The Schrodinger Equation and The Form of the Solution

One might wonder Schrodinger's equation is in the form it is in. It is called a wave equation but we are not using the wave equation. Well, I can't claim to understand what I am about to expostulate( in other words give an intuitive understanding) but rather take the cope out way and argue with a combination of mathematical and experimental reasoning.

Our story begins with an electron going through the double slit and hitting a screen somewhere behind it. The surprising result was that an interference pattern was observed as a series of these electrons passed through one at a time. The obvious explanation is that the electron has a wave property so we quickly write down our trusty sine function( or cosine if you are inclined that way). Now the next step is to get the interference pattern through our results. So we do the naive thing and write down two sine functions differing by a phase factor and add them together . One must recall that we want an interference pattern that has a maximum at the center and decreases "sinusoidally" in amplitude as we go away from it from either side.  Unfortunately when we go with our naive guess and square our wave function(addition of the sines) since E.M tells us that the intensity is proportional to wave function squared our interference is sinusoidal but does not decrease as we go away from the center. thus the interference pattern is still illusive.

In fact we realize something worse. Supposing two waves are approaching each other and let's stick to one dimension(we are modelling two electrons approaching each other). We expect a combination of destructive and constructive interference . So we re-write the sine functions in this manner  i.e \( \sin (kx - \omega t) + \sin (kx + \omega t) \); the result is this \(2 \sin kx \cos \omega t \). But this is a terrifying result, at multiples of t = \( \frac{ \pi}{2 \omega} \) the particles disappear everywhere. This sort of tom-foolery is allowed for the Cheshire cat but nor for our dear electrons. So we want waves but adding sines or cosines together does not work!

Finally we remember something about Euler's identity and it being related to waves. So we decide to represent our electrons by this wave function \( e^{(ikx - i\omega t)} \).  We add one coming from left \( e^{(ikx - i\omega t)} \) and one coming from the right  \( e^{(-ikx - i\omega t)} \) and get \( 2 e^{(-i\omega t) } \cos \omega t\). Ah! the function is still sinusoidal and never disappears , but with a weird complex amplitude. Remember our goal is to describe the interference patterns on the screen, but these patterns have real amplitudes. Before we move on we take note that we either take the wave function \( e^{(ikx - i\omega t)} \) or  \( e^{(-ikx + i\omega t)} \) but not both. Reason is if we take both and add them then we are back to cosine and sine functions which have all the wrong properties.  So we shall take the former and assume the amplitude is A.

Back to the puzzling imaginary amplitude. This is no cause for concern because remember the intensity is the square of the wavefunction, which means taking the modulus of our wave function.  First let's add two wave functions that will describe our electron (since we are looking for an interference pattern):\(A_1 e^{(ikx_1 - i\omega t)}+A_2e^{(ikx_2 - i\omega t)} \)
Now let's look at the modulus

\begin{align}
&\left(A_1 e^{(ikx_1 - i\omega t)}+A_2e^{(ikx_2 - i\omega t)}\right) \left(A_1 e^{(-ikx_1 + i\omega t)}+A_2e^{(-ikx_2 + i\omega t)}\right)\\
&\left(A_1 e^{ikx_1}+A_2e^{ikx_2 }\right) \left(A_1e^{-ikx_1}+A_2e^{-ikx_2 }\right)\\
&\text{a bit of algebra....}\\
&A_1^2 + A_2^2 + 2A_1A_2 \cos k(x_2 - x_1)
\end{align}

The formula above is quite astonishing, if we close slit 2 in our experiment and assume our wave function is the right one while holding firm to the stipulation that its modulus is the thing that matters(avoiding complex amplitudes) then we expect \(A_1^2\) namely just a mound of electrons in front of slit 1 and similarly \(A_2^2\) if we close slit 1 and leave slit 2 open. In both of these situations the cosine term is not there. Once we open both slits the interference term appears and we get our interference pattern we were looking for.

Notice the classical guess from E.M gives a cosine term squared for the intensity, which of course does not explain the places on the screen where there are no electrons hitting.

Posts on Quantum Mechanics

I recently re-looked at my Quantum Mechanics text-book by Ohanian, and I was rather impressed by his style of presentation of the material. There will be a new tab with my examination and notes on the textbook. It offers a nice complementary look to Griffiths introduction. His method avoids solving partial differential equations and instead opts for the method of operators throughout. The first post will be examining the wave picture and the matrix picture and showing their equivalence. 

