The Wave Function Exercises 01

Exercise 1 $ {<j>^2=21^2=441}$ $ {<j^2>=1/N\sum j^2N(J)=\dfrac{6434}{14}=459.6}$ Calculating for each $ {\Delta j}$ $ {j}$ $ {\Delta j=j-<j>}$ 14 $ {14-21=-7}$ 15 $ {15-21=-6}$ 16 $ {16-21=-5}$ 22 $ {22-21=1}$ 24 $ {24-21=3}$ 25 $ {25-21=4}$ Hence for the variance it follows $ {\sigma ^2=1/N\sum (\Delta j)^2N(j)=\dfrac{260}{14}=18.6}$ Hence the standard deviation is $ \displaystyle \sigma =\sqrt{18.6}=4.3$ $ {\sigma^2=<j^2>-<j>^2=459.6-441=18.6}$ And for the standard deviation it is $ \displaystyle \sigma =\sqrt{18.6}=4.3$ Which confirms the second equation for the standard deviation.

Exercise 2 Consider the first $ {25}$ digits in the decimal expansion of $ {\pi}$. What is the probability of getting each of the 10 digits assuming that one selects a digit at random. The first 25 digits of the decimal expansion of $ {\pi}$ are $ \displaystyle \{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3\}$ Hence for the digits it is $ {N(0)=0}$ $ {P(0)=0}$ $ {N(1)=2}$ $ {P(1)=2/25}$ $ {N(2)=3}$ $ {P(2)=3/25}$ $ {N(3)=5}$ $ {P(3)=1/5}$ $ {N(4)=3}$ $ {P(4)=3/25}$ $ {N(5)=3}$ $ {P(5)=3/25}$ $ {N(6)=3}$ $ {P(6)=3/25}$ $ {N(7)=1}$ $ {P(7)=1/25}$ $ {N(8)=2}$ $ {P(8)=2/25}$ $ {N(9)=3}$ $ {P(9)=3/25}$ The most probable digit is $ {5}$. The median digit is $ {4}$. The average is $ {\sum P(i)N(i)=4.72}$. $ {\sigma=2.47}$

Exercise 3 The needle on a broken car is free to swing, and bounces perfectly off the pins on either end, so that if you give it a flick it is equally likely to come to rest at any angle between $ {0}$ and $ {\pi}$. Along the $ {\left[0,\pi\right]}$ interval the probability of the needle flicking an angle $ {d\theta}$ is $ {d\theta/\pi}$. Given the definition of probability density it is $ {\rho(\theta)=1/\pi}$. Additionally the probability density also needs to be normalized. $ \displaystyle \int_0^\pi \rho(\theta)d\theta=1\Leftrightarrow\int_0^\pi 1/\pi d\theta=1 $ which is trivially true. The plot for the probability density is Compute $ {\left\langle\theta \right\rangle}$, $ {\left\langle\theta^2 \right\rangle}$ and $ {\sigma}$. $ {\begin{aligned} \left\langle\theta \right\rangle &= \int_0^\pi\frac{\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\theta d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^2}{2} \right]_0^\pi\\ &= \frac{\pi}{2} \end{aligned}}$ For $ {\left\langle\theta^2 \right\rangle}$ it is $ {\begin{aligned} \left\langle\theta^2 \right\rangle &= \int_0^\pi\frac{\theta^2}{\pi}d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^3}{3} \right]_0^\pi\\ &= \frac{\pi^2}{3} \end{aligned}}$ The variance is $ {\sigma^2=\left\langle\theta^2 \right\rangle-\left\langle\theta\right\rangle^2 =\dfrac{\pi^2}{3}-\dfrac{\pi^2}{4}=\dfrac{\pi^22}{12}}$. And the standard deviation is $ {\sigma=\dfrac{\pi}{2\sqrt{3}}}$. Compute $ {\left\langle\sin\theta\right\rangle}$, $ {\left\langle\cos\theta\right\rangle}$ and $ {\left\langle\cos^2\theta\right\rangle}$. $ {\begin{aligned} \left\langle\sin\theta \right\rangle &= \int_0^\pi\frac{\sin\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\sin\theta d\theta\\ &= \frac{1}{\pi} \left[ -\cos\theta \right]_0^\pi\\ &= \frac{2}{\pi} \end{aligned}}$ and $ {\begin{aligned} \left\langle\cos\theta \right\rangle &= \int_0^\pi\frac{\cos\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\cos\theta d\theta\\ &= \frac{1}{\pi} \left[ \sin\theta \right]_0^\pi\\ &= 0 \end{aligned}}$ We'll leave $ {\left\langle\cos\theta^2 \right\rangle}$ as an exercise for the reader. As a hint remember that $ {\cos^2\theta=\dfrac{1+\cos(2\theta)}{2}}$.

Exercise 4 In exercise $ {1.1}$ it was shown that the the probability density is $ \displaystyle \rho(x)=\frac{1}{2\sqrt{hx}}$ Hence the mean value of $ {x}$ is $ {\begin{aligned} \left\langle x \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{h}{3} \end{aligned}}$ For $ {\left\langle x^2 \right\rangle}$ it is $ {\begin{aligned} \left\langle x^2 \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\int_0^h x^{3/2}dx\\ &= \frac{1}{2\sqrt{h}}\left[\frac{2}{5}x^{5/2} \right]_0^h\\ &= \frac{h^2}{5} \end{aligned}}$ Hence the variance is $ \displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{h^2}{5}-\frac{h^2}{9}=\frac{4}{45}h^2 $ and the standard deviation is $ \displaystyle \sigma=\frac{2h}{3\sqrt{5}} $ For the distance to the mean to be more than one standard deviation away from the average we have two alternatives. The first is the interval $ {\left[0,\left\langle x \right\rangle+\sigma\right]}$ and the second is $ {\left[\left\langle x \right\rangle+\sigma,h\right]}$. Hence the total probability is the sum of these two probabilities. Let $ {P_1}$ denote the probability of the first interval and $ {P_2}$ denote the probability of the second interval. $ {\begin{aligned} P_1 &= \int_0^{\left\langle x \right\rangle-\sigma}\frac{1}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\left[2x^{1/2} \right]_0^{\left\langle x \right\rangle-\sigma}\\ &= \frac{1}{\sqrt{h}}\sqrt{\frac{h}{3}-\frac{2h}{3\sqrt{5}}}\\ &=\sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}} \end{aligned}}$ Now for the second interval it is $ {\begin{aligned} P_2 &= \int_{\left\langle x \right\rangle+\sigma}^h\frac{1}{2\sqrt{hx}}dx\\ &= \ldots\\ &=1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}} \end{aligned}}$ Hence the total probability $ {P}$ is $ {P=P_1+P_2}$ $ {\begin{aligned} P&=P_1+P_2\\ &= \sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}}+1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}}\\ &\approx 0.3929 \end{aligned}}$

