# Application of the Schrödinger Theory

# Particle Inside an Infinite Potential Well

A 1D well of length of aa with infinitely high walls,

Ep(x)={for0>x>a0for0xaE_{p}(x)=\left\{\begin{array}{l}\infty \; \text { for } \; 0>x>a \\ 0 \; \text { for } \; 0 \leq x \leq a\end{array}\right.

Since EpE_p outside (EpoE_{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.

Our task

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

不同的EE對應不同的wavefunction, 不同的wavefunction對應了不同的概率密度ψ2|\psi|^2. 在不同的能量下,檢測到粒子的分佈的概率密度是不一樣的。我們說粒子的概率分佈是有一個前提的,這個前提就是在某個能量下的概率分佈。想要更好地確定粒子的位置最好是要知道他所處的能量。

ψ(x,t)=χ(x)Γ(t)=χ(x)eiGt\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 χ\chi outside? The answer is χ=0\chi = 0. For a particle to get out, and infinite energy must be provided. This is physically impossible, and therefore χ2=0|\chi|^2 = 0 outside the well,

χ=0for0xa\chi = 0 \; \text{for} \; 0 \geq x \geq a

# Step 2 General solutions

To evaluate χ\chi inside, we must solve 22md2χdx2+Ep(x)χ=Eχ-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}+E_{p}(x) \chi=E \chi for the case Ep=0E_p = 0,

22md2χdx2=Eχ-\frac{\hbar^{2}}{2 m} \frac{d^{2} \chi}{d x^{2}}=E \chi

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

d2χdx2+2mE2χ=d2χdx2+k2χ=0\frac{d^2\chi}{dx^2} + \frac{2mE}{\hbar^2} \chi = \frac{d^2\chi}{dx^2} + k^2 \chi = 0

where k2=2mE2k^2 = \frac{2mE}{\hbar^2}. The equation is a second-order differential equation with constant coefficients. Try a solution of the form,

χ=eαx\chi = e^{\alpha x}

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

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

d2χdx2=α2eαx\frac{d^2\chi}{dx^2} = \alpha^2 e^{\alpha x}

α2eαx+k2eαx=0\alpha^2 e^{\alpha x} + k^2 e^{\alpha x} = 0

α2+k2=0α=±ik\alpha ^2 + k^2 = 0 \Rightarrow \alpha = \pm ik

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

χ(x)=aeikx+beikx=(a+b)cos(kx)+i(ab)sin(kx)\chi(x) = a e^{ikx} + b e^{-ikx} = (a + b) \cos(kx) + i(a-b) \sin(kx)

where aa and bb are arbitrary constants. Because aa and bb are arbitrary constantd, we simplify the notation by intruducing new constatns AA and BB,

χ(x)=Acos(kx)+Bsin(kx)\chi(x) = A \cos(kx) + B\sin(kx)

AA and BB 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 kk and therefore EE can take. However, as soon as we require that χ\chi be well-behaved, the restrictions on kk and EE will appear. For a well-behaved χ\chi, three requirments must be satisfied for χ\chi and dχ/dxd\chi/dx: 1. Finite 2. Single value 3. Continous.

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

At the left side of the well,

χ(0)=A×1+B×0=A=0\chi(0) = A\times 1 + B \times 0 = A = 0


χ=Bsin(kx)\chi = B\sin(kx)

At the right side of the well,

χ(a)=Bsin(ka)=0\chi(a) = B\sin(ka) = 0

There are two ways to satisfy. Either B=0B=0 or sin(kx)=0\sin(kx) = 0. If B=0B=0, the χ=0\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 sinka=0\sin ka = 0, that ka=0,π,2π,ka=0, \pi, 2\pi, \cdots or

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

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

χn=Bsin(nπax)\chi_n = B \sin(n \frac{\pi}{a} x)

By substituding kk to k2=2mE2k^{2}=\frac{2 m E}{\hbar^{2}}, we get a set of eigenvalues

k2=2mE2n2π2a2=2mE2k^2 = \frac{2mE}{\hbar^2} \Rightarrow \frac{n^2 \pi^2}{a^2} = \frac{2mE}{\hbar^2}

En=n2π222ma2=n2E0,whereE0=π222ma2E_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}


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

Left side of the well:

0nπBacos(nπax)x=0=nπBa0 \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:

0nπBacos(nπax)x=a=nπBacos(nπ)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 d2χ/dx2d^2\chi/dx^2 would be infinite at 00 and aa. From the time-independent Schrödinger Equation, this would imply that either EE or EpE_p is infinite. In this idealized example Ep=E_p = \infty.