The Summa : Euler's Rigid Body Equations

We finally arrive at the top, namely the Euler equations for rigid bodies. I mentioned the inertia tensor in previous section, but I have decided to eschew from it since there are a multitude of good resources on it. What matters to us at this point is that it is a matrix and not only that but a symmetric matrix. Given that this is the case, we can always diagonalize it so that it simply has entries on the diagonal. These entries we shall call \( I_x, I_y, I_z \). These are the so called principal axes. Now if we jump into a co-ordinate system which coincide with these axes then the angular momentum L = \( \left( I_x\omega_x, I_y \omega_y , I_z \omega_z \right) \). The angular moment as observed in the inertial reference is $$ \dot{L_{i.f}} = \dot{L_{r.f}}+ \Omega \times L = \Gamma $$ where i.f is "inertial reference frame" , r.f is "rotating reference frame" and gamma is the external torque. If we plug in what L is we arrive at the blessed equations
\begin{align}
\Gamma_x & =  \dot{L_x} - (I_y - I_z) \omega_z \omega_y\\
\Gamma_y &=   \dot{L_y}- (I_z - I_x) \omega_y \omega_z\\
\Gamma_z &=  \dot{L_z}-(I_x - I_y) \omega_x \omega_ y \\
\end{align}

Notice that the time derivate does not affect I, that is the purpose of using these principal axes. Now supposing external torque is zero, then the angular moment L should be constant. Here is the proof:
1) If external torque is zero, angular momentum is constant

Multiply with ith equation above by  \( I_i\omega_i\) and add all three equations
\begin{align}
I_x^2 \omega_x \dot{\omega_x} + I_y^2  \omega_y \dot{\omega_y}+ I_z^2  \omega_z \dot{\omega_z} &=  \left(I_x(I_y - I_z)  + (I_z - I_x)I_y + (I_x - I_y)I_z \right) \omega_z \omega_y\omega_x \\
I_x^2 \omega_x \dot{\omega_x} + I_y^2  \omega_y \dot{\omega_y}+ I_z^2  \omega_z \dot{\omega_z} &= 0\\
L_x\dot{L_x}+ L_y\dot{L_y}+L_z\dot{L_z} &= 0\\
 \frac{d}{dt}\frac{\left( L \cdot L\right)}{2} &= 0\\
\frac{d}{dt}\frac{L^2}{2} &= 0\\
\dot{L} &= 0 \hspace{20 mm} QED
\end{align}

2) If we have a lamina rotating freely(no torques) about a point O of the lamina. The component of \( \omega \) in the plane is constant.

Again since we have no torque we start from Euler's equations with no torque, but we use a slight of hand which should not look like one since we have seen it before, namely use the perpendicular axis theorem. This gets rid of the \( I_x \) and \(I_y\)  but we are getting ahead of ourselves:
\begin{align}
I_x \dot{\omega_x} &= (I_y - I_z) \omega_y \omega_z\\
I_y \dot{\omega_y} &= (I_z - I_x) \omega_z \omega_x\\
I_z \dot{\omega_z} &= (I_x - I_y) \omega_y \omega_x\\
\end{align}
Use perpendicular axis theorem

\begin{align}
\frac{I_z}{2} \dot{\omega_x} &= (\frac{I_z}{2} - I_z) \omega_y \omega_z\\
\frac{I_z}{2} \dot{\omega_y} &= (I_z - \frac{I_z}{2}) \omega_z \omega_x\\
I_z \dot{\omega_z}& = (\frac{I_z - I_z}{2}) \omega_y \omega_x\\
\end{align}
The new set of equations
\begin{align}
\dot{\omega_x}& = -  \omega_y \omega_z\\
 \dot{\omega_y} &= \omega_z \omega_x\\
\end{align}
Multiply the top equation above by \(\omega_x\) and the bottom by \( \omega_y\) and then add the two to get \(\omega_x \dot{\omega_x}+ \omega_y\dot{\omega_y} = 0 \). But this is merely \(\frac{d}{dt}\left( \omega_x^2 + \omega_y^2\right) \)= 0. Voila

A few solved problems

Since this blog is meant to concentrate on these not emphasized in textbooks or offer a slightly different way of looking at old things, we now come to place where I think a great deal of material has been written on the topic and most of it quite good. Now, we derive Euler's rigid body equations in all their glory, I found a few interesting problems, which relate to rotational motion.

1) You are to imagine a rod pivoted to a horizontal beam but free to swing in a plane perpendicular to the plane containing the beam. The rod has zero width, and a density not necessarily constant. Its moment of inertia about an axis parallel to the beam is I and the center o f mass is located distance b from the pivot.  Find the equation of motion for small swings, its period and the length of a simple pendulum that would have the same period.