Exercise 5 The probability density is $ {\rho(x)=Ae^{-\lambda(x-a)^2}}$ Determine $ {A}$. Making the change of variable $ {u=x-a}$ ($ {dx=du}$) the normalization condition is $ {\begin{aligned} 1 &= A\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &= A\sqrt{\frac{\pi}{\lambda}} \end{aligned}}$ Hence for $ {A}$ it is $ \displaystyle A=\sqrt{\frac{\lambda}{\pi}}$ Find $ {\left\langle x \right\rangle}$, $ {\left\langle x^2 \right\rangle}$ and $ {\sigma}$. $ {\begin{aligned} \left\langle x \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty ue^{-\lambda u^2}du+a\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &=\sqrt{\frac{\lambda}{\pi}}\left( 0+a\sqrt{\frac{\pi}{\lambda}} \right)\\ &= a \end{aligned}}$ If you don't see why $ {\displaystyle\int_{-\infty}^\infty ue^{-\lambda u^2}du=0}$ check this post on my other blog. For $ {\left\langle x^2 \right\rangle}$ it is $ {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right) \end{aligned}}$ Now $ {\displaystyle 2a\int_{-\infty}^\infty u e^{-\lambda u^2}du=0}$ as in the previous calculation. For the third term it is $ {\displaystyle a^2\int_{-\infty}^\infty e^{-\lambda u^2}du=a^2\sqrt{\frac{\pi}{\lambda}}}$. The first integral is the hard one and a special technique can be employed to evaluate it. $ {\begin{aligned} \int_{-\infty}^\infty u^2e^{-\lambda u^2}du &= \int_{-\infty}^\infty-\frac{d}{d\lambda}\left( e^{-\lambda u^2} \right)du\\ &= -\frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &=-\frac{d}{d\lambda}\sqrt{\frac{\pi}{\lambda}}\\ &=\frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}} \end{aligned}}$ Hence it is $ {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &= \sqrt{\frac{\lambda}{\pi}}\left( \frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}}+0+a^2\sqrt{\frac{\pi}{\lambda}} \right)\\ &=a^2+\frac{1}{2\lambda} \end{aligned}}$ The variance is $ \displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{1}{2\lambda}$ Hence the standard deviation is $ \displaystyle \sigma=\frac{1}{\sqrt{2\lambda}} $

— Mathematica file —

The resolution of exercise 2 was done using some basic Mathematica code which I'll post here hoping that it can be helpful to the readers of this blog.

// N[Pi, 25]

piexpansion = IntegerDigits[3141592653589793238462643]

digitcount = {}

For[i = 0, i <= 9, i++, AppendTo[digitcount, Count[A, i]]]

digitcount

digitprobability = {}

For[i = 0, i <= 9, i++, AppendTo[digitprobability, Count[A, i]/25]]

digitprobability

digits = {}

For[i = 0, i <= 9, i++, AppendTo[digits, i]]

digits

j = N[digits.digitprobability]

digitssquared = {}

For[i = 0, i <= 9, i++, AppendTo[digitssquared, i^2]]

digitssquared

jsquared = N[digitssquared.digitprobability]

sigmasquared = jsquared - j^2

std = Sqrt[sigmasquared]

deviations = {}

deviations = piexpansion - j

deviationssquared = (piexpansion - j)^2

variance = Mean[deviationssquared]

standarddeviation = Sqrt[variance] 

The Wave Function 02

— 1.3. Probability —

In the previous post we were introduced to the Schrodinger equation (equation 1), stated Born's interpretation of what is the physical meaning of the wave function and took a little glimpse into some philosophical positions one might have regarding Quantum Mechanics.
Since probability plays such an essential role in Quantum Mechanics it seems that a brief revision of some of its concepts is in order so that we are sure that we have the tools that allows one to do Quantum Mechanics.

— 1.3.1. Discrete variables —

The example used in the book in order to expound on the terminology and concepts of probability is of a set that consists of 14 people in a class room:

• one person has 14 years
• one person has 15 years
• three people have 16 years
• two people 22 years
• five people have 25 years

Let $ {N(j)}$ represent the number of people with age $ {j}$. Hence

• $ {N(14)=1}$
• $ {N(15)=1}$
• $ {N(16)=3}$
• $ {N(22)=2}$
• $ {N(25)=5}$

One can represent the previous data points by use of a histogram:

The the total number of people in the room is given by

$ \displaystyle N=\sum_{j=0}^{\infty}N(j) \ \ \ \ \ (2)$

Adopting a frequentist definition of probability Griffiths then makes a number of definitions of probability concepts under the assumption that the phenomena at study are discrete ones.

Definition 1 The probability of an event $ {j}$, $ {P(j)}$ is proportional to the number elements that have the property $ {j}$ and inversely proportional to the total elements ($ {N}$) under study. $ \displaystyle P(j)=\frac{N(j)}{N} \ \ \ \ \ (3)$

It is easy to see that from equation 3 together with equation 2 it follows

$ \displaystyle \sum_{j=0}^{\infty}P(j)=1 \ \ \ \ \ (4)$

After defining $ {P(j)}$ we can also define what is the most probable value for $ {j}$.