The solution of the Schrödinger Equation has given us a set of eigenfunctions χ\chi that can be used to describe a particle in the potential well. It does not tell us what particular χ\chi is associated withe the particle. Which particular χ\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=E0E = E_0.

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

χ(x)=Bsin(πax)\chi(x) = B \sin\left(\frac{\pi}{a} x\right)

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

Γ(t)=eiE0t\Gamma(t)=e^{\frac{-i E_0}{\hbar} t}

the resulting wavefunction

ψ1(x,t)=Bsin(πax)eiE0t\psi_1(x,t) = B\sin\left(\frac{\pi}{a} x\right) e^{\frac{-i E_0}{\hbar} t}


A=a+biA=a2+b2A2=a2+b2A = a + bi \Rightarrow |A| = \sqrt {a^2 + b^2} \Rightarrow |A|^2 = a^2 + b^2 A=abiAA=(abi)(a+bi)=a2+b2A^* = a - bi \Rightarrow A^*A = (a-bi)(a+bi) = a^2 + b^2. We obtain

A2=AA|A|^2 = A^* A


ψ12=ψ1ψ1=B2sin2(πax)(eiE0t)(eiE0t)=B2sin2(πax)|\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:

ψ2dx=0aψ2dx=0aB2sin2(πax)dx=1\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).

B2[x2sin2πax4πa]0a=1B^2\left[\frac{x}{2} - \frac {\sin2\frac{\pi}{a}x}{4\frac{\pi}{a}} \right]_0^a = 1

B2((a2)(0))=1\Rightarrow B^2 \left((\frac{a}{2}) - (0)\right) =1

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

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

E=E0=π222ma2E = E_{0}=\frac{\pi^{2} \hbar^{2}}{2 m a^{2}}

the eigenfunction is

χ1(x)=2asin(πax)\chi_1(x) = \sqrt{\frac{2}{a}} \sin\left( \frac{\pi}{a}x\right)

the wavefunction is

ψ1(x,t)=χ1(x)Γ1(t)=2asin(πax)eiE0t\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}


ψ1(x,t)=ψ1=2asin2(πax)|\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

x¯=ψ1xψ1dx=2a0asin2(πax)xdx=a2\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}

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

E¯=ψEψdx=ψ(iψt)dx=E0\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 mm is connected to a massless spring, with a spring constant kk, and the body is free to oscillate on the frictionless surface. At its rest, or equilibrium position, the position coordinate is x=0x=0. If the body is pushed to compress the distance x0x_0, or pulled to stretch it a distance x0x_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 FF it pulls in the opposite direction with force F-F. Thys, the force that the spring exerts on the body is kx-kx. The acceleration is not constant, we use the fundamental definition ax=d2x/dt2a_x=d^2x/dt^2. Applying Newton's second law to the body, we obtain

kx=md2xdt2(10.10)-kx = m \frac{d^2x}{dt^2} \;\;\; \text{(10.10)}

We may guess a solution.

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

Then, we obtain

dxdt=Aωcos(ωt+ϕ)\frac{dx}{dt} = A\omega\cos(\omega t + \phi)

d2xdt2=Aω2sin(ωt+ϕ)\frac{d^2x}{dt^2} = -A\omega^2\sin(\omega t + \phi)

Substitute these into the main equantion

kAsin(ωt+ϕ)=mAω2sin(ωt+ϕ)- k A \sin (\omega t + \phi) = - m A \omega^2 \sin(\omega t + \phi)

k=mω2\Rightarrow k = m \omega^2

ω=km(10.12)\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 ω=2πν\omega = 2\pi\nu, we obtain the frequency of osillation

