# 30 days of physics writing

I will be posting physics-related articles for next 30 days.

# Day 14: Common questions we ask in Quantum Mechanics

Here I will be discussing some common questions that students (like me) often ask and wonder about while learning Quantum Mechanics. I will also provide the source where I got my aha moment! This list of questions will increase as I venture in the world of quantum.

## 1. What exactly is the uncertainty principle?

We can not measure the position and momentum of a particle simultaneously and accurately. That’s what we usually take as the definition of Uncertainty Principle, but more mathematically precise and accurate definition would be,

We cannot measure any two conjugate variables, simultaneously and accurately. So, for conjugate position and momentum i.e. $x$ and $p_x$, the uncertainty principle is,

$\Delta x \Delta p_x \geq \frac{\hbar}{2}$

Where, $\Delta x$ and $\Delta p_x$ are the standard deviations.

## 2. What is the difference between state vector and wave function?

State vector is denoted as $|\psi>$ and the wave function is represented by $\psi$. Both of them are actually representing the state of the system but just in different formalism. So in wave mechanics we represent the state of systems by wave functions, which are elements of the space of wave functions.

In general scheme, the state of system is represented as state vectors which are the elements of abstract linear vector space (Hilbert space). The ket vectors $|\psi>$ are element of Hilbert space and bra vectors $< \psi|$ are elements of dual Hilbert space (Dual of Hilbert space is Hilbert space itself). Every ket vector has a corresponding bra vector.

The beauty of this notation is that it allows us to write the formulas in a very concise form. Let’s take an example. In terms of wave functions, we define the normalization as the scalar product of wave function with itself and when its integrated on some interval we should get 1.

$\int_a^b |\psi\left(x\right)|^2 = 1$

But using state vectors we say if inner product of a vector with itself is 1 then its normalized.

$\left<\psi|\psi\right> = 1$

>> Chapter 3, Quantum Mechanics – P. J. E. Peebles

# Day 13: Infinite potential well (1D)

Solving Schrödinger equation for infinite potential well:

The general method to solve the SE is by solving the TISE for $\psi$ and then multiplying it by $\phi$.

The potential function is defined as:

$V = 0$ for $\leq x \leq a$ and $V =\infty$ for $x < 0, x > a$

The idea behind taking this non-realistic potential is described quite in detail in this Wikipedia Entry. It’s clearly written that if you have a particle in a box that’s very large, the probability to find it at a certain point is fixed. But, as the box gets smaller, in the nanometer range (think about the zoom in scenes from Ant Man), all you can get is the probability distribution. The particle cannot be located exactly inside the well, but only the probability to find it somewhere inside the box can be calculated. And that’s done by Schrödinger equation.

The TISE (Time Independent Schrödinger Equation) after plugging the potential looks like this,

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

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

It’s a $2^{nd}$ order linear differential equation with constant coefficients.

Let’s simplify the equation by replacing all these constants with a single constant.

Let, $k^2 = \frac{2m}{\hbar^2} E$

Or, $k = \frac{\sqrt{2mE}}{\hbar}$

After plugging it back in the equation, we get a nice looking differential equation.

$\frac{d^2\psi}{dx^2} = -k^2 \psi$

$\frac{d^2\psi}{dx^2} + k^2 \psi = 0$

Let’s solve this equation by a general method,

First we need to make the characteristic equation. For that, we assume $\frac{d}{dx} = m$ and $\frac{d^2}{dx^2} = m^2$

Now, our equation looks like quadratic equation, which can easily be solved.

$\left(m^2 + k^2 \right)\psi = 0$

$m = \pm ik$

We have two complex roots. Let’s see how we can write the general solution for such roots.

If you have complex roots of form $a \pm ib$, then the general solution looks like this,

$y = c_1 e^{ax} sin\left(bx\right) + c_2 e^{ax} cos\left(bx\right)$

In our case, we have complex roots of the form $0 \pm ik$, so the solution will be,

$\psi\left(x\right) = A e^{0x} sin\left(kx\right) + B e^{0x} cos\left(kx\right)$

Or, $\psi\left(x\right) = A sin\left(kx\right) + B cos\left(kx\right)$

We now have the wave function for infinite potential well. But, there are still some unknown constants. So we need to evaluate the values of $A$, $B$, and $k$.