Definition 2 The most value for $ {j}$ is the one for which $ {P(j)}$ is a maximum.

Definition 3 The average value of $ {j}$ is given by $ \displaystyle <j>=\sum_{j=0}^{\infty}jP(j) \ \ \ \ \ (5)$

But what if we are interested in computing the average value of $ {j^2}$? Then the appropriate expression must be

$ \displaystyle <j>=\sum_{j=0}^{\infty}j^2P(j) $

Hence one can write with full generality that average value for some function of $ {j}$, denoted by $ {f(j)}$ is given by

$ \displaystyle <f(j)>=\sum_{j=0}^{\infty}f(j)P(j) \ \ \ \ \ (6)$

After introducing the definition of maximum of a probability distribution it is time to introduce a couple of definitions that relate t the symmetry and spread of a distribution.

Definition 4 The median is the value of $ {j}$ for which the probability of having a larger value than $ {j}$ is the same as the probability of having a value with a smaller value than $ {j}$.

After seeing a definition that relates to the the symmetry of a distribution we'll introduce a definition that is an indication of its spread.
But first we'll look at two examples that will serve as a motivation for that:

and

Both histograms have the same median, the same average, the same most probable value and the same number of elements. Nevertheless it is visually obvious that the two histograms represent two different kinds of phenomena.
The first histogram represents a phenomenon whose values are sharply peaked about the average (central) value.
The second histogram on the other hand represents a phenomenon represents a more broad and flat distribution.
The existence of such a difference in two otherwise equal distributions introduces the necessity of introducing a measure of spread.
A first thought could be to define the difference about the average for each individual value

$ \displaystyle \Delta j=j-<j> $

This approach doesn't work since that for random distributions one would expect to find equally positive and negative values for $ {\Delta j}$.
One way to circumvent this issue would be to use $ {|\Delta j|}$, and even though this approach does work theoretically it has the problem of not using a differentiable function.
These two issues are avoided if one uses the squares of the deviations about the average.
The quantity of interest in called the variance of the distribution.

Definition 5 The variance of a distribution ,$ {\sigma ^2}$, that has an average value is given by the expression $ \displaystyle \sigma ^2=<(\Delta j)^2> \ \ \ \ \ (7)$

Definition 6 The standard deviation, $ {\sigma}$, of a distribution is given by the square root of its variance.

For the variance it is

$ \displaystyle \sigma ^2=<j^2>-<j>^2 \ \ \ \ \ (8)$

Since by definition 5 the variance is manifestly non-negative then it is

$ \displaystyle <j^2> \geq <j>^2 \ \ \ \ \ (9)$

where equality only happens when the distribution is composed of equal elements and equal elements only.

— 1.3.2. Continuous variables —

Thus far we've always assumed that we are dealing with discrete variables. To generalize our definitions and results to continuous distributions.
One has to have the initial care to note that when dealing with phenomena that allow for a description that is continuous probabilities of finding a given value are vanishing, and that one should talk about the probability of a given interval.
With that in mind and assuming that the distributions are sufficiently well behaved one has that the probability of and event being between $ {x}$ and $ {x+d}$ is given by

$ \displaystyle \rho(x)dx \ \ \ \ \ (10)$

The quantity $ {\rho (x)}$ is the probability density.
The generalizations for the other results are:

$ \displaystyle \int_{-\infty}^{+\infty}\rho(x)dx=1 \ \ \ \ \ (11)$

$ \displaystyle <x>=\int_{-\infty}^{+\infty}x\rho(x)dx \ \ \ \ \ (12)$

$ \displaystyle <f(x)>=\int_{-\infty}^{+\infty}f(x)\rho(x)dx \ \ \ \ \ (13)$

$ \displaystyle \sigma ^2=<(\Delta x)^2>=<x^2>-<x>^2 \ \ \ \ \ (14)$

The Wave Function 01

— 1. The Wave Function —

The purpose of this section is to introduce the wave function of Quantum Mechanics and explain its physical relevance and interpretation.

— 1.1. The Schroedinger Equation —

Classical Dynamics' goal is to derive the equation of motion, $ {x(t)}$, of a particle of mass $ {m}$. After finding $ {x(t)}$ all other dynamical quantities of interest can be computed from $ {x(t)}$.
Of course that the problem is how does one finds $ {x(t)}$? In classical mechanics this problem is solved by applying Newton's Second Axiom

$ \displaystyle F=\frac{dp}{dt} $

For conservative systems it is $ {F=-\dfrac{\partial V}{\partial x}}$ (previously we've used $ {U}$ to denote the potential energy but will now use $ {V}$ to accord to Griffith's notation).
Hence for classical mechanics one has

$ \displaystyle m\frac{d^2 x}{dt^2}=-\dfrac{\partial V}{\partial x} $

as the equation that determines $ {x(t)}$ (with the help of the suitable initial conditions).
Even though Griffith's only states the Newtonian formalism approach of Classical Dynamics we already know by Classical Physics that apart from Newtonian formalism one also has the Lagrangian formalism and Hamiltonian formalism as suitable alternatives (and most of time more appropriate alternatives) to Newtonian formalism as ways to derive the equation of motion.
As for Quantum Mechanics one has to resort the Schrodinger Equation in order to derive the equation of motion the specifies the Physical state of the particle in study.

