m_l$ are called degenerate states.

States for which

• $l=0$ are called $s$ states
• $l=1$ are called $p$ states
• $l=2$ are called $d$ states
• $l=3$ are called $f$ states.

The Sterm-Gerlach experiment is evident that there is a magnetic dipole other than the orbital dipole that has been overlooked. What has the originial Schrödinger theory overlooked?

The electon has an intrinsic angular momentum called the spin $\bold{S}. Just as the orital angular momentum $\bold{L}$ has a magnetic dipole associated with it, so does the spin. By analogy with the behavior of $L$ and in order to explain the experimental results, G. Uhlenbeck and S. Goudsmit postulated that the magnitude of $S$ and its $z$ component were quantized as follows $S = \sqrt{s(s+1)} \hbar \;\;\text{where} \; s = \frac{1}{2}$ $S_z = m_s \hbar \;\; \text{where} \; m_s = \pm \frac{1}{2}$ Using the spin postulate, lots of experimental results can be explained. For us the main conclusion is that the state of an electron is now specified by four quantum numbers: $n$, $l$, $l_m$, $m_s$. Note that the quantum number $s$ is $\frac{1}{2}$ for all individual electrons and thus we do not need to specify it further. TIP $m_s = \frac{1}{2}$ is spin up and $m_s = -\frac{1}{2}$ is spin down. ## # Some features of the atomic wavefuctions Why don't we talk about$\Gamma(t)\$ anymore?

We have $\psi = \chi \Gamma$, why don't we leave $\Gamma$ behind.

1. State $s$ ($l=0$), $|\chi|^2$ have spherical symemetry.
2. Other states, axial symmetry, but no spherical symmetry.
3. $\sum_{m_l}\chi^*_{n,l,m_l}\chi_{n,l,m_l}$ has spherical symmetry.
4. By looking at the radial probability density $P(r)$ ($P(r)dr$ is the probability of finding electron between $r$ and $r+dr$). Given state $n$, lower $l$ more likely to be found near the the nucleus.(Lower $l \rightarrow$ lower angular momentum $L$).

WARNING