A note on the Hydrogen spectrum

February 9, 2019December 20, 2020 / 153armstrong / Leave a comment

The emission spectrum of atomic hydrogen is given by this amazing spectral series diagram:

Let’s take a closer look at only the visible portion of the spectrum i.e the Balmer series.

If a hydrogen lamp and a diffraction grating just happen to be with you, you can take a look at the hydrogen lamp through the diffraction grating, these lines are what you would see:

Source

These are known emission lines and they occur when the hydrogen atoms in the lamp return to a state of lower energy from an excited energy state.

Representation of emission and absorption using the Bohr’s model

Here’s another scenario that could also happen:

You have a bright source of light with a continuous spectrum and in between the source and the screen, you introduce a gas (here, sodium)

Source: Harvard Natural sciences

The gas absorbs light at particular frequencies and therefore we get dark lines in the spectrum.

This is known as absorption line. The following diagram summarizes what was told thus-far in a single image:

The absorption and emission spectrum for hydrogen look like so :

Stars and Hydrogen

One of the comments from the previous post was to show raw spectrum data of what was being presented to get a better visual aid.

Therefore,the following spectrum is a spectrum of a star taken from the Sloan Digital Sky Survey (SDSS)

Plot of wavelength vs median-flux

Here’s the spectrum with all the absorption lines labelled:

Source: SDSS

You can clearly see the Balmer series of hydrogen beautifully encoded in this spectrum that was taken from a star that is light-years away.

And astronomers learn to grow and love these lines and identify them immediately in any spectrum, for they give you crucial information about the nature of the star, its age, its composition and so much more.

Source: xkcd

Have a great day!

*If you squint your eyes a bit more you can find other absorption lines from other atoms embedded in the spectrum as well.

A note on Wave-functions and Fourier Transforms

December 3, 2018December 3, 2018 / 153armstrong / Leave a comment

In quantum mechanics you can denote the wave-function in the position or the momentum basis. Written in the momentum basis, it would look something like:

$|\psi(x)> = a_0 |p_0> + a_1 |p_1> + \hdots$

$|\psi(x)> = \sum\limits_{n} a_n |p_n>$

But momentum is a continuous variable and it varies from $-\infty$ to $\infty$ .

unnamed

Therefore changing to the integral representation we get that:

$|\psi(x)> = \int\limits_{-\infty}^{\infty} dp \ a_n(p) |p>$

But $a_n(p)$ is just the projection of the momentum vector on the wavefunction:

$|\psi(x)> = \int\limits_{-\infty}^{\infty} dp <p|\psi(x)>|p>$

We are also aware from our knowledge of Fourier Transform* that the wave function written in momentum space is given as :

$|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

Comparing both the above equations if we take the momentum basis as $|p> = e^{\frac{ipx}{\hbar}}$ , then:

$<p|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \tilde{\psi}(p)$

We can perform a similar analysis by expanding the wavefunction about the position basis and get

$<x| \tilde{\psi}(p)> = \frac{1}{\sqrt{2 \pi \hbar}} \psi(x)$

** Where does the $\frac{1}{\sqrt{2 \pi \hbar}}$ in the Fourier Transform come from ?

We know from Fourier transform is defined as follows:

$|\psi(x)> = \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{\infty} dk \ \psi_{1}(k) |e^{ikx}>$

Plugging in $p = \hbar k$ and rewriting the above equation we get,

$|\psi(x)> = A \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

We find that from

$\int\limits_{-\infty}^{\infty} dx <\psi(x)| \psi(x)> = 1$

that the normalization constant is not $\frac{1}{\sqrt{2 \pi}}$ but $\frac{1}{\sqrt{2 \pi \hbar}}$ . Therefore,

$|\psi(x)> = \frac{1}{\sqrt{2 \pi \hbar}} \int\limits_{-\infty}^{\infty} dp \ \tilde{\psi}(p) |e^{\frac{ipx}{\hbar}}>$

Cooking up a Lorentz invariant Lagrangian

November 23, 2018December 30, 2020 / 153armstrong / 1 Comment

Let’s consider a scalar field, say temperature of a rod varying with time i.e $T(x,t)$ . (something like the following)

w1hldkr

We will take this setup and put it on a really fast train moving at a constant velocity $v$ (also known as performing a ‘Lorentz boost’).

train-animated-gif-3

Now the temperature of the bar in this new frame of reference is given by $T^{'}(x^{'}, t^{'})$ where,

$x^{'}(x,t) = \gamma \left( x - vt \right)$

$t^{'}(x,t) = \gamma \left( t - \frac{v x}{c^{2}} \right)$

Visualizing the temperature distribution of the rod under a Length contraction.