$ \displaystyle i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi \ \ \ \ \ (1)$

— 1.2. The Statistical Interpretation —

Of course now the question is how one should interpret the wave function. Firstly its name itself should sound strange. A particle is something that is localized while a wave is something that occupies an extense region of space.
According to Born the wave function of a particle is related to the probability of it occupying a region of space.
The proper relationship is $ {|\Psi(x,t)|^2dx}$ is the density probability of finding the particle between $ {x}$ and $ {x+dx}$.
This interpretation of the wave function naturally introduces an indeterminacy to Quantum Theory, since one cannot predict with certainty the position of a particle when it is measured and only its probability.
The conundrum that now presents itself to us is: after measuring the position of a particle we know exactly where it is. But what about what happens before the act of measurement? Where was the particle before our instruments interacted with it and revealed is position to us?
These questions have three possible answers:

1. The realist position: A realist is a physicist that believes that the particle was at the position where it was measured. If this position is true it implies that Quantum Mechanics is an incomplete theory since it can't predict the exact position of a particle but only the probability of finding it in a given position.
2. The orthodox position: An orthodox quantum physicist is someone that believes that the particle had no definite position before being measured and that it is the act of measurement that forces the particle to occupy a position.
3. The agnostic position: An agnostic physicist is a physicist that thinks that he doesn't know the answer to this question and so refuses to answer it.

Until 1964 advocating one these three positions was acceptable. But on that year John Stewart Bell proved a theorem, On the Einstein Podolsky Rosen paradox, that showed that if the particle has a definite position before the act of measurement then it makes an observable difference on the results of some experiments (in due time we'll explain what we mean by this).
Hence the agnostic position was no longer a respectable stance to have and it was up to experiment to show if Nature was a realist or if Nature was an orthodox.
Nevertheless the disagreements of what exactly is the position of a particle when it isn't bening measured, all three groups of physicists agreed to what would be measured immediately after the first measurement of the particle's position. If at first one has $ {x}$ then the second measurement has to be $ {x}$ too.
In conclusion the wave function can evolve by two ways:

1. It evolves without any kind of discontinuity (unless the potential happens to be unbounded at a point) under the Schrodinger Equation.
2. It collapses suddenly to a single value due to the act of measurement.

The interested reader can also take a look at the following book from Bell: Speakable and Unspeakable in Quantum Mechanics

Hamiltonian formalism exercises

Exercise 1 Choose a set of generalized coordinates that totally specify the mechanical state of the following systems: A particle of mass $ {m}$ that moves along an ellipse. Let $ {x=a\cos\theta}$ and $ {y=b\sin\theta}$. Then the generalized coordinate is $ {\theta}$. Perhaps you might be surprised to find out that we need only a single coordinate to describe the motion of this particle since its motion is described on a plane which is a two dimensional entity. But the particle motion is restricted to be along the ellipse and that constraint decreases the particle's degrees of freedom to one instead of being two. A cylinder that moves along an inclined plane. If the cylinder rotates we need $ {x}$, the distance travelled, and $ {\theta}$, the angle of rotation. If the cylinder doesn't rotate we only need $ {x}$. The two masses on a double pendulum. The generalized coordinates are $ {\theta_1}$ and $ {\theta_2}$. Do you see why?

Exercise 2 Derive the transformation equations for the double pendulum. It is $ {x_1=l_1\cos\theta_1}$, $ {x_2=l_1\cos\theta_1+l_2\cos\theta_2}$, $ {y_1=l_1\sin\theta_1}$ and $ {y_2=l_1\sin\theta_1+l_2\sin\theta_2}$

Exercise 3 Show that $ {\dfrac{\partial \dot{\vec{r}}_\nu}{\partial \dot{q}_\alpha}=\dfrac{\partial\vec{r}_\nu}{\partial q_\alpha}}$. $ {\begin{aligned} \vec{r}_\nu&=\vec{r}_\nu(q_1,q_2,\cdots,q_n,t)\Rightarrow \\ \dot{\vec{r}_\nu}&=\dfrac{\partial\vec{r}_\nu }{\partial q_1}\dot{q}_1+\cdots+\dfrac{\partial\vec{r}_\nu }{\partial q_n}\dot{q}_n+\dfrac{\partial\vec{r}_\nu}{\partial t} \Rightarrow \\ \dfrac{\partial \dot{\vec{r}}_\nu}{\partial\dot{q}_\alpha}&=\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \end{aligned}}$

Exercise 4 Consider a system of particles that experiences an increment $ {dq_j}$ on its generalized coordinates. Derive the following expression $ {dW=\displaystyle \sum_\alpha \Phi_\alpha dq_\alpha}$ for the total work done by the force and indicate the physical meaning of $ {\Phi_\alpha}$. First note that $ \displaystyle d\vec{r}_\nu =\sum_{\alpha=1}^n \dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}dq_\alpha$ For $ {dW}$ it is $ {\begin{aligned} dW &=\sum_{\nu=1}^N\vec{F}_\nu\cdot d\vec{r}_\nu\\ &=\sum_{\nu=1}^N\left( \sum_{\alpha=1}^n \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha} \right) dq_\alpha\\ &= \sum_{\alpha=1}^n \Phi_\alpha dq_\alpha \end{aligned}}$ with $ {\displaystyle \Phi_\alpha=\sum_{\nu=1}^N \vec{F}_\nu\cdot\dfrac{\partial \vec{r}_\nu}{\partial q_\alpha}}$ being the generalized force.

Exercise 5 Show that $ {\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$. We have $ {\displaystyle dW=\sum_\alpha\frac{\partial W}{\partial q_\alpha}dq_\alpha}$ and $ {\displaystyle dW=\sum_\alpha\Phi_\alpha dq_\alpha}$. Hence $ {\displaystyle \sum_\alpha\left( \Phi_\alpha- \frac{\partial W}{\partial q_\alpha}\right)dq_\alpha=0}$. Since $ {dq_\alpha}$ are linearly independent it is $ {\Phi_\alpha=\dfrac{\partial W}{\partial q_\alpha}}$.