For that, we have to apply the boundary conditions. Only those wave functions are valid who follow these conditions. And one of these conditions is, the wave function $\psi$ and its derivative should be continuous. And, for the wave function to be continuous, $\psi\left(x=0\right) = 0$ and $\psi\left(x=a\right) = 0$. Which basically means, the wave function vanishes at the boundaries, i.e. particle cannot go out of the potential well.

Let’s apply these conditions to our solution,

At $x = 0$, $\psi\left(0\right) = A sin 0 + B cos 0 = 0$

Or, $\psi\left(0\right) = B = 0$

At $x = a$, $\psi\left(a\right) = A sin ka + B cos ka = 0$

Or, $\psi\left(a\right) = A sinka = 0$

From here we have either $A = 0$, (which doesn’t tell us anything) or $sinka = 0$. To satisfy this last case, the value of $ka$ should be an integer multiple of $\pi$.

So we have $ka = \pm \pi$, where $n$ is a Natural number.

We can now write our wave function again, and this time we plug in the values of two constants we have evaluated so far, $k =\frac{n\pi}{a}$, and $B = 0$

$\psi\left(x\right) = A sin\left(\frac{n \pi}{a} x\right)$

Or, $\psi\left(x\right) = A sin\left(kx\right)$

We have now come one step closer to our final solution. We just need $A$. And it can be done by normalization.

The process of dividing a wave function with normalization constant is called normalization. And the normalization constant can be evaluated by taking the modulus square of the wave function and integrating it in the range $0$ to $a$ and then equating it to 1. That was mathematical formula. Physically it would mean that we are trying to refine the wave functions so that the probability density remains equal to 1. Probability density is equated to one which means that the particle is got to be somewhere in a defined region or in all of space if are taking the limits $-\infty$ to $+\infty$ and not $0$ to $a$.

$\int_0^a |\psi|^2 dx = \int_0^a |A|^2 sin^2\left(kx\right) = 1$

$|A|^2 \int_0^a \frac{\left(1-cos2x\right)}{2} dx = 1$

$|A|^2 \left(\frac{a}{2}\right) = 1$

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

This is the final piece of the puzzle, let’s plug it back in.

$\psi = \sqrt{\frac{2}{a}} sin \frac{n\pi x}{a}$

The TISE has given us infinite number of solutions for the infinite potential well that we started with. Each of these wave functions have corresponding energy eigenvalues. And we can now evaluate the general expression of the nth energy level.

Here is the expression that we used previously in our derivation,

$k^2 = \frac{2 m E}{\hbar^2}$

Now that we know the possible values of $k$, we can find the expression for $E$,

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

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

It is the expression for $n^{th}$ energy level!

The wave function $\Psi = \psi\left(x,t\right)$ can now be written.

$\Psi = \psi\left(x\right) \phi\left(t\right)$

$\Psi = \sqrt{\frac{2}{a}} sin \frac{n\pi x}{a} e^{\frac{-iEt}{\hbar}}$

Reference: Introduction to Quantum Mechanics – David J. Griffiths Available to download on Archive.org

# Day 12: Solving 3D Schrödinger Equation (Part 3)

We solved the angular equation for $\phi$ yesterday. Now we have another $2^{nd}$ order linear differential equation to solve.

$\frac{1}{\Theta} \left [ sin\theta \frac{d}{d \theta} \left( sin\theta \frac{d \Theta}{d \theta} \right) \right] + l \left(l+1\right) sin^2 \theta = m^2$

This equation is also known as Associated Legendre equation, because it uses the Legendre polynomial for its solution. The only meaningful solutions of this equation are in the range $0 \leq \theta \leq \pi$ and are called associated Legendre functions of first kind.

The solution is, $\Theta \left(\theta\right) = A P^m_l \left(cos\theta\right)$ where $P^m_l$ is the associated Legendre function.

Let’s now solve the last part, the radial equation.