Temperature is a scalar field and therefore irrespective of which frame of reference you are on, the temperature at each point on the rod will remain the same on both the frames i.e

$T^{'}(x^{'}, t^{'})= T(x, t)$

Therefore we can say that Temperature (a scalar field) is Lorentz invariant. Now what other quantities can we make from T that would also be Lorentz invariant ?

Is $\nabla . T^{'}(x^{'}, t^{'})= \nabla . T(x, t) ?$

Well, let’s give it a try:

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial x} + \frac{\partial T}{\partial t^{'}} \frac{\partial t^{'}}{\partial x}$

$\frac{\partial T}{\partial x} = \frac{\partial T^{'}}{\partial x^{'}} \gamma - \frac{\partial T}{\partial t^{'}} \gamma v$

$\frac{\partial T}{\partial x} = \gamma \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T}{\partial t^{'}} v \right)$

——–

$\frac{\partial T}{\partial t} = \frac{\partial T^{'}}{\partial x^{'}} \frac{\partial x^{'}}{\partial t} + \frac{\partial T^{'}}{\partial t^{'}} \frac{\partial t^{'}}{\partial t}$

$\frac{\partial T}{\partial t} = - \frac{\partial T^{'}}{\partial x^{'}} \gamma v + \frac{\partial T^{'}}{\partial t^{'}} \gamma$

$\frac{\partial T}{\partial t} = \gamma \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right)$

Clearly, **

$\frac{\partial T}{\partial x} + \frac{\partial T}{\partial t} \neq \frac{\partial T^{'}}{\partial x^{'}} + \frac{\partial T^{'}}{\partial t^{'}}$

But just for fun let’s just square the terms and see if we can churn something out of that:

$\left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2}$

$\left( \frac{\partial T}{\partial t} \right)^{2} = \gamma^{2} \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2}$

We immediately notice that:

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \gamma^{2} \left[ \left( - \frac{\partial T^{'}}{\partial x^{'}} v + \frac{\partial T^{'}}{\partial t^{'}} \right) ^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} - \frac{\partial T^{'}}{\partial t^{'}} v \right)^{2} \right]$

$\left( \frac{\partial T}{\partial t} \right)^{2} - \left( \frac{\partial T}{\partial x} \right)^{2} = \left( \frac{\partial T^{'}}{\partial t^{'}} \right)^{2} - \left( \frac{\partial T^{'}}{\partial x^{'}} \right)^{2}$

Therefore in addition to realizing that $T$ is Lorentz invariant, we have also found another quantity that is also Lorentz invariant. This quantity is also written as $\partial_{\mu} T \partial^{\mu} T$ .

** There is a very important reason why this quantity did not work out. This post was inspired in part by Micheal Brown’s answer on stackexchange . I request the interested reader to check that post for a detailed explanation.

Ansatz to Gram-Schmidt Orthonormalization

November 18, 2018November 18, 2018 / 153armstrong / Leave a comment

The Gram–Schmidt process is a method for orthonormalising a set of vectors in an inner product space and the trivial way to remember this is through an ansatz :

Let $|v_{1}> , |v_{2}> , \hdots |v_{n}>$ be a set of normalized basis vectors but we would also like to make them orthogonal. We will call $|v_{1}^{'}> , |v_{2}^{'}> , \hdots |v_{n}^{'}>$ be the orthonormalized set of basis vectors formed out $|v_{1}> , |v_{2}> , \hdots |v_{n}>$ .

Let’s start with the first vector:

$|v_{1}^{'} > = |v_{1}>$

Now we construct a second vector $|v_{2}^{'}>$ out of $|v_{1}^{'}>$ and $|v_{2}>$ :

$|v_{2}^{'} > = |v_{2}> - \lambda |v_{1}^{'}>$

But what must be true of $|v_{2}^{'} >$ is that $|v_{1}^{'}>$ and $|v_{2}^{'}>$ must be orthogonal i.e $<v_{1}^{'}|v_{2}^{'}> = 0$ .

$<v_{1}^{'}|v_{2}^{'} > = <v_{1}^{'}|v_{2}> - \lambda <v_{1}^{'}|v_{1}^{'}>$

$0 = <v_{1}^{'}|v_{2}> - \lambda$

$\lambda = <v_{1}^{'} | v_{2}>$

Therefore we get the following expression for $v_{2}^{'}$ ,

$|v_{2}^{'} > = |v_{2}> - <v_1^{'} | v_{2} >|v_{1}>$

which upon normalization looks like so:

$|v_{2}^{'} > = \frac{|v_{2}^{'} >}{<v_{2}^{'} |v_{2}^{'} > }$

That might have seemed trivial geometrically, but this process can be generalized for any complete n-dimensional vector space. Let’s continue the Gram – Schmidt for the third vector by choosing $|v_{3}^{'} >$ of the following form and generalizing this process:

$|v_{3}^{'} > = |v_{3}> - \lambda_{1} |v_{1}^{'}> - \lambda_{2} |v_{2}^{'}>$

The values for $\lambda_{1}$ and $\lambda_{1}$ are found out to be as:

$\lambda_{1} = <v_{1}^{'}|v_{3}>$

$\lambda_{2} = <v_{2}^{'}|v_{3}>$

Therefore we get,

$|v_{3}^{'} > = |v_{3}> - <v_{1}^{'}|v_{3}>|v_{1}^{'}> - <v_{2}^{'}|v_{3}>|v_{2}^{'}>$ (or)

$|v_{3}^{'} > = |v_{3}> - \sum\limits_{j=1,2} <v_{j}^{'} | v_{3}> |v_{j}^{'}>$

$|v_{3}^{'} > = \frac{|v_{2}^{'} >}{<v_{3}^{'} |v_{3}^{'} > }$

Generalizing, we obtain:

$|v_{i}^{'} > = |v_{i}> - \sum\limits_{j=1,2,...,i-1} <v_{j}^{'} | v_{i}> |v_{j}^{'}>$

$|v_{i}^{'} > = \frac{|v_{i}^{'} >}{<v_{i}^{'} |v_{i}^{'} > }$

Now although you would never need to remember the above expression because you can derive it off the bat with the above procedure, it is essential to understand how it came out to be.

Cheers!

Example (to be added soon):

Solving the Laplacian in Spherical Coordinates (#1)

May 5, 2017June 6, 2017 / 153armstrong / Leave a comment

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

sph_coor

$x = rsin\theta cos\phi$
$y = rsin\theta cos\phi$
$z = rcos\theta$

where

$0 \leq r < \infty$
$0 \leq \theta \leq \pi$
$0 \leq \phi < 2\pi$

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

$\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0$

To solve it we use the method of separation of variables.

$f = R(r)\Theta(\theta)\Phi(\phi)$

Plugging in the value of $f$ into the Laplacian, we get that :

$\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0$

Dividing throughout by $R\Theta\Phi$ and multiplying throughout by $r^2$ , further simplifies into:

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0$

It can be observed that the first expression in the differential equation is merely a function of $r$ and the remaining a function of $\theta$ and $\phi$ only. Therefore, we equate the first expression to be $\lambda = l(l+1)$ and the second to be $-\lambda = -l(l+1)$ . The reason for choosing the peculiar value of $l(l+1)$ is explained in another post.

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0$ (1)

The first expression in (1) the Euler-Cauchy equation in $r$ .

$\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R$

The general solution of this has been in discussed in a previous post and it can be written as:

$R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}$

The second expression in (1) takes the form as follows:

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

The following observation can be made similar to the previous analysis

$\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0$ (2)

The first expression in the above equation (2) is the Associated Legendre Differential equation.

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

$sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0$

The general solution to this differential equation can be given as:
$\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)$

The solution to the second term in the equation (2) is a trivial one:

$\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi$
$\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}$

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

$R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)$

Matrix Multiplication and Heisenberg Uncertainty Principle

March 5, 2017March 5, 2017 / 153armstrong / Leave a comment

We now understand that Matrix multiplication is not commutative (Why?). What has this have to do anything with Quantum Mechanics ?

Behold the commutator operator:
$[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

where $\hat{A},\hat{B}$ are operators that are acting on the wavefunction $\psi$ . This is equal to 0 if they commute and something else if they don’t.

One of the most important formulations in Quantum mechanics is the Heisenberg’s Uncertainty principle and it can be written as the commutation of the momentum operator (p) and the position operator (x):

$[\hat{p}, \hat{x}] = \hat{p}\hat{x} - \hat{x}\hat{p} = i\hbar$

If you think of p and x as some Linear transformations. (just for the sake of simplicity).

This means that measuring distance and then momentum is not the same thing as measuring momentum and then distance. Those two operators do not commute! You can sort of visualize them in the same way as in the post.

But in Quantum Mechanics, the matrices that are associated with $\hat{p}$ and $\hat{x}$ are infinite dimensional. ( The harmonic oscillator being the simple example to this )

$\hat{x} = \sqrt{\frac{\hbar}{2m \omega}} \begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \hdots \\ \sqrt{1} & 0 &\sqrt{2} & 0 & \hdots \\ 0 & \sqrt{2} & 0 &\sqrt{3} & \hdots \\ 0 & 0 & \sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$

$\hat{p} = \sqrt{\frac{\hbar m \omega}{2}} \begin{bmatrix} 0 & -i & 0 & 0 & \hdots \\ i & 0 & -i \sqrt{2} & 0 & \hdots \\ 0 & i\sqrt{2} & 0 &\-i \sqrt{3} & \hdots \\ 0 & 0 & i\sqrt{3} & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}$

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