Exercise 6 Derive the lagrangian for a simple pendulum and obtain an equation to describe its motion. The generalized coordinate for the simple pendulum is $ {\theta}$ and the transformation equations are $ {x=l\sin\theta}$ and $ {y=-l\cos\theta}$. For the kinetic energy it is $ {K=1/2mv^2=1/2m(l\dot{\theta})^2=1/2ml^2\dot{\theta}^2}$. for the potential $ {V=mgl(1-\cos\theta)}$. Hence the Lagrangian is $ {L=K-V=1/2ml^2\dot{\theta}^2-mgl(1-\cos\theta)}$. $ {\dfrac{\partial L}{\partial \theta}=-mgl\sin\theta}$ and $ {\dfrac{\partial L}{\partial \dot{\theta}}=ml^2\dot{\theta}}$. Hence the Euler Lagrange equation is $ {\begin{aligned} \frac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}}-\dfrac{\partial L}{\partial \theta}&=0\Rightarrow\\ ml^2\dot{\theta}+mgl\sin\theta&=0\Rightarrow\\ \ddot{\theta}+g/l\sin\theta=0 \end{aligned}}$

Exercise 7 Two particles of mass $ {m}$ are connected with each other and to two points $ {A}$ and $ {B}$ by springs with constant factor $ {k}$. The particles are free to slide along the direction of $ {A}$ and $ {B}$. Use the Euler-Lagrange equations to derive the equations of motion of the particles. The kinetic energy is $ {K=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2}$. The potential energy is $ {V=1/2kx^2_1+1/2k(x_2-x_1)^2+1/2kx^2_2}$. Hence the Lagrangian is $ {L=1/2m\dot{x}^2_1+1/2m\dot{x}^2_2-1/2kx^2_1-1/2k(x_2-x_1)^2-1/2kx^2_2}$. The partial derivatives of the Lagrangian are: $ {\dfrac{\partial L}{\partial x_1}=k(x_2-x_1)}$ $ {\dfrac{\partial L}{\partial x_2}=k(x_1-x_2)}$ $ {\dfrac{\partial L}{\partial \dot{x_1}}=m\dot{x}_1}$ $ {\dfrac{\partial L}{\partial \dot{x_2}}=m\dot{x}_2}$ And the Euler-Lagrange equations are: $ {m\ddot{x}_1=k(x_2-x_1)}$ $ {m\ddot{x}_2=k(x_1-2x_2)}$

Exercise 8 A particle of mass $ {m}$ moves subject to a conservative force field. Use cylindrical coordinates to derive: The Lagrangian. The kinetic energy is $ {K=1/2m(1/2m\dot{\rho}^2+\rho^2\dot{\phi}^2+\dot{z^2})}$. The potential is $ {V=V(\rho,\phi,z)}$. The equations of motion. $ {m(\ddot{\rho}-\rho\dot{\phi}^2)=-\dfrac{\partial v}{\partial \rho}}$ $ {m\dfrac{d}{dt}(\rho^2\dot{\phi}=-\dfrac{\partial V}{\partial \phi}}$ $ {m\ddot{z}=-\dfrac{\partial V}{\partial z}}$

Exercise 9 A double pendulum oscillates on a vertical plane. Calculate: The Lagrangian. The transformation equations for the coordinates are $ {x_1=l_1\cos\theta_1}$ $ {y_1=l_1\sin\theta_1}$ $ {x_2=l_1\cos\theta_1+l_2\cos\theta_2}$ $ {y_2=l_1\sin\theta_1+l_2\sin\theta_2}$ Applying $ {\dfrac{d}{dt}}$ to the previous equations $ {\dot{x}_1=-l_1\dot{\theta}_1\sin\theta_1}$ $ {\dot{y}_1=l_1\dot{\theta}_1\cos\theta_1}$ $ {\dot{x}_2=-l_1\dot{\theta}_1\sin\theta_1-l_2\dot{\theta}_2\sin\theta_2}$ $ {\dot{y}_2=l_1\dot{\theta}_1\cos\theta_1+l_2\dot{\theta}_2\cos\theta_2}$ Hence the kinetic energy is $ \displaystyle K=1/2m_1l^2_1\theta^2_1+1/2m\left[ l^2_1\dot{\theta}^2_1+l^2_2\dot{\theta}^2_2+2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2) \right]$ And the potential is $ \begin{aligned}\displaystyle V&=m_1g(l_1+l_2-l_1\cos\theta_1)\\ &+m_2g\left[l_1+l_2-(l_1\cos\theta_1+l_2\cos\theta_2)\right]\end{aligned}$ As always the Lagrangian is $ {L=K-V=\cdots}$ The equations of motion. ${\begin{aligned}\dfrac{\partial L}{\partial \theta_1}&=-m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)\\ &- m_1gl_1\sin\theta_1-m_2gl_1\sin\theta_1\end{aligned}}$ $ {\dfrac{\partial L}{\partial \dot{\theta_1}}=m_1 l_1^2\dot{\theta}_1^2+m_2l_1^2\dot{\theta}_1+m_2l_1l_2\dot{\theta}_2\cos(\theta_1-\theta_2)}$ $ {\dfrac{\partial L}{\partial \theta_2}=m_2l_1l_2\dot{\theta}_1\dot{\theta}_2\sin(\theta_1-\theta_2)-m_2gl_2\sin\theta_2}$ $ {\dfrac{\partial L}{\partial \dot{\theta_2}}=m_2l_2^2\dot{\theta}_2^2+m_2l_1l_2\dot{\theta}_1\cos(\theta_1-\theta_2)}$ ${\begin{aligned}-(m_1+m_2)gl\sin\theta_1&=(m_1+m_2)l_1\ddot{\theta}_1+ m_2l_1l_2\ddot{_2}\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_2\sin(\theta_1-\theta_2)\end{aligned}}$ and $ {\begin{aligned}-m_2gl_2\sin\theta_2 &=m_2l_2\ddot{\theta}_2+m_2l_1l_2\ddot{\theta}_1\cos(\theta_1-\theta_2)\\ &+ m_2l_1l_2\dot{\theta}_1^2\sin(\theta_1-\theta_2)\end{aligned}}$ Assume that $ {m_1=m_2=m}$ e $ {l_1=l_2=l}$ and write the equations of motion. Left as an exercise for the reader. Write the previous equations in the limit of small oscillations. If $ {\theta\ll1}$ implies $ {\sin \theta\approx\theta}$ and $ {\cos \theta\approx1}$. Hence the equations of motion are $ {2l\ddot{\theta}_1+l\ddot{\theta}_2=-2g\theta_1}$ $ {l\ddot{\theta}_1+l\ddot{\theta}_2=-g\theta_2}$