$\frac{1}{R} \frac{d}{dr} \left(r^2\frac{dR}{dr}\right) \frac{-2mr^2}{\hbar^2} \left[ V\left(r\right) - E \right] = l \left(l+1\right)$

It can be simplified using $u\left(r\right) \equiv r R\left(r\right)$

So, now we have $R = \frac{u}{r}$

$\frac{dR}{dr} = \frac{[r\frac{du}{dr} - u]}{r^2}$

Differentiating it again we get,

$\frac{d}{dr} [r^2 \frac{dR}{dr}] = r^2\frac{d^2 u}{dr^2} +\frac{du}{dr} - \frac{du}{dr}$

Let’s now put it back in the radial equation

$r \frac{d^2u}{dr^2} - \frac{2mr^2}{\hbar^2} [V-E] = l \left(l+1\right)R$

After solving it we get,

$\frac{-\hbar^2}{2m} \frac{d^2u}{dr^2} + [V + \frac{\hbar^2}{2m} \frac{l\left(l+1\right)}{r^2}] u = Eu$

Here $[V + \frac{\hbar^2}{2m} \frac{l\left(l+1\right)}{r^2}] = V_{eff}$

References:

# Day 11: Solving 3D Schrödinger Equation (Part 2)

Today, we will start by solving the angular equation that we derived yesterday,

$\frac{1}{Y} \frac{i}{sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial Y}{\partial \theta} \right) + \frac{1}{sin^2\theta} \frac{\partial^2 Y}{\partial \phi^2} = - l \left(l+1\right)$

Let’s multiply it by $Y^2 sin^\theta$,

$sin\theta \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial Y}{\partial \theta} \right) + \frac{\partial^2 Y}{\partial \phi^2} = - l \left(l+1\right) Y sin^2\theta$

Again we use separation of variables,

$Y\left(\theta, \phi \right) = \Theta \left(\theta\right) \Phi\left(\phi\right)$

Let’s put it in the equation,

$sin\theta \frac{d}{d \theta} \left( sin\theta \frac{d \Theta\left(\theta\right)}{d \theta} \Phi\left(\phi\right) \right) + \frac{d^2 \Phi\left(\phi\right)}{d \phi^2} \Theta\left(\theta\right) = - l \left(l+1\right) \Theta \left(\theta\right) \Phi\left(\phi\right) sin^2 \theta$

Dividing throughout by $\Theta \left(\theta\right) \Phi\left(\phi\right)$

$sin\theta \frac{d}{d \theta} \left( sin\theta \frac{d \Theta}{d \theta} \frac{1}{\Theta} \right) + \frac{d^2 \Phi}{d \phi^2} \frac{1}{\Phi} = - l \left(l+1\right) sin^2 \theta$

After rearranging the equation, we get

$\left \{ \frac{1}{\Theta} \left [ sin\theta \frac{d}{d \theta} \left( sin\theta \frac{d \Theta}{d \theta} \right) \right] + l \left(l+1\right) sin^2 \theta \right\} + \frac{d^2 \Phi}{d \phi^2} \frac{1}{\Phi} = 0$

Now you can easily see that the first part of the equation is function of $\theta$ only and the second part is function of $\phi$ only. We can write them separately as follows:

$\frac{1}{\Theta} \left [ sin\theta \frac{d}{d \theta} \left( sin\theta \frac{d \Theta}{d \theta} \right) \right] + l \left(l+1\right) sin^2 \theta = m^2$

$\frac{d^2 \Phi}{d \phi^2} \frac{1}{\Phi} = - m^2$

The last equation can easily be solved. Its 2D linear differential equation.

The general solution of these equations, in terms of $\phi$ are $\phi = c_1e^{i m \phi}$ and $\phi = c_2 e^{-i m \phi}$.

Now that we have the solution to second equation its time to solve the first equation for $\Theta$. We will do that tomorrow.

# Day 10: Solving 3D Schrödinger Equation (Part 1)

Yesterday I talked about how we can solve the general 3D SE by dividing the task in different parts. And for next few days I will do exactly that. I will solve here all the parts and then finally we will have the solution $\psi\left(r,\theta,\phi\right)$. So let’s get started.