This is called a compound pendulum( contrast it with the simple pendulum where all the mass is concentrated at a point). Let \(\Gamma\) be the torque and L is the angular momentum.
\begin{align}
 \dot{L}&= \Gamma \\
  I \dot{\omega}& = - bmg\sin \theta \\
  \ddot{\theta} &= -bg \theta m \\
 \ddot{\theta} &= \frac{-bg\theta m}{I}\\
\end{align}

It should b:e obvious that the \(\omega\) or angular frequency is\( \sqrt{\frac{mbg}{I}}\) with period \(2 \pi\sqrt{\frac{I}{mbg}}\). Now the angular frequency of a simple pendulum is \( \sqrt{\frac{g}{l}}\). Setting the expressions for the different angular frequencies equal gives that the length of the simple pendulum must be \(\frac{I}{mb}\).

2) Imagine the same rod as above but this time it is hit with a horizontal force a distance c from the pivot with impulse F\(\Delta \)T = \(\xi\). What is the angular momentum delivered to the pendulum, pivot just after it has been hit? What is special distance \(b_o\) such that the impulse delivered to the pivot is zero?

Mathematically, I do not think this is a challenging problem. It is the conceptual physics involved that makes it quite challenging. So here is my solution:

We can treat the pendulum as though all its mass is centered at the center of mass. Now since it is not moving before we hit it, the initial velocity is zero. But this implies that the impulse is equal to the moment just after we hit it.(Just consider newton's second law for discrete times). Thus the angular momentum after it has been hit is simply \(b \xi\). Now the impulse delivered to the pivot will be m\(v_p\) where \(v_p\) is the velocity at the pivot. The velocity at the pivot is the velocity of the center of mass \( v_{cm} - v_{h}\) where \(v_h\) is the tangential velocity at the point  it is hit. Therefore  impulse is : \( mv_{cm} -m c\omega\) = \( \xi - \frac{mbc\xi}{I}\). Remember L = \(I\omega\),  c is the distance from the pivot to where it is hit and b is the distance from pivot to center of mass. If we want to find that special distance \(b_o\)  where impulse at the pivot is zero just set the expression for the impulse equal to zero and you will get that \( b_o = \frac{I}{mb}\)

Moments of Inertia and conclusion (Part II)

We continue our journey with moments of inertia by calculating the I for a solid cone about two different axes i.e through the axis that goes through the center of mass and the axis that goes perpendicular to its pointy edge.

5) Axis through the center of mass
\begin{align}
I_z &= \rho \left( \int r^2\, dV \right)\\
  &=  \rho \left( \int r^3 \, d\phi dr dz \right)\\
  &=  \rho  2\pi \left(\int_0^{\frac{zR}{L}} r^3 \, dr \,\int_0^ L\, dz\right)\\
 &= \frac{\pi \rho R^4 L }{10} = \frac{3MR^2}{10}
\end{align}

Axis that is perpendicular to the point edge.
Now this interesting, and it worth slowing down the pace and considering. This problem can be done two ways: One is to realize that if we put in the cone upside(not necessary) then this axis is \(I_{yy}\) in the moment of inertia tensor; now calculate away the integrals. But supposing you know nothing about the inertial tensor, is there hope? Indeed, but it requires a little foray into the perpendicular axis theorem.

Let's suppose you have a two dimensional figure(located in the xy place). Then the perpendicular axis theorem says \(I_z\) = \(I_x + I_y\). The proof is quite simple and goes as follows:
\begin{align}
 I_z &= \left(\int r^2 dm \right) = \left( \int x^2 + y^2 dm\right)\\
       &= \left(\int x^2 dm\right) + \left( \int y^2 dm\right) = I_y + I_x \hspace{20mm}QED\\
\end{align}
Its most common application is for example finding \(I_x\) or \(I_y\) of a disk. Since we know \(I_z\)  because it is easy to calculate, then finding  \(I_x\) or  \(I_y\)  is trivial. Again you do not need integrals, the theorem saves you.
 This theorem  suggest another way of calculating our desired results. Here is the strategy:
1) There is symmetry in the x- y plane so \(I_x\) and \(I_y\) are the same and we calculated \(I_z\) in the problem above(axis through center of mass)
2) Starting from the perpendicular axis theorem we can add \(I_z\) to both sides to get that \(2I = I_x+I_y+I_z\). Since we know \(I_x = I_y\) then \( 2I_y + I_z = 2I\). Now all we need to do is calculate 2I. This is a generalization of the perpendicular axis theorem.