Exercise 10 A particle moves along the plane $ {xy}$ subject to a central force that is a function of the distance between the particle and the origin. Find the hamiltonian of the system. The generalized coordinates are $ {r}$ and $ {\theta}$. The potential is of the form $ {V=V(r)}$. The Lagrangian is $ {L=1/2m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)}$. The conjugate momenta are: $ {p_r=\dfrac{\partial L}{\partial \dot{r}}=m\dot{r}}$ $ {p_\theta=\dfrac{\partial L}{\partial \dot{\theta}}=mr^2\dot{\theta}}$ For the Hamiltonian it is $ {\begin{aligned} H&=\displaystyle \sum_{\alpha=n}^np_\alpha\dot{q}_\alpha\\ &=p_r\dot{r}+p_\theta\dot{\theta}-(1/2m(\dot{r^2}+r^2\dot{\theta}^2)-V(r))\\ &=\dfrac{p_r^2}{2m}+\dfrac{p_theta^2}{2mr^2}+V(r) \end{aligned}}$ Write the equations of motion. $ {\dot{r}=\dfrac{\partial H}{\partial p_r}=\dfrac{p_r}{m}}$ $ {\dot{\theta}=\dfrac{\partial H}{\partial p_\theta}=\dfrac{p_\theta}{mr^2}}$ $ {\dot{p}_r=-\dfrac{\partial H}{\partial r}=\dfrac{p_\theta}{mr^3}-V'(r)}$ $ {\dot{p}_\theta=-\dfrac{\partial H}{\partial \theta}=0)}$

Exercise 11 A particle describes a one dimensional motion subject to a force $ \displaystyle F(x,t)= \frac{k}{x^2}e^{-t/\tau} $ where$ {k}$ and $ {\tau}$ are positive constants. Find the lagrangian and the hamiltonian. Compare the hamiltonian with the total energy and discuss energy conservation for this system. Since $ {F(x,t)= \frac{k}{x^2}e^{-t/\tau}}$ it follows $ {V=\dfrac{k}{x}e^{-t/\tau}}$. For the kinetic energy it is $ {K=1/2m\dot{x}^2}$. Hence the lagrangian is $ \displaystyle L=1/2m\dot{x}^2-\dfrac{k}{x}e^{-t/\tau}$ . Now $ {p_x=m\dot{x}\Rightarrow\dot{x}=\dfrac{p_x}{m}}$. For the Hamiltonian it is $ {H=p_x\dot{x}-L=\dfrac{p_x^2}{2m}+\dfrac{k}{x}e^{-t/\tau}}$. Since $ {\dfrac{\partial L}{\partial t}=0}$ the system isn't conservative. Since $ {\dfrac{\partial U}{\partial \dot{x}}=0}$ it is $ {H=E}$.

Exercise 12 Consider two functions of the generalized coordinates and the generalized momenta, $ {g(q_k,p_k)}$ and $ {h(q_k,p_k)}$. The Poisson brackets are defined as: $ \displaystyle [g,h]=\sum_k \left(\frac{\partial g}{\partial q_k}\frac{\partial h}{\partial p_k}-\frac{\partial g}{\partial p_k}\frac{\partial h}{\partial q_k}\right) $ Show the following properties of the Poisson brackets: $ {\dfrac{dg}{dt}=[g,H]+\dfrac{\partial g}{\partial t}}$. Left as an exercise for the reader. $ {\dot{q}_j=[q_j,H]}$ e $ {\dot{p}_j=[p_j,H]}$. Left as an exercise for the reader. $ {[p_k,p_j]=0}$ e $ {[q_k,q_j]=0}$. Left as an exercise for the reader. $ {[q_k,p_j]=\delta_{ij}}$. Left as an exercise for the reader. If the Poisson brackets between two functions is null the two functions are said to commute. Show that if a function $ {f}$ doesn't depend explicitly on time and $ {[f,H]=0}$ the function is a constant of movement.

Newtonian formalism exercises

Exercise 1 A particle of mass $ {m}$ moves in the plane $ {xy}$. Its position vector is $ {\vec{r}=a\cos \omega t \vec{i}+b\sin \omega t \vec{j}}$ with $ {a,b,\omega}$ positive constants and $ {a>b}$. Show that: The particle's trajectory is an ellipse. For $ {x=a\cos\omega\ t}$ and $ {y=b\sin\omega\ t}$ it is $ \displaystyle \left( \frac{x}{a} \right)^2+\left( \frac{y}{b} \right)^2=\cos^2\omega t+\sin^2\omega t=1 $ Which is the equation of an ellipse The particle is subject to a central force oriented to the origin It is $ {\begin{aligned} \vec{F} &= m\dfrac{d\vec{v}}{dt}\\ &= m\dfrac{d^2\vec{r}}{dt^2}\\ &= m\dfrac{d^2}{dt^2}a\cos \omega t \vec{i}+b\sin \omega t \vec{j}\\ &=-m\omega^2\vec{r} \end{aligned}}$ Which is a central force oriented to the origin (note the minus sign).