We have 3D SE in spherical coordinate,

$-\frac{\hbar^2}{2m} [\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial \psi}{\partial \theta}\right) + \frac{1}{r^2 sin^2\theta} \frac{\partial^2\psi}{\partial \phi^2}] + V \psi = E \psi$.

Let’s separate the solution, $\psi \left(r, \theta, \phi\right) = R\left(r\right) Y\left(\theta, \phi\right)$

Putting it back in the equation we get,

$-\frac{\hbar^2}{2m} [\frac{Y}{r^2} \frac{d}{dr} \left(r^2 \frac{dR}{dr}\right) + \frac{R}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial Y}{\partial \theta}\right) + \frac{R}{r^2 sin^2\theta} \frac{\partial^2 Y}{\partial \phi^2}] + V RY = E RY$

Let’s now divide the whole equation by $RY$ and multiply by $\frac{-2mr^2}{\hbar^2}$:

$\left \{ \frac{1}{R} \frac{d}{dr} \left(r^2\frac{dR}{dr}\right) \frac{-2mr^2}{\hbar^2} \left[ V\left(r\right) - E \right] \right \} + \frac{1}{Y} \left \{ \frac{i}{sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial Y}{\partial \theta} \right) + \frac{1}{sin^2\theta} \frac{\partial^2 Y}{\partial \phi^2} \right \} = 0$

The first term in the equation is function of only $r$ and the second term is only the function of $\theta$ and $\phi$. The former is called Radial equation and later Angular equation. We can write them separately as follows:

$\frac{1}{R} \frac{d}{dr} \left(r^2\frac{dR}{dr}\right) \frac{-2mr^2}{\hbar^2} \left[ V\left(r\right) - E \right] = l \left(l+1\right)$

Angular equation :

$\frac{1}{Y} \frac{i}{sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial Y}{\partial \theta} \right) + \frac{1}{sin^2\theta} \frac{\partial^2 Y}{\partial \phi^2} = - l \left(l+1\right)$

Next we will solve the angular equation to get two separate solutions $\Theta\left(\theta\right)$ and $\Phi\left(\phi\right)$. And after that we will solve the radial equation to get $R\left(r\right)$.

Last year, I read an answer on Quora by Namit Anand and I found it really helpful. It’s really good if you don’t know how to go about learning quantum mechanics. There is also an awesome video by Mithuna Yoganathan on her channel Looking Glass Universe. It’s hard to explain quantum mechanics to someone who is not familiar with the mathematics behind it. But as both Namit and Mithuna said, you first need to learn the concepts and mathematics by solving problems.

# Day 9: Solving 3D Schrödinger equation

Yesterday, I talked about solving 1D Schrödinger equation. Now, let’s see if we can generalize it to three dimensions.

We separated the SE which is 1D PDE in two ODEs. First we solved the equation having only time and then the other part which depended on position of the particle. The solution to second part depended on the potential you have chosen. Finally you multiply them and get the wave function $\psi\left(x,t\right)$ also written as $\Psi$.

Its just an overview of how the solution works. There is not any single method to solve SE for all different types of potential. They all are solved differently depending on their complexity. But the underlying idea is same.

Now, I want to show you that we can generalize this variable separation method to 3D SE as well.

One dimensional Schrödinger equation is,

$\frac{-\hbar^2}{2m} \frac{\partial^2\Psi}{\partial x^2} + V \Psi = i\hbar \frac{\partial \Psi}{\partial t}$

Its generalization to 3D is,

$\frac{-\hbar^2}{2m} \nabla^2 \Psi + V \Psi = i \hbar \frac{\partial}{\partial t}$

Here, $\nabla^2 \equiv \frac{\partial}{\partial x^2} \frac{\partial}{\partial y^2} \frac{\partial}{\partial y^2}$ is called Laplacian operator.