\begin{align}
 2I &= 2\rho \left( \int (r^2 \cos^2 \phi + r^2 \sin^2\phi + z^2 ) r dr d\phi dz \right)\\
      & = 4\pi \rho \left(\int_0^L dz \int_0^{\frac{zR}{L}}dr(r^3 + z^2r) \right)\\
      & = 4\pi \rho \left( \frac{R^4L}{20}+ \frac{L^3R^2}{10}\right)\\
    & = 12M\left( \frac{R^2}{20} + \frac{L^2}{10} \right)\\
& \text{plugging into previous results for } I_z \text{ we get } \frac{3M}{5}\left(  \frac{R^2}{4} + L^2\right)
\end{align}

 This theorem is very useful if you are in three dimensions and have symmetry in two or more axes. For example take the moment of inertia for a spherical shell, \( I_z, I_y \) and \(I_x\) are all the same so we have that \( 3I_z = 2 I\)  and I is easy to calculate. "I"here is the moment of inertia about an axis in the \( 4^{th}\) dimension!!! Poooffff ( Your mind being blown away)

Finding center of mass and moments of Inertia

The ultimate goal is still to arrive at the Euler equations for rigid body dynamics. I often find that equations are neatly derived in books and in class but they often prove to be almost useless to the student because students are not comfortable in the actual calculation of the quantities in the equations. We so far have introduced rotational motion and gone through it in some detail. Thus the idea of jumping into the rotational reference frame in order to use these equations should not be foreign. The next step is to learn how to find center of mass, and moments of inertia of various shapes. The goal is to try and make the derivations as straight forward as possible, which means using brute force methods. We shall leave the trickery and elegance for pretentious physics professors and authors. I am a simple minded man and I hope you are too.

Our first goal is to be comfortable finding the center of mass, then we shall move to finding the moment of inertia. Along this journey we shall meet interesting concepts like the parallel axis theorem and the less known perpendicular axis theorem.
First let's begin simply, we shall find the center of mass for a 1D line, cylinder, sphere and a cone. The equation is easily recalled namely: \(\int_{V}\vec{r}dm\). The important thing to remember is to take \(\vec{r}\) seriously. This is the mistake I have often made in the past, because \(\vec{r}\) is not in cartesian co-ordinates, before we take the integral this will have to be changed to Cartesian co-ordinates. This point becomes very important when we go to 3D dimensions.
1) 1D line.
  \begin{align}
   &\frac{\rho}{M}\left( \int_0^L x dx \right) \\
 &  \rho \frac{x^2}{2}|_{0}^L = \rho\left(\frac{L^2}{2}\right) = L/2\\
&\rho \text{ here is }\frac{M}{L}
 \end{align}

2) Cylinder
  \begin{align}
 &\frac{\rho}{M}\left( \int x \, dV \, \hat{x} + \int y\, dV\, \hat{y}+ \int z \, dV \, \hat{z} \right)\\
&\frac{\rho}{M}\left(\int r\,\cos\phi\, r d\phi dr dz\, \hat{x} + \int r\,\sin\phi\, rd\phi dr dz\,\hat{y}+ \int_0^L \int_0^R\, z\,r\,dr d\phi dz\, \hat{z} \right)
\end{align}
The first two integrals  will  go to zero since the integral of \(\sin \phi\) and \(\cos \phi \) over 2\(\pi\) radians is zero. So all we need  is to concentrate on the last integral.
\begin{align}
&\frac{\rho}{M}\left( \frac{R^2}{2} 2\left(\pi\right) \frac{L^2}{2}\right)\\
&\text{put in the expression for the density of a cylinder  to get} \frac{L}{2}
\end{align}
With the cylindrical example it is now clear what people mean when they say,"by symmetry, the center of mass is on this or that axis". It is important that one knows what occurs when options are eliminated by reasons of symmetry. This is why I am doing all the integrals, one has to justify one's intuitions however right they may be.
3) Spherical Shell
\begin{align}
&\frac{\rho}{M}\left( \int R\,\cos \phi \,R^2\, (\ sin\theta)^2\, d\phi d\theta\, \hat{x} + \int \, R\, \sin \phi \, (\sin\theta)^2 \,  R^2 d\theta d\phi \hat{y} + \int  R \cos \, R^2 \, \sin \theta\, d\theta d\phi\, \hat{z}\right)
\end{align}
I have already plugged in what x, y and z are in spherical co-ordinates to get the above expression
Again, the first two integrals go to zero and doing the third integral also gives you zero, i.e the center of mass is at the origin.
4)Solid Cone
\begin{align}
\frac{\rho}{M}\left( \int \, x \, r d\phi dr dz \, \hat{x}+ \int y \, r d\phi dr dz \, \hat{y}+ \int z\, r\, drd\phi dz \, \hat{z}\right)
\end{align}
Again the first two integrals disappear (could have argued this from the beginning by using symmetry arguments).  Now r depends on z the integrals over r and z can't be done completely independently, we will need to write r in terms of z. This will be done by considering the limits. If the cone is balance on its pointed edge then r and z are related in the following manner \(r = \frac{zR}{L}\) . Thus the last integral is the following $$ \frac{\rho}{M}2\pi \left(\int_0^L z\, dz \int_0 ^{\frac{zR}{L}}r\, dr \right)$$ The integral gives \( \frac{\rho \pi R^2  L^2}{4M}\) and plugging for the density gives \(\frac{3L}{4}\)