Exercise 2 A particle of constant mass $ {m}$ is subject to to a force $ {F}$. Suppose that for $ {t_1}$ e $ {t_2}$ the velocities are $ {\vec{v}_1}$ and $ {\vec{v}_2}$, respectively. Show that the work the force does on the particle equals its change in kinetic energy. $ {\begin{aligned} W &= \int_{t_1}^{t_2}\vec{F}\cdot d\vec{r}\\ &= \int_{t_1}^{t_2}\vec{F}\cdot\dfrac{d\vec{r}}{dt}dt \\ &= \int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt\\ &= \int_{t_1}^{t_2}m\dfrac{d\vec{v}}{dt}\cdot\vec{v}dt\\ &=m\int_{t_1}^{t_2}\vec{v}\cdot d\vec{v}\\ &=\dfrac{1}{2} m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})\\ &=1/2m(v_2^2-v_1^2) \end{aligned}}$

Exercise 3 For the conditions of exercise 1 calculate: The particle's kinetic energy in the positive semi-major axis and on the positive semi-minor axis. First note that $ {\vec{v}=-\omega a\sin \omega t \vec{i}+\omega b \cos \omega t \vec{j}}$. Now $ {K=1/2m(\omega^2a^2\sin^2\omega t+\omega^2b^2\cos^2\omega t)}$. Let $ {A}$ denote the semi-major axis extension and $ {B}$ the semi-minor axis extension. Hence $ {K_A=1/2m\omega ^2b^2}$ and $ {K_B=1/2m\omega ^2a^2}$ The work done by the force field when it moves the particle from the positive semi-major axis to the positive semi-minor axis. $ {\begin{aligned} W &= \int_A^B\vec{F}\cdot d\vec{r}\\ &= \int_A^B(-m\omega ^2\vec{r})\cdot d\vec{r}\\ &= -m\omega ^2\int_A^B\vec{r}\cdot d\vec{r}\\ &= -1/2m\omega ^2\int_A^B d(\vec{r}\cdot\vec{r})\\ &=1/2m\omega ^2(a^2-b^2) \end{aligned}}$ The total work done by the force field by moving the particle along the ellipse. $ {W=0}$. Do you see why?

Exercise 4 Show that if a particle is subject to $ {\vec{F}}$ and $ {\vec{v}}$ is its instantaneous velocity then the instantaneous power is $ \displaystyle \mathcal{P}=\vec{F}\cdot\vec{v} $ By definition it is $ {dW=\vec{F}\cdot d\vec{r}}$. Hence $ {\begin{aligned} \mathcal{P}&=\dfrac{dW}{dt}\\ &= \vec{F}\cdot\dfrac{d\vec{r}}{dt}\\ &= \vec{F}\cdot\vec{v} \end{aligned}}$

Exercise 5 Show that the integral $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is independent of a particle's trajectory if and only if $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Let $ {\Gamma=P_1AP_2BP_1}$ denote a closed curve and admit that $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent. Then $ {\begin{aligned} \oint \vec{F}\cdot d\vec{r}&=\int_\Gamma\vec{F}\cdot d\vec{r} \\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}$ Where the last equality follows from our assumption that $ {\displaystyle \int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}}$ is path independent. Suppose now that $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. $ {\begin{aligned} \int_\Gamma\vec{F}\cdot d\vec{r}&=\int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}}$ Where the last equality follows from our $ { \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Hence $ {\displaystyle\int_{P_1AP_2}\vec{F}\cdot d\vec{r}=\int_{P_1BP_2}\vec{F}\cdot d\vec{r}}$.

Exercise 6 A particle of mass $ {m}$ moves along $ {x}$ subject to to a conservative force field $ {V(x)}$. If the particle's positions are $ {x_1}$ and $ {x_2}$ on $ {t_1}$ and $ {t_2}$, respectively, show that, if $ {E}$ is the total energy then $ \displaystyle t_2-t_1=\sqrt{\frac{m}{2}}\int_{x_1} ^{x_2}\frac{dx}{\sqrt{E-V(x)}} $ Write $ {1/2m\left( \dfrac{dx}{dt} \right)^2+V(x)=E}$ solve in order to $ {dt}$ and integrate.

Exercise 7 Consider a particle of mass $ {m}$ that moves vertically on a resistive medium where the retarding force is proportional to the particle's velocity. Consider that particle initially moves in the downward direction with an initial velocity $ {v_0}$ from an height $ {h}$. Derive the particle's equation of motion $ {z=z(t)}$. It is $ {F=m\dfrac{dv}{dt}=-mgg-kmv}$ with $ {v<0}$ and $ {-kmv>0}$. Then solve in order to $ {dv}$ and integrate to find $ {v=\dfrac{dz}{dt}=-\dfrac{g}{k}+\dfrac{kv_0+g}{k}e^{-kt}}$ (the terminal velocity $ {v_t}$ is $ {v_t=\displaystyle \lim_{t \rightarrow +\infty} v=-g/k}$). Solving in order to $ {dz}$ and integrating and it is $ {z=h-\dfrac{gt}{k}+\dfrac{kv_0+g}{k^2}(1-e^{-kt})}$.

Exercise 8 A particle is shot vertically on a region where exists a constant gravitational field with a constant initial velocity $ {v_0}$. Show that, in the presence of resistive force proportional to the square of the particle's instantaneous velocity, the velocity of the particle when it returns to its initial position is: $ \displaystyle \frac{v_0 v_t}{\sqrt{v_0^2+v_t^2}} $ Where $ {v_t}$ denotes the particle's terminal velocity.

Quantum Mechanics Introduction

— 1. Motivation —

Unlike we did so far in the previous posts first we'll begin to study Quantum Mechanics by mentioning some of the experiments that motivated the reformulation of Classical Physics into the body of knowledge that is Quantum Mechanics. Also we'll spend more time explaining our initial formulations.

— 1.1. New results, new conceptions —

Anyone that has ever realized an experiment before knows that if one wants to know anything about a system that is being studied one needs to interact with it. In a more formal language one should say: "The act of measuring a system introduces a disturbance in the system". Thus far we've used the concept of mechanical state on our analysis of physical systems. A more critical eye to the concept of reveals the following:

1. In principle the disturbance can, in some cases, be made as small as one wants. The fact that exists a limit in our measuring apparatus lies in the nature of the apparatus and not in the nature of the theory being used.
2. There exists some disturbances whose effect can't be neglected. Nevertheless one can always calculate exactly the value of the effect of the disturbance and compensate it in the value of the quantity that is being measured afterwards.

Thus one can say that our very successful Classical Theory of Physics is casual and deterministic. Despite its many successes our Classical Theory of Physics had a few dark clouds hovering over it:

• Black body radiation.
• Photoelectric effect.
• Rydberg–Ritz combination principle.
• The existence and stability of atoms.
• Stern-Gerlach's experiment.
• Electron diffraction.
• ...