The equation is written in cartesian coordinate but we can also write it in Spherical polar coordinate system as follows

$-\frac{\hbar^2}{2m} [\frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 sin\theta} \frac{\partial}{\partial \theta} \left( sin\theta \frac{\partial \psi}{\partial \theta}\right) + \frac{1}{r^2 sin^2\theta} \frac{\partial^2\psi}{\partial \phi^2}] + V \psi = E \psi$.

It’s time independent SE and we need to find its solution $\psi\left(r, \theta, \phi \right)$. For 1D our derivation hardly reached one page. But here, we need a lot of space (8-10 pages) to finally break down the separate solutions for all these variables. So, here I am just giving an overview of how its done. Rest can be filled by just grabbing the pen and paper and following the steps of derivation given in the book. Once you see the broad-view its easy to fill in the remaining details. So let’s start.

1. You start with the 3D SE in spherical coordinate. You know that we have to deal with four variables here $\left(r, \theta, \phi, t\right)$ as opposed to only two variables in 1D case $\left(x, t \right)$. We have already seen that the solution to the time dependent part $\phi\left(t\right)$ has the same solution no matter what the potential, so we need not to consider it here. Let’s break the remaining three variables. $\psi \left(r, \theta, \phi, t\right) = R\left(r\right) Y\left(\theta, \phi\right)$. The second part is again separated $Y\left(\theta, \phi\right) = \Theta\left(\theta\right) \phi\left(\phi\right)$.
2. We now just need to put these separate variables in the 3D SE and get the equations with different variables. You can see all these equations below. As it turns out, these are not easy to understand in one go. Because they require some particular manipulations and special functions. You can only understand it by doing the derivation yourself. But you get the idea.
1. For $r$: $R\left(r\right) = A_{jl} \left(k r\right)$,
2. For $\theta$: $\Theta \left(\theta \right) = A P_l^m \left(cos\theta\right)$,
3. For $\phi$: $e^{im\phi}$
4. Using these last two equations, we finally get $Y_l^m \left(\theta, \phi\right) = \epsilon \sqrt{\frac{\left(2l+1\right)}{4\pi} \frac{ \left(l-|m|\right)!}{ \left(l+|m|\right)! }} e^{im\phi} P_l^m\left(cos\theta\right)$.
3. The solution to time independent equation looks like this, $\psi _{nlm} \left(r,\theta,\phi\right) = A_{nl} j_l \left(\beta_{nl} r/a\right) Y_l^m \left(\theta, \phi\right)$

Using this equation, you can now solve the 3D generalization of all 1D problems like infinite and finite spherical potential well. But most importantly, now you can solve the Hydrogen atom, given the Coulomb Potential. I will discuss it in more detail later this month.

# Day 8: Solving 1D Schrödinger’s equation

In this post I will briefly explain along with some extra details the Section 2.1 of Griffith’s Quantum Mechanics.

Let’s start by writing the Schrödinger’s equation,

$\frac{- \hbar^2}{2m} \frac{\partial ^2 \psi\left(x,t\right)}{\partial x^2} + V \psi \left(x,t\right) = i \hbar \frac{\partial}{\partial t} \psi\left(x,t\right)$

Here $\psi \left(x,t\right)$ is a function of $x$ and $t$. Its the true wave function because usually students confuse between $\psi \left(x,t\right)$ and $\psi \left(x\right)$. The true wave function can be separated in two parts,

$\psi \left(x,t\right) = \psi \left(x\right) \phi \left(t\right)$

$\frac{\partial \psi\left(x,t\right)}{\partial t} = \psi \left(x\right) \frac{d\phi}{dt}$

$\frac{\partial^2 \psi\left(x,t\right)}{\partial x^2} = \frac{d^2\psi \left(x\right)}{dx^2} \phi \left(t\right)$

Substituting this in the equation, we get

$\frac{- \hbar^2}{2m} \frac{d ^2 \psi\left(x\right)}{d x^2} \phi \left(t\right)+ V \psi \left(x\right) \phi \left(t\right) = i \hbar \frac{d \phi\left(t\right)}{dt} \psi\left(x\right)$

Dividing by $\psi\left(x\right) \phi\left(t\right)$

$\frac{- \hbar^2}{2m} \frac{1}{\psi\left(x\right)} \frac{d ^2 \psi\left(x\right)}{d x^2} + V = i \hbar \frac{1}{\phi\left(t\right)} \frac{d \phi\left(t\right)}{dt}$

Now, you can see the left side is function of $x$ and right side is function of $t$ only. How can both sides be equal? It’s like saying $x = t$, both are different variables, one representing position and other the time of a particle’s motion. There is only one way they can be equal, if both are equal to a constant, lets say $x = t =5$.