ν=12πkm\nu = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

and the period

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


# Quantum world

Ep=12kx2E_p = \frac{1}{2} kx^2


En=(n+12)kmE_n = \left( n + \frac{1}{2} \right) \hbar \sqrt{\frac{k}{m}}

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

En=(n+12)ω=(n+12)hνE_n = \left( n + \frac{1}{2} \right) \hbar \omega = \left( n + \frac{1}{2} \right) h \nu



  1. ω=km\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ω,ν\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 hνh\nu. The is consistent with Planck's blackbody theory.
  2. E0=12hνE_0 = \frac{1}{2} h \nu, which is not equal to 0. This is different from Plank's blackbody theory, i.e,, E=nhν,E0=0)E = nh\nu, E_0 = 0)
  3. The eigenfunction on the ground state is

χ0=Cemkx22\chi_0 = C e^{-\frac{\sqrt{mk}x^2}{2\hbar}}


This equation seems incorrect.

where CC can be decided through

χ2dx=1C=\int_{-\infty}^\infty |\chi|^2 dx = 1 \Rightarrow C =

  1. In classical mechanics, if E=12hν=12kxmax2E = \frac{1}{2} h\nu = \frac{1}{2}kx_\text{max}^2

xmax=2Ek=hνkx_\text{max} = \sqrt{\frac{2E}{k}} = \sqrt{\frac{h\nu}{k}}

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


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.




ψn(x)=12nn!π1/4Hn(x)ex2/2\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) nn, ll, mlm_l.

Because the energy

En=Z2e4m8ϵ02h21n2E_n = - \frac{Z^2e^4m}{8\epsilon_0^2h^2} \frac{1}{n^2}

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



The restrictions on the these quantum numbes are

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

l=0,1,2,...,n1l = 0, 1, 2, ..., n-1

ml=0,±1,±2,...,±lm_l = 0, \pm 1, \pm 2, ..., \pm l

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

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

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

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

Lz=ml(21.10)L_z = m_l \hbar \;\;\; (21.10)

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

Possible orientations of the angular momentum $L$ of the electron in the hydrogen atom for the case where the orbital quantum number $l$=2

Suppose l=2l=2, then mlm_l can be 2, 1, 0, -1, -2. Thus Lz=2,,0,,2L_z = 2\hbar, \hbar, 0, -\hbar, -2\hbar. By the way the magnitude to L\bold{L} can be calculated as L=2(2+1)=6L = \sqrt{2(2+1)} \hbar = \sqrt{6} \hbar. The spacial variation of χ\chi depends, on the three quntum numbers, and the wavefunciton is written with the quantum numbers sunscripts χnlml\chi_{nlm_l}. Because for a given nn 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 χ\chi habing the same energy but different values for the quantum numbers ll and mlm_l are called degenerate states.

Partial representation of the degeneracy of the eigenfunctions in the hydrogen atom.

States for which

  • l=0l=0 are called ss states
  • l=1l=1 are called pp states
  • l=2l=2 are called dd states
  • l=3l=3 are called ff 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 L\bold{L} has a magnetic dipole associated with it, so does the spin. By analogy with the behavior of LL and in order to explain the experimental results, G. Uhlenbeck and S. Goudsmit postulated that the magnitude of SS and its zz component were quantized as follows

S=s(s+1)wheres=12S = \sqrt{s(s+1)} \hbar \;\;\text{where} \; s = \frac{1}{2}

Sz=mswherems=±12S_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: nn, ll, lml_m, msm_s. Note that the quantum number ss is 12\frac{1}{2} for all individual electrons and thus we do not need to specify it further.


ms=12m_s = \frac{1}{2} is spin up and ms=12m_s = -\frac{1}{2} is spin down.

# Some features of the atomic wavefuctions

Why don't we talk about $\Gamma(t)$ anymore?

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

Source: Visualizing All Things [1]
  1. State ss (l=0l=0), χ2|\chi|^2 have spherical symemetry.
  2. Other states, axial symmetry, but no spherical symmetry.
  3. mlχn,l,mlχn,l,ml\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) (P(r)drP(r)dr is the probability of finding electron between rr and r+drr+dr). Given state nn, lower ll more likely to be found near the the nucleus.(Lower ll \rightarrow lower angular momentum LL).



[1] Youtube: Hydrogen Electron Orbital (opens new window)