All other shapes usually found in  text books can be attacked from these examples by playing around with the limits.

Moments of Inertia

Again it is quite easy to remember the general equation which is \(\int r^2 \, dm \) but as with most thing s the devil is in the details. The above form is slightly misleading if you have had vector calculus. One is led to be believe that r is from the point dm to the origin but this is wrong. The r in the equation is from the dm to axis of rotation. Do not get the impression that calculation for moment of inertia is like the calculation for center of mass except the r is squared that could not be further from the truth. This point will be illustrated when we calculate the moment of inertia for a solid sphere.

Secondly, it matters what our axis of rotation is, you have to know that before you begin writing down integrals. This point will be illustrated by deriving the parallel axis theorem.

Let \(r_i\) be the distance from some origin to a \(m_i\) and \(r_i^{'}\) be the distance from the center of mass to that same mass and lastly let \(d\) be the distance from the center of mass to the fore mentioned origin. Then the moment of inertia about this origin is clearly, \(\sum r_i^2 m_i\) which is equal to  \(\sum (r_i^{'}+d)^2 m_i\) since r = r' + d.
\begin{align}
     \sum r_i^2 m_i &= \sum  r{'}^2 m_i + \sum d^2 m_i + 2d \cdot \left( \sum r{'}m_i\right) \\
                            &= \sum r{'}^2 m_i  +d^2M +  2d \cdot \left( \sum r' m_i \right)
\end{align}
Look at the last term on the right in parentheses. It is weighted term defined in order that it be zero. Look at the definition of center of mass and  put all the terms on one side  and you will see what I mean. We thus arrive at the parallel axis theorem. Whenever you find yourself trying to calculate the moment of inertia about some weird axis. STOP! do not just write down integrals. It might be easier to do the integrals about the axis that goes through the center of mass and then use the parallel axis theorem.

1) I of line about the end and about center of mass
\begin{align}
I &= \int r^2 dm  = \rho \int r^2 dV \\
  &= \rho \int_0^L x^2 dx  = \rho\frac{L^3}{3}\\
  &= \frac{ML^2}{3}
\end{align}
Now suppose we want to find the moment of inertia about the center of mass, do we re-do the integrals. NO! we apply the parallel axis theorem: \(I = I_{cm}+ I_d\)  we just calculated I and the end is separated from the center of mass by length d. Plug and chug to get \(\frac{ML^2}{12}\).

2) I of cylinder about axis
\begin{align}
I &= \int r^2 dm  = \rho \int r^2 dV \\
  &= \rho \int r^3 \, dr d\phi dz\\
  &= \frac{\rho \pi L R^4}{2}\\
 &= \frac{MR^2}{2}
\end{align}

3) I of solid sphere about axis (any really)
  \begin{align}
  I &= \rho \int (r \sin \theta)^2 \, r^2 \sin \theta \, dr d\phi d\theta \\
     &= \rho \int (\sin \theta)^3 r^4 \, d\theta d\phi \\
     & = \frac{\rho R^5 8\pi}{15}\\
    & = \frac{2MR^2}{5}
  \end{align}
Notice the \(r \sin\theta\)  in the first integral which would not have been there if r was the usual r in spherical co-ordinates and also I have not broken up r into its components x, y and z as in the center of mass case.

4) I of a Spherical shell
   There is a rigorous way of assuming it has a finite but appreciable thickness , finding the moment of inertia then taking the limit  of the thickness dropping to zero. I will not do that instead I shall I assume it has a finite mass in an infinitesimally thin space. Hence,
\begin{align}
  I &= \rho \int r^2 dA\\
  &= \rho \int (R \sin \theta)^2\, R^2 \sin \theta \,d\phi d\theta \\
  &= \rho \int (\sin \theta)^3 R^4 \, d\theta d\phi\\
 I &= \frac{2MR^2}{3}\\
\text { remember} \rho \text{ is }\frac{M}{4\pi R^2}
\end{align}