The persistence of these experimental results and the failure of accommodating them in the classical world view indicated that a revision of concepts was needed:

• Corpuscular entities were demonstrating wavelike behavior.
• Wave entities were demonstrating pointlike behavior.
• There exists a statistical nature in atomic and subatomic phenomena that seems to be an essential characteristic of Nature.
• Nature's atomic nature (no pun intended) implies that the process of measure has to be thought over: some disturbances can't be made arbitrarily small since there exists a natural limit to how small a disturbance can be made.

— 1.2. Double-slit experiment —

In order to make our previous discussion more concrete we'll look into a particular experiment that makes it clear how disparate are the two world views that we've been discussing.

— 1.2.1. Double-slit and particles —

Suppose an experimental apparatus which consists of a wall with two opening and a second wall that will serve as target. A beam of particles is fired upon the first wall. Some of those particles will hit the wall and of some will pass through the slits and hit the second wall.

The particles that pass through slit $ {1}$ are responsible for the probability curve $ {P_1}$ while the particles that pass through slit $ {2}$ are responsible for the probability curve $ {P_2}$. The resulting probability curve, $ {P_{12}}$, results from the algebraic sum of the curves $ {P_1}$ and $ {P_2}$.

— 1.2.2. Double-slit and waves —

Now if we direct a beam of waves on the same experimental apparatus the pattern one observes in the second wall is:

 

Now what we'll study is the intensity curve. The intensity curve $ {I_1}$ is the result one would observe if only slit $ {1}$ was open, while the intensity curve $ {I_2}$ is what one would observe if only slit $ {2}$ was open. The resulting intensity is given by the mathematical expression $ {I_{12}=|h_1+h_2|^2= I_1+2I_1I_2 \cos \theta}$. The last term is responsible for the interaction between the waves that come from slit $ {1}$ with the waves that come from slit $ {2}$ which ultimately is what causes the interference pattern one observes in the second wall.

— 1.2.3. Double slit and electrons —

Now that we familiarized ourselves with the behavior of particles and waves under the double slit we'll study what one observes when we subject an electron beam through the same apparatus. Experimental results show conclusively that electrons are particles so we're expecting to see the same pattern that we saw for macroscopic particles. Nevertheless this is what Nature has for us!:

In the case of electrons Nature forces to think again in terms of curves related to wave phenomena. But unlike macroscopic wave phenomena this aren't curves of intensity but they are curves of probability. This introduces an apparent oxymoron since probability curves are a concept that is intrinsic to particles while an interference pattern is intrinsic to waves...

In order to explain what we're observing one has to assume that each probability curve $ {P_i}$ is associated to a probability amplitude $ {\phi_i}$. In order to calculate the probability one has to calculate the modulus squared of the probability amplitude, $ {P_i=\phi_i^2}$. Thus one has to first calculate the sum of probability amplitudes of an electron passing through slit $ {1}$ or passing through slit $ {2}$ and only then should we calculate the modulus squared of the probability amplitude of an electron passing through slit $ {1}$ or slit $ {2}$:

$ \displaystyle P_{12}=|\phi_1+\phi_2|^2$

— 2. Basic concepts and preliminary definitions —

After our initial discussion I suppose that you understand why physicists had the need to introduce a new paradigm that made sense of what's happening in the atomic and sub- atomic level. Now it's time to introduce our usual initial definitions.

Definition 1 An operator is a mathematical operations that carries a given function into another function.

As an example of an operator we have $ {\dfrac{d}{dx}}$ which a transforms a functions into its $ {x}$ derivative. As another example of an operator we have $ {2\times}$ that transforms a function into its double.

Definition 2 The linear momentum is represented by the operator $ \displaystyle p=\dfrac{\hbar}{i}\dfrac{d}{dx} \ \ \ \ \ (1)$

Definition 3 The energy of a particle is represented by the operator: $ \displaystyle E=i\hbar\dfrac{d}{dt} \ \ \ \ \ (2)$

Definition 4 For a free particle the following mathematical relationships are valid:$ {\begin{aligned} \label{eq:relacoesdebroglie} k &= \frac{\hbar}{p}\\ \omega &= \frac{E}{\hbar} \end{aligned}}$

— 3. Quantum Mechanics Axioms —

After this batch of initial definitions that allows us to know the entities that will take part in the construction of our quantum vision of the world it is time for us to define the rules that govern them

Axiom 1 The quantum state is defined by the specification of the relevant physical quantities and is represented by a complex function $ {\Psi(x,t)}$

Axiom 2 To every observable in Classical Physics ($ {A}$) there corresponds a linear, Hermitian operator ($ {\hat{A}}$) in quantum mechanics.

Axiom 3 Measurements made to an observable associated with the operator $ {\hat{A}}$ done on a quantum system specified by $ {\Psi(x,t)}$ will always return $ {a}$ where $ {a}$ is an eigenvalue of $ {a}$. $ \displaystyle \hat{A}\Psi(x,t)=a\Psi(x,t) \ \ \ \ \ (3)$

Axiom 4 The probability that a particle is in the volume element $ {dx}$ is denoted by$ {P(x)dx}$ and is calculated by $ \displaystyle P(x)dx=|\Psi(x,t)|^2dx \ \ \ \ \ (4)$

Axiom 5 The average value of a physical quantity $ {A}$, represented by $ {<A>}$, is $ \displaystyle <A> = \int \Psi^*\hat{A}\Psi \ \ \ \ \ (5)$

Where the integral is calculated in the relevant volume region.

Axiom 6 The state of a quantum system evolves according to the Schrodinger Equation. $ \displaystyle \hat{H}\Psi= i\hbar\frac{\partial \Psi}{\partial t} \ \ \ \ \ (6)$

Where$ {\hat{H}}$ is the hamiltonian operator and corresponds to the total energy of a quantum system.