So, we equate both sides of the equation to a constant $E$

$i \hbar \frac{1}{\phi\left(t\right)} \frac{d\phi\left(t\right)}{dt} = E$ and $\frac{- \hbar^2}{2m} \frac{1}{\psi\left(x\right)} \frac{d ^2 \psi\left(x\right)}{d x^2} + V = E$

This second equation has now become time independent and famously known as Schrödinger’s time independent equation. The first equation I wrote in the beginning is time dependent because we have used the wave function that depends both on $x$ and $t$. That’s why we had to use partial derivative there. Let’s now multiply both sides with $\psi\left(x\right)$ to make the expression look more clean.

$\frac{- \hbar ^2}{2m} \frac{d^2 \psi \left(x\right)}{dx^2} + V \psi\left(x\right)= E \psi\left(x\right)$

We started with a partial differential equation and now we have two ordinary differential equations. But it’s only for one dimensional motion.

Now, we can solve both ODEs separately to get $\psi\left(x\right)$ and $\phi\left(t\right)$ then after multiplying them we can get the true wave function which is the solution of Schrödinger’s equation.

Let’s solve for $\phi\left(t\right)$ first. Here is the equation I wrote before,

$i \hbar \frac{1}{\phi\left(t\right)} \frac{d\phi\left(t\right)}{dt} = E$

Rearranging the equation we get,

$\frac{d\phi\left(t\right)}{dt} = \frac{1}{i \hbar} E\phi\left(t\right)$

$\frac{d\phi\left(t\right)}{\phi\left(t\right)} = \frac{-i E}{ \hbar} \int dt$

$log\phi\left(t\right) = \frac{-i E}{ \hbar } t$

$\phi\left(t\right) = e^{\frac{-iEt}{\hbar}}$

This is the solution that we were looking for. Now we need $\psi\left(x\right)$. So, lets solve another equation.

$- \frac{\hbar^2}{2m} \frac{d^2\psi\left(x\right)}{dx^2} + V\psi\left(x\right) = E$

We can only solve this equation if we have the potential V given and there are a lot of potential functions in quantum mechanics that we have to deal with. Every time you need solution to the Schrödinger’s equation, you just plugin the required potential in the equation and get $\psi\left(x\right)$.

There is potential for Infinite square well, finite square well, Harmonic Oscillator etc.

So solving Schrödinger’s equation reduces to solving only the time independent equation. Then you just need to multiply it with $\phi\left(t\right)$ to get the true wave function.

Introduction to Quantum Mechanics – David J. Griffiths

# Dust of Butterfly Wings

If you ever accidentally touched the wings of a butterfly, you must have seen fine dust on your fingers. These are the scales of butterflies of Lepidoptera order. The fine dust when seen under microscope reveals that its not just ordinary dust.

Instrument used: Gemko Labwell Microscope

Magnification: 100x

Camera: 8MP

In a few days I will post a fine picture of soil particles and the photos of other tiny creatures wandering around in dirty water full of life.

# Day 7: An Impossible Electrostatic Field

One of these is an impossible electrostatic field. Which one?

$\left(a\right) \vec E = k[x y \hat{x} + 2yz \hat{y} + 3xz \hat{z}];$

$\left(b\right) \vec E = k[y^2 \hat{x} + \left(2xy + z^2 \right) \hat{y} + 2yz \hat{z}].$

Electric field $\vec E$ is a vector function whose curl is always zero. It has positive and negative divergences. So we just need to check whether the curl of one of these vectors is zero.

Problem $2.20$: Introduction to Electrodynamics